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Problems in

Chemistry

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Problems in

Chemistry Ranjeet Shahi

ARIHANT PRAKASHAN (Series), MEERUT

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Arihant Prakashan (Series), Meerut All Rights Reserved

© Author No part of this publication may be re-produced, stored in a retrieval system or distributed in any form or by any means, electronic, mechanical, photocopying, recording, scanning, web or otherwise without the written permission of the publisher. Arihant has obtained all the information in this book from the sources believed to be reliable and true. However, Arihant or its editors or authors or illustrators don’t take any responsibility for the absolute accuracy of any information published and the damages or loss suffered there upon. All disputes subject to Meerut (UP) jurisdiction only.

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ISBN : 978-93-12147-38-2

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PREFACE This work is the result of my teaching the JEE aspirants over the post decade and a half. Often I have found that the students mess up with the chemical reactions given in the problem and get themselves entangled into them. The reason is that they are not able to distinguish the relationships between the various chemical principles and concepts. This book is an effort to make them master the fundamentals of Chemistry through problems-solving, and apply them intelligently to arrive at a correct solution. The problems included in the book with their solutions, aim to give you the mastery required for solving the intricate problems asked in the exams such as JEE. Critical analysis of the situation is required whatever be format of the questions. This book just gives emphasis on this, and it is hoped that it will give a boost to all the meritorious and hard-wroking students. I shall like to thank Mr Yogesh Chand Jain, Chairman, Arihant Group, for bringing the book in this nice form.

Suggestions for further improvement of the book are welcome.

Ranjeet Shahi

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CONTENTS 1. Mole Concept

3

2. Acid-Base Titration

7

3. Redox Titration

12

4. Gaseous State

16

5. Thermochemistry

20

6. Thermodynamics

26

7. Atomic Structure

33

8. Chemical Bonding

36

9. Chemical Equilibrium

39

10. Ionic Equilibrium

47

11. Electrochemistry

57

12. Chemical Kinetics

67

13. Colligative Properties

75

14. Solid State

78

15. Surface Chemistry

80

16. Reaction Mechanism

81

17. Stereochemistry

95

18. Hydrocarbons

103

19. Alkyl Halides

120

20. Alcohols and Ethers

133

21. Aldehydes and Ketones

161

22. Carboxylic Acids and Its Derivatives

193

23. Amines

213

24. Aromatic Compounds

216

25. Carbohydrates, Amino Acids and Polymers

264

26. Miscellaneous

265

27. Qualitative Analysis

271

28. Co-odination Compounds

278

29. Representative Elements

285

30. Metallurgy

292

Solutions

295

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Problems in Chemistry

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MOLE CONCEPT PROBLEM 1 A crystalline hydrated salt on being rendered anhydrous, loses 45.6% of its weight. The percentage composition of anhydrous salt is: Al = 10.5%, K = 15.1%, S = 24.8% and O = 49.6%. Find the empirical formula of the anhydrous and crystalline salt. PROBLEM 2 How much quantity of zinc will have to be reacted with excess of dilute HCl solution to produce sufficient hydrogen gas for completely reacting with the oxygen obtained by decomposing 5.104 g of potassium chlorate? PROBLEM 3 A 1.85 g sample of mixture of CuCl 2 and CuBr 2 was dissolved in water and mixed thoroughly with 1.8 g portion of AgCl. After reaction, the solid which now contain AgCl and AgBr was filtered, dried and weighed to be 2.052 g. What was the % by weight of CuBr 2 in the mixture? PROBLEM 4 1.0 g of a sample containing NaCl, KCl and some inert impurity is dissolved in excess of water and treated with excess of AgNO 3 solution. A 2.0 g precipitate of AgCl separate out. Also sample is 23% by mass in sodium. Determine mass percentage of KCl in the sample. PROBLEM 5 A one gram sample containing CaBr 2 , NaCl and some inert impurity was dissolved in enough water and treated with excess of aqueous silver nitrate solution where a mixed precipitate of AgCl and AgBr weighing 1.94 g was obtained. Precipitate was washed, dried and shaken with an aqueous solution of NaBr where all AgCl was converted into AgBr. The new precipitate which contain only AgBr now weighed to be 2.4 g. Determine mass percentage of CaBr 2 and NaCl in the original sample. PROBLEM 6 Sulphur combines with oxygen to form two oxide SO 2 and SO 3 . If 10 g of S is mixed with 12 g of O 2 , what mass of SO 2 and SO 3 will be formed, so that neither S nor oxygen will be left at the end of reaction? PROBLEM 7 An aqueous solution of ethanol has density 1.025 g/mL and it is 8.0 M. Determine molality m of this solution. PROBLEM 8 An aqueous solution of acetic acid has density 1.12 g/mL and it is 5.0 m. Determine molarity (M). PROBLEM 9 Octane is a component of gasoline. Incomplete combustion of octane produces some CO along with CO 2 and H 2O, which reduces efficiency of engine. In a certain test run, 1.0 gallon of octane is burned and total mass of CO, CO 2 and H 2O produced was found to be 11.53 kg. Calculate efficiency of the engine, density of octane is 2.65 kg/gallon. PROBLEM 10 The formula of a hydrated salt of barium is BaCl 2 ⋅ xH 2O. If 1.936 g of this compound gives 1.846 g of anhydrous BaSO 4 upon treatment with H 2SO 4 , calculate x. PROBLEM 11 A mixture of CuSO 4 ⋅ 5H 2O and MgSO 4 ⋅ 7H 2O was heated until all the water was driven-off. If 5.0 g of mixture gave 3 g of anhydrous salts, what was the percentage by mass of CuSO 4 ⋅ 5H 2O in the original mixture? PROBLEM 12 A sample of clay contain 15% moisture, and rest are CaCO 3 and non-volatile SiO 2 . This on heating loses part of its moisture, but CaCO 3 is completely converted into CaO. The partially dried

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sample now contain 7.35% moisture and 51.5% SiO 2 . Determine mass percentage of CaCO 3 in the original sample.

PROBLEM 13 Chlorine dioxide (ClO 2 ), has been used as a disinfectant in air conditioning systems. It reacts with water according to the reaction: ClO 2 + H 2O → HClO 3 + HCl In an experiment, a 10.0 L sealed flask containing ClO 2 and some inert gas at 300 K and 1.0 atmosphere pressure is opened in a bath containing excess of water and all ClO 2 is reacted quantitatively. The resulting solution required 200 mL 0.9 M NaOH solution for neutralization. Determine mole fraction of ClO 2 in the flask.

PROBLEM 14 Potassium salt of benzoic acid (C 6 H 5COOK) can be made by the action of potassium permanganate on toluene as follows: C 6 H 5CH 3 + KMnO 4 → C 6 H 5COOK + MnO 2 + KOH + H 2O If the yield of potassium benzoate can’t realistically be expected to be more than 71%, what is the minimum number of grams of toluene needed to achieve this yield while producing 11.5 g of C 6 H 5COOK?

PROBLEM 15 Manganese trifluoride can be prepared by the following reaction: MnI 2 ( s) + F2 ( g ) → MnF3 + IF5 What is minimum number of grams of F2 that must be used to react with 12.0 g of MnI 2 if overall yield of MnF3 is no more than 75%. PROBLEM 16 A compound containing Ca, C, N and S was subjected to quantitative analysis and formula mass determination. A 0.25 g of this compound was mixed with Na 2CO 3 to convert all Ca into 0.16 g CaCO 3 . A 0.115 g sample of compound was carried through a series of reactions until all its S was changed into SO 2– 4 and precipitated as 0.344 g of BaSO 4 . A 0.712 g sample was processed to liberate all of its N as NH 3 and 0.155 g NH 3 was obtained. The formula mass was found to be 156. Determine the empirical and molecular formula of the compound. PROBLEM 17 A 0.2 g sample, which is mixture of NaCl, NaBr and NaI was dissolved in water and excess of AgNO 3 was added. The precipitate containing AgCl, AgBr and AgI was filtered, dried and weighed to be 0.412 g. The solid was placed in water and treated with excess of NaBr, which converted all AgCl into AgBr. The precipitate was then weighed to be 0.4881 g. It was then placed into water and treated with excess of NaI, which converted all AgBr into AgI. The precipitate was then weighed to be 0.5868 g. What was the percentage of NaCl, NaBr and NaI in the original mixture. PROBLEM 18 A mixture of NaI and NaCl when heated with H 2SO 4 produced same weight of Na 2SO 4 as that of original mixture. Calculate mass percentage of NaI in the original mixture. PROBLEM 19 Ammonia is manufactured by the reaction of N 2 and H 2 . An equilibrium mixture contains 5.0 g of each N 2 , H 2 and NH 3 . Calculate mass of N 2 and H 2 present initially and maximum amount of NH 3 that can be produced. PROBLEM 20 Consider the following reactions: XeF2 + F2 → XeF6 and XeF6 + —( CH 2 —CH 2— ) n → —( CF2 —CF2 — ) n + HF + XeF4 Determine mass of F2 ( g ) required for preparation of 1.0 kg fluorinated polymer.

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PROBLEM 21 2.5 g of a sample containing Na 2CO 3 ; NaHCO 3 and some non-volatile impurity on gentle heating loses 12% of its weight. Residue is dissolved in 100 mL water and its 10 mL portion required 15 mL 0.1 M aqueous solution of BaCl 2 for complete precipitation of carbonates. Determine mass percentage of Na 2CO 3 in the original sample. PROBLEM 22 2.0 g of a sample containing NaCl, NaBr and some inert impurity is dissolved in enough water and treated with excess of AgNO 3 solution. A 3.0 g of precipitate was formed. Precipitate on shaking with aqueous NaBr gain 0.76 g of weight. Determine mass percentage of NaCl in the original sample. PROBLEM 23 Based on the following information, determine value of x and y: AgNO3

(CH 3 ) x AlCl y → xCH 4 ( g ) + yCl – + Al 3+ → AgCl( s) 0.643 g

0.222 g

0.996 g

PROBLEM 24 An organic compound containing C, H, O, N and Cl was analyzed and 0.15 g of sample on combustion produced 0.138 g of CO 2 and 0.0566 g of H 2O. All the nitrogen in different 2.0 g sample of compound was converted into NH 3 which was found to weigh 0.238 g. Finally the chlorine in a 0.125 g sample of compound was converted to Cl – and by reacting with AgNO 3 , 0.251 g AgCl was obtained. Deduce the empirical formula of the starting organic compound. PROBLEM 25 A 5.0 g sample of felspar containing Na 2O, K 2O and some inert impurity is dissolved in dilute HCl solution and NaCl and KCl formed are separated by fractional crystallization. During crystallization some less soluble impurities also comes out. Mass of NaCl, KCl and impurity accompanying these salts was found to be 6.47 g. Solid crystal was then re-dissolved and required 300 mL of 0.3 M AgNO 3 for complete precipitation of chlorides. The precipitate thus, obtained was found to contain 4.23% insoluble impurity. Determine mass percentage of Na 2O and K 2O in the original sample. PROBLEM 26 Potassium chlorate (KClO 4 ) is made in the following sequence of reactions: Cl 2 ( g ) + KOH → KCl + KClO + H 2O KClO → KCl + KClO 3 KClO 3 → KClO 4 + KCl What mass of Cl 2 is needed to produce 1.0 kg of KClO 4 ? PROBLEM 27 Titanium oxide (TiO 2 ) is heated in stream of hydrogen to give water and a new oxide Ti x O y . If 1.598 g TiO 2 produces 1.438 g Ti x O y , what is the formula of new oxide. PROBLEM 28 A solution of copper sulphate that contain 15% CuSO 4 by weight has a density of 1.169 g/mL. 25 mL portion of this solution was reacted with excess of ammonia solution to form a dark blue solution. When cooled, filtered and dried, 6.127 g of dark blue solid was obtained. A 0.195g solid was analyzed for ammonia and required 30.63 mL of 0.1036 M HCl solution to reach the equivalence point. In a separate analysis, 0.200 g was heated at 110°C to drive off water, producing 0.185 g of anhydrous material. Deduce formula of the compound crystallized out from blue solution assuming that it contain only one copper atom per formula unit. Also determine the percentage yield of crystallization process. PROBLEM 29 0.1152 g of a compound containing carbon, hydrogen, nitrogen and oxygen are burned in oxygen. The gases produced are treated further to convert nitrogen containing product into N 2 . The resulting mixture of CO 2 , H 2O and N 2 is passed through a CaCl 2 drying tube, which gains 0.09912 g. The gas stream was then bubbled through water where the CO 2 forms H 2CO 3 . Titration of this solution required 28.8 ml 0.3283 M NaOH solution to reach the phenolphthalein end point. The excess O 2 was

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removed by reaction with copper metal and the N 2 was collected in a 225 mL measuring bulb where it exerted a pressure of 65.12 mm of Hg at 25°C. In a separate analysis, the molar mass of this compound was found to be 146 g mol –1 . Deduce molecular formula of the starting compound.

PROBLEM 30 Pb(NO 3 ) 2 and KI reacts in aqueous solution to form an yellow precipitate of PbI 2 . In one series of experiments, the masses of two reactants varied, but the total mass of the two was held constant at 5.0 g. What maximum mass of PbI 2 can be produced in the above experiment ? PROBLEM 31 An element X react with hydrogen leading to formation of a class of compounds that is analogous to hydrocarbons. 5 g of X forms 5.628 g of a mixture of two compounds of X, XH 4 and X 2 H 6 in the molar ratio of 2 : 1. Determine molar mass of X. PROBLEM 32 The mineral Argyrodite is a stoichiometric compound that contain silver, sulphur (– 2) and an unknown element Y ( + 4). The mass-ratio of silver and Y in the compound is, m( Ag) : m(Y ) = 11.88 Y forms a reddish brown lower sulphide on heating the mineral in stream of H 2 ( g ), in which Y is in + 2 state. The residue are Ag 2S and H 2S. To convert 10 g Argyrodite completely, 0.295 L of H 2 ( g ) measured at 400K and 1.0 atmosphere is required. Determine molar mass of Y and empirical formula of mineral.

PROBLEM 33 Uranium is isolated from its ore by dissolving it as UO 2 (NO 3 ) 2 and separating it as solid UO 2 (C 2O 4 ) ⋅xH 2O. A 1.0 g sample of ore on treatment with nitric acid yielded 1.48 g UO 2 (NO 3 ) 2 which on further treatment with 0.4 g Na 2C 2O 4 yielded 1.23 g UO 2 (C 2O 4 ) ⋅ xH 2O. Determine weight percentage of uranium in the original sample and x. PROBLEM 34 When iodine was added to liquid chlorine in cold condition, orange crystal of a compound separate out. The amount of chlorine in a sample of crystal was determined by precipitating AgCl. A 0.467 g sample of crystal gave 0.861 g of AgCl. Deduce empirical formula of the crystal. PROBLEM 35 Urea is manufactured on large scale by passing CO 2 (g ) through ammonia solution followed by crystallization. CO 2 for the above reaction is prepared by combustion of hydrocarbons. If combustion of 236 kg of a saturated hydrocarbon produces as much CO 2 as required for production of 1000 kg of urea, deduce molecular formula of hydrocarbon. PROBLEM 36 Sodium bicarbonate can be purified by dissolving it in hot water (at 60°C), filtering to remove insoluble impurities, cooling to 0°C to precipitate solid NaHCO 3 and the filtering to remove the solid leaving soluble impurities in solution. Some NaHCO 3 that remain in solution is not recovered. The solubility of NaHCO 3 in water at 60°C is 164 g/L. Its solubility in cold water at 0ºC is 69 g/L. If a 250 g impure sample of NaHCO 3 was purified by this method by dissolving first in 250 mL water at 60°C and then crystallizing NaHCO 3 from 100 mL water at 0°C, 150 g NaHCO 3 was recovered. Determine percentage purity of original sample. PROBLEM 37 A 100 g solution was prepared by dissolving 46 g CuSO 4⋅xH 2O in 54 g of water and mole fraction of CuSO 4 in solution was found to be 0.05. Determine x. PROBLEM 38 An ore of iron contain FeS and non-volatile impurity. Roasting of this ore converts all FeS into Fe 2 O3 and a 4% loss in weight was observed. Determine mass percentage of FeS in ore. PROBLEM 39 Optical measurement is a very efficient method of determining molar mass of unknown material. In one experiment, 3.0 g of an unknown polymeric material was dissolved in 100 mL of CCl 4 and transmittance of this solution was found to be 72%. Transmittance of a 0.001 M standard solution in

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the same solvent, under identical experimental condition was 60%. Determine molar mass of unknown polymer.

PROBLEM 40 shown below,

A crystalline polymer molecule is uniform prismatic in shape with dimensions as

300 Å 100Å

If density of this polymer is 1.2 g/cm 3 , determine molar mass.

PROBLEM 41 A mother cell disintegrate into sixty identical cells and each daughter cell further disintegrate into 24 smaller cells. The smallest cells are uniform cylindrical in shape with diameter of 120 Å and each cell is 6000 Å long. Determine molar mass of the mother cell if density of the smallest cell is 1.12 g/cm 3 . PROBLEM 42 A sample of rock taken for analysis weigh 1.0 g on air dried basis. After drying for one hour at 110°C, the sample weigh 0.9437 g. The calcium is precipitated as oxalate but weighed as CaSO 4 ; that weigh 0.5g. The magnesium is precipitated as MgNH 4 PO4 which finally ignited to 0.5 g Mg 2 P2O 7 . Find the percentage of CaO and MgO on oven dried basis and percentage of them and H 2O on air dried basis. PROBLEM 43 A sample is a mixture of Mohr’s salt and (NH 4 ) 2 SO 4 . A 0.5 g sample on treatment with excess of BaCl 2 solution gave 0.75 g BaSO 4 . Determine percentage composition of the salt mixture. What weight of Fe 2O 3 would be obtained if 0.2 g of the sample were ignited in air? PROBLEM 44 A chloride mixture is prepared by grinding together pure BaCl 2⋅2H 2O, KCl and NaCl. What is the smallest and largest volume of 0.15 M AgNO 3 solution that may be used for complete precipitation of chloride from a 0.3g sample of the mixture which may contain any one or all of the constituents?

ACID-BASE TITRATION PROBLEM 45 A 1.5 g sample containing oxalic acid and some inert impurity was dissolved in enough water and volume made up to 250 mL. A 20 mL portion of this solution was then mixed with 30 mL of an alkali solution. The resulting solution was then treated with stoichiometric amount of CaCl 2 just needed for precipitation of oxalate as CaC 2O 4 . Solution was filtered off and filtrate was finally titrated against 0.1 M HCl solution. 8.0 mL of acid was required to reach the equivalence point. At last, the above neutral solution was treated with excess of AgNO 3 solution and AgCl obtained was washed, dried and weighed to be 0.4305 g. Determine mass percentage of oxalic acid in the original sample. PROBLEM 46 A 1.5 g sample containing P2O 3 and some inert impurity was dissolved in enough water and boiled gently where P2O 3 disproportionated quantitatively into PH 3 and H 3 PO 4 . The solution was

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further boiled for some time to let-off all PH 3 ( g ) and finally cooled to room temperature and diluted to 100 mL. A 10 mL portion of this solution was then mixed with 20 mL 0.3 M NaOH solution. Excess alkali required 11.0 mL 0.05 M H 2SO 4 solution for back titration. Determine mass percentage of P2O 3 in the original sample.

PROBLEM 47 2.5 g of a mixture containing CaCO 3 , Ca(HCO 3 ) 2 and NaCl was dissolved in 100 mL water and its 10 mL portion required 10 mL 0.05 M H 2SO 4 solution to reach the phenolphthalein end point. An another 10 mL portion of the same stock solution required 32.35 mL of the same acid solution to reach the methyl orange end point. Determine mass percentage of CaCO 3 and Ca(HCO 3 ) 2 in the original mixture. PROBLEM 48 A solution contain both Na 2CO 3 and NaHCO 3 . 10 mL portion of this solution is mixed with few drops of phenolphthalein indicator and titrated against 0.08 M H 2SO 4 solution. 7.0 mL of acid was required to reach the end point A 5.0 mL portion of this solution was then taken for further analysis and a few drops of methyl orange was added to it and finally titrated against same acid solution. 3.53 mL of acid was required to reach the end point. Determine mass of Na 2CO 3 and NaHCO 3 per litre of solution. Ignore volume change due to addition of indicator. PROBLEM 49 A mixture was known to contain both KNO 3 and K 2SO 3 . To 0.486 g of the mixture, dissolved in enough water to give 50 mL solution, was added 50 mL of 0.15 M HCl solution. The reaction mixture was heated to expel all SO 2 and then 25 mL of the reaction mixture was titrated with 0.1 M KOH. The titration required 13.11 mL of the base. Calculate mass percentage of K 2SO 3 in the mixture. PROBLEM 50 An amino acid isolated from a piece of animal tissue was believed to be glycine. A 0.05 g sample was treated in such a way that all nitrogen in it was converted into ammonia. This ammonia was added to 50 mL of 0.05 M HCl solution. The excess acid remaining in the solution required 30.57 mL 0.06 M NaOH solution for complete neutralization. What was the percentage by mass of nitrogen? How does this mass compare with percentage mass of nitrogen calculated from glycine (H 2 NCH 2COOH)? PROBLEM 51 In a reaction, calcium orthophosphate on heating with magnesium produced calcium phosphide, magnesium metaphosphate, calcium oxide and oxygen gas. Phosphide on hydrolysis produces PH 3 gas. The PH 3 gas is burnt completely to P2O 5 using air, which contains 21%, by volume of oxygen. Calculate the volume of air at STP required for combustion, if 2.4 g Mg was initially reacted with calcium orthophosphate. All volumes are measured at STP. PROBLEM 52 9.3 g of a mixture containing Li 2CO 3 , NaHCO 3 , Na 2CO 3 on strong heating produced 7.37 g of solid residue. The residue is dissolved in 200 mL water. A 10 mL portion of this solution is mixed with 15 mL of a normal HCl solution. The excess acid required 12 mL 0.5 N NaOH solution to reach the equivalence point. Determine the mass percentage of NaHCO 3 and Na 2CO 3 in the original mixture. Li = 7, Na = 23. PROBLEM 53 4.0 g of a monobasic, saturated carboxylic acid is dissolved in 100 mL water and its 10 mL portion required 8.0 mL 0.27 M NaOH to reach the equivalence point. In an another experiment, 5.0 g of the same acid is burnt completely and CO 2 produced is absorbed completely in 500 mL of a 2.0 N NaOH solution. A 10 mL portion of the resulting solution is treated with excess of BaCl 2 to precipitate all carbonate and finally titrated with 0.5 N H 2SO 4 solution. Determine the volume of the acid solution that would be required to make this solution neutral. PROBLEM 54 5.0 g of a mixture containing NaHCO 3 , NaCl and Na 2CO 3 is dissolved in 500 mL water and its 10 mL portion required 12.4 mL 0.1 M HCl solution to reach the equivalence point. In an another

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experiment, 10 mL portion of the same stock solution is mixed with 10 mL 0.15 M NaOH solution. Excess NaOH required 12.6 mL 0.1 M HCl solution for back titration. Determine the mass percentage of each component in the original mixture.

PROBLEM 55 6.4 g of a pure monobasic organic acid is burnt completely in excess of oxygen and CO 2 evolved is absorbed completely in one litre of an aqueous solution of NaOH. A 10 mL portion of this solution required 14.5 mL of a normal HCl solution to reach the phenolphthalein end point. An another 10 mL portion of the same solution required 18 mL of the same HCl solution to reach the methyl orange end point. If the organic acid contains 25% oxygen by weight, deduce the empirical formula of this acid and strength of original NaOH solution. PROBLEM 56 A complex of cobalt with ammonia is analyzed for determining its formula, by titrating it against a standardized acid as follows: Co(NH 3 ) x Cl 3 ( aq ) + HCl → NH +4 ( aq ) + Co 3+ ( aq ) + Cl – ( aq ) A 1.58 g complex required 23.63 mL 1.5 M HCl to reach the equivalence point. Determine formula. If the reaction mixture at equivalence point is treated with excess of AgNO 3 solution, what mass of AgCl will precipitate out?

PROBLEM 57 One litre solution of alkali is prepared by dissolving impure solid of alkali which contain 5% Na 2CO 3 and 8% CaCO 3 and 10% NaCl. A 10 mL portion of this solution required 9.8 mL of a 0.5 M H 2SO 4 solution for neutralization. Calculate weight of alkali dissolved initially. PROBLEM 58 40 g of a sample of caustic soda containing NaOH, Na 2CO 3 and inert impurity is dissolved in water to prepare 1.0 litre solution. A 25 mL portion of this solution required 23.15 mL 1.022 N HCl for complete neutralization. To 25 mL another solution, excess of BaCl 2 is added, and resulting solution required 22.55 mL HCl of same strength to reach the end point. Calculate mass percentage of NaOH and Na 2CO 3 in the original sample. PROBLEM 59 1.5 g of a sample containing Na 2CO 3 and NaHCO 3 is dissolved in 100 mL of water. A 25 mL portion of this solution required 22.45 mL 0.202 N HCl using methyl orange as indicator. In a separate analysis, 25 mL portion of the same stock solution is mixed with 30 mL 0.204 N NaOH and then excess of BaCl 2 is added resulting in precipitation of all carbonate as BaCO 3 . Filtrate required 9.98 mL HCl of same strength. Calculate mass percentage of Na 2CO 3 and NaHCO 3 in the mixture. PROBLEM 60 One gram sample of a saturated hydrocarbon is burned completely and liberated CO 2 was absorbed in a 1.0 L 0.2 N NaOH solution. To the resulting solution, excess of BaCl 2 crystals was added and the solution was filtered off to free from BaCO 3 . A 10 mL portion of the extract required 12 mL 0.025 M H 2SO 4 solution for neutralization. Determine molecular formula of the hydrocarbon. PROBLEM 61 2.0 g of a saturated, monobasic carboxylic acid was burned and liberated CO 2 was passed through a concentrated solution of NaOH. The resulting solution was separated into two equal half and analyzed. One half required 71.72 mL 1.0 N HCl to reach the end point in presence of phenolphthalein indicator. The other half required 123.44 mL 1.0 N HCl to reach the end point in presence of methyl orange indicator. Deduce formula of acid and determine mass of NaOH present initially. PROBLEM 62 2.5 g of a mixture containing NaHCO 3 , Na 2CO 3 and NaCl is dissolved in 100 mL water and its 50 mL portion required 13.33 mL 1.0 N HCl solution to reach the equivalence point. On the other hand its other 50 mL portion required 19 mL 0.25 M NaOH solution to reach the equivalence point. Determine mass percentage of each component.

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PROBLEM 63 2.0 g of a crystal of CaCO 3 is dissolved in 50 mL water and then mixed with 50 mL of a HCl solution. The resulting solution is boiled to remove all CO 2 and its 10 mL portion required 8.0 mL of a NaOH solution to make the solution neutral. Also 20 mL of original HCl solution is equivalent to 96 mL of NaOH solution. Determine molarity of both NaOH and HCl solution. PROBLEM 64 2.725 g of a mixture of K 2C 2O 4 , KHC 2O 4 and H 2C 2O 4 ⋅ 2H 2O is dissolved in 100 mL H 2O and its 10 mL portion is titrated with 0.1 N HCl solution. 20 mL acid was required to reach the equivalence point. In another experiment, 10 mL portion of the same stock solution is titrated with 0.1 N KOH solution. 20 mL of base was required to reach the equivalence point. Determine mass percentage of each component in the mixture. PROBLEM 65 A 1.0 g sample containing NH 4 NO 3 , (NH 4 ) 3 PO 4 and some inert impurity was dissolved in 100 mL water its 10 mL portion required 15 mL 0.1 M NaOH solution to reach the equivalence point. In a separate experiment, 10 mL of the same stock solution was treated with excess of BaCl 2 solution and 0.077 g of barium phosphate precipitate was obtained. Determine mass percentage of ammonium nitrate in the original sample. PROBLEM 66 10.38 mg of a diprotic acid (containing (C, H and O) is burned completely and all CO 2 was absorbed in 100 mL of alkali solution. The resulting solution is separated into two-half and one-half required 55 mL 0.005 M H 2SO 4 solution to reach the phenolphthalein end point. Other half was titrated in presence of methyl orange indicator and 80 mL H 2SO 4 solution of same strength was required to reach the end point. In a separate analysis, 0.168 g of the same acid required 16.18 mL 0.125 M NaOH solution to reach the end point. Deduce formula of the acid and determine molarity of alkali solution used initially. PROBLEM 67 A 3.0 g sample containing Na 2CO 3 , NaHCO 3 , NaCl and some inert impurity was dissolved in 100 mL of water and its 10 mL portion was titrated against 0.1 M HCl solution using phenolphthalein indicator. 11.32 mL of acid solution was required to reach the end point. The resulting solution was then mixed with excess of AgNO 3 solution resulting in formation of 0.306 g of AgCl precipitate. The solution was filtered-off and filtrate was again titrated, but now against 0.05 M NaOH solution. 42.64 mL of alkali was required to reach the end point. Determine mass percentage of Na 2CO 3 , NaHCO 3 and NaCl in the original sample. PROBLEM 68 In neutralization titration of Na 3 PO 4 , if phenolphthalein is used as indicator, end point is indicated only when Na 3 PO 4 is converted into Na 2 HPO 4 while, if methyl orange is used as indicator, end point appear only when Na 3 PO 4 is converted into H 3 PO 4 . In an experiment a 4.0 g mixture containing Na 3 PO 4 , Na 2 HPO 4 and NaH 2 PO 4 is dissolved in 50 mL water and its 10 mL portion required 24.4 mL 0.1 M HCl solution to reach the end point using phenolphthalein indicator. In a separate analysis, 10 mL portion of the same stock solution required 23.572 mL 0.5 M HCl solution to reach the end point using methyl orange as indicator. Determine mass percentage of all components in the mixture. PROBLEM 69 A mixture containing LiHCO 3 , NaCl and Na 2CO 3 on gentle heating loses 26.5% of its weight. 5.0 g of this mixture was heated gently and residue was dissolved in 100 mL water. A 10 mL portion of this solution was then treated with 20 mL 0.2 M H 2SO 4 solution. A 10 mL portion of the resulting solution required 3.86 mL 0.1 M NaOH solution to reach the end point. Determine mass percentage of each component in the mixture. PROBLEM 70 A mixture containing LiHCO 3 , NaHCO 3 and CaCO 3 on gentle heating loses 48.4% of its weight. In an experiment, 5.0 g of this mixture was dissolved in 100 mL water and its 10 mL portion was treated with 10 mL 0.5 M NaOH solution. The resulting solution was then treated with excess of

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BaCl 2 solution resulting in precipitation of all carbonates as BaCO 3 . Precipitate was separated out by filtration and filtrate required 15.3 mL 0.1 N HCl solution to reach the end point. Determine mass percentage of all components present in the mixture.

PROBLEM 71 5.0 g of a mixture containing NaCl, NaHCO 3 , Na 2CO 3 and CaCO 3 on gentle heating reduces to 4.25 g of solid residue. In a separate experiment, 1.0 g of the same mixture required 10 mL 0.2 M NaOH to reach the end point. In a 3rd experiment, 1.0 g of the same mixture was dissolved in 100 mL water and required 10 mL 1.053 M HCl solution to reach the end point. Determine mass percentage of each component in the mixture. PROBLEM 72 2.0 g of a sample of CaCO 3 , NaHCO 3 and some volatile, inert impurity, was heated strongly where CaCO 3 and NaHCO 3 , were decomposed into CaO and Na 2CO 3 respectively and all CO 2 gas produced in decomposition was absorbed in a 50 mL NaOH solution. NaOH was little less than the stoichiometric requirement therefore, CO 2 during reaction with NaOH, produced Na 2CO 3 and some NaHCO 3 . The resulting solution was titrated first in presence of phenolphthalein indicator and 5.0 mL 1.0 M HCl was required to reach the phenolphthalein end point. Methyl orange was then added and titration continued with HCl of same strength where 15 mL HCl was required to reach the final end point. On the other hand, the residue obtained after heating of the original sample was dissolved in water and treated with excess of BaCl 2 , giving 0.985 g of BaCO 3 precipitate. Determine mass percentage of CaCO 3 and NaHCO 3 in the original sample. PROBLEM 73 A one gram sample containing NaOH as the only basic substance and some inert impurity was left exposed to atmosphere for a very long time so that part of NaOH got converted into Na 2CO 3 by absorbing CO 2 from atmosphere. The resulting sample was dissolved in water and volume made upto 100 mL. A 100 mL portion of this solution required 16 mL 0.25 M HCl solution to reach the equivalence point when methyl orange was used as indicator. In a separate analysis, 20 mL portion of the same solution was taken alongwith phenolphthalein indicator and mixed with 50 mL of 0.1 M HCl solution. An additional 9.00 mL 0.1 M Ba(OH) 2 solution was required to just restore the pink colour of solution. Determine mass percentage of NaOH in the original sample and mass percentage of Na 2CO 3 in the sample after exposure to atmosphere. PROBLEM 74 The monochloroacetic acid (ClCH 2COOH) preservative in a 100 mL of carbonated beverage was extracted by shaking with dimethyl ether and then returned to aqueous solution as ClCH 2COO – by extraction with 1.0 M NaOH. This solution was acidified and treated with 50 mL 0.0452 M AgNO 3 solution where the following reaction occurred: ClCH 2COOH + AgNO 3 + H 2O → HOCH 2COOH + H + + NO –3 + AgCl( s) After filtering the AgCl, titration of filtrate required 10.43 mL of an NH 4SCN solution. Titration of a blank taken through the entire procedure used 22.98 mL of same NH 4SCN solution. Calculate weight in mg, of ClCH 2COOH in the beverage sample. PROBLEM 75 2.0 g of a sample containing sodium oxalate, oxalic acid dihydrate and some inert impurity was dissolved in 100 mL water and its 20 mL portion required 23.34 mL 0.04 M acidified permanganate solution to reach the equivalence point. In a separate analysis, 20 mL portion of the same stock solution required 26.67 mL 0.1 N NaOH solution to reach the end point. Determine mass percentage of Na 2C 2O 4 and H 2C 2O 4⋅2H 2O in the original sample. PROBLEM 76 A 1.5 g sample containing (NH 4 ) 2 SO 4 , NH 4 NO 3 and some inert impurity was dissolved in water and volume made upto 100 mL. A 20 mL portion of this solution was mixed with 50 mL 0.1 M NaOH solution. A 30 mL aliquot of this resulting solution required 9.00 mL 1/28 M H 2SO 4 solution for complete neutralization. In a separate analysis, 32 mL of the original stock solution on

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treatment with excess of BaCl 2 solution produced 0.466 g BaSO 4 precipitate. Determine mass percentage of NH 4 NO 3 and (NH 4 ) 2 SO 4 in the original sample.

PROBLEM 77 A 1.0 g impure sample containing [Zn(NH 4 ) 4 ]Cl 2 and some inert impurity was treated with 15 mL of 1 M NaOH solution where all complex is converted into Na 2 [Zn(OH) 4 ] . The excess base 1 required 10 mL M HCl solution for back titration. 6 (a) Determine percentage purity. (b) If the last solution obtained after neutralization was treated with excess of AgNO 3 , what weight of AgCl would have been produced? PROBLEM 78 1.2 g of a salt with their empirical formula K x H y (C 2O 4 ) z was dissolved in 50 mL of water and its 10 mL portion required 11.00 mL of a 0.1 M HCl solution to reach the equivalence point. In a separate analysis, 15 mL of the stock solution required 20 mL 0.2475 M KOH to reach the equivalence point. Determine empirical formula of the salt. PROBLEM 79 Impure phosphoric acid for use in the manufacture of fertilizer is produced by the reaction of sulphuric acid on phosphate rock of which a principal component is Ca 3 (PO 4 ) 2 and rest are silica and other inert impurity. In an analysis, 2.0 g of a sample of rock salt was dissolved in 100 mL H 2SO 4 solution. Excess sulphuric acid left in 20 mL of this solution required 40 mL 0.02 M NaOH for back titration. In a separate analysis 20 mL of the above solution required 50 mL 0.04 M NaOH for complete neutralization. Determine mass percentage of Ca 3 (PO 4 ) 2 in rock-sample. PROBLEM 80 A 10 g sample of ammonium perchlorate containing some inert impurity was mixed with 3 g Al powder where all perchlorate reacted to produce Al 2O 3 , N 2 , HCl and H 2O. All HCl was absorbed in 100 mL 1 M NaOH solution. Determine percentage purity of perchlorate sample and volume of 0.5 M HCl required to neutralize the above solution. PROBLEM 81 Potassium superoxide (KO 2 ) is utilized in closed system breathing apparatus to remove CO 2 and water from exhaled air. The removal of H 2O generate oxygen gas and KOH and this KOH in the subsequent step remove CO 2 as KHCO 3 . 5.0 kg of an impure sample of KO 2 is just sufficient to remove all CO 2 and H 2O from a closed room of dimension 10 m × 5 m × 3m. Determine mass of this KO 2 required to neutralize a 100 mL 0.1 M H 2SO 4 solution in a separate analysis. Assume room conditions to be at 1.0 atmosphere and 300 K and mole fraction of CO 2 in that room is 0.01. PROBLEM 82 3.25 g of a saturated, tribasic carboxylic acid required 68.4 mL of a 0.750 M NaOH solution to reach the equivalence point. Determine molecular formula of acid.

REDOX TITRATION PROBLEM 83

A sample of chrome-vanadium steel weighing 2.0 g was dissolved in a mixture of

sulphuric acid and just sufficient oxidant was added to raise the oxidation state of iron to Fe 3+ , the chromium to Cr 2O 72– , vanadium to VO –3 and Mn to MnO –4 . The solution was then treated with HCl and – resulting solution still containing Fe 3+ , Cr 2O 2– 7 and VO 3 then treated with 25 mL of 0.101 M FeSO 4 . This resulted in reduction of dichromate and VO –3 to Cr 3+ and VO 2+ in the solution respectively. Fe 2+ and VO 2+ in the solution was then titrated with 0.02236 M KMnO 4 and required 12.6 mL to reach the

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equivalence point. A small amount of Fe 2+ was then added to again reduce the VO –3 produced by KMnO 4 back to VO 2+ and this then titrated directly with 0.02236 M KMnO 4 , a process requiring 0.86 mL to reach the equivalence point. Calculate the following quantities: (a) Moles of Fe 2+ in 25 mL sample of standard FeSO 4 solution. (b) Moles of Fe 2+ titrated with 12.6 mL of standard KMnO 4 . (c) Moles of Fe 2+ consumed by Cr 2O 2– 7 . (d) Percentage of V and Cr in the steel [Atomic weight of V = 51, Cr = 52]

PROBLEM 84 A sample of crude uranium oxide is known to be contaminated with iron. To determine the extent of contamination, the crude oxide were dissolved and reduced with Zn to yield a solution containing U 4+ and Fe 2+ . A 20 mL aliquot of this solution was treated with cupferron which precipitated all uranium and the resulting precipitate on ignition yielded 423.3 mg of U 3O 8 . A further 20 mL sample was treated with 0.024 M KMnO 4 solution and consumed 27.23 mL. Calculate mass percentage of contamination if the iron were present as Fe 2O 3 in a sample of crude oxide containing 100 g of U 3O 8 . KMnO 4 solution oxidised Fe 2+ to Fe 3+ and U 4+ to UO 2+ 2 . Atomic mass of U = 238. PROBLEM 85 A 5.0 g sample containing Pb 3O 4 , PbO 2 and some inert impurity is dissolved in 250 mL dil. HNO 3 solution and 2.7 g of Na 2C 2O 4 was added so that all lead converted into Pb 2+ . A 10 mL portion of this solution required 8.0 mL 0.02 M KMnO 4 for titration of excess of oxalate. In an another experiment, 25 mL of solution was taken and excess oxalate was removed by extraction, this required 10 mL of a permanganate solution for oxidation of Pb 2+ to Pb 4+ . 10 mL this permanganate solution is equivalent to 4.48 mL 5 V H 2O 2 solution. Calculate mass percentage of PbO 2 and Pb 3O 4 in the original sample. [Atomic mass of Pb = 207] PROBLEM 86 An unknown cupric salt with formula Cu x (CO 3 ) y (OH) z is analyzed to determine the exact formula. A 1.7225 g sample of salt was dissolved in 100 mL of pure water. A 50 mL portion of this solution required 10 mL 1.0 N H 2SO 4 solution to reach the equivalence point if phenolphthalein was used as indicator. Another 50 mL portion was titrated using methyl orange as indicator and 15 mL acid of same strength was required. Deduce the formula of the salt. PROBLEM 87 Both CaCl 2 and NaCl are used to melt ice and snow on roads in winter. A certain company was marketing a mixture of these two compounds for this purpose. A chemist, wishing to analyze the mixture, dissolved 1 g of it in water and precipitated the calcium by adding sodium oxalate. The calcium oxalate was then carefully filtered, dissolved in dilute sulphuric acid, and titrated with 0.1 M KMnO 4 solution. The titration required 22 mL of the KMnO 4 solution. Calculate freezing point of an aqueous solution which is 5% (w/V) of the above mixture. K f of water is 1.86 K kg mol –1 . PROBLEM 88 A 4.25 g sample containing CaC 2O 4 , Na 2C 2O 4 and some inert impurity is heated gently so that CaC 2O 4 decomposed as: CaC 2O 4 → CaO + CO( g ) + CO 2 ( g ) All gaseous products were passed through a NaOH solution where following reaction occurred quantitatively: 2NaOH + CO 2 ( g ) → Na 2CO 3

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The resulting solution is separated into two equal part (by volume) and one part required 30 mL 0.5 M HCl to reach the phenolphthalein end point while the other half solution required 50 mL 0.5 M HCl solution to reach the methyl orange end point. In a separate experiment same mass of the same sample is dissolved into 100 mL dilute HCl solution and its 10 mL portion required 10 mL 0.1 M K 2Cr 2O 7 solution. Determine the mass percentage of Na 2C 2O 4 in the original sample.

PROBLEM 89 In acidic solution, 45 mL KMnO 4 solution is required to react with 50 mL 0.25 N Na 2C 2O 4 solution. How many mL of this same KMnO 4 solution would be required to oxidise 25 mL 0.1 N K 2C 2O 4 solution in alkaline medium where KMnO 4 is reduced to MnO 2 . PROBLEM 90 A sample weighing 0.3 g containing K 3 [Fe(C 2O 4 ) 3 ] ⋅ 3H 2O,FeCl 3 ⋅ 6H 2O and inert impurity is dissolved in dilute sulphuric acid and volume made up to 100 mL. A 20 mL portion of this solution required 3.75 mL of 0.005 M acidified KMnO 4 solution to reach the equivalence point. In an another experiment, 50 mL sample of the same stock solution is treated with Zn-amalgum and the resulting solution required 17.5 mL of permanganate solution of same strength. Determine mass percentage of FeCl 3 ⋅ 6H 2O in the original sample. PROBLEM 91 A 6.1 g sample containing oxalic acid dihydrate, sodium oxalate and NaHC 2O 4 and is dissolved in 100 mL of water and its 10.0 mL portion required 16 mL 0.25 M HCl to reach the equivalence point. In another experiment 10.0 mL portion of the same stock solution required 24 mL 0.25 M NaOH to reach the equivalence point. Determine the mass percentage of all components in the original mixture. PROBLEM 92 A 0.127 g of an unsaturated oil was treated with 25 mL of 0.1 M ICl solution. The unreacted ICl was then treated with excess of KI. Liberated iodine required 40 mL 0.1 M hypo solution. Determine mass of I 2 that would have been required with 100.0 g oil if I 2 were used in place of ICl. PROBLEM 93 Alkali metal nitrate on heating decomposes to metal nitrite and oxygen whereas alkaline earth metal on heating decomposes into metal oxide, NO 2 and oxygen. In an experiment 15 g mixture of NaNO 3 and Mg(NO 3 ) 2 was heated until no more gas were evolved. The water soluble part of residue was used for analysis and dissolved in 1.0 litre water. 10 mL portion of this solution was reacted with 20 mL 0.02 M acidified KMnO 4 solution. The excess reagent required 10.00 mL 0.05 M oxalic acid solution. Determine mass percentage of each nitrate in the mixture. Also determine the molar ratio of oxygen to NO 2 in the gaseous products given off. PROBLEM 94 The mass percentage of MnO 2 in a sample of mineral is determined by reacting with As 2O 3 in acid solution. A 0.225 g sample of mineral is ground and boiled with 75 mL 0.0125 M As 2O 3 solution. After the reaction is complete, the solution is cooled and titrated with 2.28 ×10 –3 M acidified KMnO 4 solution. 16.34 mL of the oxidizing agent solution was required to reach the end point. Determine mass percentage of MnO 2 in the sample. PROBLEM 95 A driver is arrested and asked to pass “breath analyzer” test. A sample consisting 56.5 mL of exhaled air is then bubbled into a spectrometer cell containing 3 mL 0.025% (w/V) K 2Cr 2O 7 solution. The transmittance of the solution was 41.5% initially and 43.4% after bubbling the sample through the reaction cell. It is known that the alcohol content in blood stream is 2300 times higher than in exhaled air and that the legal limit is 80 mg of alcohol per 100 mL of blood. Determine the concentration of alcohol in the blood and state whether or not the driver should be charged with drunk driving.

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PROBLEM 96 A sample of 0.3657 g powder containing only Ba(NO 3 ) 2 and Ca(NO 3 ) 2 are dissolved in 50 mL water. Ammonia is added to the solution to raise the pH than an excess of Na 2C 2O 4 is added to precipitate the metals. The precipitate is then filtered, washed with 1.0 L of water and transferred to a beaker containing 50 mL H 2O. The solution is acidified to solublise the precipitate and finally titrated with 0.05 M KMnO 4 solution. A total of 13.94 mL of oxidizing agent solution was required to reach the end point. Find the composition of the initial mixture. K sp. of BaC 2O 4 = 1.5 × 10 –8 and of CaC 2O 4 = 2.34 × 10 –9 .

PROBLEM 97 1.0 g sample containing KO 2 and some inert impurity is dissolved in excess of aqueous HI solution and finally diluted to 100 mL. The solution is filtered off and 20 mL of filtrate required 15 mL 0.4 M Na 2S 2O 3 solution to reduce the liberated iodine. Determine mass % of KO 2 in the original sample. PROBLEM 98 Cuprous ion is known to disproportionate quantitatively in acid medium. A 3.0 g sample of Cu 2O is dissolved in dilute H 2SO 4 solution. The solution is filtered off and 8.3 g pure KI crystal is added to filtrate. This caused precipitation of CuI with evolution of I 2 . The solution is filtered off and filtrate is boiled till all I 2 is expelled off. Now, excess of an oxidizing agent is added to filtrate and liberated iodine required 10 mL 1.0 N Na 2S 2O 3 solution. Calculate mass percentage of Cu 2O in the original sample. PROBLEM 99 To a 10 mL 1.0 M aqueous solution of Br 2 , excess of NaOH is added so that all Br 2 disproportionated to Br – and BrO –3 . The resulting solution is freed from bromide ion by extraction and excess of OH – neutralized by acidifying the solution. The resulting solution is just sufficient to react with 1.5 g of an impure CaC 2O 4 sample. Calculate percentage purity of oxalate sample. PROBLEM 100 One gram of an impure sample of NaCl was dissolved in water and treated with excess of AgNO 3 solution. The precipitate AgCl thus, formed undergo decomposition into Ag and Cl 2 ( g ) and latter disproportionate into chlorate (V) and chloride ions and chloride is re-precipitated due to presence of excess of AgNO 3 . If the original precipitate was 60% decomposed and final precipitate weigh 1.5 gram, determine mass percentage of NaCl in original sample. PROBLEM 101 0.4 g of a sample containing CuCO 3 and some inert impurity was dissolved in diute sulphuric acid and volume made up to 50 mL. To this solution was added 50 mL 0.04 M KI solution where copper precipitate as CuI and iodide ion is oxidized into I –3 . A 10 mL portion of this solution is taken for analysis, filtered, made free from I –3 and treated with excess of acidic permanganate solution. Liberated iodine required 20 mL 2.5 m M sodium thiosulphate solution to reach the end point. Determine mass percentage of CuCO 3 in the original sample. PROBLEM 102 One gram of an unknown sample of NaCN is dissolved in 50 mL 0.33 M alkaline solution of KMnO 4 and refluxed so that all cyanide is converted into cyanate (OCN – ). The reaction mixture was cooled and its 5.0 mL portion was acidified by adding excess of sulphuric acid solution and finally titrated with 19.0 mL 0.1 M FeSO 4 solution. Determine mass percentage of NaCN in the original sample. PROBLEM 103 5.0 mL of a pure liquid toluene is dissolved in 100 mL of dilute alkaline KMnO 4 solution and refluxed so that all toluene is oxidized into benzoic acid and a dark brown precipitate of MnO 2 is formed. Solution is filtered off and filtrate and precipitate were analyzed separately. Precipitate

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was re-dissolved into 100 mL 1.0 M acidified solution of Na 2C 2O 4 and excess of oxalate required 50.73 mL 0.1 M acidic dichromate solution for back titration. On the other hand 10.5 mL of filtrate was acidified by adding excess of sulphuric acid and titrated with 0.1 M acidified solution of Na 2C 2O 4 . A 38 mL of oxalate solution was required to reach the end point. Determine density of liquid toluene and molarity of original permanganate solution.

PROBLEM 104 A 2.0 g sample containing CaOCl 2 and NaOCl is dissolved in 100 mL water and its 10 mL portion was titrated against 0.15 M acidified solution of Na 2C 2O 4 . 10 mL of oxalate solution was required to reach the end point. Titrated solution was then treated with excess of aqueous solution of AgNO 3 where all chloride precipitates as AgCl and weighed to be 0.287 g. Determine mass percentage of CaOCl 2 and NaOCl in original sample.

GASEOUS STATE PROBLEM 105 6.0 g of He having average velocity 4 × 10 2 ms –1 is mixed with 12.0 g of Ne 20 having the same average velocity. What is the average kinetic energy per mole in the mixture? PROBLEM 106 The valve of a commercial cylinder of N 2 gas was left slightly open so that small amount of gas leaked into the laboratory. The leak rate was proportional to the pressure difference (internal pressure one atm). If the initial leak rate was found to be 1 g s –1 and initial pressure inside the 7.28 m 3 tank was 17180 kPa, what would be the pressure inside the tank after 10 days assuming temperature of the lab to be 27°C. PROBLEM 107 Calculate pressure exerted by 22.0 g of CO 2 in 0.5 L bulb at 300 K assuming it to be real gas with a = 363 kPaL2 mol –2 and b = 42.67 cc/ mol. PROBLEM 108 Molar volume of He at 10.1325 Mpa and 273 K is 0.011075 times its molar volume at 101.325 kPa. Calculate radius of He atom assuming negligible ‘a’. PROBLEM 109 A gas mixture containing 5% by mass of butane and 95% by mass of Ar (40) is to be prepared by allowing gaseous butane to fill an evacuated 40 L cylinder at 1.0 atm and 27°C. Calculate mass of Ar that gives the desired composition and total pressure of the final mixture. PROBLEM 110 Cl 2O 7 gas decomposes as: Cl 2O 7 → Cl 2 + O 2 A partially decomposed gaseous mixture is allowed to effuse through a pin-hole and the gas coming out initially was analyzed. The mole fraction of the O 2 was found to be 0.60, determine the degree of dissociation.

PROBLEM 111 Proportion of a lighter isotope in a gaseous mixture containing both heavier and lighter isotopes is increased by successive effusion of the gas mixture. A sample of neon gas has 22 Ne = 90% and 20 Ne = 10% by moles. In how many stages of successive effusion, 25% enrichment of 20

Ne would be achieved?

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PROBLEM 112 The density of vapour of a substance at 1.0 atm and 500 K is 0.35 k/ m 3 . The vapour effuses through a small hole at a rate of 1.33 times faster than oxygen under similar condition. (a) Determine (i) Molecular weight (ii) Molar volume (iii) Compression factor (Z) of the vapour (iv) Which forces among the gas molecules are dominating, the attractive or repulsive? (b) If the vapour behaves ideally at 1000 K, determine the average translational kinetic energy possessed by a molecule. PROBLEM 113 Using van der Waals’ equation of state, calculate the pressure correction factor for two moles of a gas confined in a four litre flask that exert a pressure of 11 atmosphere at 300 K. b = 0.05 L mol –1 . PROBLEM 114 For a van der Waals’ gas Z (compressibility factor) was found to be 1.5 at 273 K and one atmosphere and TB of the gas is 107 K. Determine value of a and b. PROBLEM 115 A flask containing 2.0 moles of He gas at 1.0 atm and 300K is connected to another flask containing N 2 ( g ) at the same temperature and pressure by a narrow tube of negligible volume. Volume of the nitrogen flask is three times volume of He-flask. Now the He-flask is placed in a thermostat at 200 K and N 2 -flask in another thermostat at 400 K. Determine final pressure and final number of moles in each flask. PROBLEM 116 In a spherical glass flask A of radius 1.0 m, containing 300 g H 2 (g ), there was a rubber balloon B containing some N 2 ( g ). Inside B, there was another rubber balloon C containing some oxygen gas. At 27°C, it was found that the balloon B had radius 60 cm and of C was 30 cm. Calculate the total weight of the gas inside the flask. Now 50 g H 2 ( g ) is further added to A, what would be the volume of B and C. PROBLEM 117 A partially decomposed PCl 5 (g ) along with its dissociation product is subjected to diffusion study and the gases coming out initially collected in an another flask. The rate of effusion of collected gaseous mixture was found to be 0.45 times rate of effusion of pure oxygen gas. Determine the degree of dissociation of PCl 5 ( g ) in the original sample. PROBLEM 118 One mole of a monoatomic gas confined in a 22.5 litre flask at 273 K exert a pressure of 0.98 atm, whereas expected pressure was 1.0 atm has the gas behaved ideally. Determine the van der Waal’s constants ‘a’ and ‘b’ and Boyle’s temperature (TB ). PROBLEM 119 One litre of a gas at 300 atm and 473 K is compressed to a pressure of 600 atm and 273 K. The compressibility factors found to be 1.072 and 1.375 respectively at the initial and final states. Calculate the final volume. PROBLEM 120 Calculate the van der Waal’s constants for ethylene. TC = 282 K,PC = 50 atm. PROBLEM 121 The second Virial coefficient of an imperfect gas is 2 × 10 –2 (L/ mol) 2 . Calculate the volume of a gm mole of the gas at 27°C and 5 atmosphere pressure. PROBLEM 122 The van der Waal’s constant ‘b’ of a gas is 4.42 centilitre/mol. How near can the centres of the two molecules approach each other? PROBLEM 123 For carbon dioxide, critical density is 0.45 g/cc and its TC = 300 K. Determine its van der Waal’s constants.

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PROBLEM 124 The Virial equation for ethane gas is given by PV = RT + BP . At 0°C, B = – 0.1814 L/ mol. Calculate volume of one mole of ethane at 10 atm, and ‘a’. PROBLEM 125 An unknown gas (X) at 2.0 atmosphere and Ar (40) at 1.0 atmosphere were injected simultaneously from the two ends of a 1.0 metre long glass tube and the first collision between X and Ar occurred at a distance of 38 cm from Ar-end.Determine the molar mass of X assuming that gases were injected at same temperature and through the pin-hole of identical geometry. PROBLEM 126 Using van der Waals’ equation of state, calculate pressure developed by 100 g of CO 2 contained in a volume of 5.0 litre at 40°C. Also compare this value with that calculated using ideal gas law and determine the percentage deviation from ideality. a = 3.6 atm L2 mol –2 , b = 44 cm 3 mol –1 . PROBLEM 127 An equation of state for a non-ideal gas can be written as: PVm = A + BP + CP 2 ; where Vm is the molar volume and P is the gas pressure in atmosphere. B = – 2.879 × 10 –2 and C = 14.98 × 10 –5 in litre atmosphere unit. Under the experimental condition, determine the pressure at which PV-P curve will attain minimum.

PROBLEM 128 A modified form of van der Waal’s equation of state for 1.0 mole of gas is given as: α   P +  (V – β) = RT  TV 2  Deduce expression for the first Virial coefficient (B) and Boyle’s temperature in term of α and β if Virial equation of state is: PV B C = 1 + + 2 +… RT V V PROBLEM 129 Assuming that dry air contain 79% N 2 and 21% O 2 by volume, calculate the density of moist air at 25°C and 1.0 atmosphere when the relative humidity is 60%. The vapour pressure of water at 25°C is 23.76 mm of Hg. PROBLEM 130 At what temperature, three moles of SO 2 will occupy 10 litre at a pressure of 15.0 atm if it is a van der Waal’s gas with a = 6.71 atm L2 mol –2 and b = 56.4 cm 3 mol –1 . PROBLEM 131 Pressure of He gas confined in a steel chamber drops from 4.0 to 1.0 atmosphere in 4.0 hours due to diffusion through a pin-hole in the steel chamber. If an equimolar mixture of He and methane gas at 20 atmosphere and the same temperature are confined in the same chamber, what will be the partial pressure of He and methane after 1.0 hour. Assume rate of diffusion to be linear function of gas pressure and inverse function of square root of molar masses. PROBLEM 132 One mole of a van der Waal’s gas at 0°C and 600 atmosphere occupies 0.075 L. If b = 0.024 L mol –1 , determine compressibility factor (Z) and predict the type of force dominating among the gas molecule. PROBLEM 133 A one litre flask containing NH 3 (g ) at 2.0 atmosphere and 300 K is connected to another 800 mL flask containing HCl(g) at 8.0 atmosphere and 300 K by means of a narrow tube of negligible volume and gases were allowed to react quantitatively as: NH 3 ( g ) + HCl( g ) → NH 4Cl( s); ∆H = – 43kJ/ mol

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If heat capacity of HCl(g) CV is 20 JK –1 mol –1 , determine final pressure inside the flask assuming negligible heat capacity of flask and negligible volume of solid NH 4Cl.

PROBLEM 134 A long cylindrical glass tube, equipped with a porous disc at the centre, contain methane gas at 5.0 atmosphere on one side and He gas at 2.0 atmosphere on the other side of the disc as shown in the diagram below: Disc is permeable to both gases and rate of diffusion is directly proportional to the gas pressure and inversely proportional to square root of molar masses as: CH4 5.0 atm.

–

He 2.0 atm.

dp P where, k is a constant. =k dt M

If k for the diffusion of methane gas is 2.5 ×10 –2 second –1 , determine time after which pressure of methane chamber will drop to 4.0 atmospheres.

PROBLEM 135 At a given condition of temperature, rate of change of r.m.s. of He gas is twice the rate of change of absolute temperature. Determine rms of He in the given condition. PROBLEM 136 1.6 moles of ammonia gas at 300 K is taken in a 2.0 litre flask, sealed and heated to 500 K. At this temperature, ammonia is partially decomposed into N 2 and H 2 and a pressure measurement at this point gave 48.5 atmosphere. Determine number of moles of each component present at 500 K. PROBLEM 137 Decomposition of KClO 3 produces oxygen gas and KCl solid. In a typical experiment, some KClO 3 was decomposed and 36.00 mL oxygen gas was collected over water at 23°C. The laboratory barometer reads 751 mm and vapour pressure of water at 23°C is 21.1 mm of Hg. Find the volume of the dry oxygen at 0°C and 1.0 atmosphere. PROBLEM 138 A narrow tube of negligible volume connects two evacuated bulb of 1.0 litre capacity each. One bulb is placed in a 200 K thermostat bath and other in a 300 K thermostat bath and then 1.0 mole of an ideal gas is injected into the system. Find the pressure in the two flasks. PROBLEM 139 Isothermal compressibility (κ ) of a gas is defined as: 1  ∂V  κ =–   V  ∂P  T , n Determine isothermal compressibility for an ideal gas at 1.0 atmosphere.

PROBLEM 140 What will be the temperature difference needed in a hot air balloon to lift 1.0 kg weight. Assume that the volume of balloon is100 m 3 , the temperature of atmosphere is 25°C and pressure is 1.0 atmosphere. Average molar mass of air is 29 amu. PROBLEM 141 Using van der Waals’ equation of state, find pressure at which the PV vs P curve acquires minima for 1.0 mole of oxygen gas at 0°C. a =1.36 L2 atm mol –2 , and b = 32 cm 3 mol –1 . PROBLEM 142 The van der Waals’ constant ‘a’ is a correction factor to the ideal gas law for intermolecular force of attractions within the substance. Match the following values of ‘a’ ( L2 atm mol –1 ): 0.2107, 5.464, 18.00 and 24.06 with gases benzene, toluene, Ne and steam.

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PROBLEM 143 The van der Waals’ constant ‘b’ is a correction factor to the ideal gas law for the intrinsic volume of the molecule. Match the following values of ‘b’(L mol –1 ): 0.017, 0.0305, 0.1154 and 0.1463, with the gases: toluene, benzene, Ne and steam.

THERMOCHEMISTRY PROBLEM 144 The specific heat capacity of water is 4.18 J(° C) –1 g –1 and that of copper is 0.38 J( ° C) –1 g –1 . Calculate the heat that must be supplied to a 500 g copper kettel containing 450 g of water to raise its temperature from 25°C to the boiling point of water. What percentage of heat is used to raise the temperature of the water? PROBLEM 145 How much heat can be produced from a reaction mixture of 50 g of iron (III) oxide and 25 g of aluminium in the thermite reaction: Fe 2O 3 ( s) + 2Al( s) → Al 2O 3 ( s) + 2Fe( s); ∆H = – 851.5 kJ/ mol

PROBLEM 146 Calculate the reaction enthalpy for the hydrogenation of ethyne to ethane, given standard enthalpy of combustion of ethyne, ethane and hydrogen; – 1300, – 1560 and – 286 kJ/mol respectively. PROBLEM 147 Calculate the reaction enthalpy for the synthesis of HCl(g) from the following data: NH 3 ( g ) + HCl( g ) → NH 4Cl( s) ∆H = – 176 kJ N 2 ( g ) + 3H 2 ( g ) → 2NH 3 ( g ) ∆H = – 92.22 kJ N 2 ( g ) + 4H 2 ( g ) + Cl 2 ( g ) → 2NH 4Cl( s) ∆H = – 628.86 kJ PROBLEM 148 An important reaction that occurs in the atmosphere is NO 2 ( g ) → NO( g ) + O( g ) Which is brought about by the sunlight. How mucy energy the sun to cause it must supply? Given, dissociation energy of oxygen = 498 kJ/ mol and NO( g ) + O 3 ( g ) → NO 3 ( g ) + O 2 ( g ) ∆H = – 200 kJ 3O 2 ( g ) → 2O 3 ( g ) ∆H = 285.4 kJ PROBLEM 149 Using reaction a, b and c determine the enthalpy change of this reaction: 3 CH 4 ( g ) + O 2 ( g ) → CO( g ) + 2H 2O( g ) 2 (a) CH 4 ( g ) + 2O 2 ( g ) → CO 2 ( g ) + 2H 2O( g ) ∆H ° = – 802 kJ/ mol (b) CH 4 ( g ) + CO 2 ( g ) → 2CO( g ) + 2H 2 ( g ) ∆H ° = + 206 kJ/ mol (c) CH 4 ( g ) + H 2O( g ) → CO( g ) + 3H 2 ( g ) ∆H ° = + 247 kJ/ mol PROBLEM 150 The bond energy of H 2 (g ) is 436 kJ/mol and that of N 2 (g ) is 941.3 kJ/mol. Calculate the average bond energy of an N—H bond in ammonia if ∆H ° f of ammonia is – 46 kJ/mol.

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PROBLEM 151 The heat of formation of PCl 3 and PH 3 are 306 kJ/mol and 8 kJ/mol respectively, and the heats of atomization of phosphorus, chlorine and hydrogen are 314, 121 and 216.5 kJ/mol respectively. Calculate P-Cl and P-H bond energy. PROBLEM 152 At 25°C, the molar heat of formation of SO 2 and H 2O are – 296.81 and – 285.83 kJ respectively. Using the information from the following reactions, 2H 2S( g ) + Fe( s) → FeS 2 + 2H 2 ( g ) 3 H 2S( g ) + O 2 ( g ) → H 2O( l) + SO 2 ( g ) 2 Calculate heat of formation of H 2S( g ) and FeS 2 ( s) at 25°C.

∆H ° = – 137 kJ/ mol ∆H ° = – 562 kJ/ mol

PROBLEM 153 The standard molar enthalpy of formation of cyclohexane (l) and benzene (l) at 25°C are – 156 and + 49 kJ/ mol respectively. The standard enthalpy of hydrogenation of cyclohexene (l) at 25°C is – 119 kJ/ mol. Use this data to estimate the magnitude of the resonance energy of benzene. PROBLEM 154 For the reaction cis-2-butene → trans-2-butene and cis-2-butene → 1-butene, ∆H = – 950 and +1771cal/ mol respectively. The heat of combustion of 1-butene is – 649.8 kcal/mol. Determine the heat of combustion of trans-2-butene. Also calculate the bond energy of C==C bond in trans-2-butene. Given B.E of C==O =196, O—H =110, O==O = 118, C—C = 80 and C—H = 98 kcal/mol respectively. ∆H v (H 2O) = 11 kcal / mol. PROBLEM 155 Using the data (all values are in kJ/mol at 25°C) given below: (i) Enthalpy of polymerization of ethylene = – 72. (ii) Enthalpy of formation of benzene(l) = 49 (iii) Enthalpy of vaporization of benzene(l) = 30 (iv) Resonance energy of benzene(l) = – 152 (v) Heat of formation of gaseous atoms from the elements in their standard states H = 218, C = 715. Average bond energy of C—H = 415. Calculate the B.E. of C—C and C==C. [ A : 331 and 590 kJ/mol] PROBLEM 156 Calculate energy of aromatization of cyclohexane according to the following reaction, both cyclohexane and benzene are in liquid state: Given, bond energies: C—C = 348, C—H = 415, C==C = 600, H—H = 436 kJ/mol respectively, sublimation energy of C(gr) is 717 kJ/mol, resonance energy of C 6 H 6 ( l) = – 152 kJ/ mol, ∆H v° of benzene = 30.8 and of cyclohexane is 33 kJ/mol. + 3H2(g)

PROBLEM 157 Enthalpy of polymerization of ethylene and acetylene into corresponding polymers are – 86 kJ/ mol and – 148 kJ/mol respectively. Enthalpy of hydrogenation of ethylene is – 132 kJ/mol, determine C==C bond energy. B.E. of H 2 ( g ) is 436 kJ/mol and of C—H = 415 kJ/ mol. PROBLEM 158 ∆HComb of methane and ethane are – 210 kcal/mol and – 368 kcal/mol respectively. Determine ∆HComb of decane. PROBLEM 159 Determine resonance energy of benzene [C 6 H 6 (l)] from the following information : ∆H °f of C 6 H 6 ( l) = + 49 kJ; ∆H °f of C 2 H 2 ( g ) = + 75 kJ ∆H °v of C 6 H 6 ( l) = + 45 kJ B.E. C ≡≡C = 930 kJ/ mol; C==C = 615 kJ/ mol; C  C = 348 kJ/ mol

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PROBLEM 160 Consider the following thermodynamic data: Enthalpy of formation of CaC 2 ( s) = – 60 kJ/ mol; Enthalpy of sublimation of Ca( s) = 179 kJ/ mol; Enthalpy of sublimation of C( s) = 718 kJ/ mol; First ionization energy of Ca(g) = 590 kJ/ mol; Second ionization energy of Ca(g) 1143 kJ/mol; Bond energy of C 2 ( g ) = 614 kJ/ mol; First electron affinity of C 2 ( g ) = – 315 kJ/ mol; Second electron affinity of C 2 ( g ) = + 410 kJ/ mol. Draw a clear Born-Haber cycle and determine lattice energy of CaC 2 ( s). PROBLEM 161 Normal L.P.G. contains 90% propane and 10% methane by weight. If combustion of L.P.G. produces acetylene, CO(g) and H 2O( l), calculate the heat evolved by combustion of 100 g of L.P.G. Given: enthalpy of combustion of methane = – 890 kJ/ mol, C 3 H 8 = – 2220 kJ/ mol, C 2 H 2 = – 1300 kJ/ mol and CO( g ) = – 285 kJ/ mol. PROBLEM 162 A swimmer breaths 20 times in one minute when swimming and inhale 200 mL of air in one breath. Inhaled air contain 20% O 2 by volume and exhaled air contain 10% O 2 by volume. If all oxygen are consumed in combustion of glucose in the body and 25% of energy obtained from combustion is available for muscular work. Determine the maximum distance this swimmer can swim in one hour if 100 kJ energy is required for 1.0 km swimming. Standard molar enthalpy of combustion of glucose is – 2880 kJ/ mol and body temperature is 37°C. PROBLEM 163 Standard molar enthalpies of formation of H 2O(l) and H 2O 2 (l) are – 285 and – 200 kJ/mol respectively and their molar enthalpies of vaporization are 41 and 60 kJ respectively. If enthalpy of atomization of O 2 ( g ) is 298 kJ/mol, determine bond energy of O—O bond. PROBLEM 164 Determine resonance energy of 1,3-butadiene using the following information: Enthalpy of combustion : 1,3-butadiene = – 2841 kJ/ mol, C(gr ) = – 394 kJ/ mol, H 2 ( g ) = – 285 kJ/ mol Bond enthalpy : C—C = 348 kJ/ mol, C==C = 615 kJ/ mol. Also standard enthalpy of formation of cyclobutene =130 kJ/ mol, PROBLEM 165 Standard molar enthalpy of formation of hydrazine liquid (N 2 H 4 ) is 50 kJ/mol, NH 3 ( g ) = – 46 kJ/ mol. Average N—H and H—H bond energies are 393 and 436 kJ/mol respectively. If enthalpy of vaporization of N 2 H 4 is 18 kJ/mol, determine N—N bond energy in N 2 H 4 . PROBLEM 166 Using following standard enthalpies: ∆H °f HF( aq ) = – 329 kJ/ mol ∆H ° f H 2O( l) = – 285 kJ/ mol, ∆H ° f F – ( aq ) = – 320 kJ /mol and H + ( aq ) + OH – ( aq ) → H 2O( l), ∆H = – 56 kJ/ mol. Determine enthalpy of neutralization of HF against a strong base.

PROBLEM 167 From the following reactions and thermal information at 25°C: 3 2Fe( s) + O 2 → Fe 2O 3 ( s) ∆H ° = – 821.4 kJ 2

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1 2FeO( s) + O 2 → Fe 2O 3 ( s) 2

∆H ° = – 284 kJ/ mol

Fe( s) + 2H + ( aq.) → Fe 2+ ( aq.) + H 2 ( g )

∆H ° = – 87.8 kJ

1 H 2 ( g ) → H + ( aq.) 2 1 H 2 ( g ) + O 2 ( g ) → H 2O( l) 2 Calculate ∆H ° for the reaction :

∆H ° = 0 ∆H ° = – 285 kJ

FeO( s) + 2H + ( aq ) → H 2O( l) + Fe 2+ ( aq )

PROBLEM 168 A 150 cc portion of 0.4 N HCl is neutralized with excess of NH 4OH in a bomb calorimeter which results in a temperature rise of 2.36°C. If the heat capacity of calorimeter content is 1316.7 J/°C, calculate heat of neutralization of HCl Vs NH 4OH. PROBLEM 169 Determine S—S bond energy; Given ∆H ° f of (C 2 H 5 ) 2 S( g ) = – 147 kJ/ mol, ∆H ° f of (C 2 H 5 ) 2 S 2 ( g ) = – 202 kJ/ mol and ∆H °Sublimation of S( s) = 223 kJ/ mol. PROBLEM 170 Given the following standard molar enthalpies: of ∆H °Sublimation ∆H ° f of CH 3CN( g ) = 88 kJ/ mol, ∆H ° f of C 2 H 6 = – 84 kJ/ mol, C(gr) = 717 kJ/ mol, bond dissociation energy of N 2 ( g ) and H 2 ( g ) are 946 and 436 kJ/mol respectively, B.E. (C—H) = 410 kJ/mol. Determine C—C and C ≡≡N bond energies. PROBLEM 171 Determine standard state enthalpy of the following reaction: CH 3COOH( l) → CH 4 ( g ) + CO 2 ( g ) Given ∆H °Combustion CH 4 = – 860 kJ/ mol. Bond energies in kJ/mol. C—H = 410, C—C = 348, C==O = 728, C—O = 352, O—H = 463, O==O = 498. ∆H °Vaporization of acetic acid and water are 52 and 41 kJ/mol respectively. PROBLEM 172 ∆H ° f of enthalpy of combustion of C 2 H 5OH( l) = – 66 kcal/ mol, CH 3OCH 3 ( g ) = – 348 kcal/ mol, ∆H ° f of water is – 68 kcal/mol and ∆H ° f of CO 2 ( g ) = – 94 kcal/ mol. Determine enthalpy of the following isomerization reaction: C 2 H 5OH( l) → CH 3OCH 3 ( g )

PROBLEM 173 The standard enthalpies of formation of BH 3 (g ) and B2 H 6 (g ) are 100 kJ and 36 kJ per mol respectively and the enthalpies of formation of B( g ) and H( g ) are 563 kJ mol –1 and 218 kJ mol –1 respectively. Determine mean B H bond enthalpies in each case. Assume terminal B—H bonds have same strengths, estimate enthalpies of the three centre B H  B bonds in B2 H 6 . Which bonds would you expect to be longer-terminal or bridged one? PROBLEM 174 Enthalpy of combustion of C 6 H 6 (l), C 6 H12 (l) (cyclohexane) H 2 (g ) are –3268, –3920 and –289 kJ/mol respectively. If enthalpy of hydrogenation of cyclohexane is –120 kJ/mol, determine resonance energy of benzene( l). PROBLEM 175 Assuming that mileage of an automobile gets is directly proportional to the heat of combustion of fuel, calculate how many times farther an automobile could be expected to go on one litre gasoline than on 1.0 litre ethanol.

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Assume gasoline to be pure n-octane (ρ = 0.7025 gmL–1 ). Density of ethanol is 0.7893 gmL–1 . ∆H f° of ethanol and octane are −278 kJ mol –1 and –208.4 kJ mol –1 . ∆H f° of CO 2 ( g ) and H 2O( l) are −394 kJ mol –1 and –286 kJ mol –1 respectively.

PROBLEM 176 10 g of propane was burnt in air at 30°C and 1.0 atm pressure. Assume air to be 21.00% O 2 , determine volume of air required for combustion process. If all the heat produced from combustion of 10 g of propane was transferred to 8.00 kg of water at 30°C, determine final temperature of water. C P of water is 418 . Jg –1 K –1 . Also, ∆H f° of propane, CO 2 and H 2O are: −104 kJ mol –1 , – 394 kJ mol –1 and –286 kJ mol –1 respectively.

PROBLEM 177 With the following informations, determine standard state Gibb’s free energy of formation of N 2O 4 ( g ). ∆G° = 86.6 kJ

…(i)

NO( g ) + 12 O 2 ( g ) → NO 2 ( g )

∆G° = − 34.82 kJ

…(ii)

2NO 2 ( g ) → N 2O 4 ( g )

∆G° = − 5.77 kJ

…(iii)

1 2

N 2 ( g ) + 12 O 2 ( g ) → NO( g )

Standard state enthalpy of formations of CO( g ) and CO 2 ( g ) are −111 kJ mol –1 and –394 kJ mol –1 respectively. Bond dissociation energies of O 2 ( g ) and C==O( g ) are 498 kJ mol –1 and 743 kJ mol –1 respectively. Supposing that there is a double bond in CO as two double bond in CO 2 , determine the enthalpy of combustion of CO( g ) and compare it with actual value. Also, explain the difference in calculate and observed value of enthalpies.

PROBLEM 178

PROBLEM 179 One mole of N 2 (g ) and 3.0 mole of H 2 (g ) taken in a flask at 25°C and heated to 450°C. Now pressure was applied on the gaseous mixture which results in conversion of 0.1 mole of N 2 into NH 3 . The gases are then cooled rapidly back to 25°C. Determine the net heat change in this process given the following bond enthalpies: N 2 ( g ) = 944 kJ mol –1 ; H 2 = 436 kJ mol –1 and average N  H bond energy = 388 kJ mol –1 . PROBLEM 180 A 150 cc portion of 0.4 NHCl is neutralized with an excess of NH 4OH in a Dewar vessel with a resulting rise in temperature of 2.36°C. If the heat capacity of Dewar and its contents after the reaction is 1316.7 J/ ° C, calculate heat of neutralization. PROBLEM 181 At 25°C, the heat of solution of anhydrous CuSO 4 in a large volume of water is −66.044 kJ mol –1 , while that of CuSO 4 ⋅ 5H 2O is –11.495 kJ. Determine heat of reaction: CuSO 4 ( s) + 5H 2O → CuSO 4 ⋅ 5H 2O( s)

PROBLEM 182 The integral heats of solution at 25°C, for the various solid modification of CaCl 2 in the indicated quantities of water are shown below: CaCl 2 ( s) + 400 H 2O( l) → CaCl 2 (400 H 2O)

∆H1° = − 4.3 kJ

CaCl 2 ⋅ 2H 2O + 398 H 2O( l) → CaCl 2 (400 H 2O) ∆H 2° = − 41.925 kJ

…(i) …(ii)

CaCl 2 ⋅ 4H 2O + 396 H 2O → CaCl 2 (400 H 2O)

∆H 3° = − 7.65 kJ

…(iii)

CaCl 2 ⋅ 6H 2O + 394 H 2O → CaCl 2 (400 H 2O)

∆H 4° = +19.06 kJ

…(iv)

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Problems in Chemistry

Determine enthalpies of the following hydration reactions: (a) CaCl 2 (s) + 2H 2O → CaCl 2 ⋅ 2H 2O (b) CaCl 2 ⋅ 2H 2O + 2H 2O → CaCl 2 ⋅ 4H 2O (c) CaCl 2 ( s) + 6H 2O → CaCl 2 ⋅ 6H 2O

PROBLEM 183 The enthalpy of following reactions at 25°C are: (i) Na( s) + 12 Cl 2 ( g ) → NaCl( s) ∆H ° = − 410.6 kJ (ii)

H 2 ( g ) + S( s) + 2O 2 ( g ) → H 2SO 4 ( l)

∆H ° = − 810.54 kJ

(iii) 2Na( s) + S( s) + 2O 2 ( g ) → Na 2SO 4 ( s) ∆H ° = −1381.5 kJ (iv)

H 2 ( g ) + 12 Cl 2 ( g ) → HCl( g )

1 2

∆H ° = − 92.21 kJ

From the above thermal data, determine enthalpy of the following reaction: 2NaCl( s) + H 2SO 4 ( l) → Na 2SO 4 ( s) + 2HCl( g ).

PROBLEM 184 Given the following standard state enthalpies of reaction, calculate the standard molar heat of formation of AgCl. (i) Ag 2O ( s) + 2HCl( g ) → 2AgCl(g ) + H 2O( l) (ii) 2Ag( s) + 12 O 2 ( g ) → Ag 2O( s) (iii)

1 2

H 2 ( g ) + 12 Cl 2 ( g ) → HCl( g )

(iv) H 2 ( g ) + 12 O 2 ( g ) → H 2O( l)

∆H ° = − 324.4 kJ ∆H ° = − 30.56 kJ ∆H ° = − 92.21 kJ ∆H ° = − 394 kJ

PROBLEM 185 Draw Lewis structures of hypothetical molecule N 6 (g ) consisting of a six membered ring of nitrogen atom. If its standard enthalpy of formation is 1072 kJ mol –1 , predict the most likely structure. Given: B.E. N 2 = 944 kJ mol –1 , N  N = 163 kJ mol –1 and N == N = 409 kJ mol –1 . PROBLEM 186 A male burns 2000 kJ of energy while jogging for 1.0 hour. If the standard heat of combustion of a typical fat is 38 kJ g –1 and only 70% energy is available for muscular activity. What minimum hours would he need to jog if he wished to lose 0.5 g fat? PROBLEM 187 Strong sunshine bombards the Earth with about 1 kJ m –2s –1 . If a beaker containing ethanol, is placed in sunlight for 10 minutes, 3.24 g of liquid was vaporized. Assuming that all the heat is used for vaporization, not to increase temperature, determine surface area of beaker. Enthalpy of vaporization is 42.6 kJ mol –1 . PROBLEM 188 “Benzene”.

From the following enthalpies values, determine resonance energy of C 6 H 6 ( l)

(i)

+ H2

∆H° = – 38 kJ

(ii)

+ 2H2

∆H° = – 170 kJ

Also, given the resonance energy of 1,3-cyclohexadiene is 70 kJ mol –1 .

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THERMODYNAMICS PROBLEM 189 Suppose that a gas obeys the modified van der Waals’ equation P × (Vm – b) = RT and b = 0.02 L mol –1 . If 0.5 mol of the gas is reversibly compressed from an initial volume of 2 dm 3 to a final volume of 0.5 dm 3 , how much work is done on the system at 27°C. PROBLEM 190 One mole of a monoatomic, ideal gas confined in a 5 L piston fitted cylinder at 300 K is heated such that its temperature increased to 400 K but at the same time gas also expanded to a volume of 8 L. Calculate change in enthalpy of the system. PROBLEM 191 One mole of a monoatomic ideal gas confined in a 5 L, piston fitted cylinder at 300 K is heated to 800 K as well as allowed to expand to a volume of 8 L simultaneously. Calculate change in enthalpy of the system. PROBLEM 192 100 g of nitrogen gas at 300 K are held by a piston under 30 atmosphere. The pressure is suddenly released to 10 atmosphere and gas is allowed to expand adiabatically. If CV = 20.8 JK –1 mol –1 , calculate ∆S System . PROBLEM 193 Calculate entropy change when 0.5 L of an ideal gas (CV =12.6 JK –1 mol –1 ) at 300 K and one atmosphere is allowed to expand to double its volume and simultaneously heated to 373 K. PROBLEM 194 10 g of ice at 0°C are added to 20 g water at 90°C in a thermally insulated flask of negligible heat capacity. The heat of fusion of ice is 6 kJ/mol. What is the final temperature, ∆S System ? C p = 75.42 JK –1 mol –1 . PROBLEM 195 One mole of a supercooled liquid water at – 10°C and one atmosphere turns into ice at – 10°C. Calculate entropy change for the system. C p for liquid water and ice are 75.42 and 37.2 JK –1 mol –1 respectively. PROBLEM 196 In an open beaker at 27°C and one atm pressure, 100 g of zinc are caused to react with dilute sulphuric acid. Calculate the work done by the liberated hydrogen gas assuming it to behave ideally. What would be the work done if the reaction took place in a sealed vessel? PROBLEM 197 A balloon is 0.5 m in diameter and contains air at 25°C and 1 bar pressure. It is then filled with air isothermally and reversibly until the pressure reaches to 5 bar. Assume that pressure is proportional to the diameter of the balloon, calculate (a) final diameter and (b) work done in the process. PROBLEM 198 One mole of an ideal gas initially at 10 bar and 300 K is allowed to expand against a constant external pressure of 2.00 bar to a final pressure of 2 bar. During this process, the temperature of this gas falls to 250 K. Construct a reversible path connecting this initial and final state as a combination of reversible isothermal expansion followed by reversible adiabatic expansion so that the final state is attained and calculate work done by the system in attaining the final state. C vm = 3/ 2 R . PROBLEM 199 With the temperature maintained at 0°C, 2 mole of an ideal gas are allowed to expand against a piston that supports 2.0 bar pressure. The initial pressure of the gas is 10 bar and the final pressure 2 bar. (a) How much energy is transferred to the surrounding during the expansion? (b) What is the change in internal energy and enthalpy of the gas? (c) How much heat the gas has absorbed?

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PROBLEM 200 A gas behaving ideally was allowed to expand reversibly and adiabatically to twice its volume. Its initial temperature was 25°C and C vm = (5 / 2) R , calculate ∆E m and ∆H m . PROBLEM 201 One mole of a gas at 300 K is compressed isothermally and reversibly from an initial volume of10 dm 3 to a final volume of 0.2 dm 3 mol –1 . Calculate work done on the system if the equation of state of the gas is (Vm – b) P = RT with b = 0.03 dm 3 mol –3 . PROBLEM 202 One mole of a gas at 100 K is compressed isothermally from an initial volume of 20 dm 3 to a final volume of 5 dm 3 . Calculate the work done on the system if the equation state is:  a   P + 2  Vm = RT Vm  

where a = 0.384 m 6 Pa mol –1

PROBLEM 203 Find q,W, ∆U and ∆H if 2.0 g of He undergoes a reversible isobaric expansion from 20 to 40 L at 0.8 atm pressure followed by reversible isochoric heating till pressure reaches to 1.0 atm. Depict the change of state on a P-V diagram. PROBLEM 204 One mole of liquid water at 30°C is adiabatically compressed, pressure increasing from 1.0 atm to 10.0 atm. Since, liquids and solids are rather incompressible, it is a fairly approximastion to take V constant. Calculate q, ∆U and ∆H for the process. C p of H 2O( l) = 75.42 JK –1 mol –1 . PROBLEM 205 For a perfect gas, C v = 2.5 R . 2.0 moles of this gas undergoes following change of state: (a) A reversible isobaric expansion from 1.0 atm, 20 L to 1.0 atm, 40 L. (b) A reversible isochoric change from 1.0 atm, 40 L to 0.5 atm, 40 L (c) A reversible isothermal compression from 0.5 atm, 40 L to 1.0 atm, 20 L. Sketch each process on the P-V diagram and calculate : q, W, ∆U and ∆H. PROBLEM 206 A sample of an ideal gas underwent an adiabatic expansion from 298 K, 15 bar to 2.5 bar against a constant external pressure of 1.0 bar. What is the final temperature of the system and work done by the system, assume C vm = 2.5 R? PROBLEM 207 A gas behaves ideally and its C v is given by: C v = 21.52 + 8.2 × 10 –3 T (all parameters in SI unit). A sample of this gas is initially at T1 = 300 K, P1 = 10 bar andV1 = 1 L. It is allowed to expand until P2 = 1 atm and V2 = 10 L. What are ∆U and ∆H for this process? Could the process be carried out adiabatically. Calculate C p at 300 K. PROBLEM 208 The entropy change of argon is given to a good approximation by the expression: Sm / JK –1 mol –1 = 36.36 + 20.79 ln T Calculate change in Gibb’s free energy of one mole of argon gas if it is heated at constant pressure from 25°C to 50°C.

PROBLEM 209 Initially at 300 K and 10 atm pressure, 1.0 mole of an ideal gas is allowed to expand adiabatically against a constant pressure of 4.0 atm until equilibrium is established. Assume the gas to be ideal with: C p = 28.58 + 1.76 × 10 –2 T Calculate ∆U , ∆H, and ∆S.

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PROBLEM 210 An ideal gas expand against a constant external pressure of 2.0 atmosphere from 20 L to 50 L and absorb 20 kJ of energy from surrounding. What is the change in internal energy of the system? 10 atm, where V is volume of V gas at each stage of expansion. Further during expansion from volume 10 L to 100 L, the gas undergoes a change in internal energy of 420 J. How much heat is absorbed by the gas during expansion?

PROBLEM 211 A gas expands against a variable pressure given by P =

PROBLEM 212 Three moles of an ideal gas is heated at constant pressure of one atmosphere from 27°C to 127°C. If C v is expressed as: C v = 30 + 14 × 10 –3 T JK –1 mol –1 , determine W, ∆E and q. PROBLEM 213 One mole of an ideal gas at state A (500 K, 5.0 atm) is cooled at constant volume to B (300 K) and then expanded isothermally and reversibly to C and finally compressed adiabatically to A. 5 Sketch the change on a P-V diagram and determine the net work done in this cyclic process. γ = . 3 PROBLEM 214 One mole of an ideal gas is subjected the following change of state: Reversible Isothermal expansion

A(5.0 atm, 500 K) → B Isochoric cooling

Reversible Adiabatic compression

B → C (300 K) → A Depicting the above mentioned change on a P-V diagram, determine the net work done in the cyclic process. C v =1.5 R .

PROBLEM 215 One mole of an ideal gas initially at A (300 K and 5.0 bar) is heated at constant pressure to double its volume (B). The gas is then expanded isothermally and reversibly to a new state C. The gas is then cooled at constant pressure to another new state D (200 K) and finally compressed adiabatically and reversibly to A. Depicting on a P-V diagram, determine the net work done in the above cyclic process. C v =1.5 R . PROBLEM 216 One mole of an ideal gas at A (500 K, 5.0 bar) is expanded isothermally and reversibly to a new state B and then cooled at constant pressure to C (250 K) and finally compressed adiabatically and reversibly to A. Depicting on a P-V diagram, determine the net work done in the cyclic process. C v =1.5 R . PROBLEM 217 One mole of an ideal gas at 500 K and 10 bar, defined by state A is allowed to expand isothermally and does a work equal to 4200 J. Construct a combination of initial irreversible expansion upto 2.0 bar followed by reversible expansion so that final state is reached and determine final pressure of the gas. PROBLEM 218 One mole of an ideal gas at 500 K and 10 bar is allowed to expand till the final pressure falls to 1.0 atmosphere and final temperature falls to 250 K. Construct a combination of reversible path of: (a) initial adiabatic expansion followed by isothermal expansion so that final state is reached and determine the total work done. C v =1.5 R . (b) Determine the work by reversing the order of combination in (a) and compare the two work done.

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Problems in Chemistry

PROBLEM 219 One mole of an ideal gas at 300 K and 1.0 atmosphere is heated to 500 K and expanded simultaneously to 36 litre. Determine ∆H assuming heat capacity to be independent of temperature and C v =1.5 R . PROBLEM 220 One mole of a gas initially at 300 K is heated to 500 K. Determine the Gibb’s free energy change, ∆G if S = 1.5 + 3 × 10 –3 T [JK –1 mol –1 ] and C v =1.5 R . PROBLEM 221 One mole of an ideal gas at 300 K and 1.0 atmosphere is heated as well as expanded simultaneously to 500 K and 2.0 atmosphere. Determine ∆S if C v = 2.5 R . PROBLEM 222 One mole of an ideal gas is taken in a one litre sealed flask at 300 K and heated till the pressure becomes equal to 40 atmosphere. If C v = 12 + 28 × 10 –3 T (in SI unit), determine ∆S. PROBLEM 223 A gaseous reactant A forms two different product in a parallel reaction B and C as follows: A → B ;

∆H ° = – 3 kJ, ∆S ° = 20 JK –1

A → C ;

∆H ° = – 3.6 kJ, ∆S ° = 10 JK –1

Discuss the relative stability of B and C on the basis of Gibb’s free energy change at 27°C.

PROBLEM 224 One mole of an ideal gas contained in a sealed flask at 1.0 bar is heated from 27°C to 127°C. Determine ∆G if: S (JK –1 ) = 10 +12 × 10 –3 T . PROBLEM 225 Two moles of NO 2 is heated at constant volume from 27°C to 127°C and C p (JK –1 mol –1 ) = 28 + 31 × 10 –3 T Determine ∆S.

PROBLEM 226 Two moles of an ideal gas is expanded isothermally and irreversibly at 27°C from volumeV1 to 2.5 V1 and 4.17 kJ heat is absorbed from surroundines. Determine ∆S sys , ∆S surr and ∆S univ . PROBLEM 227 One mole of He(g) is mixed isothermally and reversibly with 2.0 mole of Ne(g). Determine ∆S. PROBLEM 228 C vm for an ideal gas is 2.5 R and it is independent of temperature. If 2.0 moles of this gas is subjected to the following change of state : Reversible

A (1.0 bar , 20 L ) → B (1.0 bar, 40 L) Isobaric heating

Isochoric cooling

Reversible Isothermal compression

B → C (0.5 bar, 40L) → A Representing the above change of states on a P-V diagram, determine the net work done.

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PROBLEM 229 An ideal gas has C vm = a + bT where a = 25 JK mol and b = 0.03 JK mol –1 . If 3.0 moles of this gas is subjected to a thermodynamic change of state from A (300 K, 2.0 bar) to B (500 K, 3.0 bar), determine q, W, ∆E, ∆H and ∆S. –1

–2

–1

PROBLEM 230 One mole of an ideal gas defined by state A (400 K, 5.0 bar) is heated at constant pressure to B (500 K) and then cooled at constant volume to C. The gas is then expanded isothermally and reversibly to a new state D (1.0 bar) and finally compressed adibatically to A. Depicting on a P-V diagram, determine the net work done in this cyclic process. C v =1.5 R .. PROBLEM 231 One mole of an ideal gas is subjected to the following change of state: Reversible Isothermal expansion

Isochoric cooling

A(500 K, 5.0 bar ) → B → C (250K, 1.0 bar) Reversible

C → D (3.0 bar); C v =1.5 R Adiabatic compression

Depicting the above change on a P-V diagram, determine the net work done.

PROBLEM 232 The entropy of vaporization of benzene is 85 JK –1 mol –1 . (a) Estimate the enthalpy of vaporization of benzene at its normal boiling point of 80°C. (b) Determine the entropy change of surroundings when 100 g benzene vaporizes at its normal boiling point. PROBLEM 233 The entropy of vaporization of acetone is 85 JK –1 mol –1 . (a) Estimate enthalpy of vaporization of acetone at its normal boiling point 56°C. (b) Determine entropy change of surrounding if 100 g of acetone condenses at its boiling point. With the help of following reduction reactions:

PROBLEM 234

at

1000

K,

given

TiO 2 ( s) + 2C( s) → Ti( s) + 2CO( g )

…(i)

TiO 2 ( s) + C( s) → Ti( s) + CO 2 ( g )

…(ii)

∆Gf° (CO) = − 200 kJ mol –1 ,

∆Gf° (CO 2 ) = − 396

kJ mol –1

and

∆G f° TiO 2 ( s) = − 762 kJ mol –1 , determine which will be the predominant mode of reduction of TiO 2 ( s) at 1000 K.

PROBLEM 235

The reaction for the production of synthetic fuel ‘water gas’ from coal is: C( gr ) + H 2O( g ) → CO( g ) + H 2 ( g )

Standard molar entropies of C( gr ), H 2O( g ), CO( g ) and H 2 ( g ) are 5.7, 70, 190 and 131 JK –1 mol –1 respectively. Also, standard enthalpy of formations of H 2O and CO are –242 and –111 kJ mol –1 respectively. Determine the standard reaction free energy of reaction at 27°C. Also, predict about the spontaneity and effect of temperature on direction of reaction.

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Problems in Chemistry

PROBLEM 236 follows:

The thermodynamic informations for isomerization of alkene (C 4 H 8 ) at 300 K are as cis-2-butene

H 3C

trans-2-butene

CH 3 C==C

H

H 3C

1

H

∆G ° = 66 kJ mol –1 f ∆H ° = − 7 kJ mol –1 f

H C==C

H

2

3

H

CH 3

∆G ° = 63 kJ mol –1 f ∆H ° = −11.2 kJ mol –1 f

CH 3 C==C

H

CH 3 2-methylpropene ∆G °f = + 58 kJ mol

If the temperature of the above system is increased to 400 K and equilibrium was allowed to re-establish, mole percentage of trans-2-butene at new equilibrium was 18. Determine ∆H ° and ∆S ° for the isomerization reactions below: cis-2-butene

2-methylpropene

trans-2-butene

2-methylpropene

PROBLEM 237 At a temperature above 65 K, decarboxylation of acetic acid, (i.e., loss of CO 2 ) becomes spontaneous. If ∆H f° of CH 3COOH, CO 2 ( g ) and CH 4 are −484.5, − 394 and − 74.8 kJ/mol respectively, determine standard state entropy change ( ∆S ° ) for the decarboxylation reaction. What is the driving force for getting this reaction to proceed? PROBLEM 238

For the reaction:

H 2 ( g ) + CO 2 ( g ) H 2O( g ) + CO( g ) ∆G at 2000 K is 2540 J, where partial pressures of the species are PH = 0.25, PCO2 = 0.78, 2 PH O( g ) = 0.66 and PCO = 1.2 atm respectively. Determine equilibrium composition of the gaseous 2 mixture.

PROBLEM 239

Consider the thermal decomposition of solid CaCO 3 as:

CaCO 3 ( s) CaO( s) + CO 2 ( g ). The equilibrium vapour pressure of CO 2 at 700°C and 950°C are 22.6 and 1830 mm of Hg. Calculate ∆H ° and ∆S ° for the reaction.

PROBLEM 240 A certain reaction is spontaneous at 72°C. If the enthalpy change for the reaction is 19 kJ, what is the minimum value of ∆S for the reaction? PROBLEM 241 The internal engine of a 1200 kg car is designed to run on octane whose enthalpy of combustion is 5510 kJ/mol. If the car is moving up a slope, calculate the maximum height to which the car can be driven on 2.0 gallon of the fuel. Assume the cylinder temperature is 2200°C and the exit temperature is 760°C and ignore all form of friction. The mass of 1.0 gallon of fuel is 3.1 kg. PROBLEM 242 One gram sample of oxygen undergoes free expansion from 0.75 L to 3.0 L at 298 K. Calculate ∆S , q, W , ∆H and ∆E.

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PROBLEM 243 A 550 ml sample of an ideal gas at 300 K exerts 3 atm. The thermodynamic state of the system changes in a process. In the final state, P = 3.5 atm and V = 730 mL. Calculate ∆S and ∆E and ∆H, C vm = (5 / 2) R . PROBLEM 244 A sample of 0.0133 mole of an ideal gas, initially at 5.00 atm, expands isothermally and reversibly from 3.00 L to 10 L. Calculate ∆S , ∆G and ∆H. PROBLEM 245 One mole of an ideal gas originally at a volume of 8.00 Lit. at 1000 K, is allowed to expand adiabatically until final volume is 16.00 Lit. For the gas C v =1.5 R . Calculate values of ∆S for the process when: (a) The expansion takes place reversibly. (b) The expansion takes place against a constant pressure of 3.00 atm. (c) The change in volume involves a free expansion. PROBLEM 246 One mole of an ideal gas at 0°C and 1.0 atm pressure is mixed adiabatically with one mole of a different gas at 100°C and 1.0 atm to yield a mixture. If C P for each gas is (5 / 2) R, determine ∆S (mixing). PROBLEM 247

For chloroform gas C PM is expressed as: C PM = 24.9 + 14.8 × 10 −2 T − 9 × 10 −5 T 2 JK –1 mol –1 .

Assuming this gas to be ideal, determine entropy change involved in heating 2.0 mole of gas from volume 100 L at 500 K to a volume of 70 Lit. at 700 K.

PROBLEM 248

For N 2 ( g ), entropy function as a function of temperature is expressed as: S = 25.1 + 29.3 ln T

Determine Gibb’s free energy change ∆G of one mole of nitrogen if it is heated from 298 K to 348 K at 2.0 atm pressure.

PROBLEM 249 One mole of an ideal gas initially at 400 K and 10 atm, is adiabatically expanded against a constant pressure of 5.0 atm until equilibrium is attained. If CV = 18.8 + 0.021T JK –1 mol –1 , determine ∆E , ∆H and ∆S . PROBLEM 250 Molar volume of C 6 H 6 (l) is 89 c.c. at 27°C and 1.0 atm pressure. Assuming the volume to be constant, determine ∆G for compression of 5.00 moles of liquid benzene from 1.0 atm to 100 atm. PROBLEM 251 One mole of an ideal gas at 25°C is subjected to a reversible isoentropic expansion until final temperature reached to 75°C. If the initial pressure was 1.0 atm, determine final pressure CV = (3 / 2) R . PROBLEM 252 A flask containing 1.00 mol of N 2 at 4.00 atm and 298 K was connected to a flask containing 1.00 mol of N 2 gas at 2.00 bar and 298 K. The gases were allowed to mix isothermally. Determine the entropy change for the system. PROBLEM 253 One mole of solid iron was vaporized in an oven at 3500 K. If iron boils at 3133 K and enthalpy of vaporization is 349 JK –1 mol –1 , determine ∆S system , ∆S surroundings and ∆S universe .

Problems

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ATOMIC STRUCTURE PROBLEM 254 Determine the number of revolutions made by an electron in one second in the 2nd Bohr orbit of H-atom. PROBLEM 255 (a) What electronic transition in He + ion would have the same wavelength as the first Lyman line of H-atom? (b) Which electronic transition in Li 2+ ion would have same wavelength as 3rd transition in Lyman series of H-atom? PROBLEM 256 Energy required for the excitation of H-atom from its ground state to the 2nd excited state is 2.67 times smaller than dissociation energy of H 2 ( g ). If H 2 ( g ) placed in a 1.0 litre flask at 27°C and 1.0 bar is to be excited to their 2nd excited state, what will be the total energy consumption? PROBLEM 257 A H-like species is in their excited state (A) and absorb a photon of 3.868 eV and get excited into a new state B. The electron from B in returning back to lower energy orbit, can give a maximum of 10 emission lines with different wavelengths, some lines having energy greater than 3.868 eV, some lines with energy equal to 3.868 eV but two lines with energy less than 3.868 eV. Determine the orbit number of states A and B and ionization energy of the species. PROBLEM 258 What is the speed of an electron whose de Broglie wavelength is 0.1 nm. By what potential difference, must have such an electron be accelerated from an initial speed zero. PROBLEM 259 Consider a colloidal particle of mass 6 × 10 –16 kg. Suppose that its position is measured to 1 nm accuracy, calculate the uncertainty in its velocity. PROBLEM 260 The wavelength of a photoelectric threshold of a metal is 230 nm. Determine the kinetic energy of photoelectron ejected from the surface by UV radiation emitted from the second largest wavelength of transition of electron in Lyman series of atomic spectrum of H-atom. PROBLEM 261 A hydrogen like species with atomic number Z is in higher excited state ‘n’ and emits photons of energy 25.7 and 8.7 eV when makes a transitions to 1st and 2nd excited state respectively. Determine ‘n’ and ionization energy of the species. PROBLEM 262 Light from a discharge tube containing H-atoms in some excited state, falls on the surface of sodium metal. The kinetic energy of the fastest photoelectron was found to be 10.93 eV. If He + ions were present in the same excited state, the kinetic energy of the fastest photoelectron would have been 49.18 eV. Determine the excited state orbit number and work function of Na. PROBLEM 263 Certain sun glasses have small crystals of AgCl incorporated in the lenses. When the lenses are exposed to light of appropriate wavelength, the following reaction occurs: AgCl → Ag + Cl The Ag atoms formed produce a uniform gray colour that reduces the glare. If ∆H for the reacton is 248 kJ, what maximum wavelength would be required to induce this process?

PROBLEM 264 Ozone in the stratosphere absorbs the harmful radiation from the sun by undergoing decomposition: O 3 → O + O 2

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Calculate the maximum wavelength of photons that possess energy to cause the decomposition of O 3 . Standard enthalpy of formation of ozone is 142 kJ/mol and dissociation energy of oxygen is 498 kJ/mol.

PROBLEM 265 The retina of human eye can detect light when radiant energy incident on it is at least 4.0 × 10 –17 J. For light of 600 nm wavelength, how many photons must incident on retina for clear vision. PROBLEM 266 An electron in the excited state in a hydrogen atom can return in its ground state in two different ways : (a) via a direct transition in which a photon of wavelength λ 1 is emitted. (b) via an intermediate excited state reached by emission of a photon of wavelength λ 2 . This intermediate excited state then decays to the ground state by emitting another photon of wavelength λ 3 . Derive an equation that relates λ 1 , λ 2 and λ 3 . PROBLEM 267 Calculate wavelength of He atom whose speed is equal to the r.m.s. at 20°C. PROBLEM 268 How many photons at 600 nm must be absorbed to melt 500 g of ice? On average how many H 2O molecules does one photon convert from ice to liquid water? Standard enthalpy of fusion of ice is 6.0 kJ mol –1 . PROBLEM 269 The energy needed for the reaction Li(g ) → Li 3+ (g ) + 3e is 19612 kJ/mol. If the first ionization energy of lithium is 520 kJ/mol, determine its second ionization energy. PROBLEM 270 Alveoli are the tiny sacs of air in the lungs whose average diameter is 50 µm. Consider an oxygen molecule trapped within a sac. Calculate the uncertainty in the velocity of oxygen molecule. (Hint : The maximum uncertainty in the position of molecule is given by the diameter of sac.) PROBLEM 271 The sun is surrounded by a white circle of gaseous material known as Corona, which becomes visible during total eclipse of the sun. The temperature of Corona is in million degree Celsius which is high enough to breakup molecules and remove some or all electrons from the atom. One line having wavelength equal to the emission line of Si14+ has been observed in this Corona. On this observation what could be the estimated temperature of Corona. PROBLEM 272 A H-like neutral species is in some excited state (A) and on absorbing a photon of energy 3.066 eV gets promoted to a new state B. When the electron from state B return back, photons of a maximum ten different wavelengths can be observed in which some photons of energy smaller than 3.066 eV. Some of the equal energy and only four photon having energy greater than 3.066 eV. Determine the orbit number of states A and B and ionization energy. PROBLEM 273 Calculate average energy of an oscillator with frequency1014 Hz at 1000 K. Boltzmann constant K = 1.38 × 10 −23 JK –1 . PROBLEM 274 The threshold wavelength for photoelectric emission in tungsten is 2300 Å. What wavelength of light must be used to eject electrons with a maximum energy of 1.5 eV? PROBLEM 275 A beam of X-rays is scattered by loosely bound electrons at 45° to the direction of beam. The wavelength of scattered beam is 0.22 Å. What is the wavelength of the incident beam? PROBLEM 276 By what potential difference, an electron at rest must be accelerated to have a de-Broglie wavelength of 0.38 Å? PROBLEM 277 Determine the de-Broglie wavelength associated with an electron in the 3rd Bohr’s orbit of He + ion.

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Problems

PROBLEM 278 If uncertainty (maximum) in location of an electron in a hydrogen atom is 0.1 mm, what would be the minimum uncertainty in its speed if measured simultaneously? PROBLEM 279 An automobile of mass 500 kg is moving with speed of 50 ± 0.001km hr –1 . Determine uncertainty in position of moving automobile and interprete the result. PROBLEM 280 If a photon of wavelength 200 pm strikes an atom and one of the inner bound electrons is ejected out with a velocity of 2 × 10 7 ms –1 , calculate the energy with which it is bound to nucleus? PROBLEM 281 When radiation of wavelength 253.7 nm falls on a copper surface, electrons are ejected. Calculate work function if the stopping potential is 0.5 V. PROBLEM 282 The size of an atomic nucleus is 10 −14 m. Calculate uncertainty in momentum of an electron if it were to exist inside the nucleus. PROBLEM 283 Calculate de-Broglie wavelength of a hydrogen atom with translational energy corresponding to a temperature of 27°C. Calculate the electrostatic potential energy of two electrons separated by 3 Å in

PROBLEM 284 vacuum?

The wave function for electron in ground state of hydrogen atom is

PROBLEM 285 1 − ( πa 03 ) 2

−

r a0

, where “a 0 ” is radius of Bohr’s orbit. Calculate the probability of finding the ψ1s = e electron somewhere between 0 and 2a 0 . The normalized wave function of the hydrogen atom for the 1s orbital is

PROBELM 286 ψ1s =

( πa 03 )

−

1 2

PROBLEM 287 −

1 2

−

e

r a0

. Show that in such a state the most probable distance from nucleus to electron is a 0 .

The wave function for the electron in the ground state of H-atom is − r/a

ψ1s = ( πa 03 ) e 0 . What is the probability of finding an electron somewhere inside a small sphere of radius 10 −12 m centered on nucleus?

PROBLEM 288

Determine de-Broglie wavelength associated with He atom at room temperature 25°C.

PROLBEM 289 At what temperature the translational kinetic energy of atomic hydrogen equal that for n2 = 1 to n2 = 2 transition. PROBLEM 290 A hypothetical element “positronium” consists of an electron moving in space around a nucleus consisting a positron. Using the Bohr’s atomic model, determine the first Bohr radius. Positron is a subatomic particle similar to electron in all respect except possessing a positive charge.

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CHEMICAL BONDING PROBLEM 291 Discuss the bonding of the following species: – + + + CO 2 , COS, CO 2– 3 , SiCl 4 , PCl 5 , NH 4 , PH 4 , PCl 4 , PCl 6 , SF6 , IF7 . PROBLEM 292 Discuss the bonding of the following species: SO 2 , NH 3 , H 2O, PCl 3 , NCl 3 . PROBLEM 293 Discuss the relative bond angles in the following species: (a) NH 3 , PH 3 , AsH 3 , SbH 3 and BiH 3 . (b) H 2O, H 2S, H 2Se, H 2Te. PROBLEM 294 Arrange the following species in the increasing order of their bond angles, explaining the reason for your order: H 2O, NH 3 , H 2Se, PH 3 , AsH 3 .

PROBLEM 295 Discuss the bonding of: ClF3 , COCl 2 , Cl 2O, OF2 , ICl –4 , IF5 , IOCl –4 , IOCl +4 .

PROBLEM 296 Discuss the bonding of: N –3 , I –3 , BrF5 , IF4– , BF4– , ICl +2 , ICl –2 and PF5 . PROBLEM 297 The energy necessary to break similar bond is not always equal, it varies from molecule to molecule as: NCl 3 → NCl 2 + Cl ∆H = 375 kJ/ mol and NOCl → NO + Cl Discuss the difference.

∆H =158 kJ/ mol

PROBLEM 298 Draw the shape of the following species: – – – AsF5 , AsF2+ , SnCl –3 , NOF, SO 2– 3 , TeF5 , GeF3 , SCl 2 , SbCl 6 . PROBLEM 299 Discuss the bonding of following: ClF4+ , FClO 3 , F2ClO + , SeO 2– 3 . PROBLEM 300 Discuss the relative polarity of the following species: (a) SCl 2 , BF3 , ICl 3 , POCl 3 , PCl 5 . (b) XeF2 , SF4 , XeO 3 , SnCl 4 , PCl 4 F. PROBLEM 301 Arrange the following compounds in the increasing order of their melting points explaining reason for your order: Li 2O, LiF, Li 3 N.

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Problems

PROBLEM 302 Arrange the following in the increasing order of their thermal stability with suitable explanations: (a) CaCO 3 , BaCO 3 , MgCO 3 , Na 2CO 3 , Al 2 (CO 3 ) 3 , BeCO 3 . (b) CaCO 3 , CaSO 4 and CaSO 3 . (c) MgSO 4 , BaSO 4 , Al 2 (SO 4 ) 3 , SrSO 4 . (d) CaC 2O 4 , K 2C 2O 4 , FeC 2O 4 and CaCO 3 . PROBLEM 303 The molecule XeF4 has two lone pairs at the central atom, instead the bond angle is 90° as expected from its geometry, explain. PROBLEM 304 Arrange the following species in the increasing order of their ionic character: NaCl, CaCl 2 , AlCl 3 , BaCl 2 , MgCl 2 and GaCl 3 . PROBLEM 305 In the following pairs of molecules, select one, which has greater bond angle, and explain the reason for your answer. (a) NH 3 or NF3 (b) PH 3 or PF3 (c) AsH 3 or AsF3 (d) AsF3 or AsCl 3 (e) H 2O or F2O PROBLEM 306 Arrange the following sets of molecules in increasing order of bond angles providing appropriate explanation for your order: (a) H 2O, H 2S, NH 3 and PH 3

(b) CH –3 , SiH –3 and GeH –3

PROBLEM 307 Draw the shape of the following species indicating bond angles and distortion (if present): (e) ICl –2 (d) ICl +2 (c) PCl 2 BrF2 (b) PF2Cl 3 (a) PF3Cl 2 PROBLEM 308 Considering the molecules in question 223, arrange them in the increasing order of their dipole moments. PROBLEM 309 Arrange the following molecules in the increasing order of their polarity: CH 3Cl, CH 2Cl 2 , CHCl 3 and CCl 4 . PROBLEM 310 Discuss the bonding in the following molecules with respect to hybridisation of central atom, shape and bond angles: (a) ClF3O 2 (d) I +3 (b) XeOF4 (c) IOCl –4 PROBLEM 311 Discuss the bonding of PCl 5 in gas phase and in solid phase. PROBLEM 312 Dimethyl ether has tetrahedral geometry of hybrid orbital at central atom whereas disilyl ether has triangular planar geometry of hybrid orbital at central atom. Explain. PROBLEM 313 Draw the shape of the following molecules: (a) XeF2

(b) XeF4

(c) XeF6

(d) XeOF2

(e) XeOF4

(f) XeO 2 F2

(g) XeO 3 F2

(h) I 2Cl 6

(i) I 2 Br 2Cl 4

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PROBLEM 314 Draw the shape of the following molecules: (b) (CH 3 ) 3 P(CF3 ) 2 (a) (CH 3 ) 2 P(CF3 ) 3 PROBLEM 315 In sp 3 d-hybridized phosphorus atom in trigonal bipyramidal molecule, will the atom have a greater electronegativity when bonding through equatorial or axial orbitals? Explain. PROBLEM 316 B—F bond distance in BF3 is shorter than the same in BF4– , explain. PROBLEM 317 In CCl 4 , C—Cl bond length is the sum of the covalent radii of carbon and chlorine whereas in SiCl 4 , Si—Cl bond distance is smaller than the sum of the covalent radii of Si and Cl, explain. PROBLEM 318 Arrange the following in increasing order of their Lewis acid strength with proper reasoning. BCl 3 , BI 3 , BF3 and BBr 3 . PROBLEM 319 In gas phase, N(CH 3 ) 3 acts as a good Lewis base but N(SiH 3 ) 3 doesn’t, explain. PROBLEM 320 The molecule CHBrCHBr can have two different structures in which one is polar and other is non-polar. Draw the structures labelling them as polar and non-polar. PROBLEM 321 Arrange the following compounds in order of increasing dipole moment: Cl

Cl Cl

Cl

Cl

(a)

Cl

Cl

Cl

Cl

Cl

Cl

(b)

(c)

(d)

PROBLEM 322 Although both carbon and silicon are in the same group of periodic table, very few Si==Si bonds are known. Account for the instability of Si==Si in general. PROBLEM 323 Molecule N 2 F2 can acquire two different structures in which one is polar and other is non-polar. Draw them labelling polar and non-polar. PROBLEM 324 Compound 1,2-dichloro ethane is a non-polar whereas cis-1,2-dichloro ethene is polar, explain the difference. PROBLEM 325 Discuss the dipole moment of the following molecules in view of chemical bonding: Cl H Cl H and C==C==C==C C==C==C H Cl H Cl PROBLEM 326 Draw all possible structures for the molecule C 2 H 2Cl 2 and rank them in increasing order of their dipole moment.

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PROBLEM 327 Draw shape of a hypothetical molecule N 6 in which nitrogen atoms are part of a six membered ring and has: (a) two pi-bonds, (b) three pi-bonds. PROBLEM 328 Sodium chloride (NaCl) and sodium fluoride (NaF) both crystallizes in same type of unit cell. Which is expected to have higher lattice energy and why? PROBLEM 329 The bond energy in NO is 632 kJ mol –1 and that of each N  O bond in NO 2 is 469 kJ mol –1 . Explain. PROBLEM 330

In the air, NO can react with NO 2 . What is the most likely structure of product?

PROBLEM 331 Draw the Lewis structure of the following species: (a) SO 2Cl + , (b) S 2 F4 (contain S—S bond). PROBLEM 332 The heteronuclear diatomic ion CN – has an orbital structure similar to that of N 2 . How will the fact C has an electronegativity different from that of N affect the energy level diagram. PROBLEM 333 From the following pair of molecules, select one which will be more soluble in a polar solvent and explain the reason for your choice. (a) SiF4 and PF3 , (b) SF6 and SF4 , (c) IF5 and AsF5 PROBLEM 334 In addition to forming σ- and π-bonds similar to those formed by p-orbitals, d-orbitals may overlap in δ-bonds with two nodal planes cutting through the internuclear axis. Draw the overlap diagrams showing how d-orbitals can overlap in these three ways.

CHEMICAL EQUILIBRIUM PROBLEM 335 An equilibrium mixture of: XeF2 ( g ) + OF2 ( g )

XeOF2 ( g ) + F2 ( g )

was found to contain 0.6 mole of XeF2 ( g ), 0.3 mole of OF2 ( g ), 0.1 mole of XeOF2 ( g ) and 0.4 mole of F2 in a one litre container. How many moles of OF2 must be added to increase [XeOF2 ] to 0.2 M?

PROBLEM 336 A compound HB is formed from H and B according to the following reaction: H+B HB. A solution was prepared by dissolving 0.1 mole of H and 0.1 mole of B in enough water to make the total volume equal to one litre. After equilibrium had been reached, it was found that 20% of H had been reacted. What are the equilibrium concentration of H, B and HB? What is the equilibrium constant, K c for this reaction? PROBLEM 337 2.0 g of ammonium chloride was heated in a one litre flask to 300°C. From the measurement of pressure, it was found that 98% of ammonium chloride was dissociated. If to this flask 2.0g of dry ammonia was added, what would be the percentage dissociation?

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PROBLEM 338 H 2 (g ) + I 2 (g ) 2HI( g ) If the system is equilibrated at [H 2 ] = 0.5, [I 2 ] = 0.5 and [HI] = 1.23 in a one litre flask and suddenly 0.6 mole of HI is removed, what will be the new equilibrium composition when the equilibrium is re-established? PROBLEM 339 The equilibrium constant K p for the gas phase decomposition of ter-butyl chloride is 3.45 at 500 K: (CH 3 ) 3 CCl( g ) (CH 3 ) 2 C = CH 2 ( g ) + HCl( g ) Calculate molar concentration of reactants and products in the equilibrium mixture obtained by heating 1.0 mole of ter-butyl chloride in a 5.0 L container at 500 K. PROBLEM 340 A 79.2 g chunk of dry ice and 30 g of graphite were placed in an empty 5.0 L container, and the mixture was heated to achieve equilibrium. The reaction is: CO 2 ( g ) + C( s)

2CO( g )

What is the value of K p at 1000 K if the gas density at this temperature is 16.3 g/L. What is the value of K p at 1100 K if the gas density at this temperature is 16.9 g/L. Also calculate the enthalpy of the reaction.

PROBLEM 341 The reaction: N 2 + 3H 2 2NH 3 is started with an equal number of moles of N 2 and H 2 . Calculate mole fraction of NH 3 at 723 K and a total pressure of 0.5 atm given K p = 6 × 10 –3 . 1 H2 + O2 2 is 8.7 ×10 –11 . For a significant production of hydrogen, which may be used as fuel, at least 15% of water should be dissociated. To what temperature should you heat the water? The enthalpy of reaction is 57.8 kcal. For a rough estimate assume that enthalpy is independent of temperature and P =1.0 atm.

PROBLEM 342 At 1000 K, the equilibrium constant for the dissociation of water, H 2O

PROBLEM 343 In a study of the equilibrium: H 2 + I 2 2HI, 1.0 mole of H 2 and 3 mole of I 2 gave rise at equilibrium to x mole of HI. Addition of a further 2 mole of H 2 gave an additional x mole of HI. What is x? What is K at the temperature of experiment? PROBLEM 344 The equilibrium constant for the reaction: H 2 (g ) + I 2 (g ) 2HI ( g ) is 20 at 40°C, and vapour pressure of solid iodine is 0.1 bar at that temperature. If 12.7 g of solid iodine are placed in a 10 L vessel at 40°C, what is the minimum amount of hydrogen gas that must be introduced in order to remove all solid iodine? PROBLEM 345 A standard solution of I 2 in water contains 0.33 g of I 2 in one litre flask. More than this dissolve in a KI solution because of the following equilibrium: I2 + I–

I –3

A 0.1 M KI solution actually dissolve 12.5 g of iodine per litre, most of which is converted to I –3 . Assuming that concentration of I 2 in all saturated solution is same, calculate the equilibrium constant for the above reaction. What is the effect of adding water to a clear saturated solution of I 2 in KI solution?

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Problems

PROBLEM 346 The following gaseous equilibrium was obtained by heating 0.46 moles of A in a 5.0 L vessel. The equilibrium pressure at 300 K was 3.0 atm. The equilibrium pressure changed to 3.6 atm when temperature was raised to 320 K. Calculate percentage change in degree of dissociation of A at 350 K with respect to that of A at 300 K. A(g )

B ( g ) + Cg )

PROBLEM 347 At 400°C, a 1 : 3 mixture of N 2 and H 2 reacts to form an equilibrium mixture of N 2 , H 2 and NH 3 . The total pressure at equilibrium was found to be 350 bar and mole fraction of NH 3 = 0.5. Calculate K p for the reaction: N 2 ( g ) + 3H 2 ( g ) 2NH 3 PROBLEM 348 Show that the maximum yield of product at equilibrium occurs when stoichiometric amount of reactants are used. Use the reaction: Cl 2 ( g ) + Br 2 ( g )

2BrCl( g )

PROBLEM 349 A flask containing 0.06 mole of F2 (g ) is allowed to equilibrate with F(g) at 1000 K. If the total pressure of the gases at equilibrium is 2.07 bar, calculate mole fraction of each gases at equilibrium. Given:

F2 ( g )

2F( g ), K p (1000 K) = 9.5 × 10 −3

PROBLEM 350 A mixture of air at 1.0 bar and 2000 K was passed through an electric arc to produce nitric oxide as N 2 ( g ) + O 2 ( g ) 2NO( g ), K 2000 = 4 × 10 –4 . What are the partial pressure of gases once the equilibrium has been established.

PROBLEM 351 At a pressure of one bar, an equilibrium exist at 2000 K between 0.25 mole of Br 2 (g ), 0.75 mole of F2 ( g ) gas and 0.497 mole of BrF3 ( g ). What will be the amounts of each gas after the pressure on the system has been increased to 2.0 bar and equilibrium at 2000 K re-established? Br 2 ( g ) + 3F2 ( g ) 2BrF3 ( g ) PROBLEM 352 Would 1.0% CO 2 in air be sufficient to prevent any loss in weight when Ag 2CO 3 is dried at 120°C Ag 2CO 3 ( s)

Ag 2O ( s) + CO 2 ( g ),

K p = 0.0095 at 120°C

PROBLEM 353 In a gas phase reaction 2 A + B 3C + 2D, it was found that when 1.0 mol of A, 2.0 mol of B, and 1.0 mol of D were mixed in one litre flask and came to equilibrium, the resulting mixture contained 0.9 mole of C. Calculate equilibrium constant K c . PROBLEM 354 A container whose volume is V constains an equilibrium mixture that consists of 2 mole each of PCl 5 , PCl 3 and Cl 2 and the equilibrium pressure is 30.4 kPa at T. A certain amount of Cl 2 ( g ) is now introduced keeping P and T constant, until equilibrium volume is 2V. Calculate amount of Cl 2 ( g ) added and K p .

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PROBLEM 355 A 250 ml flask containing NO(g) at 0.46 atm is connected to a 100 mL flask containing oxygen gas at 0.86 atm by means of a stop cock at 350 K. The gases are mixed by opening the stop cock where the following equilibrium established: 2NO + O 2 → 2NO 2 N 2O 4 . The first reaction is complete while the second is at equilibrium. Calculate K p if the final total pressure is 0.37 atm. PROBLEM 356 Equal moles of F2 (g ) and Cl 2 (g ) are introduced into a sealed container and heated to certain temperature where following equilibria established: Cl 2 ( g ) + F2 ( g )

2ClF( g ), K p = 3.2

Cl 2 ( g ) + 3F2 ( g ) 2ClF3 ( g ) If partial pressure of ClF and ClF3 at equilibrium are 0.2 and 0.04 atm respectively, calculate K p for the second equilibrium reaction. PROBLEM 357 0.2 mole of each Cl 2 (g ) and F2 (g ) are introduced in a sealed flask and heated to 2000 K where following equilibrium established. Cl 2 ( g ) + F2 ( g ) 2ClF( g ); and at equilibrium, moles of ClF = 0.267. At this stage, 0.1 mol of Br 2 is added and equilibrium is re-established as: Cl 2 ( g ) + F2 ( g )

2ClF( g )

Cl 2 ( g ) + Br 2 ( g ) 2BrCl( g ) Now moles of ClF found to be 0.25. Calculate K c for the second equilibrium reaction.

PROBLEM 358 A glass bulb initially contains mixture of N 2 and NO at a total pressure of one atm. Some Br 2 ( g ) is added such that the pressure would have been 2.25 atm had not the equilibrium: 2NO + Br 2 2NOBr been established, as a result, the actual pressure after adding Br 2 ( g ) was 2.12 atm. A second addition of Br 2 was made such that total pressure would have been 22.75 atm had no reaction at all been occurred, while the actual pressure found to be 22.5 atm. Calculate K p for the reaction assuming N 2 to be inert. PROBLEM 359 When 0.1 mole of Cl 2 (g ) and an excess of solid iodine are placed in 5 L vessel at 300 K, the total pressure is 0.767 atm. The vapour pressure of solid iodine is 0.03 atmosphere at this temperature. After this 1st equilibrium, 0.1 mole of bromine vapour is introduced into the reaction vessel. It is found that the weight of excess solid iodine diminished by an amount corresponding to 0.03 mole when new equilibrium is established. It is also found that partial pressure of bromine in new equilibrium mixture is 0.2 atm. Calculate K p for the following equilibria at 300 K: Cl 2 ( g ) + I 2 ( g )

2ICl( g )

Cl 2 ( g ) + Br 2 ( g )

2BrCl( g )

I 2 ( g ) + Br 2 ( g )

2IBr( g )

PROBLEM 360 The equilibrium constant of a reaction A2 (g ) + B 2 (g ) 2 AB ( g ) at 100°C is 50. If a one litre flask containing one mole of A2 ( g ) is connected to a 2.0 L flask containing two moles of B 2 ( g ), how many moles of AB will be formed at 373 K. PROBLEM 361 0.96 g of HI was heated till equilibrium 2HI(g ) H 2 + I 2 ( g ) is reached. The reaction mixture was suddenly cooled and the amount of iodine produced required 15.7 mL N/10 hypo solution. Calculate K p for the above equilibrium reaction.

Problems

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PROBLEM 362 Hydrazine was taken in a constant volume container at 27°C and 0.3 atm and equal moles of oxygen gas was injected, sealed and finally heated to 1000 K where the following equilibria established: N 2 H 4 + 3O 2 N 2 H4

2NO 2 + 2H 2O N 2 + 2H 2

K p1 = 3

K p2 = ?

N 2 H4 + H2 2NH 3 K p3 = ? If the gaseous mixture at equilibrium is passed through moisture absorbent, a decrease of 360 mm in the equilibrium pressure was observed. Now if the dried gaseous mixture is passed through ammonia absorber a further decrease of 20 mm in the equilibrium pressure was observed. Calculate K p 2 and K p 3 .

PROBLEM 363 Pure nirtosyl chloride (NOCl) gas was heated to 240°C in a 1.0 L container. At equilibrium, the total pressure was found to be 1.00 atmosphere and partial pressure of NOCl was 0.64 atm. Now some Cl 2 ( g ) is added at constant pressure and equilibrium was allowed to re-establish. At new equilibrium, the volume of the container was 1.5 L. Determine the moles of Cl 2 ( g ) added at first equilibrium. PROBLEM 364 N 2O 4 gas decomposes partially as N 2O 4 2NO 2 . In an experiment, some N 2O 4 is taken in a flask and heated to 300 K, where the above equilibrium is established. At equilibrium pressure of 1.0 atmosphere, density of the equilibrium mixture was found to be 2.33 g/L. Now, the gas is compressed till the density of new equilibrium mixture reaches to 5.08 g/L. Determine the new equilibrium pressure and the density of equilibrium mixture if the equilibrium pressure in this case is 1.5 atm. PROBLEM 365 H 2S dissociates according to the following reaction: H 2S H 2 + S( g ). At 1000 K and total pressure of 1.0 atmosphere, degree of dissociation was found to be 0.3. Determine the degree of dissociation if the gas is compressed isothermally to a new equilibrium pressure of 2.0 atmosphere. PROBLEM 366 (a) PCl 5 dissociates as: PCl 5 PCl 3 + Cl 2 . When 0.03 mole of PCl 5 was brought to equilibrium at 500 K and 1.0 atmosphere, the equilibrium volume was 2.09 L. Calculate degree of dissociation. (b) What will be the degree of dissociation when 0.2 mole of PCl 5 is brought to equilibrium in a 3 L flask at 500 K. PROBLEM 367 0.1 mole of hydrogen gas and 0.2 mole of CO 2 (g ) are introduced in an evacuated flask at 723 K and the following reaction occurs to give an equilibrium pressure of 50.67 kPa. H 2 ( g ) + CO 2 ( g ) H 2O( g ) + CO( g ) Analysis of the mixture shows that it contains 10 mol % of H 2O( g ). A mixture of CoO(s) and Co(s) is then introduced such that the additional equilibria: CoO( s) + H 2 ( g )

Co( s) + H 2O( g )

CoO( s) + CO( g )

Co( s) + CO 2 ( g )

are established. Analysis of the new equilibrium mixture thus, obtained is found to contain 30% (mol) of H 2O( g ) are present. Determine K c for the three reactions.

PROBLEM 368 At 25°C, the equilibrium : 2NOBr 2NO + Br 2 is readily established. When 1.1 g of NOBr is present in 1.0 litre flask at 25°C, total pressure in the flask was found to be 0.335 atmosphere.

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Determine the pressure if 5.0 g of NOBr is placed in a two litre evacuated flask at 25°C, sealed and allowed to attain the decomposition equilibrium to establish.

PROBLEM 369 A gaseous mixture containing equimolar amount of HCl and O 2 was taken in a flask, sealed and heated to 500°C where the following equilibrium is established: 4HCl + O 2 ( g ) 2Cl 2 + 2H 2O( g ) If the initial gas pressure was 1.0 atmosphere and mole fraction of HCl reacted before the equilibrium was established, was 0.76. Determine K p .

PROBLEM 370 When N 2O 5 is heated, it dissociates as: N 2O 5 N 2O 3 + O 2 K c = 4.5. At the same time, N 2O 3 also decomposes as: N 2O 3 N 2O + O 2 . If initially 4.0 moles of N 2O 5 are taken in a 1.0 litre flask and allowed to attain equilibrium, concentration of O 2 was found to be 4.5 M. Determine equilibrium concentration of other species and K c for the second equilibrium. PROBLEM 371 At 25°C, a mixture of NO 2 and N 2O 4 are in equilibrium in a cylinder fitted with a movable piston. The concentration of species present at equilibrium are as follows: [ N 2O 4 ] = 0.487, [NO 2 ] = 0.0475. Now, the piston is pushed to half the volume where the equilibrium was re-establish. Determine concentration of gases present at new equilibrium. PROBLEM 372 Gaseous nitrosyl chloride (NOCl) and N 2 are taken in a flask, sealed and heated to some temperature where the total pressure would have been 1.0 bar had not the following equilibrium been established: 2NOCl 2NO + Cl 2 The actual pressure was found to be 1.2 bar. Now into the equilibrium mixture, some Cl 2 gas was introduced so that the total pressure would have been 9.0 bar had no further reaction occurred but the actual pressure was found to be 8.9 bar. Determine the equilibrium constant for the decomposition equilibrium under the given experimental condition. PROBLEM 373 20.85 g of PCl 5 (g ) is introduced in a vessel washed with a nonvolatile solvent (B.Pt. = 350 K, M. Pt. =154). The equilibrium is established at 523 K when PCl 5 ( g ) is 52% dissociated and a total pressure was found to be 5.5 bar. If K p for the decomposition reaction: PCl 5 PCl 3 + Cl 2 is 1.78. Calculate the weight of solvent left in the vessel during washing. PROBLEM 374 Cl 2 (g ) and O 2 (g ) are taken in the molar ratio of 2 : 7 where the following equilibrium was established: 2Cl 2 + 7O 2 2Cl 2O 7 At equilibrium, mole fraction of Cl 2O 7 was found to be 0.1 when the total pressure was 100 bar. In an another experiment, two gases were taken in equimolar amount under identifical condition of temperature and mole fraction of Cl 2O 7 at equilibrium was found to be 0.06. Determine the equilibrium pressure in the new flask. PROBLEM 375 PCl 5 (g ) is taken in a flask at 1.0 atmosphere, sealed and allowed to attain the following equilibrium: PCl 5

PCl 3 + Cl 2

The equilibrium mixture was then allowed to pass through a pin hole and the gases coming out of pin-hole initially, was collected, analyzed and mole fraction of Cl 2 was found to be 0.53. Determine equilibrium constant ( K p ) for the decomposition reaction.

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Problems

PROBLEM 376 A solid substance A decomposes into two gaseous products B and C as: A ( s) 2B ( g ) + C ( g ) If at equilibrium, some C(g) at 1.0 atmosphere is added in constant volume condition, 10% of B(g) solidified before the equilibrium was re-established. Determine total pressure at final equilibrium. PROBLEM 377 Toxic level of Pb 2+ in human blood is reduced by forming a stable Pb 2+ EDTA complex which is excrete able through kidneys. Formation constant ( K f ) for this complex is 1018 . The ligand is administered by infusion of a solution of Na 2 [Ca-EDTA] (K f = 5 × 1010 ). In the blood stream, exchange of Ca 2+ for Pb 2+ occurs. The level of Pb 2+ in a patient blood was found to be 4µΜ. To this patient a mixed solution containing Ca(NO 3 ) 2 and Na 2 [Ca-EDTA] was administered so that their initial concentration in their blood were 2.5 µM and 1.0 µM respectively. Determine the ratio [Pb-EDTA ]2– /[Pb 2+ ] in the patient blood at equilibrium.

PROBLEM 378 NH 4 HS(s) is an unstable solid, decomposes as:NH 4 HS NH 3 + H 2S and the following thermodynamic informations are available: ∆H ° f ( kJ/ mol) : NH 4 HS = −157, NH 3 = – 46 and H 2S = – 20.5 ∆S f° (JK –1 mol –1 ) : NH 4 HS = 113.5, NH 3 = 193 and H 2S = 206. Suppose 1.0 mole of solid NH 4 HS is introduced into an empty 25 L flask, calculate the equilibrium pressure at 27°C. PROBLEM 379 N 2O 4 decomposes as N 2O 4 2NO 2 and at 300 K, ∆G ° f of N 2O 4 ( g ) and NO 2 ( g ) are 98 and 52 kJ/mol respectively. Starting with one mole of N 2O 4 at one bar and 300 K, calculate the fraction of N 2O 4 decomposed when equilibrium is established at 1.0 bar and 300 K. Also determine percentage volume change if decomposition is carried out in a cylinder fitted with a mass-less, frictionless piston. PROBLEM 380 A gaseous substance AB 2 (g ) convert to AB(g) in presence of solid A(s) as: AB 2 ( g ) + A ( s) 2 AB ( g ) The initial pressure and equilibrium pressure are 0.7 and 0.95 bar. Now the equilibrium mixture is expanded reversibly and isothermally till the gas pressure falls to 0.4 bar. Determine volume percentage of AB(g) and AB 2 ( g ) at the final equilibrium. PROBLEM 381 One mole of N 2 (g ), three moles of H 2 (g ) and one mole of H 2S(g ) were taken in a one litre flask, sealed and heated to 700 K where the following equilibria were established: N 2 ( g ) + 3H 2 ( g ) H 2S( g ) + NH 3 ( g )

2NH 3 ( g )

NH 4 HS( g )

K p = 8 × 10 –3 atm –1

At equilibrium, concentration of ammonia gas was found to be 0.9 M. Determine K p for the first equilibrium and equilibrium concentrations of H 2S. PROBLEM 382 N 2O 4 (g ) is taken in a cylinder equipped with movable piston and heated first at constant volume where the following equilibrium is established and the gas was 30% dissociated : N 2O 4 ( g ) 2NO 2 ( g ) Now the gaseous mixture was expanded isothermally till volume was doubled. Determine the percentage of N 2O 4 dissociated at this stage.

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PROBLEM 383

Consider the following reaction:

2NO 2 + O 3 N 2O 5 ( g ) + O 2 ( g ) ∆H °f (O 3 ) = 143 kJ mol , ∆H °f (N 2O 5 ) = 11 kJ mol –1 and ∆H °f (NO 2 ) = 33 kJ mol –1 . –1

The

above reaction is spontaneous at lower temperature but turned non-spontaneous as temperature approaches to 1175 K. Assuming ∆H ° and ∆S ° to be independent of temperature, determine K p at 500 K.

PROBLEM 384

Consider formation of N 2O 5 ( g ) according to the reaction below : 2NO 2 + 12 O 2

N 2O 5

∆H ° = − 55 kJ; ∆S ° = − 227 JK –1

Also, ∆H °f (NO 2 ) = + 33.2 kJ mol –1 , S ° ( NO 2 ) = 240 JK –1 , S ° (O 2 ) = 205 JK –1 . (a) Determine ∆H °f N 2O 5 , S ° (N 2O 5 ), ∆G ° at 25°C. (b) State and explain, whether this reaction is spontaneous at 25°C. (c) How the relative amounts of reactants and products would be affected at equilibrium?

PROBLEM 385 Cis-2-butene when heated to 500 K, it isomerizes into trans-2-butene and 2-methylpropene. If standard state Gibb’s free energy of formations of cis-2-butene, trans-2-butene and 2-methyl-propene are 65.85 kJ mol –1 , 62.77 kJ mol –1 and 58.07 kJ mol –1 respectively, determine the equilibrium composition. PROBLEM 386 A vessel containing 0.015 mol N 2 (g ) and 0.02 mol PCl 5 (g ) is heated to 227°C where the pressure was found to be 1.843 atm`. Also at 227°C, K p for : PCl 5 ( g ) PCl 3 ( g ) + Cl 2 ( g ) = 0.4 atm. Assuming N 2 to be inert gas, determine, volume of the vessel. PROBLEM 387

The ∆G° for the reaction: SO 2 + 12 O 2

SO 3 is −22600 + 21T

Determine the temperature at which mixture is 80% (by mole) in SO 3 , if initial mixture contain 15% SO 2 and 20% O 2 at one atmosphere total pressure. Assume total pressure is maintained constant at one atmosphere throughout.

PROBLEM 388 A 2.0 L flask, initially containing one mol of each CO and H 2O, was sealed and heated to 700 K, where the following equilibrium was established : CO( g ) + H 2O( g ) CO 2 ( g ) + H 2 ( g ); K ( 700) = 9 Now the flask was connected to another flask containing some pure CO 2 ( g ) at same temperature and pressure, by means of a narrow tube of negligible volume. When the equilibrium was restored, moles of CO was found to be double of its mole at first equilibrium. Determine volume of CO 2 ( g ) flask. PROBLEM 389

Consider the reaction:

H 2 (g ) + I 2 (g ) 2HI ( g ) If we start with 0.5 moles of each H 2 ( g ) and I 2 ( g ) at 700 K in a flask, equilibrium concentration of HI was observed to be 0.15 M. What would have been the concentration of HI at equilibrium had the reaction been started with 0.8 mole of each H 2 and I 2 ( g ) at same temperature and in the same vessel?

PROBLEM 390 For the reaction C(s) + CO 2 (g ) 2CO( g ), K p = 63 atm at 1000 K. What will be the total pressure of the gases above an equilibrium mixture if PCO = 10 PCO2 .

Problems

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IONIC EQUILIBRIUM PROBLEM 391 Addition of 100 mL 0.1 M HCl to certain volume of a slightly weak monobasic (0.1 M) acid solution decreases its pH from 2.0 to 1.7. Determine ionization constant of the weak acid and its volume. PROBLEM 392 Calculate pH of 0.02 M succinnic acid solution taking into account both ionization. K a 1 = 7 × 10 –5 , K a 2 = 3 × 10 –6 . PROBLEM 393 K a of formic acid at 27°C is1.7 ×10 –4 . What will be the pH of a 0.1 M aqueous solution of formic acid at 47°C. Standard enthalpy of neutralization of formic acid and HCl versus NaOH are – 42 kJ and – 57 kJ respectively. PROBLEM 394 Determine exact pD of a 10 –7 M DCl solution is D 2O. Ionic product of D 2O is 1.35 ×10 –15 .

PROBLEM 395 Calcium hypochlorite [Ca(OCl) 2 ] is used as a disinfectant for swimming pools. The recommended pH of a swimming pool is 7.8. Calculate the percentage of HClO and ClO – in the swimming pool. K a for HOCl is 3 × 10 –8 . PROBLEM 396 In the vapour phase acetic acid molecule associate to the some extent to form dimmers. At 50°C, the pressure of a certain acetic acid vapour is 0.0342 atm in a 360 mL flask. The vapour is condensed and neutralized with 13.8 mL 0.0568 M NaOH. Calculate the degree of dissociation of the dimmer. PROBLEM 397 Henery’s law constant for CO 2 at 38°C is 2.28 ×10 –3 mol/ L. Determine pH of a solution of CO 2 at 38°C in equilibrium with the gas at a partial pressure of 3.2. For CO 2 : K1 = 4.2 × 10 –7 , K 2 = 4.8 × 10 –11 . PROBLEM 398 Determine pH of a 5% aqueous solution of NaOCl (by weight). Density of the solution is 1.0 g/cc. K a = 3 × 10 –8 . PROBLEM 399 An ammonia-ammonium chloride buffer has a pH value of 9 with [NH 3 ] = 0.25. By how much the pH will change if 75 mL of 0.1 M KOH be added to 200 mL buffer solution. K b = 2 × 10 –5 . PROBLEM 400 A solution is prepared by dissolving 15 g of acetic acid and 25 g sodium acetate in 750 mL of water. (a) What is the pH of this solution? (b) What would the pH of the solution be after 25.00 mL 0.25 M NaOH is added? (c) What would be the pH is 25 mL 0.4 M HCl is added into (b)? K a = 2 × 10 –5 .

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PROBLEM 401 Determine freezing point of a 0.5 M solution of dichloroacetic acid (K a = 0.05). Assume density of the solution to be 1 g/cc and K f for water =1.86. PROBLEM 402 pK w of heavy water at 20°C and 30°C are 15.05 and 14.7 respectively. Determine pD of a pure heavy water at 50°C. PROBLEM 403 Suppose that two hydroxides MOH and M ′ (OH) 2 , both have K sp =10 –12 and that initially both cations are present in a solution at concentrations of 0.001 M. Which hydroxide precipitate first and at what pH when a NaOH solution is added dropwise? PROBLEM 404 Determine molar solubility of Fe(OH) 2 at pH − 8 and at pH = 6. K sp = 1.6 × 10 –14 . PROBLEM 405 Determine solubility of CaF2 (K sp = 4 × 10 –11 ) (a) at pH = 7,

(b) at pH = 5.

K a = 3.45 × 10 –4 .

PROBLEM 406 Determine the molar solubility of BaF2 (K sp = 1.7 × 10 –6 ) at (a) pH = 7, (b) pH = 4. PROBLEM 407 When 10 mg sodium barbiturate are dissolved in 250 mL of water to form a solution, the resulting pH was found to be 7.71. The molar mass of the salt is 150. Determine percentage protonation of barbiturate ion and K a of the acid. PROBLEM 408 Determine pH of a 0.024 M hydroxyl amine hydrochloride solution. K b of hydroxyl amine =10 –8 . PROBLEM 409 A 25 mL 0.2 M oxalic acid is titrated with 0.2 M NaOH solution. Determine pH at the first and second equivalence point. K a 1 = 6 × 10 –2 and K a 2 = 6.5 × 10 –5 . PROBLEM 410 Determine molar solubility of ZnS (K sp = 1.6 × 10 –24 ) in a saturated solution of H 2S (0.1 M) buffered at pH = 7. What would be the solubility in a saturated H 2S solution buffered at pH =10. For H 2S K1 = 10 –7 and K 2 = 10 –14 . PROBLEM 411 Determine simultaneous solubility of Ag 2CO 3 and Ag 2CrO 4 . K sp (Ag 2CO 3 ) = 8.1 × 10 –12 and K sp ( Ag 2CrO 4 ) = 1.2 × 10 –12 . PROBLEM 412 To a certain volume of a weak monobasic acid, when 20 mL of NaOH solution is added, pH of the resulting solution was found to be 3.7 whereas when 30 mL of the same NaOH is added to the same volume of the acid from same stock, pH was found to be 4.18. Determine K a of the acid. PROBLEM 413 Determine mass of sodium dihydrogen phosphate and volume of 1.5 M HCl solution required for preparation of a 100 mL buffer solution of pH 2.42 with concentration of H 2 PO 4– be 0.15 M. K1 = 7.6 × 10 –3 , K 2 = 6.2 × 10 –8 and K 3 = 2 × 10 –13 .

PROBLEM 414 Carbondioxide gas from a steel cylinder is bubbled for some time through pure water placed in a jar. When the supply of gas is terminated, pH of the solution was found to be 3.83. What is the pressure of gas in the cylinder if Henery’s constant of CO 2 is 2.3 ×10 –2 . For CO 2 , K1 = 4.2 × 10 −7 and K 2 = 4.8 × 10 −11 .

Problems

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PROBLEM 415 Determine pH of a 0.1 M Fe(NO 3 ) 2 solution. Given Fe(OH) 2

Fe(OH) + + OH –

K1 = 10 –4

Fe(OH) +

Fe 2+ + OH –

K 2 = 2.5 × 10 –6

PROBLEM 416 H 2S is bubbled into a 0.2 M NaCN solution which is 0.02 M in each Ag(CN) –2 and –50 , K sp of CdS = 7.1 ×10 –28 . K d Cd(CN) 2– 4 . Determine which sulphide precipitate first. K sp ( Ag 2S) =10 –18 . Ag(CN) –2 = 10 –20 and Kd(Cd(CN) 2– 4 ) = 7.8 × 10

PROBLEM 417 Over what range of concentration of hydrogen ion concentration (pH) is it possible to separate Cu 2+ from Ni 2+ when both metal ions are present at 0.01 M concentration and solution is made saturated with 0.1 M H 2S? K sp of CuS and NiS are 6 × 10 –37 and 4 × 10 –20 respectively. PROBLEM 418 A solution contains calcium nitrate and nickel nitrate, each at concentration of 0.1 M. CO 2 is bubbled into make its concentration equal to 0.03 M. What pH range would make possible for the selective precipitation of metal carbonates? K sp of CaCO 3 and NiCO 3 are 4.5 ×10 –9 and 1.3 ×10 –7 respectively.

PROBLEM 419 How many moles AgBr can dissolve in 1.0 L of 1.0 M NH 3 solution. K sp of AgBr is 5 × 10 –13 and K f for Ag(NH 3 ) +2 is 1.5 ×10 7 . PROBLEM 420 A salt whose formula is of the form MX has a value of K sp equal to 3.2 ×10 –10 . Another sparingly soluble salt MX 3 must have what value of K sp if the molar solubilities of the two salts are to be identical? PROBLEM 421 A salt having formula of the type M 2 X 3 has K sp = 2.2 × 10 –20 . Another salt, M 2 X , has to have what K sp value if M 2 X has twice the molar solubility of M 2 X 3 ? PROBLEM 422 Suppose Na 2SO 4 is gradually added to 100 mL of a solution that contains both Ca 2+ and Sr 2+ ions each at 0.15 M concentrations. (a) What will the Sr 2+ ion concentration be when CaSO 4 just begin to precipitate. (b) What percentage of Sr 2+ ion has precipitated when CaSO 4 just begin to precipitate? K sp (CaSO 4 ) = 2.4 × 10 –5 , K sp (SrSO 4 ) = 3.2 × 10 –7 . PROBLEM 423 How many grams of solid NaCN have to be added to 1.2 L of water to dissolve 0.11 mole of Fe(OH) 3 in the form of [ Fe(CN) 6 ]3– ? [K sp of Fe(OH) 3 = 1.6 × 10 –39 , K f =10 31 ] PROBLEM 424 Silver ion forms a complex with thiosulphate ion with their formula [Ag(S 2O 3 ) 2 ]3– . K f for this complex is 2 × 1013 . How many grams of AgBr ( K sp = 5 × 10 –13 ) will dissolve in 125 mL of a 1.20 M Na 2S 2O 3 solution.

PROBLEM 425 A sample of hard water was found to contain 278 ppm of calcium ion. Into 1.00 L of this water, 1.06 g of Na 2CO 3 was added. What is the new concentration of Ca 2+ ion in ppm. Density of all solution be 1.0 g/cc and K sp of CaCO 3 is 4.5 ×10 –9 .

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PROBLEM 426 What are the concentration of Pb 2+ , Br – and I – in an aqueous solution that is in contact with both PbI 2 and PbBr 2 ? K sp (PbBr 2 ) = 2 × 10 –6 , K sp (PbI 2 ) = 8 × 10 –9 . PROBLEM 427 An indicator has its standard ionization constant 9 × 10 –9 . The acid colour of the indicator is yellow while its basic colour is red. The yellow colour is visible when the ratio yellow/red is 30 and red colour becomes visible when the ratio red/yellow is 2. What is the working pH range of this indicator? PROBLEM 428 (a) At what maximum concentration of hydroxide ion, 1.0 m mole Zn(OH) 2 will go –16 into 1.0 L solution as Zn(OH) 2– and K f = 2 × 10 20 . 4 ? K sp = 3 × 10 (b) At what maximum concentration of hydroxide ion will 1.0 m mole Zn(OH) 2 will go into 1.0 L solution as Zn 2+ ?

PROBLEM 429 Determine

the

simultaneous

solubility

of

AgCN

( K sp = 2.2 × 10 –16 )

and

AgCl( K sp = 1.6 ×10 –10 ) in 1.0 M ammonia solution. K f [ Ag(NH 3 ) +2 ] = 1.5 × 10 7 .

PROBLEM 430 What must be the minimum concentration of acetic acid in a one litre buffer solution of acetic acid acetate ( pH = 4.74) if the pH changes by not more than 0.1 unit on addition of 50 mL 1 M HCl. K a = 1.8 × 10 –5 . PROBLEM 431 How much 1.0 M HCl must be added to 1.00 L of 0.10 M sodium lactate (NaC 3 H 5O 3 ) to make a buffer of pH = 3.77. K a = 1.4 × 10 –4 . PROBLEM 432 Potas alum is KAl(SO 4 ) 2 ⋅ 12 ⋅ H 2O. As a strong electrolyte, it is considered to be 100% 3+ ionized into K + , Al 3+ and SO 2– 4 . The solution is acidic because of the hydrolysis of Al , but not so acidic as might be expected, because the sulphate ion can spong-up some of the H + by forming HSO –4 . Given a solution is made by dissolving 11.4 g of alum in enough water to make 100 mL of the solution. Calculate pH considering the following hydrolysis reactions: Al 3+ + H 2O HSO –4

Al(OH) 2+ + H + ; H + + SO 2– 4 ;

K h = 1.4 × 10 –5 K a 2 = 1.26 × 10 –2

PROBLEM 433 Determine molar solubility of AgCN in water considering hydrolysis of CN – ion. Given K sp (AgCN) = 2.3 ×10 –16 and K a ( HCN ) = 5 × 10 –10 . PROBLEM 434 To what volume a 50 mL 6.0 M ammonia solution be diluted to double the degree of protonation? K b = 2 × 10 –5 . PROBLEM 435 Molar solubility of PbS is 6.7 ×10 –10 . Sulphide ion hydrolysis completely into HS – but the further hydrolysis of HS – is insignificant. Determine K sp of PbS. K a 1 (H 2S) = 10 –7 and K a 2 (H 2S) = 10 –14 .

PROBLEM 436 Lactic acid is formed in the muscles during intense activity (anaerobic metabolism). In

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the blood, lactic acid is neutralized by reaction with hydrogen carbonate. Lactic acid written HL is monoprotic, and the acid dissociation constant is K (HL) = 1.4 ×10 –4 . The acid dissociation constants for carbonic acid are: K a 1 = 4.5 × 10 –7 and K a 2 = 4.7 × 10 –11 . All carbon dioxide remains dissolved during the reactions. (a) Calculate pH in a 3.00 ×10 –3 M solution of HL. (b) Calculate the value of the equilibrium constant for the reaction between lactic acid and hydrogen carbonate. (c) 3.00 ×10 –3 mol lactic acid (HL) is added to 1.00 L of 0.024 M solution of NaHCO 3 (no change in volume, HL completely neutralized). (i) Calculate the value of pH in the solution of NaHCO 3 before HL is added. (ii) Calculate the value of pH in the solution after the addition of HL. (d) pH in the blood of a person changed from 7.40 to 7.00 due to lactic acid formed during physical activity. Let an aqueous solution having pH = 7.40 and [ HCO –3 ] = 0.022 M represent blood in the following calculation. How many moles of lactic acid have been added to 1.00 L of this solution when its pH has become 7.00? (e) In a saturated aqueous solution of CaCO 3 ( s), pH is measured to 9.95. Calculate the solubility of calcium carbonate in water and show that the calculated value for the solubility product constant K sp is 5 × 10 –9 . (f) Blood contains calcium. Determine the maximum concentration of “free” calcium ions in the solution (pH = 7.40, [HCO –3 ] = 0.022 M) given in d).

PROBLEM 437 The pH recorded when 1.7 g of a monoprotic acid (K a = 2 × 10 –10 ) was added to 100 mL of NaOH of 0.1 M strength was 10.2. Determine molar mass of the acid. PROBLEM 438 A saturated solution of Mg(OH) 2 has pH value 10.5. A mixture of 10.0 g of Mg(OH) 2 and 100 mL 0.1 M HCl is stirred magnetically for some time at 25°C. Calculate pH of the liquid phase when equilibrium is reached. PROBLEM 439 The concentration of CO 2 is expected to rise 440 ppm in year 2020. What would be the pH of rainwater in the year 2020 if molar solubility of CO 2 is 0.0343 L–1 bar –1 . K a 1 of CO 2 = 4.2 × 10 –7 . PROBLEM 440 Nitrogen in agricultural materials is often determined by the Kjeldahl method. The method involves a treatment of the sample with hot concentrated sulphuric acid, to convert organically bound nitrogen to ammonium ion. Concentrated sodium hydroxide is then added, and the ammonia formed is distilled into hydrochloric acid of known volume and concentration. The excess hydrochloric acid is then back-titrated with a standard solution of sodium hydroxide, to determine nitrogen in the sample. (a) 0.2515 g of a grain sample was treated with sulphuric acid, sodium hydroxide was then added and the ammonia distilled into 50.00 mL of 0.1010 M hydrochloric acid. The excess acid was back-titrated with 19.30 mL of 0.1050 M sodium hydroxide. Calculate the concentration of nitrogen in the sample, in per cent by mass. (b) Calculate the pH of the solution which is titrated in (a), when 0 mL, 9.65 mL, 19.30 mL and 28.95 mL of sodium hydroxide has been added. Disregard any volume change during the reaction of ammonia gas with hydrochloric acid. K a for ammonium ion is 5.7 ×10 –10 . (c) Draw the titration curve based on the calculations in (b).

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(d) What is the pH transition range of the indicator which could be used for the back titration. (e) The Kjeldahl method can also be used to determine the molecular weight of amino acids. In a given experiment, the molecular weight of a naturally occuring amino acid was determined by digesting. 0.2345 g of the pure acid, and distilling the ammonia into 50.00 mL of 0.1010 M hydrochloric acid. A titration volume of 17.50 mL was obtained for the back titration with 0.1050 M sodium hydroxide. Calculate the molecular weight of the amino acid based on one and two nitrogen groups in the molecule, respectively.

PROBLEM 441 In the precipitation titration of KCl against AgNO 3 , K 2CrO 4 is used as an indicator since, AgCl is white coloured. End point is detected by appearance of deep yellow coloured precipitate of Ag 2CrO 4 . Determine the minimum concentration of chromate ion required for detection of end point. K sp of AgCl = 2.5 × 10 –10 and K sp of Ag 2CrO 4 = 1.8 × 10 –12 . PROBLEM 442 50 mL of a10 –4 M aqueous solution of La(NO 3 ) 3 is mixed with 50 mL of an NH +4 /NH 3 buffer that is originally 0.2 M in NH +4 and 0.4 M in NH 3 . What percentage of La 3+ has been precipitated at equilibrium. K sp of La(OH) 3 = 10 –19 and K b of NH 3 = 2 × 10 –5 . PROBLEM 443 Aluminium phosphate is more soluble than its expected solubility due to hydrolysis of both Al 3+ and PO 3– 4 as: Al 3+ + H 2O PO 3– 4 + H 2O

Al(OH) 2+ + H +

K a =10 –5

– HPO 2– 4 + OH

K b = 5 × 10 –2

Determine the actual solubility of AlPO 4 . K sp of AlPO 4 = 10 –20 .

PROBLEM 444 0.2 moles of MgCl 2 is mixed with 0.2 moles of NaOH in a total volume of 1.0 litre. (a) Calculate pH of this solution. K sp of Mg(OH) 2 is 1.6 ×10 –12 . (b) Also determine pH when 0.04 moles of NaOH is added in the solution (a). (c) Also determine pH of solution obtained after adding 0.04 mole of HCl to solution (a). PROBLEM 445 How many grams of PbS will precipitate from a 1.0 L saturated PbSO 4 solution if the solution is made saturated with H 2S (0.1 M) and pH adjusted to 0. K sp of PbSO 4 = 1.6 × 10 –8 and K sp of PbS = 2.5 × 10 –27 . Also for H 2S; K a 1 = 10 −7 and K a 2 = 10 −14 .

PROBLEM 446 A typical wine sample contain 2.3% by mass of the base urea. K b for urea is1.5 ×10 –14 . If the density of urea solution is 1.06 g/mL and it is buffered at pH 6.35, determine the equilibrium concentration of protonated urea. PROBLEM 447 At 25°C K sp of AgCl is 2 × 10 –10 . Using the following standard molar Gibb’s free energies of formations: AgCl( s) = – 110 kJ/ mol, Ag 2SO 4 ( s) = – 618.5 kJ/ mol, Cl − ( aq ) = − 131 kJ/mol and SO 24− ( aq ) = − 742 kJ/mol determine solubility of Ag 2SO 4 in a 1.0 m M Na 2SO 4 solution at 25°C.

PROBLEM 448 pH of a 100 mL HOCN solution is 2.75. Addition of 100 mL 0.05 M HOCN to the above solution lowers the pH to 2.5. Determine concentration of the original acid solution (assume degree of ionization to be negligible in comparison to one).

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PROBLEM 449 1.0 metric ton coal containing 2.5% sulphur is burned and SO 2 produced in the combustion was dissolved in rainwater of volume equivalent to 2.0 cm rain fall over 2.6 km 2 area. Determine pH of rainwater if K a of H 2SO 3 = 1.5 × 10 –2 . PROBLEM 450 1.0 g impure sample of Mg is burnt completely in N 2 atmosphere and nitride is dissolved in 1.0 litre of water. pH of the resulting solution was found to be 10.75. Determine percentage purity of the sample ignoring solubility of Mg(OH) 2 . K b of NH 3 = 2 × 10 –5 . PROBLEM 451 Bromine in excess is added drop wise to a 0.01 M solution of SO 2 . All SO 2 is oxidized to sulphuric acid. Excess of bromine is removed by flushing with gaseous N 2 . Determine pH of the resulting solution assuming K a 1 of H 2SO 4 to be very large and K a 2 = 10 –2 . PROBLEM 452 In laboratory class, a student is given three flasks that are labeled Q, R and S . Each flask contains one of the following solutions : 1.0 M Pb(NO 3 ) 2 , 1.0 M NaCl and 1.0 M K 2CO 3 . The student is also given two flasks that are labeled X and Y . One of these flasks contains 1.0 M AgNO 3 and other contains 1.0 M BaCl 2 . (a) When the student combined a sample of solution Q with a sample of solution X , a precipitate formed. A precipitate also formed when samples of solution Q and Y were combined. Identify Q. (b) When solution Q is mixed with solution R, a precipitate forms. However, no precipitate forms when solution Q is mixed with solution S. Identify R and S . (c) After knowing the identity of Q, R and S , employ them to identify X and Y . PROBLEM 453 You are given two burettes. One is filled with 0.15 M acetic acid and other is filled with 0.10 M sodium acetate. How much of each would you mix together to produce a total of 20 mL of solution with a pH of 4.7. You may not add any water to solution. K a of acetic acid is 2 × 10 −5 . PROBLEM 454 You are given two burettes. One is filled with 0.15 M acetic acid and other with 0.1 M NaOH solution. How much of each would you mix together to produce a total volume of 50 mL with a pH = 4.7 K a of acetic acid = 2 × 10 −5 . PROBLEM 455 1.6 g solid Ba(NO 3 ) 2 and 0.3 g H 3 PO 4 are taken in a flask and volume made up to 2.0 litre by adding sufficient water. Determine pH of the final solution assuming Ba 3 (PO 4 ) 2 to be insoluble in water. PROBLEM 456 A solution contains the cations Mn 2+ , Co 2+ and Ag + at an original concentration of 0.01 M each. Determine the pH range of this precipitation when the solution is saturated in H 2S(0.1 M). K a 1 and K a 2 of H 2S are 10 −7 and 10 −14 respectively. MnS K sp = 2.5 × 10 –10 ,

CoS : K sp = 4 × 10 −21 , Ag 2S : K sp = 6.3 × 10 –50 .

PROBLEM 457 A saturated hydrogen electrode (SHE) with [H + ] = 1.0 M is connected to a silver electrode placed in a saturated solution of silver oxalate Ag 2C 2O 4 via a salt bridge and e.m.f was found °+ to be 0.589 V. Calculate solubility product constant K sp for given oxalate. E Ag = 0.80 V. /Ag PROBLEM 458 If 150 mg of zinc hydroxide crystals are mixed with 250 mL of a 0.1 M NaOH solution, what mass of Zn(OH) 2 crystals will remain undissolved? K sp = 2.1 × 10 −16 and K f for 15 Zn(OH) 2– 4 = 2.8 ×10 .

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PROBLEM 459

K sp for AgCl = 2 × 10 –10 and K f for Ag(NH 3 ) +2 = 2 × 10 7 . Calculate molar solubility of

AgCl in an aqueous solution that is 0.1 M in KCl and 0.1 M in NH 3 .

PROBLEM 460

values

K sp

of

silver

halides

are :

AgCl = 1.8 ×10 –10 ,

AgBr = 5 × 10 –13 ,

AgI = 8.4 ×10 –17 . State the order in which halide ions should be added to a concentration of 0.1 M so that each precipitate will form from the [Ag + ] in equilibrium with the previous precipitate. Support your answer with appropriate calculations.

PROBLEM 461

Phosphoric acid ionizes according to the equations : H 3 PO 4 ( aq )

H + ( aq ) + H 2 PO 4– ( aq ) K1 = 7 × 10 –3

H 2 PO 4– ( aq )

–8 H + ( aq ) + HPO 2– 4 ( aq ) K 2 = 6 × 10

HPO 2– 4 ( aq )

H + ( aq ) + PO 3– 4 ( aq )

K 3 = 4.5 × 10 –13

(a) If you are asked to prepare a buffer with a pH = 7.00, identify the species that should be used in the solution and calculate their ratio. (b) Assume 50 mL of the buffer prepared in part ‘a’ is available in which more abundant species has a concentration of 0.10 M. If to this solution 20 mL 0.1 M NaOH is added further, what will be the new pH?

PROBLEM 462 A 0.50 gram mixture containing NaHCO 3 , Na 2CO 3 and KCl was dissolved in 50.00 mL water and titrated with 0.115 M HCl resulting in the following titration curve. 14 12 10

(11.00 mL)

8 6 4

(35 mL)

2 0

10

20

30

40

Determine weight percentage of NaHCO 3 , Na 2CO 3 and KCl in the mixture.

PROBLEM 463

K sp of Ca(H 2 PO 4 ) 2 = 10 –3 . What minimum pH must be maintained in a 0.25 M

H 3 PO 4 solution to prevent precipitation of Ca(H 2 PO 4 ) 2 if the solution is 0.15 M in calcium ion? K a 1 of H 3 PO 4 is 7 × 10 –3 .

PROBLEM 464 Fe 3+ ion forms a blood red complexes with thiocyanate ion, SCN – . Following equilibria exist on adding SCN – to an aqueous solution of Fe 3+ ion : (i) Fe(OH) 3 ( s) Fe 3+ ( aq ) + 3OH – K sp = 4 × 10 –38 (ii) Fe 3+ + 2SCN –

Fe(SCN) +2

(iii) SCN – + H 2O

HSCN + OH –

K f = 2.3 × 10 3 K a = 71

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Problems

Discuss the effect of followings on concentration of Fe 3+ ion in solution : (a) Addition of HNO 3 , (b) Addition of NaOH, (c) Addition of NaSCN, (d) Addition of Fe(OH) 3 .

PROBLEM 465 What will be the effect of adding a strong acid to an aqueous solution of Cu(CN) 2– 4 , –10 HCN is a weak acid with K a = 6 × 10 ? PROBLEM 466 What concentration of NH 3 must be present in a 0.1 M AgNO 3 solution to prevent AgCl from precipitating when 4.0 g of NaCl are added to a 250 mL of this solution. K sp = 2 × 10 –10 , K f [Ag(NH 3 ) +2 ] = 10 7 .

PROBLEM 467 It is possible to keep Co(OH) 3 from precipitating from a 0.01 M CoCl 3 solution by buffering the solution at pH 9.1 with a buffer that contain NH 3 and NH +4 ion. How much of 6 M NH 3 and 6 M HCl must be added per litre of this solution to prevent Co(OH) 3 from precipitating? 35 [NH 3 : K b = 2 × 10 –5 , Co(OH) 3 ( s) : K sp =10 –43 , Co(NH 3 ) 3+ 6 : K f = 2 × 10 ] PROBLEM 468 Will Co(OH) 3 precipitate from a solution that is initially 0.1 M in Co 3+ and 1M in SCN – if this solution is buffered at pH = 7. K sp of Co(OH) 3 = 1.6 × 10 –44 and K f [Co(SCN) –4 ] = 10 3 ? – PROBLEM 469 Calculate the concentration of CO 2– 3 in a 0.1 M HCO 3 solution buffered with equal 2– + number of moles of NH 3 and NH 4 . Is this CO 3 concentration large enough to precipitate BaCO 3 when the solution is mixed with equal volume of 0.1 M Ba 2+ ion. [K sp = 5 × 10 –9 , H 2CO 3 : K a 1 = 4.5 × 10 –7 ,

K a 2 = 4.7 × 10 –11 , NH 3 : K b = 2 × 10 –5 ]

PROBLEM 470 Enough carbonate ion CO 2– 3 can be leached out of clay to buffer ground water at pH = 8. Assume that total concentration of HCO –3 and CO 2– 3 ions in this solution is 0.10 M. Calculate the 60 maximum concentration of Co-isotope that could leach into ground water if clay were used as a barrier to store this radioactive isotope. K sp of CoCO 3 and Co(OH) 3 are1.4 ×10 –13 and1.6 ×10 –44 respectively. For H 2CO 3 K a 1 = 4.5 × 10 –7 and K a 2 = 4.7 × 10 –11 .

PROBLEM 471 Determine pH and degree of hydrolysis of a 10 –3 M C 6 H 5 NH 3Cl solution. K b = 5 × 10 –10 . PROBLEM 472 A solution is made by dissolving 0.001 mol Ca(OI) 2 in 100 mL of water and to this solution 100 mL 0.01 M acetic acid solution was added. Determine pH of this solution. K a (acetic acid) = 2 × 10 –5 , K a ( HOI) = 2 × 10 –11 . PROBLEM 473

Determine pH of a 0.01 M ClC 6 H 4 NH 3Cl, K b = 4 × 10 –12 .

PROBLEM 474 The indicator dinitrophenol has K a = 1.1 × 10 –4 . In a10 –4 M solution, it is colourless in acid medium and yellow in basic medium. Calculate pH range over which it goes from 25% (ionized) to 75% (ionized). PROBLEM 475

Determine concentration of ammonium ion that is required to prevent the precipitation

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of Mg(OH) 2 in a solution with [Mg 2+ ] = 0.1 M and [NH 3 ] = 0.1 M. K sp [ Mg(OH) 2 ] = 1.5 ×10 –11 and K b ( NH 3 ) = 2 × 10 –5 .

PROBLEM 476 Determine molar solubility of BaSO 4 in a 0.25 M solution of NaHSO 4 . K a for HSO –4 = 1.2 × 10 –2 . Also determine the factor by which solubility is decreased as compared to solubility in pure water K sp =10 –10 . 1 M HCl will have to be added to 3 500 c.c. of 0.1 M Na 2CO 3 solution in order to adjust the pH to 10? K a 1 of CO 2 = 4 × 10 –7 and

PROBLEM 477

If the final volume is to be 1 litre, how many mL of

K a 2 = 5 × 10 –11 .

PROBLEM 478 Formation constant of [Ag(CN) 2 ]– is 2.5 ×1018 . Determine concentration of Ag + (aq ) in a solution which was originally 0.1 M in KCN and 0.03 M in AgNO 3 . PROBLEM 479 The solubility product of CaF2 at 18°C is 3.4 ×10 –11 while that of CaCO 3 is 95 . × 10 –9 . (a) What will be the nature of 1st precipitate when a solution of CaCl 2 is added to a solution which is 0.05 M in NaF and 0.02 M in Na 2CO 3 . (b) In a 0.02 M solution of Na 2CO 3 , what is the minimum concentration of NaF at which both CaF2 and CaCO 3 will precipitate simultaneously? PROBLEM 480 The solubility product of PbI 2 is 7.47 ×10 –9 at 17°C and1.4 ×10 –8 at 27°C. Calculate (a) the molar heat of solution of PbI 2 (b) the solubility of PbI 2 at 77°C. PROBLEM 481 The equivalence conductivity of a solution containing 2.54 g CuSO 4 /L is 91 cm 2 Ω –1 equ –1 . Calculate the specific conductivity of a solution. Also determine the resistance of a cm 3 of this solution when placed between two electrodes 1.00 cm apart, each having an area of 1.00 cm 2 ? PROBLEM 482 The equivalent conductance of a 0.01 M ammonia solution is 10 cm 2 Ω –1equ –1 . The equivalent conductance of ammonia at infinite dilution is 238 cm 2 Ω –1equ –1 . Determine ionization constant K b of ammonia. PROBLEM 483 Equivalent conductance at infinite dilution for acetic acid is 390 cm 2 Ω –1 at 25°C. Determine specific conductance and equivalent conductance of a 0.01 M acetic acid solution if K a of acetic acid = 2 × 10 –5 . PROBLEM 484 A conductivity cell is filled with 0.1 M KCl solution at 27°C. The measured resistance R = 25 Ω. Calculate cell constant if specific conductance of 0.1 M KCl solution is 0.0116 Ω –1cm –1 . If the same cell is filled with acetic acid, the cell resistance is 1982 Ω. Determine molar conductance of 0.1 M acetic acid solution. PROBLEM 485 The conductivity of a 0.1 M NaOH solution is 0.022 Ω –1cm –1 . When an equal volume of 0.1 M HCl solution is added, the conductivity decreases to 0.0056 Ω –1cm –1 . A further addition of HCl solution, the volume of which is equal to that of first portion added, conductivity increases to 0.017 Ω –1cm –1 . Calculate molar conductance of NaOH, NaCl and HCl solutions in the given conditions. PROBLEM 486 A sample of water from a large swimming pool has a resistance of 9200 Ω at 25°C when placed in a certain conductance cell. When filled with 0.02 M KCl solution, the cell has a

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resistance of 85 Ω at 25°C. Five hundred grams of NaCl was then dissolved in swimming pool, stirred throughly and a sample of this solution gave a resistance of 7600 Ω. Calculate the volume of water in the pool. Given molar conductivity of NaCl solution at that concentration is126.5 Ω –1cm 2 mol –1 and molar conductivity of KCl solution at 0.02 M is 138.3 Ω –1cm 2 mol –1 .

PROBLEM 487 The molar conductivity of a 0.05 M aqueous solution of MgCl 2 is 195 Ω –1cm 2 mol –1 at 27°C. A cell whose electrodes are of1.5 cm 2 surface area and placed at a distance of 0.5 cm, is filled with 0.05 M MgCl 2 solution. How much current will flow if the potential difference across the electrodes is 5 V? PROBLEM 488 A conductivity cell when filled with 0.02 M KCl gives resistance of 85 Ω. The same conductivity cell, when filled with a saturated solution of PbCl 2 , gives resistance of 70 Ω. If the equivalent conductance of 0.02 N KCl solution and saturated solution of PbCl 2 , under identical conditions are 138.3 and 106 cm 2 Ω –1eq –1 respectively, determine solubility product of PbCl 2 .

ELECTROCHEMISTRY PROBLEM 489 (a) When a current of 150 mA is used for 8.0 hr, what volume of fluorine gas at N.T.P. can be produced from a molten mixture of potassium and hydrogen fluoride? (b) If the same current is passed for same interval of time, how many litre of oxygen gas could be produced from electrolysis of water? PROBLEM 490 An aqueous solution of Na 2SO 4 was electrolysed for 30 min; 25 mL of O 2 was produced at the anode over water at 22°C at a total pressure of 722 torr. Determine the strength of current that was used to produce the oxygen gas. Vapour pressure of water at 22°C is 19.83 torr. PROBLEM 491 A piece of copper metal is to be electroplated on all side with silver to a thickness of 1 micrometre. If the metal strip measures 50 mm × 10 mm × 1 mm, how long must the solution, which consists the Ag(CN) –2 ion, be electrolysed, using a current of 100 mA? The density of silver metal is 10.5 g/cc. PROBLEM 492 19 g of fused SnCl 2 was electrolysed using inert electrode when 0.119 g of Sn was deposited at cathode. If nothing was given out during electrolysis, calculate the ratio of weight of SnCl 2 and SnCl 4 in fused state after electrolysis. PROBLEM 493 Electrolysis of molten KBr generates bromine gas, which can be used in industrial bromination process. How long will it take to convert a 500 kg batch of phenol to monobromo phenol using a current of 20 kA? PROBLEM 494 The cell Pt | H 2 (1 bar), H + || KCl (1.0 M saturated) | Hg 2Cl 2 | Hg was used to measure the pH of a solution of 0.01 M acetic acid in 0.0358 M sodium acetate. Calculate the cell potential expected at 25° C [K a = 1.81 × 10 –5 for acetic acid; E ° (Hg 2Cl 2 / Hg, Cl – ) = 0.28 V]

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PROBLEM 495 The voltage required to electrolyse certain solutions changes as the electrolysis proceeds because the concentration in the solution is changing. In an experiment, 500 mL of a 0.1 M solution of copper (II) bromide was electrolysed until 2.827 g of Cu was deposited. Calculated the theoretical minimum voltage required to sustain the electrolysis reaction at the beginning and at the end of experiment. E °Cu 2+ /Cu = 0.34 V and E ° Br / Br – = 1.07 V. 2

PROBLEM 496 Calculate the concentration of I −3 in a standard solution of iodine in 0.5 M KI, making use of the following standard electrode potentials: E ° (I 2 /I – = 0.5355 V) and E ° (I –3 /I – = 0.5365 V).. The molarity of I 2 in the standard solution can be assumed to be 0.5 M.

PROBLEM 497 Write both electrode reaction and overall cell reaction for the cell Tl | TlCl( s) || CdCl 2 (0.01 M) | Cd and calculate EMF of the cell if E ° (Tl + /Tl = – 0.34 V), E ° (Cd 2+ /Cd = – 0.40 V) and the solubility product of TlCl is 1.6 × 10 −3 . PROBLEM 498

The meal M forms a soluble nitrate and a very slightly soluble chloride. The cell M | M + (0.1 M ), HNO 3 (0.2 M ) || H 2 (1 bar), Pt

Has a measured EMF = – 0.4 V. When sufficient solid KCl is added to make the solution of the cell 0.2 M in K + , the EMF changes to – 0.05 V. Calculate the solubility product of MCl.

PROBLEM 499 The voltage of the following cell is 0.987 V Pt(H 2 , 1.0 atm ) | HOCN(1.4 × 10 –3 M ) || Ag + (0.8 M ) | Ag ( s) Calculate K a of HOCN, E ° Ag + / Ag = 0.8 V

PROBLEM 500 For the galvanic cell: Ag | AgCl( s), KCl(0.2 M ) || KBr(0.001 M ), AgBr ( s) | Ag Calculate the e.m.f. generated and assign correct polarity to each electrode for a spontaneous process after taking account of the cell reaction at 25°C. Given K sp of AgCl = 2.8 ×10 –10 and of AgBr = 3.3 × 10 –13 .

PROBLEM 501 Zinc granules are added in excess to 750 mL of 1.5 M Ni(NO 3 ) 2 solution at 30°C untill the equilibrium is reached. If emf of the cell Zn ( s) | Zn 2+ (1 M ) || H + (1 M ) | H 2 ( g ) 1.0 atm, | Pt is 0.76 V and that of cell

Ni | Ni 2+ (1 M ) || H + ( pH = 0) | H 2 ( g, 1 atm ) | Pt is 0.24 V, then calculate [ Ni 2+ ] at equilibrium. Also find the amount of zinc consumed.

PROBLEM 502 An electrochemical cell consist of a silver electrode in contact with 346 mL of 0.10 M AgNO 3 solution and a magnesium electrode in contact with 288 mL 0.1 M Mg(NO 3 ) 2 solution. (a) Calculate E for the cell at 25°C. (b) A current is drawn from the cell until 1.2 g of silver is deposited at the silver electrode. Calculate E for the cell at this stage of operation. E °(Ag + /Ag) = 0.8 V and E °( Mg 2+ /Mg) = – 2.37 V.

Problems

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PROBLEM 503 An electrochemical cell is constructed by immersing a piece of copper wire in 25 mL of a 0.2 M CuSO 4 solution and a zinc strip in 25.0 mL of a 0.2 M ZnSO 4 solution. (a) Calculate the e.m.f. of the cell at 25°C and predict what would happen if a small amount of concentrated NH 3 solution were added to (i) the CuSO 4 solution and (ii) to ZnSO 4 solution? (b) In a separate experiment 25 mL of 3.0 M NH 3 are added to the CuSO 4 solution. If the e.m.f. of the cell is 0.68 V, calculate formation constant of Cu(NH 3 ) 24+ . E °Cu 2+ /Cu = 0.34 V, E ° Zn 2+ / Zn = − 0.76 V PROBLEM 504 The voltage of a certain cell at 25°C and 20°C are 0.3525 V and 0.3533 V respectively. If the number of electrons involved in the overall reactions are two, calculate ∆G°, ∆S ° and ∆H ° at 25°C. PROBLEM 505 For the cell: Pt | H 2 ( g ) | sol X || KCl ( sat ) | Hg 2Cl 2 ( s) | Hg | Pt the observed e.m.f. at 25°C was 612 mV. When solution X was replaced by a standard phosphate buffer whose assigned pH is 6.86, the e.m.f. was 741 mV. Find pH of the solution X. PROBLEM 506 Use the following data to calculate solubility product of PbF2 . A galvanic cell consists of a hydrogen gas electrode ( p(H 2 ) = 1 bar , [ H + ] = 1 M ) and a Pb electrode in a 1.0 M KF solution in equilibrium with PbF2 ( s). The cell potential is 0.348 V and E ° (Pb2+ / Pb) = − 0.13 V. PROBLEM 507 A concentration cell has Zn electrodes. The electrolyte in each of the half-cells is a solution of ZnCl 2 dissolved in water. One has a freezing point of – 2.0°C and the other has a freezing point of – 0.9°C. What is the potential of the cell at 25°C? PROBLEM 508 A lead storage battery is allowed to discharge until 23.92 g of PbO 2 have been reduced. Determine the mass of PbSO 4 formed and time required to recharge the battery to its original state using a current of strength 3.0 A. PROBLEM 509 The chlorate ion ClO –3 can disproportionate in basic solution according to the reaction 2ClO –3 ( aq )

ClO –2 ( aq ) + ClO –4 ( aq )

What is the equilibrium concentration of the ions resulting from a solution initially at 0.1 M in chlorate ion? E ° (ClO –4 /ClO –3 ) = 0.36 V and E ° (ClO –3 /ClO –2 ) = 0.33 V.

PROBLEM 510 A platinum electrode is in contact with a solution containing Fe 2+ and Fe 3+ ions. The total concentration of iron ions is 0.033 M. What is the ratio of the concentrations if the potential difference between solution and metal surface is 35 mV and E ° (Fe 3+ / Fe 2+ ) = 0.77 V ? PROBLEM 511 Calculate Gibb’s free energy change and equilibrium constant for the reaction Ni 2+ + 2H 2O → Ni(OH) 2 + 2H + Given that and

NiO 2 + 4H + + 2e → Ni 2+ + 2H 2O NiO 2 + 2H 2O + 2e → Ni(OH) 2 + 2OH

E ° =1.678 V –

E ° = – 0.49 V

PROBLEM 512 The standard electromotive force of the cell: Fe | Fe 2+ ( aq ) || Cd 2+ | Cd is 0.0372 V The temperature coefficient of e.m.f. is – 0.125 V K –1 . Calculate the quantities ∆G°, ∆H ° and ∆S ° at 25°C.

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PROBLEM 513 The concentration of Ca 2+ in sea water is determined using a Ca 2+ selective electrode. A 10 mL sample of sea water is diluted to 100 mL and its 50 mL aliquot is placed in a beaker with Ca 2+ ion selective electrode and a reference electrode. The e.m.f. was found to be – 0.053 V. A 1.0 mL aliquot of 0.05 M Ca 2+ ion is added and emf was found to be – 0.0422 V. What is the concentration of Ca 2+ in sea water? Ignore volume change due to addition of second aliquot. PROBLEM 514 Use the van der Waals’ equation of state to calculate the pressure at 35°C that would result if the chlorine produced from electrolysis of 10 kg molten NaCl is compressed into a tank with a volume of 75 L. Assume 96% efficiency of the electrolysis. a = 6.5 atm L2 mol –2 , b = 0.0562 L mol −1 . PROBLEM 515 The pH of the solution in the cell Pt | H 2 ( g ) | HCl(aq.) || AgCl(s) | KCl(1M) | Ag is 0.65 and E ° (Ag + | Ag) = 0.8 V. Calculate e.m.f. K sp AgCl = 1.6 × 10 −10 . PROBLEM 516 The e.m.f. of the following cell: Ag(s) | AgCl(saturated), KCl( M = 0.05) | KNO 3 | AgNO 3 ( M = 0.1) | Ag ( s) is 0.4312 V. Calculate solubility product of AgCl. PROBLEM 517 The e.m.f. of cell: H 2 (g ) | Buffer || Normal calomal electrode is 0.6885 V at 40°C when the barometric pressure is 725 mm of Hg. What is the pH of the solution. E ° calomal = 0.28. PROBLEM 518 The cell : Ag ( s) | AgCl( s), HCl(0.1 N ) | Glass | Buffer || Saturated calomal electrode gave an e.m.f. of 0.112 V when the pH of the buffer used was 4.00. When a buffer of unknown pH was used, the potential was 0.3865 V. What is the pH of unknown buffer? PROBLEM 519 One hundred mL of 0.01 N KCl are titrated with 0.1 N AgNO 3 . Calculate potential of a silver electrode in the solution at equivalence point. The solubility product of AgCl is 1.56 ×10 –10 ? [E ° Ag + / Ag = 0.8 V] PROBLEM 520 A 2.0 V battery is used in an electrolysis in which an e.m.f. of 1.45 V is developed. If the resistance of the entire circuit is 10 ohms, determine magnitude of current and quantity of heat produced by the current flow per faraday of electricity. PROBLEM 521 A solution is made originally 0.1 M in Ag + ions and 0.25 M in KCN. If the dissociation constant of the complex Ag(CN) –2 is 3 × 10 –19 , what will be the concentration of Ag + ions in this solution and what will be the deposition potential of Ag? [ E ° (Ag + / Ag) = 0.8 V]

PROBLEM 522 A solution is 0.1 M in Au + ions and 0.1 M in Ag + ions. The standard reduction potential of Au + is 1.68 V, while the dissociation constant of Au(CN) –2 is 5 × 10 –39 . What concentration of NaCN will have to be maintained in the given solution in order to deposit Au and Ag simultaneously? [E ° Ag + / Ag = 0.8 V]. PROBLEM 523 A Galvanic cell consist of Zn anode dipped into a 1.0 L 0.2 M Zn(NO 3 ) 2 solution and a silver cathode dipped into a 1.0 L 0.1 M AgNO 3 solution at 25°C and has e.m.f. =1.52 V. Now, KCl(s) is added to the cathode chamber resulting in precipitation of AgCl and a change in e.m.f. After addition, e.m.f. was found to be 1.04 V and [ K + ] = 0.3 M. Determine K sp of AgCl.

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Problems

PROBLEM 524 (a) Calculate E° for: Ag 2S + 2e

2Ag( s) + S 2– if K sp of Ag 2S = 6 × 10 –50 and E ° Ag + / Ag = 0.8 V

(b) Consider an electrochemical cell having an indicator electrode Ag/Ag 2S coupled with a SCE (E° = 0.244 V) calculate [S 2– ] if the overall voltage of this cell is 0.766 V. – + PROBLEM 525 An I 2 (s)/ I (0.1 M ) half cell is connected to a H /H 2 (1.0 atm) half cell and e.m.f. found to be 0.755 V. If E °I 2 /I – is 0.535 V, determine pH of H + / H 2 half cell.

PROBLEM 526 To an aqueous solution of a weak acid (HA), 20 mL of a NaOH is added and connected to a reference electrode Ag/AgCl/Cl – (0.1 M ) electrode. E.m.f. of the cell was found to be 0.47 V. To the resulting solution, 30 mL of NaOH of the same strength was further added and new e.m.f. was found to be 0.5 V. Determine K a of the weak acid. E° (AgCl/Ag, Cl – ) = 0.23 V. PROBLEM 527 A direct current of 1.25 A was passed through 200 mL of 0.25 M Fe 2 (SO 4 ) 3 solution for a period of 1.1 hour. The resulting solution in cathode chamber was analyzed by titrating against acidic KMnO 4 solution. 25 mL permanganate solution was required to reach the end point. Determine molarity of KMnO 4 solution. PROBLEM 528 A 80 mL sample solution of KI was electrolyzed for 3.0 minute, using a constant current. The I 2 produced required 0.25 M, 37.2 mL sodium thiosulphate solution and unreacted KI required 36.3 mL 0.02 M acidic solution of KMnO 4 . Determine current strength and original molarity of KI solution. PROBLEM 529 K sp for AgBr = 8 × 10 –13 . What is the e.m.f. of the cell: Ag, AgNO 3 (1.0 M) || KBr (1.0 M), AgBr, Ag

PROBLEM 530 E.m.f. of the cell: Hg, Hg 2Cl 2 , KCl(1.0 M) || R 2 NH 2+ Cl – (0.1 M) | H 2 , Pt is – 0.52 V Calculate ionization constant of the base R 2 NH ⋅ E ° ( Hg 2Cl 2 | Cl – ) = 0.28 V. PROBLEM 531

An electrochemical cell is constructed with an open switch as shown below : V

Sn

Salt Bridge

0.5 M Sn2+

x

0.1 M Xn+

When the switch is closed, mass of tin-electrode increases. If E ° (Sn 2+ /Sn) = −0.14 V and for E ° (X n + /X) = − 0.78 V and initial emf of the cell is 0.65 V, determine n and indicate the direction of electron flow in the external circuit.

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PROBLEM 532

Consider the following electrochemical cell : V

Ag

Salt Bridge

Zn E°Ag+/Ag = 0.8 V E°Zn2+/Zn = –0.76 V Zn(NO3)2

0.1 M AgNO3

(a) (b) (c) (d)

Write a balanced net ionic equation for the spontaneous reaction that take place in the cell. Calculate the standard cell potential E ° for the cell reaction. If the cell emf is 1.6 V, what is the concentration of Zn 2+ ? How will the cell potential be affected if KI is added to Ag + half-cell?

PROBLEM 533 You have available a 0.001 M Al(NO 3 ) 3 , 0.1 M Cu(NO 3 ) 2 and 1.0 M Fe(NO 3 ) 2 solutions, Al, Cu and Fe metal strips. Construct a galvanic cell that would have greatest cell potential at 25°C. E ° 3+ = −176 . V, E ° 2+ = − 0.44 V, and E ° 2+ is +0.34 V. Al

/Al

Fe

/Fe

Cu

/Cu

PROBLEM 534 Magnesium metal is produced commercially by isolation of MgCl 2 from seawater followed by electrolysis of molten salt. MgCl 2 → Mg + Cl 2 (a) What mass of Mg can be produced if a current of 430 A is passed for 1.0 hour? (b) If a current of 500 A is used, how many hours will be required to convert all the 1000 kg MgCl 2 into Mg metal?

PROBLEM 535 Commercial production of Al-metal involves electrolysis of Al 2O 3 obtained from bauxite ore. The balanced net equation is : 2Al 2O 3 + 3C → 4Al + 3CO 2 (a) How many moles of Al are produced if 12 moles of electrons are passed through this cell? (b) How many kg of Al metal are produced if a current of 1250 A is passed for 1.25 hours? PROBLEM 536 Given :

Consider the following redox reaction : – + 2IrCl 3– 6 + 3HCOOH → 2Ir + 3CO 2 +12Cl + 6H CO 2 + 2H 3O + + 2e → HCOOH + 2H 2O IrCl 3– 6

E ° = – 0.20 V

–

+ 3e → Ir + 6Cl E ° = 0.77 V (a) Determine standard state emf of cell. (b) Is this reaction thermodynamically spontaneous as written? Briefly explain.

PROBLEM 537 An electrical source which might be used on a spacecraft should be light weight. Yet deliver high voltage and maximum useful energy. A galvanic cell using aluminium, one of the lightest metal, might be considered. Al | Al 3+ || Ce 4+ | Ce 3+ | Pt

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Problems

The cell is constructed with 2.0 L 4 M Al 2 (SO 4 ) 3 and 2.0 L 6 M Ce(SO 4 ) 2 . If E ° (Al 3+ /Al) = − 1.76 V and E ° (Ce 4+ /Ce 3+ ) = + 1.443 V. (a) Determine standard state emf and emf of the cell in the given conditions, if the cathode compartment is initially 1.0 M in Ce 3+ ion. (b) What is the maximum amount of useful electrical energy which could be derived from this cell? (c) The instruments on spacecraft require a reasonably steady voltage to operate correctly. What will be the cell voltage when the concentration of Ce 4+ ion has dropped to half of its original concentration? Assume this galvanic cell operate under a reversible condition at 25°C.

PROBLEM 538 A Daniell cell was constructed from 2.00 L 4 M aqueous CuSO 4 solution and 2 Lit 4 M aqueous ZnSO 4 solution. Also, given

e

Cu + ( aq ) → Cu( s)

E ° = + 0.522 V

Cu

2+

+

( aq ) + e → Cu ( aq ) E ° = 0.158 V

Zn

2+

( aq ) + 2e → Zn ( s)

E ° = – 0.76 V

(a) Calculate standard cell potential for this Daniell cell. (b) Calculate the cell potential when the concentration of Cu 2+ ( aq ) dropped to 0.4 M. Assume current was drawn slowly, so that this galvanic cell operated under reversible conditions at 25°C.

PROBLEM 539

The half-reactions for a typical rechargable nickel-cadmium (“nicad”) battery are : Cd(OH) 2 ( s) + 2e → Cd( s) + 2OH – ( aq )

E ° = – 0.81 V

Ni(OH) 3 ( s) + e → Ni(OH) 2 ( s) + OH – ( aq ) E ° = + 0.49 V (a) Identify cathode and write a balanced equation that shows the reaction that occurs spontaneously as the cell discharges. (b) If E° for electrolysis of molten MgCl 2 into Mg and Cl 2 ( g ) is –3.74 V, can this nicad battery be used to drive decomposition of molten MgCl 2 into Mg metal and Cl 2 gas? (c) If the nicad battery can deliver 0.1 A for 12 hour, how many grams of Mg can be produced from MgCl 2 by electrolysis? (d) The voltage delivered by a nicad battery doesnot change significantly as the battery discharges and the reaction reaches equilibrium. Using Nernst’s equation, briefly explain, why this is true?

PROBLEM 540 Lead storage battery is used in automobiles. In order to reduce the amount of hazardous lead and compounds that end up in landfills when these batteries are dumped, it has been suggested that the lead be replaced with carbon monoxide. The unbalanced cell reaction when the battery discharge is : PbO 2 ( s) +CO( g ) + H 2SO 4 ( aq ) → PbSO 4 ( s) + CO 2 ( g ) + H 2O E ° (PbO 2 /PbSO 4 ) = +1.685 V and

E ° (CO 2 ( g ) /CO( g )) = − 0.12 V

(a) Calculate E° for this lead-storage battery. (b) If density of sulphuric acid in battery decreases from 1.3 g/ml (40% H 2SO 4 by wt.) to 1.2 g/ml (20% H 2SO 4 by wt.) during its used, determine the ampere-hour for which battery has been used. Volume of battery is 3.0 Lit. (c) Voltage of such battery is maintained constant during its used. Very briefly explain, why this is so?

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PROBLEM 541 Diabetic must monitor their levels of blood glucose to determine the proper insulin dose. An electronic device that does this uses the reaction below: OH

OH HO HO

O

OH [Fe(CN) ]3– 6 H3O+

O

HO HO

O + [Fe(CN)6]4– OH

OH

gluconolactone (gln)

glucose (glu)

E ° (gln/glu) = + 0.29 V and

E ° {[Fe(CN) 6 ]3− / [Fe(CN) 6 ]4– } = + 0.69 V

(a) Balance the cell reaction and determine standard state cell potential. (b) Is this reaction spontaneous as written? (c) When the concentration of glucose is increased, what happens to the cell voltage?

PROBLEM 542

The diagram below shows the experimental setup for a typical Zn–Ni galvanic cell : e

e–

e–

Salt-Bridge

E ° (Zn 2+ /Zn) = – 0.76 V E ° (Ni 2+ /Ni) = – 0.25 V MSO 4 = 0.01 M XSO 4 = 1.00 M M2+ SO42–

X2+ SO42–

(a) Identify M and X and determine cell potential at 25°C. (b) If concentration of M 2+ ion changes to 1.0 M during its uses, what would be the new cell voltage? (c) Describe, what would happen to cell voltage if salt bridge was removed.

PROBLEM 543

At 25°C, H 2O 2 decomposes according to the following equation :

2H 2O 2 ( aq ) → 2H 2O( l) + O 2 ( q ) E ° = 0.55 V (a) Determine the value of K eq for decomposition reaction. 4e

(b) If standard reduction potential for O 2 ( g ) + 4H + → 2H 2O is 1.23 V, using this information in addition to the information provided above, determine the value of E° for O 2 ( g ) + 2H + + 2e → H 2O 2

PROBLEM 544 In an electrolytic cell, Cu(s) is produced by electrolysis of CuSO 4 (aq ). Calculate maximum mass of Cu( s) that can be deposited by a direct current of 100 A passed through 2 Lit 2 M CuSO 4 ( aq ) solution for a period of 1.5 hours. Also determine the final concentration of CuSO 4 in solution and volume of O 2 ( g ) measured at STP, produced at anode.

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Problems

(a) Using the following E° values :

PROBLEM 545

°+ ° 2+ + = 0.16 V, = 0.52 V, ECu ECu /Cu /Cu

determine K for : Cu( s) + Cu 2+ ( aq ) 2Cu + ( aq ). (b) If E° for the reaction Cu( s) + Cu 2+ ( aq ) + 2Br – ( aq )

2CuBr( s) is 0.5147 V, determine K sp

of CuBr at 25°C.

PROBLEM 546

Consider the galvanic cell Zn( s) Zn 2+ (0.04 M)Cl – (5 × 10 −3 M )Cl 2 (0.1 atm ) Pt( s)

(a) Determine emf of cell and equilibrium constant for the net cell reaction. (b) Now, if NH 3 is added to anode chamber at 1.0 M concentration, what would be the new emf. K f ° 2+ = – 0.76 V and E ° – = 1.36 V. for [Zn(NH 3 ) 4 ]2+ = 7.8 × 10 8 , E Zn Cl /Cl /Zn 2

PROBLEM 547

A galvanic cell was constructed using Ag/Cu electrode as shown below : CuCu 2+ ( aq ) Ag + (0.01 M) Ag

The cell emf was measured to be 0.382 V. Now some CaCl 2 was added to 250 ml electrolyte present in cathode chamber such that all Ag + precipitate out as AgCl and finally voltage drop to 0.01 V. If [Cu 2+ ] was 0.02 M, determine K sp of AgCl.

PROBLEM 548 A solution of M(NO 3 ) 2 is electrolyzed using a current of 2.5 A and 3.06 g metal was deposited in 35 minutes. Determine molar mass of the metal. PROBLEM 549 Consider the galvanic cell : PtSn 4+ ( aq, 0.010 M), Sn 2+ ( aq, 0.10 M)O 2 (g, 1.0 atm); H + (pH = 4)C( gr ) E ° 4+ 2+ = 0.15 V and E ° + = 1.23 V. Sn

/ Sn

O 2 , H /H 2 O

Assuming that the cathode chamber is buffered at constant pH = 4, determine (a) Cell-potential, (b) Concentrations of Sn 4+ and Sn 2+ when cell potential dropped to 0.8 V during its uses at 25°C.

PROBLEM 550 When a pH-meter was standardized with a basic-acid borate-buffer with a pH 9.4, the cell potential was 0.06 V. When the buffer was replaced by a solution of unknown hydronium ion, the cell potential was 0.22 V. Determine pH in this solution. PROBLEM 551 A 200 ml CuSO 4 solution was electrolyzed using a current of strength 4.0 A for 30 min. Determine pH of analyte at the end of electrolysis assuming initial pH = 7. PROBLEM 552 A solution of Mn(NO 3 ) 2 (0.15 M) and Fe(NO 3 ) 2 (0.1 M) is placed in a beaker and buffered at pH = 5. Two platinum electrodes are inserted and current is passed through the solution in order to plate out the metals. ° 2+ = − 0.44 V, E ° 2+ = − 1.18 V, and E Fe /Fe Mn /Mn

EO° , H + /H O = 1.23 V 2

2

Which metal will be deposited first and what minimum voltage would be required for the onset of electrolysis?

PROBLEM 553 Consider the electroplating of a metal +1 cation from a solution of unknown concentration according to the half reaction : M + + e → M with a standard potential of E°. When the half-cell is connected to an appropriate oxidation half-cell and current is passed, M + begins to

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plateout at a potential of E1 . To what value ( E 2 ) must the applied potential be adjusted, relative to E1 , if 99.99% of the metal is to be removed from the solution?

PROBLEM 554 A steel pot containing acidic tomato sauce is covered with the aluminium foil; the foil is in contact with both the pot and sauce. Latter a large excess of aluminium foil has dissolved. It is believed that the following electrochemical reaction caused aluminium to dissolve : 3Fe 2+ ( aq ) + 2Al

3Fe( s) + 2Al 3+ ( aq )

At some instant, emf was found to be 1.2 V. Determine the area of holes created in the Al-foil at that ° 3+ = − 1.66 V and E ° 2+ = − 0.44 V. Density of Al metal is 2.7 g/cc, thickness of instant. E Al Fe /Fe /Al aluminium foil and volume of sauce was 100 ml and [Fe 2+ ] = 5 × 10 –4 M.

PROBLEM 555 A balloon is being filled with hydrogen produced by the electrolysis of an aqueous solution of acid. How long will it take to generate enough hydrogen to lift 1.5 kg by using a current of 8.5 A. Average molar mass of air = 28.8 amu. PROBLEM 556 A galvanic cell was constructed by dipping a zinc electrode in 0.1 M Zn(NO 3 ) 2 and a Pt electrode in 0.1 M HIO 3 solution and emf of this cell was found to be 0.72 V at 25°C. ° 2+ = − 0.76 V. Determine acid dissociation constant (K a ) of HIO 3 . Assume partial pressure of E Zn /Zn gaseous species to be 1.0 atm.

PROBLEM 557 A battery was used to supply a constant current of what was believed to be exactly 0.4 A as read on a meter in the external circuit. The cell was based on the electrolysis of a 100 ml 0.1 M aqueous copper sulphate solution. After a 30 minute duration, concentration of electrolyte dropped to 0.0528 M. Determine the extent to which meter was incorrect? PROBLEM 558 A galvanic cell with a measured potential of 0.11 V at 25°C, contain a Pt electrode in a solution which is 0.0135 M in Cr +2 and 2.16 × 10 −4 M in Cr +3 . In second compartment a Ni electrode dips into a solution of Ni(NO 3 ) 2 . Determine concentration of Ni 2+ second compartment. E ° (Cr 3+ /Cr 2+ ) = − 0.41 V and E ° ( Ni 2+ /Ni) = − 0.25 V. PROBLEM 559 The electrolysis of Na 2SO 4 (aq ) is conducted into two separate half-cells joined by a salt bridge, also containing Na 2SO 4 . The cell diagram for the electrolysis is Pt Na 2SO 4 ( aq ) Na 2SO 4 ( aq ) Pt Phenolphthalein indicator is added to each half-cells. (a) Describe any colour-change occurring in half-cells as electrolysis progresses. (b) After electrolysis is stopped, the solution of the two half-cells are mixed. Describe and explain any colour change that occur. (c) In an experiment 10 ml HCl is added to cathode chamber alongwith phenolphthalein. Electrolysis is carried out with a 25 mA current and the solution colour becomes pink after 8.00 minutes. What is the molarity of HCl? 2+ ° PROBLEM 560 Determine formation constant of complex HgI 2– /Hg) = 0.85 V and 4 . E (Hg 2– – . E ° (HgI 4 /Hg, I ) = − 0.04 V

Problems

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CHEMICAL KINETICS PROBLEM 561 A sample of pitch blend has amount of 206 Pb equivalent to one-fifth of the amount of 238 U (by mass). If the disintegration constant of 238 U is 1.54 ×10 –10 Y –1 and all Pb is supposed to be obtained from uranium, what will be the age of mineral? PROBLEM 562 The gaseous decomposition of Cl 2O 7 (g ) according to following reaction follow 1st order kinetics: 7 Cl 2O 7 ( g ) → Cl 2 ( g ) + O 2 ( g ) 2 When the decomposition is carried out in a closed container, pressure measured after 15 min and after a very long time were found to be 240 and 900 mm of Hg respectively. Calculate the pressure that would be observed in the flask after 140 min under identical conditions of temperature. PROBLEM 563 The isomerization of cyclobutene to 1,3-butadiene follow first order kinetics and the rate constant is 2 × 10 –4 s –1 at 150°C in a 530 mL flask. Determine the partial pressure of cyclobutene and its concentration after 30 min if an isomerization reaction was carried out at 150°C with an initial pressure of 55 mm of Hg. PROBLEM 564 The hydrolysis of sugar sucrose to sugars glucose and fructose C12 H 22O11 + H 2O → C 6 H12O 6 + C 6 H12O 6 follow first order kinetics. (a) In a neutral solution, k = 2.1 × 10 –11 s –1 at 27°C and 8.5 ×10 –11 s –1 at 37°C. Determine activation energy, frequency factor and rate constant at 47°C. (b) When a solution of sucrose with an initial concentration of 0.15 M reaches equilibrium, the concentration of sucrose is 1.65 ×10 –7 M. How long will it take the solution to reach equilibrium at 27°C? PROBLEM 565 Fluorine-18 is a radioactive isotope that decays by positron emission to form oxygen-18 with a half-life of 109.7 min. Physician use 18 F for the study of brain by injecting a quantity of fluoro substituted glucose into the blood of a patient. The glucose accumulates in the regions where the brain is active and needs nourishment. (a) What is the rate constant for the decomposition of fluorine-18? (b) If a sample of glucose that contains 18 F is injected into the blood, what percentage will remain after 5.6 hours? (c) How long does it take for 99% decay of 18 F ? PROBLEM 566 The nucleidic ratio of 1 H 3 to 1 H1 in a sample of water is 8.0 ×10 –18 :1. Tritium undergoes decay with a half-life period of 12.3 years. How many tritium atoms would a 10 g sample of water contains 40 year after the original sample is collected. PROBLEM 567 The decomposition of N 2O 5 according to the following reaction follow first order kinetics: 2N 2O 5 → 4NO 2 + O 2 After 30.0 min from start of decomposition in a closed vessel, the total pressure found to be 284.5

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mm of Hg and after a very long time, the total pressure found to be 584.5 mm of Hg. Calculate the rate constant and total pressure after 1.0 hour.

PROBLEM 568 A certain radioactive isotope z X A (half-life =10 days) decays to give z – 2 X g of X is kept in a sealed vessel, find the volume of He accumulated at S.T.P. in 20 days.

A–4

. If 1.0

PROBLEM 569 A reaction rate increases by a factor of 500 in the presence of a catalyst at 37°C. If the reaction is carried out in presence of a catalyst at 127°C, by what factor, compared to original rate in absence of catalyst at 27°C, the rate will increase? Activation energy of original pathway is 106 kJ/mol. PROBLEM 570 A gaseous mixture containing N 2 and an unknown gas A2 (g ) are present in a sealed vessel at 700 mm of Hg. Gas A2 decomposes as A2 ( g ) → 2 A ( g ) with progress of time and decomposition follow first order kinetics. If the total pressure inside the flask measured after 30 min and 60 min are 760 and 800 mm of Hg, calculate t1/ 2 and total pressure after 75 min. PROBLEM 571 If the rate constant for the decomposition of COCl 2 (g ) according to the following reaction: COCl 2 ( g ) → CO + Cl 2 is expressed as log k ( min –1 ) = 15 –

∆H =103 kJ/ mol

11420 . Calculate activation energy for the formation of COCl 2 ( g ) at T ( K)

the same temperature.

PROBLEM 572 Gaseous decomposition of N 2O 5 follows first order kinetics. Pure N 2O 5 gas is taken in a flask, sealed and finally heated to 100°C where decomposition occurs as follows: 1

N 2O 5 → 2NO 2 + 2 O 2 After 15 minutes, a leak was developed in the flask. On analysis of the gaseous mixture coming out initially, mole fraction of NO 2 was found to be 0.5. Calculate half-life for the decomposition reaction.

PROBLEM 573 The half-life of a drug is 3.0 hours. A patient is administered a 5 mg dose of this drug at an interval of 6.0 hour. Calculate the mass of drug remaining in the patient ’s body just after 6th dose. PROBLEM 574 0.01 moles of sulphuryl chloride, SO 2Cl 2 (g ) is taken in a sealed tube and heated to 400 K where it decomposes following first order kinetics according to the following reaction: SO 2Cl 2 ( g ) → SO 2 ( g ) + Cl 2 ( g ) The tube is broken after 4.0 hour and gas passed through a 15 mL of an acidified, 1.0 N iodine solution where all SO 2 is oxidised to SO 2– 4 . The resulting solution required 7.0 mL 1 M hypo solution. Calculate the total pressure inside the flask just before it broken down. If the volume of the flask is 200 mL, also calculate its half-life. PROBLEM 575 A reactant A decomposes to produce two different products simultaneously, but at different rate as shown below: k2

k1

E2

E1

C ← A → B The order of reaction is unity for both reaction and their respective rate constants k1 and k 2 are 1.2 ×10 –2 and 3 × 10 –2 per second respectively. Calculate activation energy for the overall reaction if the activation energy for individual steps E1 and E 2 are 180 and 200 kJ/mol respectively.

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Problems

PROBLEM 576 The reaction cis - Cr(en) 2 (OH) +2

K1 K2

trans - Cr(en) 2 (OH) 2+

is first order in both directions. At 25°C, the equilibrium constant is 0.16 and rate constant k1 is 3.3 ×10 –4 s –1 . In an experiment starting with the pure cis form, how long would it take for half the equilibrium amount of the trans-isomer to be formed?

PROBLEM 577

40

K decays into

40

Ca and

40

Ar simultaneously as:

t 1/ 2 = 1.5 × 109 y

→ 20 K

20Ca

40

+

–1β

0

40 t 1/ 2 = 1.2 × 1010 y

→ 18 Ar 40 + +1β 0 If the atomic concentration of Ar in a sample of rock is 10% that of potassium, calculate the age of the rock sample considering the only source of Ar is radioactive disintegration of potassium.

PROBLEM 578 2,3-di methyl-3-chloro pentane on heating with methanol undergoes substitution as well as elimination according to the following reaction: Cl CH3 CH 3 OMe     CH3OH C 2 H 5 —C— CH —CH 3 → C 2 H 5 —C== C— CH 3 + C 2 H 5 —C—CH—CH 3 ∆     CH 3CH 3 CH 3 CH 3 A 100 mL 0.1 M solution of dextro-rotatory sample of 2,3-di methyl-3-chloro pentane is heated with excess of methanol and optical-rotation were measured at different intervals as: t (min) 0 10 ∞ Optical-rotation 40° 30° 5° Calculate the optical-rotation after 20 minute and volume of 0.15 M aqueous solution of bromine required after 20 minute for bromination of alkene produced if the molar ratio of alkene to ether is 6 : 4.

PROBLEM 579 The activation energy for the reaction H + CH 4 → H 2 + CH 3 has been found to be 49.8 kJ/mol. Some estimates of enthalpies of formations, ∆H f° are : H = 218, CH 4 = – 74.8 and CH 3 = 139.5 kJ/ mol. Estimate activation energy for the reverse reaction. PROBLEM 580 The half-life for the thermal de-naturation of haemoglobin, a first order process, has been found to be 3460 s at 60°C and 530 s at 65°C. Calculate the enthalpy of activation and entropy of activation at 60°C assuming the Arrhenius equation to apply. PROBLEM 581 For a parallel reaction: A

→ B E a 1 = 150 kJ/ mol → C E a 2 = 80 kJ/ mol → D E a 3 = 30 kJ/ mol

Mole percentage of B, C and D after one hour was found to be 10, 25 and 15 respectively. Determine the overall activation energy for this reaction.

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PROBLEM 582 You need to convert 1.0 mole of X into product. The reaction follow first order kinetics. At 1.00 p.m. you start the reaction at 27°C with one mole of X. At 2.0 p.m. you find that 0.8 mole of X remains. You immediately increase the temperature to 37°C. At 3.0 p.m. you discover that 0.4 mole of X are still remaining. You want to be able to finish the reaction by 4.0 p.m. but you cannot stop the reaction until only 0.1 mole of X remains. You decided to increase the temperature again. What is the minimum temperature required to complete the reaction by 4.0 p.m.? PROBLEM 583 Gaseous decomposition of A2 (g ) into A(g) follow first order kinetics as: A2 ( g ) → 2 A ( g ) If the decomposition is carried out in a sealed flask at a constant temperature, the total pressure measured after 20 minute and 40 minute from the beginning are 2.5 atm and 2.875 atm respectively. Calculate the total pressure inside the flask after 1.0 hour. PROBLEM 584 A gaseous reactant A decomposes to produce gaseous products B and C in a parallel reaction, both by first order, as follows: K = 2 × 10–3 min –1

1 B (g ) →

A(g ) → C (g ) –2 –1 K 2 = 3 × 10

min

If the decomposition is carried out in a sealed flask, partial pressure of B(g) at the end of reaction was found to be 40 mm of Hg. Determine partial pressure of C after 1.0 hour.

PROBLEM 585 Isomerization of cyclobutene into 1,3-butadine follow first order kinetics as: (g )

(g)

The kinetic study was performed by taking same amounts of cyclobutene in three sealed flasks. First flask was broken after 20 minute and the reaction mixture was absorbed completely in bromine solution. 16.0 mL 1.0 M bromine solution was required. The second flask was broken after a very long time and the reaction mixture required 20 mL bromine solution of the same strength. If the third flask was broken after 30 minute, what volume of bromine solution of same strength would have been required?

PROBLEM 586 Racemization of an optically pure enantiomer in its aqueous solution follow first order kinetics and optical rotation found after 10 minute and 20 minute from beginning were 40° and 24° respectively. Determine the optical-rotation after 40 minute. PROBLEM 587 Decomposition of both A2 (g ) and B 3 (g ) follows 1st order kinetic as : K1

A2 ( g ) → 2 A ( g ) K2

B 3 ( g ) → 3B ( g )

K1 ( hr

−1

K 2 (hr

−1

2

−

14000 RT

3

−

20000 RT

) = 10 e ) = 10 e

If one mole of each A2 ( g ) and B 3 ( g ) are taken in a 10 L evacuated flask and heated to some temperature so that they start decomposing at the same rate, determine total pressure in the flask after 1.0 hr.

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Problems

PROBLEM 588 Decomposition of a non-volatile solute “ A ” into another non-volatile solute B and C, when dissolved in water follow first order kinetics as: A → 2B + C When one mole of A is dissolved in 180 g of water and left for decomposition, vapour pressure after 12 hours was found to be 20.69 mm of Hg. Determine V.P. of the solution after 30 hours. Assume constant temperature through out to be 27°C and V.P. of pure water at 27°C is 24 mm of Hg.

PROBLEM 589 For the reaction A + B + C → P , following rate informations are available : Rate (mol L–1 hr –1 )

[A] L mol −1

[B] L mol −1

[C] L mol −1

2.5 × 10 −3

0.1

0.2

0.3

10 −2

0.1

0.2

0.6

1.67 × 10 −3

0.1

0.3

0.2

1.78 × 10 −2

0.2

0.4

0.4

Deduce the rate law.

PROBLEM 590 A sample of 0.42 mg 233 UF6 shows an activity of 9.88 ×10 4 count per second. Find t1/ 2 . PROBLEM 591 The uranium present in earth today is 99% 238 U and 1% 235 U. The half-lives are 4.5 ×10 9 years and 7 × 10 8 years respectively. How long ago, this mineral was 50% in each isotope? PROBLEM 592 Mole fraction of A as a function of time in the following reversible first order reaction A B are: Time (hr)

0

1

Infinity

Mole % of A

100

75

30

Determine mole percentage of A after four hour from the beginning.

PROBLEM 593 Isomerization of N-chloroacetanilide to p-chloro acetanilide follow first order kinetics and isomerization is achieved by adding KI solution to N-chloroacetanilide solution. Liberated iodine is titrated with standard thiosulphate solution for kinetic study and following results were observed: Time (hr)

0

1

Vol. of 0.1 M S 2O 32– required

50

35

Determine volume of thiosulphate that would be required after 8.0 hour assuming only N-chloroacetanilide reacts with KI.

PROBLEM 594 Decomposition of phosgene gas as COCl 2 with:

Kf Kb

CO + Cl 2 follow first order kinetics

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PROBLEM 595 Thermal decomposition of H 2O 2 follows first order kinetics. Three flask, each containing 25 mL 2.5 V H 2O 2 were taken for kinetic study. Into the first flask, 20 mL of a KI solution was added after 1.0 hour, heated gently to expel-off liberated iodine and finally titrated against 0.4 M chromic acid solution. 10.4 mL of chromic acid was required to reach end point. After three hour from beginning, 20 mL of KI solution of same strength was added to the second flask, heated gently and finally titrated against chromic acid solution of same strength. After a very long time, similar experiment was repeated with the third flask and required 16.67 mL chromic acid solution. What was the volume of chromic acid required for second experiment? PROBLEM 596 The acid catalyzed hydrolysis of ethyl acetate is first order in ester and overall rate constant for the hydrolysis reaction is 0.11 L mol –1 s –1 . Determine time required for 30% hydrolysis of ester at pH 2.5. PROBLEM 597 A mixture of two different substance A and B undergo simultaneous reaction to produce the same product as: K = 2 × 10−2 min −1

1 A 

K = 5 × 10−3 min −1

2 B 

C

After 30 minute from beginning, mole percentage of C was found to be 25. Determine its mole per cent after 50 minute assuming initially there were no C.

PROBLEM 598 Isotopes of oxygen with mass number less than 16 undergo β + emission. Assuming an equimolar mixture of O14 and O15 , determine time required for nuclei ratio O14 /O15 to become equal to 0.25. t1/ 2 for O14 and O15 are 71 and 124 second respectively. PROBLEM 599 A first order chemical reaction has t1/ 2 = 200 hr at 27°C. Adding a catalyst lowers the half-life to 20 hour. How long the reaction would take to complete 75% at 127°C in presence of this catalyst. Activation energy of the uncatalyzed pathway is 80 kJ. PROBLEM 600 The compound Xe(CF3 ) 2 is unstable and decomposes by 1st order reaction as: Xe(CF3 ) 2 ( g ) → Xe ( g ) + C 2 H 4 ( g ) + F2 ( g ), t1/ 2 = 30 minute at 300 K, If 1.345 g of Xe(CF3 ) 2 ( g ) is taken in an evacuated 100 mL flask, sealed and left for some time at 300 K, pressure inside the flask was found to be 2.63 bar. How long did the gas stay in flask? PROBLEM 601 The nuclide Ac 227 decay by β − emission (90%) and α-emission (10%). If 100 g of pure actinium-227 is taken in a sealed evacuated 1.0 L flask, pressure inside the flask after 5.0 hour was found to be 0.16 bar at 300 K. Determine half-life of the decay process. PROBLEM 602 Nitrogen oxides NO x (a mixture of NO and NO 2 ) plays an important role in the

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Problems

production of photochemical smog. The half-lives for the removal of NO and NO 2 in smoke stack emission are 1.5 hr. and 2.5 hr. respectively. Starting with 1.5 g of mixture in an experiment, 0.225 g of NO x was left after 5.0 hour. Determine composition of original mixture assuming both gases are removed by first order reaction. How many hours of light must elapse to decrease amount of NO x to 2.5 × 10 −6 g. K

PROBLEM 603 For a chemical reaction A → product, following mechanism has been proposed : K

E a =190 kJ

K

E a = 80 kJ

1 A → A* 2 A * → B*

K3

B * → Product

E a = 40 kJ

K K  Also, overall rate constant K is related to individual rate constants by the equation: K =  1 2   K3  Determine activation energy for the overall reaction.

2/ 3

.

PROBLEM 604 The first order decomposition of a coloured chemical species X into colourless products is monitored with a spectrophotometer by measuring changes in absorbance over time. The data from the experiment are given below: [ X ] (M)

Absorbance

Time (min)

?

0.600

0.00

0.200

35.0

0.150

44.2

4.00 × 10

−5

3 × 10 −5

Calculate initial concentration of the coloured species and time when absorbance will fall to 0.0075. The graph below show results from a study of the decomposition of H 2O 2 .

–1.0

ln [H2O2]

PROBLEM 605

–2.0 –3.0 –4.0 0

800

1600 Time (min)

2400

4800

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(a) Write rate law for the reaction justifying your answer. (b) Determine value of rate constant and [H 2O 2 ] after 2000 minute. PROBLEM 606 The gas phase reaction: 2Cl 2O + 2N 2O 5 → 2NO 3Cl + 2NO 2Cl + O 2 has rate law: Rate = K [N 2O 5 ]. If the reaction was carried out in a sealed flask with equimolar amounts of reactant, the total pressure measured after 10 minute and after a very long time were 650 and 750 mm of Hg respectively. What was the total pressure after 30 minute from beginning? Assume constant temperature throughout the experiment. PROBLEM 607 For a chemical reaction A → P, the ratio of amount of reactant left unreacted to the initial amount ([A] / [A]0 ) were measured after 400 seconds and 964 seconds and were found to be 0.5 and 0.25 respectively. Determine the order of reaction and rate constant if initial moles of reactant was 0.6. PROBLEM 608 Considering the following concentration of reactant left unreacted Vs time graph for a first order chemical reaction, determine rate of reaction at 10 sec.

0.4

Molarity

0.3 0.2 0.1 0

20

10

30

40

50

Time in sec

PROBLEM 609 For a first order chemical reaction, variation of rate constants vs temperature is depicted in the following graph:

–1

log K (s )

3

2

1 x=1 (0, 0)

1

x 2

3

4

5

6

7

8

9

4 –1 1 × 10 (K ) T

Determine the time required for 80% completion of reaction at 500 K.

PROBLEM 610 Rate of a first order chemical reaction measured after 10 min and 30 min were 10.64 × 10 −3 and 8.37 × 10 −3 mol L–1 min –1 . What was the rate after 2 hr. assuming temperature was constant throughout?

Problems

PROBLEM 611

Join @unacademyplusdiscounts 75 A reversible chemical reaction: Kf

A

Kb

P

is first order in both A and P and has equilibrium constant of 2 at 300 K. If rate constant for the forward direction is 2.5 ×10 –2 min –1 at 300 K, determine the time required to form half of equilibrium amount of P assuming that initially there was no P.

PROBLEM 612 A 1st order chemical reaction was carried out for 1.0 hour in absence of a catalyst and 20% reaction was complete. A catalyst was then added and reaction was allowed to continue for next 30 min when 60% reaction was complete. A second catalyst was then added at this time and reaction was allowed to continue for further 10 min. when 90% reaction was complete. If activation energy of the original path was 80 kJ, determine activation energies of catalyzed pathways. Assume constant temperature throughout to be 300 K. PROBLEM 613 The half-life of 14 C isotope is 5730 years. The activity of carbon in a living being is 12.5 counts per minute per gram of carbon. (a) Determine the percentage of carbon in living being that is 14 C. (b) Determine activity of carbon from remains of an organism that died 50,000 years ago. (c) Determine age of Egyptian tomb that shows an activity of 7.0 counts per minute per gram of carbon.

COLLIGATIVE PROPERTIES PROBLEM 614 A certain aqueous solution of a non-electrolytic solute freezes at – 0.6°C. For this solution estimate (a) normal boiling point (b) the vapour pressure at 25°C. (c) the osmotic pressure at 20°C. The vapour pressure of pure water at 25°C is 23.75 torr. K f =1.86 and K b = 0.52. PROBLEM 615 At 25°C, vapour pressure of pure benzene and pure toluene are 93.4 and 26.9 torr respectively. A solution is prepared by mixing 60 g of benzene and 40 g of toluene. What pressure should be maintained in the flask containing this solution so that it start boiling at 25°C. PROBLEM 616 A urea solution in 250 g of water freezes at – 0.744°C. This solution was cooled to some temperature where some ice is formed. Solution was decanted-off and heated to 100°C where the vapour pressure was found to be 757.7 mm of Hg. Determine mass of ice formed and temperature to which solution was cooled. K f of water is 1.86 kg/mol. PROBLEM 617 At 27°C, 12 L of pure N 2 measured at 1.0 atmosphere are passes through an aqueous solution of a non-volatile solute, whereby the solution loses 0.25 g in weight. If vapour pressure of pure water is 23.75 mm of Hg and K b is 0.52 kg/mol, determine boiling temperature of this solution. PROBLEM 618 An aqueous solution containing 10 g mixture of urea and glucose boils at 100.58°C. Addition of a further 6.0 g glucose to the above solution causes it to boil at 100.77°C. Determine mass percentage of urea in the original mixture. PROBLEM 619 A one litre solution is prepared by dissolving some solid lead-nitrate in water. The solution was found to boil at 100.15°C. To the resulting solution 0.2 mole NaCl was added. The resulting

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solution was found to freeze at – 0.83°C. Determine solubility product of PbCl 2 . Given K b = 0.5 and K f =1.86. Assume molality to be equal to molarity in all case.

PROBLEM 620 Vapour pressure of a volatile substance A is 400 mm of Hg at 298 K. When some another volatile solvent B (V.P. = 500 mm at 298 K) is added to A, vapour pressure of the resulting solution was found to be 470 mm of Hg. To the resulting solution a third volatile solvent C (V.P. = 600 mm of Hg at 298 K) is added so that the vapour pressure of the resulting solution becomes equal to 496 mm of Hg. Determine mole fraction of each A, B and C in the vapour phase in equilibrium with the final solution. PROBLEM 621 A solution, prepared by dissolving 38.2 g of Na 2 B4O 7 ⋅ xH 2O in 250 g water has boiling point 100.582°C. Deduce formula of the salt. K b of the water is 0.52 kg mol −1 . [Atomic mass of B =11, Na = 23] PROBLEM 622 A mixture of NaCl and sucrose of combined mass 10.2 g is dissolved in enough water to make up a 250 mL solution. The osmotic pressure of the solution is 7.32 atm at 23°C. Calculate the mass percentage of NaCl in the mixture. PROBLEM 623 A certain organometallic complex contain carbon, hydrogen and a metal from the first transition series (one metal per molecule). In a non-protic complexing solvent like diethyl ether, this compound dissociate into ions. In a non-polar solvent like benzene, this compound does not dissociate. Below are data obtained from freezing point depression and boiling point elevation experiment carried out on the sample of this compound. (a) Freezing point of benzene solution containing 1.5 g of complex in 36 g benzene is 4.35°C. (fr. pt of C 6 H 6 is 5.5°C and K f of benzene is 5.12) (b) The b.pt of diethyl ether solution containing 0.48 g of the complex in 10 g of the solvent is 36.05°C (b.pt. of pure ether is 34.5°C, K b = 2.02) (c) Elemental analysis : 64.88% C and 5.45 %H. In how many ions does the complex dissociated when dissolved in the complexing solvent? Write the molecular formula of the complex. PROBLEM 624 A glucose solution in 100 g of water boils at 100.26°C. If this solution is heated to 101°C, determine the mass of water left at equilibrium. PROBLEM 625 A 100 g 10% by weight urea solution is placed together with a 200 g 10% by weight glucose solution in a closed jar and allowed to attain the equilibrium. Determine mass percentage of urea in its solutions at equilibrium. PROBLEM 626 A 100 g 10% by weight urea solution is placed together with a 200 g glucose solution in a big jar and left for a very long time. The vapour pressure in the jar at equilibrium was found to be 23.6 mm of Hg. If the vapour pressure of pure water at experimental temperature is 24 mm of Hg, determine mass percentage of glucose in its solution. PROBLEM 627 The vapour pressure of a solution of methanol and ethanol at 20°C was found to be 70 mm of Hg. Adding 10 g of urea to 80 g of this solution lowers the vapour pressure to 64.6 mm of Hg. Determine composition of the original solution. PROBLEM 628 Vapour pressure of an equimolar mixture of benzene and toluene was found to be 80 torr. If the vapour above the liquid phase is condensed in a beaker, vapour pressure of this condensate was found to be 100 torr. Determine vapour pressure of pure benzene and pure toluene in the given condition.

Problems

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PROBLEM 629 Vapour pressure of methanol and ethanol are 94 and 44 mm Hg at 20°C. To a 50 g mixture containing equal weight of both methanol and ethanol, 11.85 g of a mixture of NaCl and urea was dissolved and vapour pressure of the resulting solution was found to be 58 mm of Hg. Determine mass percentage of urea and NaCl in the mixture. PROBLEM 630 Determine vapour pressure of water at 50°C if enthalpy of vaporization of water is 40.6 kJ/mol. PROBLEM 631 An aqueous solution of canesugar ( MW = 342) has osmotic pressure equal to 1.5 atm at 18°C. What will be the V.P. of this solution at 40°C. If 100 g of this solution is cooled to – 2°C, what mass of ice will be separated out. (V.P. of H 2O = at 40°C = 55.324 mm of Hg, K f =1.86, density =1.0 g/ mL) PROBLEM 632 Two elements A and B forms compounds having molecular AB 2 and AB 4 . When dissolved in 20 g of benzene, 1.0 g of AB 2 lowers freezing point by 2.3 K whereas 1.0 g of AB 4 lowers the freezing point by 1.3 K. Determine atomic masses of A and B. The molal depression constant for benzene is 5.1 K kg mol −1 . PROBLEM 633 1.0 g of a monobasic acid when dissolved in 100 g of water lowers the freezing point by 0.168°C. 0.2 g of the same acid when dissolved and titrated, required 15.1 mL of N/10 alkali. Calculate degree of dissociation of the acid. K f for water is 1.86. PROBLEM 634 A complex is represented as CoCl 3 ⋅ xH 2O. Its 0.1 m solution in aqueous medium shows ∆T f = − 0.558 K. K f for water is 1.86 and assume 100% ionization of complex having coordination number 6, deduce the formula of the complex. PROBLEM 635 10 g of a weak monobasic acid in 100 g of benzene freezes at 3.35°C. Assume that acid undergoes complete association in benzene. Find the degree of dissociation of 10 g of acid in 100 g of water whose freezing point found to be – 1.75°C. K f for water =1.86. K f for the benzene = 5.12 and freezing point of benzene = 5.51° C. PROBLEM 636 A saturated solution of a sparingly soluble salt MCl 2 has a vapour pressure of 31.78 mm of Hg at 30°C, while pure water exert a pressure of 31.82 mm of Hg at the same temperature. Calculate solubility product of the compound at this temperature. PROBLEM 637 A protein has been isolated as sodium salt with their molecular formula Na x P (this notation means that xNa + ions are associated with a negatively charged protein P −x ). A solution of this salt was prepared by dissolving 0.25 g of this sodium salt of protein in 10 g of water and ebulliscopic analysis revealed that solution boils at temperature 5.93 × 10 −3 ° C higher than the normal boiling point of pure water. K b of water is 0.52 Kg mol −1 . Also elemental analysis revealed that the salt contain 1% sodium metal by weight. Deduce molecular formula and determine molecular weight of acidic form of protein H x P.

PROBLEM 638 A non-volatile organic compound X was used to make-up two solution. Solution A contains 5.0 g of X in 100 g of water and solution B contains 2.0 g of X in 100 g of benzene. Solution A has vapour pressure of 754.5 mm of Hg at normal boiling point of water and solution B has the same vapour pressure at the normal boiling point of benzene. Assuming X undergo partial dimerization in benzene, determine percentage of X dimerized in benzene solution.

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PROBLEM 639 1.32 g of a mixture of cyclohexane and naphthalene is dissolved in 20 g of benzene and solution freezes at 2.2°C. Calculate composition of the mixture, given K f of benzene is 5.12 kg mol −1 and freezing point of benzene is 5.51°C.

PROBLEM 640 When 1.0 of urea is dissolved in 200 g of an unknown solvent X, the X freezing point is lowered by 0.25°C. When 1.5 g of an unknown, non-electrolytic solute Y is dissolved in 125 g of same solvent X, freezing point is lowered by 0.2°C and vapour pressure is lowered by 1%. If freezing point of X, is 12°C, determine molar enthalpy of fusion of X. PROBLEM 641 Chloroform boils at 62°C. For a solution containing 0.4 g of naphthalene in 25 g of chloroform, boiling point is elevated by 0.45 K. If chloroform is to be distilled at 50°C in order to avoid any decomposition during distillation, what maximum pressure can be maintained in the flask? PROBLEM 642 Phenol undergoes partial dimerization in bromoform. When 2.5 g of phenol is dissolved in 100 g bromoform, freezing point is lowered to 6.4°C. Pure bromoform freezes at 8.4°C and its K f is 14 Kkg mol −1 . Determine K c (equilibrium constant) for the following dimerization reaction if the density of solution is 0.88 g/mL; 2C 6 H 5OH

(C 6 H 5OH) 2

PROBLEM 643 6.0 g of a mixture of anthracene (C14 H10 ) and naphthalene (C10 H 8 ) when dissolved in 300 g of benzene, freezing point is lowered by 0.7°C. Normal freezing point of benzene is 5.5°C and its K f is 5.12 Kkg mol −1 . Determine percentage composition of mixture and mass of solid benzene that will be produced if the above solution is cooled to 4.5°C.

PROBLEM 644 A 1.0 m aqueous solution of HF freezes at – 1.9°C. K f of water is 1.86 Kkg mol −1 . If the density of solutions is 1.12 g/mL, determine pH of solution and ionization constant of HF.

SOLID STATE PROBLEM 645 A body centre cubic lattice is made up of two different type of atom A and B. Atom A occupying body centre and B occupying the corner positions. One of the corner is left unoccupied per unit cell. Deduce the empirical formula of such solid. PROBLEM 646 A face centred cubic solid is made up of two different type of atoms A and B. Atom A occupying the corner and B are at the face centres. One of the face centre is left unoccupied per unit cell. Deduce the formula of the solid. PROBLEM 647 A face centred cubic lattice is made up of a single type of atom and one of its corner is left unoccupied per unit cell. Calculate packing fraction of such solid. PROBLEM 648 A body centred cubic lattice is made up of hollow spheres whose inner radius is one-fourth of its outer radius. Calculate packing fraction of such solid. PROBLEM 649 A hexagonal close packed atomic solid has some defect and its one corner and one face centre is left unoccupied per unit cell. Calculate packing fraction of such solid.

Problems

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PROBLEM 650 Al crystallizes in cubic shape unit cell and with edge length 405 pm and density 2.7 g/cc. Predict the type of crystal lattice. PROBLEM 651 CdO has NaCl structures with density 8.27 g/cc. If the ionic radius of O 2− is 1.24 Å, determine ionic radius of Cd 2+ . PROBLEM 652 A compound of mercury chlorine has a crystal structure with two formula weights per unit cell. The unit cell edges are 4.47 Å, 4.47 Å and 10.89 Å respectively and the unit cell angles are all 90Å. The density of substance is 7.15 g/cc, determine formula of compounds. PROBLEM 653 Sulphide of cobalt metal has a cubic structure with four formula weights per unit cell. If density is 4.269 g/cc and edge length of unit cell is 6.93 Å, determine mass of sulphur required to produce 1.5 kg of this compound. PROBLEM 654 Graphite has HCP arrangements of carbon atoms and the parallel planes are 3.35 Å apart. Determine density of graphite. PROBLEM 655 An atomic solid crystallizes in hexagonal closest packing. Calculate packing fraction. PROBLEM 656 An ionic solid crystallizes in hexagonal closest packing in which cations occupies the middle layer of unit cell. Calculate packing fraction of the ionic solid. PROBLEM 657 An atomic solid crystallizes in a body centre-cubic lattice and the inner surface of the atoms at the adjacent corner are separated by 60 pm. If the atomic weight of A is 48, calculate density of the solid. PROBLEM 658 FeO crystallizes in NaCl type of crystal lattice. The crystal however are always non-stoichiometric and deficient in iron. Some cation sites are vacant and some contains Fe 3+ ions but the combination is such that structures is electrically neutral. The formula approximate to Fe 0.95O. (a) What is the ratio of Fe 2+ to Fe 3+ ion in the solid. (b) What percentage of cation sites are vacant? PROBLEM 659 Cadmium oxide crystallizes in NaCl type of crystal lattice. The compound is however usually non-stoichiometric with approximate formula CdO 0.95 . The defect arises due to some cationic positions are occupied by neutral Cd-atom instead of Cd 2+ ions and equivalent numbers of anion sites are vacant: (a) What percentage of anion sites are vacant? (b) If the edge length of the unit cell is 470 pm, what would be the density of perfect solid? (c) What is the density of non-stoichiometric solid. PROBLEM 660 A hexagonal closed packed solid is made-up of cylindrical atom. Calculate packing fraction of this solid. PROBLEM 661 In an atomic BCC, what fraction of edge is not covered by atoms? PROBLEM 662 CaO crystallizes in face-centred cubic lattice with unit cell edge length of 4.8 Å. Calculate the closest distance between oppositely charged ions. PROBLEM 663 In an atomic FCC, all the positions are occupied by A and the body centred octahedral hole is appropriately occupied by B, without disturbing the FCC of A. Calculate packing fraction of this solid.

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PROBLEM 664 What is the packing fraction of the two dimensional unit cell shown in figure?

PROBLEM 665 What fraction of the Ca atoms lies on the surface of a cubic crystal that is 1.0 cm in length. Radius of Ca atoms is 96 pm and it crystallizes in BCC arrangement. PROBLEM 666 TlAl(SO 4 ) 2 ⋅ xH 2O is BCC with ‘ a’ =1.22 nm. If the density of the solid is 2.32 g/cc, determine x. PROBLEM 667 The density of CaF2 is 3.18 g/cc. Calculate the unit cell edge length for CaF2 . PROBLEM 668 Ar crystallizes in FCC arrangements and density of solid and liquid Ar are 1.59 and 1.42 g/cc respectively. Find percentage of empty space in liquid Ar. PROBLEM 669 An atomic BCC is made-up of A type of atom and a guest atom B is inserted in the unit cell of A. If radius of A is R, what could be the maximum radius of B without disturbing the unit cell dimension of A. Also determine packing fraction of this solid. PROBLEM 670 Na crystallizes in BCC arrangement with the interfacial separation between the atoms at the edge 54 pm. Determine density of this solid. PROBLEM 671 Polonium crystallizes in simple cubic crystal lattice. If the size of largest atom that can be placed at the body centre of unit cell is 120 pico meter, determine density. M ( Po) = 209. PROBLEM 672 Density of one form of CaCO 3 solid is 2.93 g/cc and it has orthorhombic unit cell with unit cell parameter : a = 4.6 Å, b = 5.7 Å and c = 57 Å. Calculate number of Ca 2+ ions per unit cell. PROBLEM 673 Determine packing fraction of NaCl solid in the following cases: (a) Ions along an axis connecting opposite face centres are absent. (b) Ions along an axis connecting opposite edge centres are absent. (c) Ions along an edge diagonals are absent.

SURFACE CHEMISTRY PROBLEM 674 In the adsorption of hydrogen gas over a sample of copper powder, 1.36 cm 3 of H 2 (g ) measured over S.T.P. was found to absorb per gram of copper powder. Considering only mono-layer adsorption, determine specific surface area of copper. Density of liquid H 2 is 0.07 g/cc.

PROBLEM 675 A 0.016 M of an acid solution in benzene is dropped on a water surface, the benzene evaporates and the acid forms a monomolecular film of solid type. What volume of the above solution would be required to cover a 500 cm 2 surface area of water with monomolecular layer of acid? Area covered by single acid molecule is 0.2 nm 2 .

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Problems

PROBLEM 676 One gram of activated charcoal has a surface area of 1000 m 2 . Considering complete coverage as well as monomolecular adsorption, how much ammonia in mL at S.T.P. would be adsorbed on the surface of 25 g charcoal? Diameter of a ammonia molecules is 0.3 nm. PROBLEM 677 Calculate surface area of a catalyst of which adsorb 100 cm 3 of N 2 (g ) at S.T.P. per gram in order to form a monomolecular layer of N 2 ( g ). The effective surface area occupied by nitrogen molecule on the surface of catalyst is 0.16 nm 2 . PROBLEM 678 Discuss how extent of adsorption varies on (a) Increasing temperature of system. (b) Increasing surface area of adsorbent. (c) Increasing the gas pressure. PROBLEM 679 Effect produced on extent of adsorption by changing either temperature or pressure are consistent with Le-Chatelier’s principle. Justify. PROBLEM 680 Classify adsorption of gases on solids on the basis of force of interaction between adsorbent and adsorbate. What are their main characteristics? PROBLEM 681 What are surface active substances? Explain why the surface tension of a liquid is very much modified in the presence of a surface active substance?

REACTION MECHANISM PROBLEM 682 Rank the following sets of intermediates in increasing order of their stability giving appropriate reasons for your choice. (a) •• CH 2 , •• C(C 6 H 5 ) 2 , CH 3CH •• , C 6 H 5CH •• , O 2 NCH 2CH •• (b) C 6 H 5+ , p-NO 2 (C 6 H 4 ) + , p-CH 3 —(C 6 H 4 ) + , p-Cl—C 6 H +4 + +

+

(c) + +

(d) O

O

+

O

O O O O O      − − − − (e) CH 2 —C—CH 3 , CH 2 —C—H, CH 2 —C—OCH 3 ,CH 2 —C—NH 2 , CH 2 —C—CH 2 —OCH 3 , −

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PROBLEM 683 Reorder the following sets of compounds according to increasing pK a . (a) 1-butyne, 1-butene, 1-butanol, H 2O (b) 1-butanamine, aniline, butanamide (c) Acetic acid, oxalic acid, 1,3-propandioic acid. (d) Protonated forms of pyrrole, pyridine, N-methyl pyrrole. (e) Phenol, cyclohexanol, cyclohexan carboxylic acid, p-nitrophenol. PROBLEM 684 2-butenal has sufficient acidic character despite the fact that it lacks enolizable hydrogen α-to carbonyl group. Explain? PROBLEM 685 For the following compounds, arrange the labelled proton in increasing order of their ease of deprotonation : C≡≡C—H3

O

(b)

(a) H2

CH2

O

H

(c) C H

N

1

H

1

H H

2

H2C H

H1

O

3

H (d)

CH2

H

1

3

2

2

O

SO3H

3

CH2 O

O

PROBLEM 686 In the following pair of compounds, A is insoluble in water but soluble in dilute sulphuric acid while B is insoluble in both water and dilute acid but soluble in aqueous solution of strong base, explain : O

O

O

NH2

A

NH2

B

PROBLEM 687 Draw all proton tautomers of each of the following structures and arrange them in the increasing order of their enol content. O NH2 (a)

N

(b) N H

H

O (c)

O

O

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Problems

PROBLEM 688 K a 1 of phthalic acid is greater than K a 1 of isophthalic acid while reverse is true for K a 2, explain. PROBLEM 689 Benzene sulphonic acid is a stronger acid than benzoic acid, explain. PROBLEM 690 Which is stronger acid and why-phenol or thiophenol? PROBLEM 691 Compare the acid strength of benzoic acid with three isomeric hydroxy benzoic acid. PROBLEM 692 Orthohydroxy benzoic acid is much stronger than ortho methoxy benzoic acid, justify. PROBLEM 693 Which is a stronger acid, A or B and why? COOH

COOH

NO2

NO2

A

B

PROBLEM 694 Which is a stronger base, A or B and why? NH2

NH2 or NO2

NO2

A

B

PROBLEM 695 Discuss the mechanism of following reactions : (a)

H

(b) CH 3CH 2CH •2 → CH 3CH==CH 2 + CH 3CH 2CH 3

PROBLEM 696 Discuss the following observations : (a) C—Cl bond in vinyl chloride is stronger than in chloroethane. (b) Carbon-carbon bond length in ethene is shorter than in CH 2 == CHOCH 3 . (c) CH 3SH is stronger acid than CH 3OH. (d) CH 3CH 2 NH 2 is stronger base than CH 2 == CHNH 2 . PROBLEM 697 Although trimethyl amine has greater electron density at nitrogen atom than methyl amine, later is stronger Lewis-base in water than former. Explain.

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PROBLEM 698 A cyclobutandicarboxylic acid exist in two stereo-isomeric forms in which one is polar but non-resolvable while other is non-polar but resolvable into enantiomers. Deduce structures of all these compounds . PROBLEM 699 Which of the following rearrangement is thermodynamically favoured and why? +

+

—CH2 A +

CH2

+

B

PROBLEM 700 An alkyne is more reactive than alkene in catalytic hydrogenation reaction while reverse is true for electrophilic addition reaction, justify. PROBLEM 701 Unlike naphthalene, azulene have significant zwitter ionic character and dissolves in aqueous acid. Provide a mechanistic explanation.

Azulene

Naphthalene

PROBLEM 702 Label the following 1-4 in order of decreasing acidity. Explain your answer with the aid of mechanism: OH OH OH OH

NO2 B

PROBLEM 703

OCH3

CH3O

A

NO2 C

D

Discuss the basic strength of two nitrogens in benzimidazole. N N H Benzimidazole

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Problems

PROBLEM 704 Number the following compounds in order of increasing acidity of indicated proton giving mechanistic reasoning: CH2—H

O—H

CH2—CH2—H

II

I

O—H

III

IV

PROBLEM 705 For each of the following acid-base reactions, predict the side of the equilibrium that will be favoured. Briefly explain the choice by describing the features of the species involved that bias the equilibrium to the chosen side: O

O +

H

+

NH + NH4

N H

O

O

+ CH3—C

+

…(i)

+ NH3

N

+ CH3—C

N

+

…(ii)

NH

N

H

CF3

CF3

CF3

H

+

+

CF3 +

+

N

N

H

H

…(iii) N

N H

PROBLEM 706 2,6-di-t-butyl pyridine is more basic the pyridine but later is a stronger nucleophile than former. Rationalize. PROBLEM 707 With the help of structure, discuss the positions of Br  C  H and Cl  C  H planes in 1-bromo-3-chloro-1,2-propadiene. BrHC==C==CHCl (1-bromo-3-chloro-1,2-propadiene)

PROBLEM 708 Taking into account hybridization and resonance effects, rank the following C  C bonds in order of decreasing length: I

III

II

IV +

+

+

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Which side is favoured at equilibrium, provide quantitative explanation:

PROBLEM 709 O

O

+ H2 O

O

O

O

O

–

+ HO

O

O

–

pKa = 13

pKa = 16

Which of the following reaction would proceed to the right? Explain reason for your

PROBLEM 710 choice?

+

NH3

(a)

+

NO2

NH2

s

NO2 + H

pKa = 9.33

pKa = 10

+

NH3 +

(b)

pKa = 10.71

PROBLEM 711 acid?

NO2

NO2

NH2 +

s

H

For the following pairs of compounds, predict with clear reasoning, which is stronger

+ N—H or

(a)

—NO2 or HO—

(b) HO—

—NO2

N+ H

O (c)

O

O

O (d)

or

MeO

OMe

O

O

COOH

(e)

or

(f)

COOH

or

or N+

N+

H

H

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Problems

or

(g)

(h) PhCOOH or PhCO3H OH

OH (i)

or NO2 NO2

PROBLEM 712

From the following pair, select the stronger acid providing clear reasoning: OH OH

(a) O N 2

COOH

or

COOH

or

(b) COOH

NH3Cl (c)

COOH

NH3Cl or

PROBLEM 713 From the following pair, select the stronger base: (a) p-methoxy aniline or p-cyanoaniline (b) pyridine or pyrrole (c) CH 3CN or CH 3CH 2 NH 2 CH3

N

O2N

CH3

NH2 O2N

NO2

(d)

NO2

or

NO2

NO2

PROBLEM 714 In each of the following pair of compounds, which is more basic in aqueous solution? Give an explanation for your choice: (a) CH 3 NH 2 or CF3 NH 2 NH  (b) CH 3CONH 2 or H 2 N NH 2

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(c) PhNH 2 or CH 3CN (d) C 6 H 5 N(CH 3 ) 2 or 2,6-dimethyl-N-N-dimethylaniline (e) m-nitroaniline or p-nitroaniline

PROBLEM 715

(a) Arrange the following in increasing order of acid-strength: +

(i) H 2SO 4 (ii) CH 3CHCH 3 (iii) CH 3CH 2 O H 2 (iv) CH 3CH 2CH 3  SH (b) What are the major components at equilibrium if 1.0 mol of each of the four components below are mixed together? +

(i) HSO –4 (ii) CH 3CHCH 3 (iii) CH 3CH 2 OH 2 (iv) CH 3CH 2CH 3  S–

PROBLEM 716

Which is the better site of protonation in the following compound and why? (a)

PROBLEM 717 nitrogen?

l l

—N

N

l l

CH3 (b) CH3

In the following structure, which is better site of protonation and why-oxygen or

O

N—

H+

PROBLEM 718 Rank the following from greatest (most) to smallest (least): (a) Proton acidity: O

(i)

H

H

H

;

;

;

C

B

A

H

D

O O—H

(ii)

N—H

A

C B

CH3

O—H

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Problems

(b) Basic strength: – CH2

– NH O

O– , O– C

, D

C

B

A

C

,

,

–

–

O , F

E

(c) Heat of combustion:

A

B

D

C

H

H

E

(d) Rate of reaction with tertiary butanol:

HCl,

HBr , B

A

HI,

HF.

C

D

(e) Rate of reaction with HCl: OH OH CH3OH,

OH

and

A C

B

D

(f) Relative stability: +

+

CH2 ,

A

B +

+

, C

D

,

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(g) Rate of reaction with NaCN: Br

I

,

OMe

, C

B

A

F

,

,

D

Cl

E

(h) Heat of hydrogenation

(i) C

B

A

D

,

,

,

(ii)

B

A

C

,

E

D

(i) Rate of reaction with potassium t-butoxide:

A

F,

OTs ,

OMe B

C

Cl D

,

Br E

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Problems

(j) Rate of reaction in SN-2 reaction: Br O ,

Br ,

MeBr , B

Br

C

A

,

Br

D

E

PROBLEM 719 Draw the important resonance structures that contribute significantly to the resonance hybrid of the following molecules. If there is more than one important resonance contributor, indicate the major one. CH3

CH3

N

O

(i)

B

(ii)

– N l l

l l

(iii)

O

PROBLEM 720

Compare the C  N bond-length in the following species: H2 N

H2 N

O (i)

PROBLEM 721 choice.

(ii)

(iii)

Rank the followings in increasing order of acidic strength: H +

PROBLEM 722 choice:

H2 N

giving reasons for your

H

H

N

N

I

II

N+

N +

III

H

H

Rank the followings in increasing order of basic strength, explaining reason for your H

H

NLi+

N

N

I

II

III

–

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Arrange the following in increasing order of electrophilic strength:

PROBLEM 723

+

+

+

II

I

III

PROBLEM 724 In which direction would the following equilibria lie predominantly: (i) NH 3 + OH – NH 2 – + H 2O (ii) CH 3 NH 2 + (CH 3 ) 3 NH + CH 3 NH 3 + + (CH 3 ) 3 N (iii) CH 3 NH 2 + H 2O CH 3 NH 3+ + OH –

Lists the following groups of cations in increasing order of acidity explaining reasons

PROBLEM 725 for your order:

O +

+

+

NH2

(a)

and why?

N

A

NH2

H

O +

+

NH2

NH2

(b)

+

NH2

B

A

C

B

C

NH3+ +

(c)

Ph2NH2

PROBLEM 726

C

B

A

+

Ph3NH

Discuss the enhanced basic strength of the following base: NH N

N N

H2 N

N

I

NH2 III

II

PROBLEM 727 For the following compounds, draw all the resonance structures in which every atom has complete valence shell. CH3 CH3

N

+

N—CH3

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Problems

PROBLEM 728 Answer the following questions: (i) Which of the indicated H is abstracted rapidly by bromine radical and why? Ph

Hc

Hb Ha

(ii) One of the indicated proton H a or H b , is approximately 10 30 times more acidic than other, which is more acidic and why? H

H

Ha

Hb

(iii) Which proton is more rapidly abstracted by ethoxide ion and why? O

O

Br

Hb

Ha

(iv) Which compound is more rapidly attacked by a nucleophile and why? O

O OCH3

or (b)

(a)

PRLOBLEM 729

Discuss the relative acid strength of the following substituted phenols: —OCH3 HO—

HO— (i)

PROBLEM 730

—Cl

—CHO

HO— (iii)

(ii)

(iv)

The basic amino acid agrinine has the following structural formula: NH NH2 H2N

N H

HO—

COOH

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It has three pKa values. pKa1 = 2.18, pKa 2 = 9.09 and pKa 3 = 13.2. Give the formula of agrinine as it exist in aqueous solution at pH: (a) below 2.18, (b) between 2.18 − 9.09, (c) between 9.09 − 13.2.

PROBLEM 731

Discuss the relative acidity of the following compounds:

H

NH2

NH2

N and

(b)

and

(a)

N

O

NO2

H

PROBLEM 732

Amino acid shown below has three pKa values: O pKa1 = 1.94 H2 N

OH NH2

pKa2 = 8.65 pKa3 = 10.76

Give formula of above amino acid in aqueous solution when pH is: 1.9–8.7 (c) between 8.7–10.8 (d) above 10.8.

PROBLEM 733 for your order:

(a) below 1.9 (b) between

Rank the following acids in increasing order of their pKa value, explaining reasons

COOH COOH

COOH

H2N

(1)

(2)

COOH

NO2

(3)

NO2 (4)

PROBLEM 734

The γ-proton of crotonaldehyde is sufficiently acidic, justify.

PROBLEM 735 The two ammonium ions shown below differ in their pKa value by more than four units, explain the difference. +

—CH3

H3N— A

+

—CH2—NH2—CH3

and B

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Problems

STEREOCHEMISTRY PROBLEM 736 How many stereoisomers are possible for 1,3-dichlorocyclopentane. Draw structural formulae and discuss the stereochemical relationship among them. PROBLEM 737 Draw structures of all stereoisomers of 1-bromo-2- chlorocyclopentane and discuss the stereochemical relationship among them. PROBLEM 738 For each of the following pair, deduce the stereochemical relationship, i.e., whether they are enantiomers, diastereomers or identical. Br Br (a)

C Et

H CH3

and

C H

Et CH3

H H CH3 OH (b) CH3 C—C and CH3 C—C CH H 3 OH OH OH H H CH (c) 3 Cl

Cl Cl and C2 H 5

H——C2H5 H——Cl

H

(d)

CH3 H

CH3 H C==C

C

CH3 and H Cl

CH3 H

Cl C==C

C

H H CH3

PROBLEM 739 (+) 2-butanol has specific rotation of +13.9° when measured in pure form. A sample of 2-butanol was found to have an optical rotation of –3°. What is the stereomeric composition of this mixture? PROBLEM 740 (a) Draw structures of all stereoisomers of C 2 HClBrF. (b) Draw structures of all stereoisomers of 1,3,5-trichloro-1,4-pentadiene and discuss the stereochemical relationship among them. (c) How may isomers (including stereoisomers) can be obtained on monochlorination of 2-methyl butane? Draw structures of all isomers. PROBLEM 741 A hydrocarbon A(C 6 H 8 ) is resolvable and it decolourise brown colour of bromine water. A, on treatment with H 2 /Pt yields C 6 H14 ( B ) which can’t be resolved into enantiomers. Deduce structures of A and B. PROBLEM 742 An organic compound A has molecular formula C10 H17 Br and it can be resolved into enantiomers. Also, A decolourises cold, dilute and alkaline solution of KMnO 4 . A on treatment with alcoholic solution of KOH yields two organic compounds, B and C both have molecular formula C10 H16 . Only B is resolvable but C is non-resolvable. Reducing either B or C yields the same 4-methyl-1-isopropylcyclohexane. Deduce structures of A to C.

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PROBLEM 743 An organic compound A has molecular formula C 4 H10O and it can be resolved into enantiomers. A on treatment with chromic acid yields B (C 4 H 8O) which is non-resolvable. B on treatment with CH 3 NH 2 yields C (C 5 H11 N) which exhibit stereoisomerism. C on treatment with H 2 | Pt yields D(C 5 H13 N) which is a racemate. Deduce structures of A to D. PROBLEM 744 Bromination of cis-3,4-dimethyl-3-hexene yields a racemic mixture of vicinal dibromide while catalytic hydrogenation of the same alkene yields a meso compound, explain. PROBLEM 745 An optically active organic compound A(C 8 H13Cl) does not decolourise bromine water solution. A on treatment with alcoholic KOH can produce two products in principle but infact only one product B (C 8 H12 ) is obtained. B on treatment with ozone followed by work-up with H 2O 2 yields C (C 8 H12O 4 ) which cannot be resolved into enantiomers. C on heating with NaOH/CaO yields D(C 6 H12 ) which on monochlorination yields C 6 H11Cl as single isomer. Deduce structures of A to D. PROBLEM 746 An optically active hydrocarbon A has molecular formula C 6 H12 . A on treatment with H 2 / Pt gives inactive B. B on monochlorination yield four products C, D, E and F, all having molecular formula C 6 H13Cl. C is enantiomeric while D is diastereomeric. Both E and F are non-resolvable. Treatment of E with alcoholic solution of KOH yields two possible alkene G and H while similar treatment with F gives only H. Deduce structures of A to H. PROBLEM 747 For each of the following molecules, indicate, whether they are chiral, achiral or meso compound: CH3

H HO A B

CH3 OH CH3 H

CH3

H H

H3C

H

H

Br E

CH3 H

H

CH3

I

H

G

F

CH3

H

H3C

CH3

Br

H

D

C

CH3 CH3

H

PROBLEM 748 For each of the following pair of structures, indicate, if the compounds are identical, constitutional isomers, enantiomers, distereomers, different. OH

(a)

OH

and HO OH

H

OH

H

H

(b)

H

CH 3

H 3C

OH H

and

CH 3

OH

HO

H CH 3

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Problems

PROBLEM 749 Discuss the optical activity of the following two compounds and also label them as polar and non-polar. Cl Cl

Cl H H

H

H

Cl II

I

PROBLEM 750

Discuss the origin of optical activity in 1,3-dichloropropdiene.

PROBLEM 751 From the following four structures select: (a) The optically active isomers, (b) Optically inactive isomers, (c) Enantiomer pairs, (d) Distereomer pairs.

PROBLEM 752

Cl

Cl

Cl

Cl

I

II

III

IV

Consider the following six structures: OH H N CH3

CH3

HO

O

I

OH

H N

CH3

HO

O CH3 II

OH H N

HO

O CH3

CH3 III

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CH3

HO

O CH3 IV

OH H N CH3

CH3

HO

O

V

OH

H N

HO

CH3

O CH3 VI

Establish the stereochemical relationship between: and V, (e) IV and VI.

PROBLEM 753 molecule :

(a) I and II, (b) III and IV, (c) II and III, (d) I

Draw structures of all the stereoisomers produced on addition of HCl to the following CH3 HCl

PROBLEM 754 Draw the energy vs dihedral angle diagram for rotation around C 2  C 3 bond of 2,3-dimethyl butane. PROBLEM 755 For the following two molecules, draw structures of one set of functional, one set of positional and one set of stereoisomers. C 5 H 6O 2 I

and C 5 H11 NO II

PROBLEM 756 How many stereomers are formed upon hydration of the following compound (in kinetic condition) : H3C

CH3

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Problems

PROBLEM 757 Indicate the stereo centres in the following molecule and total number of stereomers in the following molecule. Also draw the structures of pair of distereomers. H N O

O O

I

PROBLEM 758 Draw the structures of stable configuration obtained after acidic hydration of the following unsaturated compounds: (exclude rearranged products)

CH3 I

CH3

CH3

CH3 III

II

IV

PROBLEM 759 N-methylethenamine as such does not show any stereoisomerism but one of its resonance form exhibit stereoisomerism. Explain. PROBLEM 760 Draw the Newman projection formula of important conformers around C 2  C 3 bond of 2,3-dimethylpentane and label them according to the increasing order of stability. PROBLEM 761

Select the compound which is optically active. Draw the structure of stereoisomers:

CH2CH3

O (b)

(a)

(c)

(d)

OH

OH

PROBLEM 762 From the following set of compounds, select: (a) enantiomer (b) distereomer pairs in which both are optically active, (c) label them as D or L sugar.

pairs,

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CHO HO

HOH 2C

H

H H

OH OH

HO

CH 2OH

H H H H

HO HO HO

H CH 2OH

H

OH

HO H H

H OH

OH

OH

CHO

CHO

II

I

III

CHO

CHO OH

H

H H

HO HO

CH 2OH

H

H

HO HO HO HO

H H H CH 2OH

OH IV

V

How many stereoisomers are possible for the molecule shown below–draw their

PROBLEM 763 structures:

O

COOH OH

Draw resonance structure of the amides shown below and select them which are

PROBLEM 764 stereoisomeric: O (a)

(b)

H2 N

H

CH3

(c)

N H

N H

PROBLEM 765

Discuss the type of isomerism exhibited by the following pairs: N N N

N (a)

N CH3

O (d)

O

O

and

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CH3

(b)

CH3

and

OH

OH

CH3

CH3

(c)

and O

O

OH OH

OH OH

O CH3 and

(d)

OH

H3C

PROBLEM 766 For the following enol tautomers, write structures of corresponding keto tautomers and other enol tautomers if possible and label them as chiral/achiral. OH OH

I

II

H (a) I

F (b)

F Cl F

Cl CH3

H II

I

Br C2H5

H3C Cl I

HO (c)

III

Establish the stereochemical relationship in the given sets of compounds:

PROBLEM 767

H3C

OH

I

H3C

C2H5

Br II

COOH

OHC HO I

H

F CHO

H

H

Cl

H

HO HO II

COOH

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H

H 3C

CH3

CH3

OH

H

H3C

OH

H3C

(d) H OH

H

CH3

H

OH

I

OH

OH

H

II

III

PROBLEM 768 Both of the quaternary ammonium salts shown below are resolvable into enantiomers but one when dissolved in water racemizes. Identify that one and explain the reason for its racemization in aqueous solution. C6H5 H H3C

N+

CH3CH2

N+ CH2CH

H3C

CH2

I

PROBLEM 769

II

CH C2H5

CH2

Discuss the optical isomerism of the following molecule:

O

O

H3CH2C

PROBLEM 770 Both cis and trans isomers of 3-methoxycyclohexanol are chiral. However, one distereomer reacts with a base and then with methyl iodide to give an optically inactive compound while the other distereomer gives a racemic mixture of optically active compound. Explain. PROBLEM 771 conformers:

Indicate, whether each of the following structures is isomers, resonance structures, Cl O

OH and

(a)

(b)

N

N

and Cl

H (c)

B

O

H

and

sB

O +

O (d)

O and

H

Problems

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HYDROCARBONS PROBLEM 772 A hydrocarbon A has molecular formula C10 H10 and decolourise brown colour of bromine water. A on treatment with HgSO 4 / H 2SO 4 produces two isomeric compounds B and C, both having molecular formula C10 H12O. Compound B forms an yellow coloured precipitate on treatment with alkaline solution of I 2 while C does not. Also A on treatment with O 2 followed by work-up with H 2O 2 yields D(C 8 H 8O 2 ) as one of the product. Heating D with soda lime (NaOH/CaO) yields E (C 7 H 8 ). E on treatment with chlorine in presence of sunlight yields F (C 7 H 7Cl) as single isomer while E on treatment with Cl 2 in dark but in presence of AlCl 3 produces two isomeric compounds with their molecular formula C 7 H 7Cl. Deduce structures of A to F. PROBLEM 773 An organic compound A has molecular formula C 8 H12O and it evolve a colourless gas on treatment with Na-metal. Also A forms a white precipitate with Tollen’s reagent. A on treatment with HgSO 4 / H 2SO 4 yields B (C 8 H14O 2 ) which of reduction with NaBH 4 yields a resolvable compound C (C 8 H16O 2 ). A on boiling with aqueous KMnO 4 gives D(C 7 H12O 3 ). D can also be obtained by treatment of aqueous KCN with cyclohexanone followed by hydrolysing the product with dilute H 2SO 4 . Deduce structures of A to D. PROBLEM 774 A hydrocarbon A has molecular formula C10 H10 and decolourises brown colour of bromine water. A on treatment with ozone followed by work-up with dimethyl sulphide (CH 3 ) 2 S yields B (C10 H10O 2 ). B gives positive iodoform test as well as positive Tollen’s test. A on boiling with acidic KMnO 4 yield C (C 8 H 6O 4 ) as one of the product. C on heating with P2O 5 dehydrate to yield D(C 8 H 4O 3 ). C does not decolourise brown colour of bromine water. Identify A to D. PROBLEM 775 A hydrocarbon A has molecular formula C11 H18 and it decolourise bromine water solution. A has a chiral centre and structure of four stereo-isomers can be drawn for A. A on treatment with H 2 / Pt yields B (C11 H 22 ) which contain two chiral centre. A on treatment with O 3 followed by work-up with Zn – H 2O yields the following compound and ethanol. Deduce structures of A and B and O O   draw structures of all stereo-isomers of A. CH 3CH 2 —CH—CH 2 —C—CH 2 —C—CH 3 .  CHO PROBLEM 776 A hydrocarbon A has molecular formula C10 H18 and it decolourise purple colour of cold, dilute and alkaline permanganate solution. A on treatment with HBr yields B (C10 H19 Br) which can’t be resolved into enantiomers while A on treating with HBr in presence of a peroxide yields, C an isomer of B, but C can be resolved into enantiomers. Also A on treatment with acidic KMnO 4 produces D(C 7 H12O 2 ) as one of the product which on heating with NaOH/CaO yields E (C 6 H12 ). E on treating with Cl 2 ( g ) in presence of sunlight yields C 6 H11Cl as single mono-chloro derivative. Deduce structures of A to E. PROBLEM 777 An organic compound A has molecular formula C13 H10O. A decolourises Bayer’s solution but does not evolve any gas on treatment with Na-metal. A on controlled hydrogenation yields B (C13 H14O) which does not decolourise brown colour of bromine water solution. A on treatment with O 3 followed by work-up with H 2O 2 yields C (C 8 H 8O 2 ) and D(C 5 H 4O 3 ). Heating C with NaOH/CaO yields

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E (C 7 H 8 ) while similar treatment on D yields furan. E on heating with Br 2 yields F (C 7 H 7 Br) F on treatment with aqueous solution of KCN followed by hydrolysis of product with dilute H 2SO 4 yields C. Deduce structures of A to F.

PROBLEM 778 An optical active organic compound A has molecular formula C13 H18O and decolourise brown colour of bromine water solution but does not give any gas on heating with sodium metal. A on hydroboration-oxidation yields B (C13 H 20O 2 ) which is still resolvable. B on refluxing with dilute solution of H 2SO 4 yields C (C 7 H 8O) and D(C 6 H14O 2 ) neither of them can be resolved into enantiomers. C on treating with chromic acid solution gives benzoic acid while D on heating with concentrated solution of H 2SO 4 yields E (C 6 H12O). E does not evolve any gas on heating with sodium metal. Deduce structures of A to E. PROBLEM 779 Propose mechanism of the following reactions : Br

CH2 (a)

Br

+ HBr

(b) OH

+

H2SO4

O

CCl3 (c)

+ CHCl3

Peroxide

OH

CH3

H2SO4

(d)

CH3

N (e)

CH3CH2—C

CH2 + CH2N2

CH3

N H3C CH2CH3

PROBLEM 780 Write major product in the following reactions : (a) F3C—CH==CH 2 + HI → Peroxide

(b) C 6 H 5 —CH==CHCH 3 + HBr →

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CH 3  Peroxide (c) CH 3CH==C—CH 3 + CHBr 3 → H 2O 2

(d) C 6 H 5 —CH==CH 2 + B2 H 6 → NaOH

HCl 50°C

(e) CH 3 —CH==CH—CH==CH 2 →

PROBLEM 781 Select one from the following pair of isomer that has higher heat of combustion, justifying your choice : (a)

and

(c)

(b)

or

(d)

and

or

PROBLEM 782 Select one from the following pair of isomers, that has higher heat of hydrogenation, justifying your choice. (a)

or

or

(b)

or

(c)

(d)

or

(e)

or

(f)

or

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PROBLEM 783 Propose mechanism of the following reactions :

(a)

+ CHCl3

C2H5OK C2H5OH

Cl (b) N

+ CHCl3

Cl

Cl

C2H5OK C2H5OH

N

H

PROBLEM 784 Write mechanism of chlorination of an alkane using sulphuryl chloride (SO2Cl2) and a free radical initiator (R2O2). PROBLEM 785 An organic compound A has molecular formula C8H18 which on monobromination produced three isomeric products B, C and D. B and C are enantiomeric whereas D is achiral. Treatment of either C or D with alcoholic KOH solution produced the same product E while B on similar treatment produced F(C8H16), an isomer of E as major product. Establish structures of A to F. PROBLEM 786 A hydrocarbon A(C8H16) does not decolourise Br2-water. A on monochlorination produced four isomeric products B, C, D and E, among which only C and D are enantiomeric. Compound B is resistant to dehydrohalogenation reaction. Treatment of C with alcoholic solution of KOH produced F(C8H14) as only product while similar treatment on E produced G, an isomer of F. D on treatment with alcoholic solution of KOH produces both F and G in comparable amount. Deduce structures of A to G. PROBLEM 787 A hydrocarbon A(C8H16) does not decolourise Br2 water. A on monochlorination produced four isomeric products B, C, D and E among which only D is enantiomeric. Compound B is resistant to dehydrohalogenation reaction. Treatment of either C or D with alcoholic solution of KOH produced the same alkene F(C8H14) as major product which does not rotate the plane polarized light while similar treatment on E produced G, an isomer of F which is enantiomeric. Deduce structures of A to G. PROBLEM 788 Predict major addition products in the following reactions: (a) (CH 3 ) 3 CCH==CH 2 + ICl → (b) C 6 H 5  CH==CHCH 3 + BrCl → CH3  (c) C 6 H 5  CH==CHCH 3 + IBr → (d) C 6 H 5  C==C(C 6 H 5 ) 2 + IBr → PROBLEM 789 Propose mechanism of formation of the indicated products in the following reactions:

Cl (a)

+ HCl

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(b)

+ HCl Cl

PROBLEM 790 Show how would you convert 1-methylcyclopentanol to 2-methylpentanol. PROBLEM 791 Predict major products of the following reactions: (a) Propene + BH 3 /THF → (b) The product of part (a) + H 2O 2 / NaOH → H O /NaOH

2 2 (c) 2-methyl-2-pentene + BH 3 /THF →

H2O2 /NaOH

(d) 1-methylcyclohexene + BH 3 /THF →

PROBLEM 792 Show how would you accomplish the following conversion: (a) 1-butene → 1-butanol (b) 1-butene → 2-butanol (c) 2-bromo-2,4-dimethylpentane → 2,4-dimethyl-3-pentanol. PROBLEM 793 A hydrocarbon A(C6H10) on reduction first gives B(C6H12) and finally C(C6H14). A on ozonolysis followed by work-up with Zn-H2O gives two molecules of aldehydes C2H4O(D) and one molecule of aldehyde E(C2H2O). Oxidation of B with acidified KMnO4 gives an acid F(C4H8O2). Determine structures of A to F with proper reasoning. PROBLEM 794 A hydrocarbon exist in two stereomeric forms (A) and (B) with their molecular formula C8H16. A on treatment with cold, dilute and alkaline solution of KMnO4 produces C8H18O2(C) which is a meso form. B, on the other hand, on similar treatment produces racemic mixture which is isomeric to C. Either A or B on treatment with O3 followed by work with H2O2 produces D(C4H8O) as the only product. Identify A to D. PROBLEM 795 An organic compound A(C13H23Cl) exist as diastereomers and decolourise bromine water. A on treatment with ethanolic solution of KOH produces isomeric B and C with their molecular formula C13H22. Treatment of either B or C with Rany Nickel produces 4-isopropyl-1-tertiarybutyl cyclohexane. A on oxidative ozonolysis gives acetone as one product. Identify A, B and C considering C to be enantiomeric. PROBLEM 796 Bring about the following conversions:

(a)

OH OH

(b)

(c)

COOH

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H (e) H D

(f)

Br

(g) Br

O O

(h) BrCH2

O

CH2Br

PROBLEM 797 (a) Propose mechanism of formation of products in the following reaction: HO H3O

OH

+

+

+ O

OH

OH

(b) Arrange the following alkenes in increasing order of their heat of hydrogenation explaining reason for your choice:

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(II)

(I)

(IV)

(III)

(V)

PROBLEM 798 Draw the structures of compounds A to C: (C 2 H 5 ) 3 N •• + •• CCl 2 → A (an unstable adduct) A → B + C 2 H 4 HO

2 B → C

PROBLEM 799 Predict mechanism of formation of products in the following reactions: Ph

(a)

Ph

+ HCl Cl

(b)

+ HCl O

O

Cl

Cl  (c) C 6 H 5  CH==CH  CH 3 + HOCl → C 6 H 5  C H  CH  CH 3  OH

PROBLEM 800 An optically active hydrocarbon A has molecular formula C8H18. A on monochlorination gives five alkyl halide B to F with their molecular formula C8H17Cl. B does not undergo dehydrohalogenation on treatment with alcoholic solution of KOH. Treatment of either C or D with alcoholic KOH yields same alkene G(C8H16) which on ozonolysis followed by work-up with Zn-dimethyl sulphide gives an optically inactive compound C6H12O and ethanal. Also C is enantiomeric whereas D is diastereomeric. E on dehydrohalogenation yields an alkene, which on reductive ozonolysis yields H(C7H14O) which is optically inactive. H on treatment with LiAlH4 yields I(C7H16O) which can be resolved into enantiomers. F on dehydrohalogenation yields an alkene (C8H16) which on reductive ozonolysis yields J(C7H14O) which is optically active and have same configuration as that of A. Identify A to J explaining the reactions involved. PROBLEM 801 A hydrocarbon (A) contains five carbon atoms, reacts with cold, dilute and alkaline solution of KMnO4 to form B(resolvable). B on oxidation with hot concentrated KMnO4 solution forms two compounds (C) and a neutral (D). C has molecular formula C2H4O2 and turns blue litmus paper red. Compound D neither reacts with Fehling solution nor with Tollen’s reagent and its formula is C3H6O. A on treatment with Cl2 in CCl4 yields another compound E(C5H10Cl2). E on treatment with alcoholic solution of KOH yields a stable compound F(C5H8). Identify A to F.

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PROBLEM 802 An organic compound A(C8H14) on treatment with H2/Pt gives C8H18. Ozonolysis of A followed by work-up with Zn-dimethyl sulphide yields three products B, C and D. B reduces Tollen’s reagent and gives a yellow precipitate on treatment with NaOH/I2. C doesn’t reduces Tollen’s reagent but gives yellow precipitate with NaOH/I2. D has molecular formula C2H2O2 and on treatment with concentrated solution of NaOH followed by acidification of product yields E(C2H4O3). Deduce structures of A to E. PROBLEM 803 An organic compound (A) has molecular formula C5H10, and decolourise bromine-water solution. When A is treated with cold concentrated solution of sulphuric acid followed by refluxing of product, B(C5H12O) was formed. B turns the orange colour of chromic acid solution to blue-green, converting itself into C(C5H10O). B and C both reacts with alkaline solution of iodine producing yellow precipitate and salt of isobutyric acid. Identify A to C. PROBLEM 804 A hydrocarbon A(C8H12) is optically active and on treatment with H2/Pt gives (C8H18) which does not rotate plane polarized light. A does not give any precipitate on treatment with ammonical silver nitrate solution. A on treatment with Pd/BaSO4 gives C(C8H14) which is also optically inactive but A on treatment with Na/liq NH3 gives D, an isomer of C, which is able to show enantiomerism. Also A on treatment with ozone followed by hydrolysis gives C4H6O3(E) which is optically active. Identify A to E representing their stereochemical structure. PROBLEM 805 A hydrocarbon A has molecular formula C10H18. A on treatment with H2/Pt gives B(C10H22). B on monochlorination gives two isomer C and D with molecular formula C10H21Cl. Out of C and D, only C can undergo dehydrohalogenation with alcoholic solution of KOH as well as only C can be resolved into enantiomers. A on addition of one equivalent of Cl2 in CCl4 gives a stereomer E(C10H18Cl2). E on treatment with Bayer’s reagent yields a racemic mixture of molecular formula (C10H20Cl2O2). Also A on partial hydrogenation with Pd/BaSO4/quinone gives F(C10H20) which on treatment with Bayer’s reagent gives a meso compound G(C10H22O2). Deduce the structures of A to G. PROBLEM 806 A hydrocarbon A(C10H12) has no chiral carbon. A gives a white precipitate with ammonical solution of silver nitrate. A on treatment with H2/Pt gives B(C10H20). A on ozonolysis gives C(C8H12O4) as one product which on heating with soda lime gives D(C6H12). D on monochlorination with Cl2/hν gives C6H11Cl as sole isomer. Identify A to D. PROBLEM 807 An organic compound A(C8H6) gives brick red precipitate with ammonical solution of Cu2Cl2 . A on treatment with acidic solution of HgSO4 yields B(C8H8O). B gives positive iodoform test. B can also be prepared by the reaction of benzene with acetic anhydride in presence of AlCl3. Identify A and B. PROBLEM 808 Potassium hydroxide is mixed with 2,3-dibromohexane, the mixture is sealed in a fusion tube and heated to 200oC for one hour. The product mixture (A) is mixed with Cu(I)-ammonia complex and a precipitate forms. The precipitate (B) and liquid phase (C) are separated. The precipitate is acidified and the product (D) is distilled at 71oC. Product D is treated with NaNH2, followed by acetone and then with dilute acid to give alcohol (F). The liquid phase (C) is distilled and the product alkyne is treated with NaNH2 at 150oC for 1.0 hour and product mixture is distilled to give a pure alkyne (E) of boiling point 71 oC. Identify A to F.

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PROBLEM 809 Propose structures for A to L: 1-bromobutane KOH/H2O

Mg/Ether

(K)

Pentanal H3O+

(E)

H2SO4 /heat

Na/heat CH3Br

( J) (C)

O3 /(CH3)2S

(F)

Butanal + Pentanal

Br2

(butanone/H+)

(A)

(B)

(D)

(G)

(l) NaNH2 HBr Peroxide (i) NaNH2/heat

1-nonyne

(ii) H2O HBr (two mole)

fused KOH

(H) mixture

(L)

PROBLEM 810 An optically active compound A has molecular formula C7H11Br. A reacts with HBr, in absence of peroxide to yield isomeric products, B and C with molecular formula C7H12Br2. Compound B is optically active; C is not. Treating A with one mole of potassium-t-butoxide yields D(C7H10). Subjecting one mole of D to ozonolysis followed by treatment with Zn and water yield two moles of formaldehyde and one mole of 1,3-cyclopentandione. A on addition of HBr in presence of H2O2 produces another isomer of B which is E, and shows diastereomerism. Deduce structure of A to E. PROBLEM 811 An alkane (A) with formula C6H14 reacts with Cl2 to yield three compounds with formula C6H13Cl, B, C and D of these only C and D undergo dehydrohalogenation with C2H5ONa/EtOH to produce alkene (E). Also C can be resolved into enantiomers but D is non-resolvable. E on addition of HCl produces ‘B’ which on treatment with Zn-acetic acid produces A. Deduce structure of A to E. PROBLEM 812 Complete the following: (only major product) (a) Cyclohexene + CHCl 3 (50% NaOH/ H 2O) → OH CHBr2 (b)

+ CH2I2

Zn/CuCl

+ 50% NaOH + H O 2

(c)

PROBLEM 813 An organic compound A has molecular formula C10H16 and is known not to contain any triple bond. On catalytic hydrogenation of A, a new compound B (C10H22) is formed. Ozonolysis of A followed by treatment with zinc-H2O yields two mole of CH2O, one mole of acetone and a third compound C(C5H6O3). Deduce the structure of A,B and C considering C being optically active. PROBLEM 814 Identify A to E (Provide structures) C10H16Br2(A) C10H15Br(C)

Br2/CCl4

E(C10H18O2)

H3O+

C10H16O(D)

PhCO3H

C10H16

H2SO4/∆

Br2/H2O

C10H17OBr(B)

O

O3/Zn-H2O

O

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PROBLEM 815 An organic compound A (C10H16) on catalytic hydrogenation produces 1-isopropyl-4-methylcyclohexane. Ozonolysis of A followed by work-up with dimethylsulphide produces formaldehyde and O O   CH 3  C  CH  CH 2  CH 2  C  CH 3  CH2CHO Deduce structure of A.

PROBLEM 816 Three isomeric hydrocarbons A, B and C have molecular formula C6H10. All three compounds readily decolourise bromine in CCl4, compound A gives precipitate with AgNO3/NH3 solution but compound B and C donot. Compound A and B both yield hexane when treated with excess of H2/Pt. Under these conditions, C absorbs only one molar equivalent of H2 and give product with formula C6H12. When A is oxidised with hot basic KMnO4 and resulting solution is acidified, the only organic product that can be isolated is CH3(CH2)3COOH. Similar oxidation of B gives only CH3CH2COOH and similar treatment of C gives only HO2C(CH2)4COOH. Identify A, B and C. PROBLEM 817 Propose mechanism: OH (a)

H2SO4 Heat

Br (b)

Br2, H2O

Br

NaCl

Br

+

OH

Br

+ Cl

PROBLEM 818 An organic compound A has molecular formula C9H14 and on treatment with H2/Pt gives C9H20. A exist in four stereomeric forms. A on treatment with O3 followed by work-up with Zn-H2O produced CO2, B (C4H6O3) and C (C4H8O). B rotates plane-polarized light but C doesn’t. C changes the colour of acidic dichromate solution converting itself into an another optically inactive compound D (C4H8O2). D on treatment with I2/Red-P produced an optically inactive compound E(C4H7O2I) as the only isomer. Deduce structures of A to E and draw the structures of all four stereomers of A. PROBLEM 819 An optically active compound A has molecular formula C9H10. A does not decolourise aqueous solution of Br2. A on treatment with NBS produced another optically active compound B (C9H9Br). B on hydrolyzing in aqueous KOH solution gave C (C9H10O) which doesn’t change the colour of acidic dichromate solution. B on treatment with C2H5ONa/C2H5OH produced D (C9H8) which is optically inactive. D on oxidative cleavage produced E (C9H8O3) which is optically inactive. E on heating with soda lime gave C8H8O, which can also be obtained by Friedel Craft’s acylation. Deduce the structures of A to E describing the reactions involved in each step. PROBLEM 820 (a) Reagent (two possible isomers) + KMnO 4 / H + → 3,4-diketo hexandioic acid. Provide structures of the reagent which will lead to the above product.

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(b) Propose mechanism: H

(i)

+

Br Bu-Li

(ii)

C Cl

CHCl3/(CH3)3COK

(iii) N

N

H

PROBLEM 821 (a) An organic compound A(C12H20) on catalytic hydrogenation yields B(C12H24). A on ozonolysis followed by work-up with H2O2 yields acetone and the following compound. Deduce the structures of A and B. O

O

O

OH

HO

(b) Predict product in the following sequence of reaction: EtOOC

LiAlH4

COOEt

X + 2 EtOH Conc. H2SO4

NaI

Z

NBS

Y

PROBLEM 822 An organic compound A(C13H21Br) is optically active and decolourise Bayer’s reagent. A on treatment with EtONa/EtOH yields two and only two isomers B and C both having molecular formula C13H20 with B as major product. Catalytic hydrogenation of either B or C gave the same product C13H24 which is optically active. B is stereomeric whereas C is non-stereomeric. C on oxidative cleavage with acidic KMnO4 produced D(C12H18O5) as one product which on simple heating gave E(C11H18O3). E gives yellow precipitate with NaOH/I2. E on treatment with NaBH4 gives F(C11H20O3), an optically active compound. F on heating with catalytic amount of H+ gave the following compound: O

O

Deduce structure of A to F.

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PROBLEM 823 Write structures of single major product of each of the following reactions. If two or more major products are formed, write structures of both. Show stereochemistry of products where appropriate. H Br

O (a)

CH3ONa

CH3

CH3OH/E2

CH3

H

(b)

H

H3 C

Br2 H 2O

H

CH 3  (d) CH 3 —C== CH—CH 3 + CH 3CO 3 H →

(i) BH3

(c)

(ii) H2O2/NaOH

50% H SO

2 4 (e) HO—CH 2CH 2 —CH==CH 2 →

Cold, dilute KMnO4

(f)

NaOH

H 2O

CCl4

+ Br2

(g)

Cl2

+ CH3OH

(h)

PROBLEM 824 Hydroboration oxidation of trans-2-( p-anisyl)-2-butene yielded an alcohol A whose melting point is 60°C while similar treatment on cis isomer yielded an isomeric alcohol which is a liquid at room temperature. Suggest resonable structures for A and B. PROBLEM 825 An optically active alkyne A contain 89.25% C and 10.48% H. After hydrogenation with H 2 /Pt, it is converted into 1-methyl-4-propyl cyclohexane. A gives no gas with CH 3 MgBr. Treatment of A with acidified KMnO 4 solution yields B which gives positive iodoform test and a salt C. Acidification of salt C yields an optically active compound D. Identify A to D. Discuss the stereochemistry of addition of Br 2 to the following alkenes:

PROBLEM 826 H (a)

Br H

H C

(b)

C

C

H3C

H C

C

H

CH3

Br

C

CH3

H

CH3

Br H (c) H C 3 Br

C

C

C

C

H

Br

H

H

H H CH3

(d) Br H

C C

C CH3

C H

CH3

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Problems

PROBLEM 827 (a)

A

Provide the structures of missing components: CH3

HO

Br2 H 2O

HgSO4

(b)

H2SO4/H2O

B

Br

OH Br CH ONa 3

(c)

Excess

H2SO4

C (d)

∆

OH

Br (e)

H3C Ph

PROBLEM 828

H CH3

B2H 6 H2O2/NaOH

E

Propose mechanism of the following reaction: Br2

OH

O

CCl4

Br H

PROBLEM 829

Bring about the following transformations: OH CH3

(a)

(b)

Br OMe

OH

Br (c) COOH COOH

D

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PROBLEM 830 (a)

Complete the following reactions: C2H5ONa

NBS hν

CH3

Bu3SnH

(b)

C2H5OH

AIBN

Br

[AIBN : Azobis isobutyronitrile]

PROBLEM 831

Discuss the relative heat of hydrogenation of following alkenes:

PROBLEM 832

C

B

A

Propose mechanism of the following acid catalyzed reaction:

H2SO4

H2SO4

OH

I

PROBLEM 833

II

Complete the following reactions:

OMe Br (a)

tBuOK tBuOH

Br2

(b)

MeOH

OH RCO3H

(c)

HO– H2O

H2SO4

(d)

Heat

H2/Pd

(e)

BaSO4

PROBLEM 834

Convert: Br

OMe

O

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Problems

Provide structural formula for the major product in the following reactions:

PROBLEM 835 O

Zn(Hg) HCl

(a)

DCl

(b)

0°C

O Ph3P

(c)

BuLi

Br

PROBLEM 836 The reaction of the diene shown below with dry HCl can lead to four products. Provide structural formula of all the products. HCl One-mole

PROBLEM 837 A hydrocarbon A has molecular formula “C14 H 22 ” and it can be resolved into enantiomers. A on hydrogenation with H 2 /Pt yields B (C14 H 30 ) which can’t be resolved into enantiomers. Also A on partial hydrogenation with H 2 in presence of Pd/BaSO 4 /PbO yields C (C14 H 26 ) which is still resolvable and decolourise brown colour of bromine water solution. A on treatment with alkaline permanganate solution yielded D(C 9 H18O 2 ), oxalic acid and propanoic acid. D can also be resolved into enantiomers. Another organic compound E (C 8 H16O) forms yellow precipitate on treatment with alkaline iodine solution. E on treatment with NaBH 4 yielded a resolvable organic product F (C 8 H18O). F on treatment with p-toluenesulphonyl chloride followed by workup with aqueous NaCN yielded G(C 9 H17 N). Hydrolysing G with dilute sulphuric acid produced D. Identify A to G. PROBLEM 838 An organic compound A has molecular formula C12 H 22 and can’t be resolved into enantiomers. A on ozonolysis followed by work-up with Zn/H 2O yields B (C12 H 22O 2 ),. B neither reduces Tollen’s reagent nor produced any yellow precipitate with alkaline iodine solution. B on exhaustive oxidation with hot, concentrated, acidic permanganate solution produced C (C 6 H10O 4 ) as one of the product. C on heating with sodalime yielded isobutane. Identify A, B and C. PROBLEM 839 An organic compound C 5 H 6O 2 exist in three isomeric forms A, B and C of which only B can be resolved into enantiomers. All three evolves a gas on heating with sodium metal but only A and B evolves gas with NaHCO 3 . Also B and C formed white precipitate with ammonical silver nitrate while A didnot. A on heating with soda-lime yielded D (C 4 H 6 ) which did not yield any precipitate with ammonical silver nitrate. Hydrolysing C with aqueous alkali followed by heating with sodalime produced C 3 H 4 . Identify A to D. PROBLEM 840

Propose synthetic routes to accomplish the indicated transformations: CH3

OH CH3

(a) O

O CH3

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H Cl

(b) H

Cl

OH  (c) H—C ≡≡C—H → CH 3 —CH —CH 2 —C ≡≡C—CH 3 CH3 C—CH3

(d) H—C

CH3

Propose mechanism to account for the following transformations:

PROBLEM 841

H2SO4

O

(a)

O

OH

(b)

O

NH3

HO

OH Br2

(c)

PROBLEM 842

N

OH

O Br

Propose mechanism: H2SO4

PROBLEM 843

Propose mechanism of the following reactions: Cl

(a)

Cl2 CH3OH

H OCH3 CH3

Br Br2

(b)

H

KOH, H2O

H

H O

O

OH O

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Problems

H 2O

(c)

H2SO4

OH

O

O

OH

PROBLEM 844 An organic compound A has molecular formula C10 H12O. A forms salt with aqueous NaOH but does not exhibit stereo-isomerism. A on treatment with HBr yields B (C10 H13OBr) which can be resolved into enantiomers. On the otherhand, if A is treated with HBr in presence of a peroxide, it gives C, an isomer of B but C can’t be resolved into enantiomer. C on treating with potassium tertiary butoxide in tertiary butanol yields D, an isomer of A, which is still non-resolvable. D on hydration in presence of alkaline H 2O 2 catalyzed by B2 H 6 yields E (C10 H14O 2 ) which can be resolved into enantiomers. Refluxing E with dilute sulphuric acid yields F (C10 H12O) which does not forms any salt with aqueous NaOH. Also, A on boiling with alkaline solution of potassium permanganate yields 2-hydroxybenzoic acid. Identify A to F. PROBLEM 845

Propose mechanism of the following reactions: H3C

I

I

CH3

CH3 +

+ HI

B

A

PROBLEM 846

Propose mechanism: Br + HBr

Br +

+ Br

PROBLEM 847 The hydrocarbon shown below is optically active and can be resolved into enantiomers. A pure enantiomer of this compound on refluxing with conc. H 2SO 4 for sometime isomerizes into an optically inactive isomer. Explain the observations with the help of mechanism. CH3 CH CH3

PROBLEM 848 A hydrocarbon A has molecular formula C 7 H12 . A on treatment with B2 H 6 followed by work-up with H 2O 2 /NaOH produced B (C 7 H14O) as the only product. Reacting B with torylchloride followed by workup with KOH yielded C, an isomer of A, in addition to other olefinic products. Ozonolysis of C followed by work-up with Zn  CH 3COOH produced 2-methyl hexan-1,6-diol. Identify A to C.

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PROBLEM 849

Suggest the most likely site of protonation explaining reason:

ALKYL HALIDES PROBLEM 850 Considering that following reaction proceed by S N 2 mechanism, select the faster reaction of each pair and explain your reasoning : (a) Reaction of cyanide ion with n-iodoheptane or n-chloroheptane. (b) Reaction of ethanol or sodium ethoxide with n-butyl bromide (c) Reaction of azide ion with 1-butyltosylate or 2-butyltosylate. (d) Reaction of tertiary butoxide with ethyl bromide or reaction of ethoxide with tertiary butyl bromide. (e) Reaction of cyanide ion with 2-bromopentane or CH 3CH 2 CH—CH(CH 3 ) 2  CH 2 Br PROBLEM 851 Explain the following observations : NH2

CH3I

NHCH3

Slower

NH2 and

CH3I

NHCH3

Faster

PROBLEM 852 An organic compound A has molecular formula C 7 H15Cl and it can be resolved into enantiomers as well as diastereomers. A on treatment with alcoholic solution of KOH yields three isomeric alkenes B, C and D having molecular formula C 7 H14 , all of them can be resolved into enantiomers. Treating either B or C with acidic solution of KMnO 4 yield E (C 5 H10O 2 ) and ethanoic acid. E is still resolvable. Heating E with soda lime (NaOH/CaO) yields C 4 H10 which is non-resolvable. Identify A to E. PROBLEM 853 A organic compound A has molecular formula C 6 H13Cl and it can be resolved into enantiomers. A on treatment with alcoholic solution of KOH yield four alkenes B, C, D and E, none of them can be resolved into enantiomers. B and C are stereoisomers and treatment of either B or C with Cl 2 yields racemic mixture of C 6 H12Cl 2 . On the other hand D and E are stereoisomers and D on treatment with Cl 2 yields a meso compound C 6 H12Cl 2 while E on similar treatment yield a racemic mixture of C 6 H12Cl 2 . Also B has greater heat of hydrogenation than C. Deduce structures of A to E.

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Problems

PROBLEM 854 Give reason: (a) Hydrolysis of methylbromide takes place much faster in presence of NaI. (b) When tertiary butyl chloride undergo hydrolysis in aqueous NaOH, the rate of formation of tertiary butyl alcohol does not increase appreciably as OH – concentration is increased, however a marked decrease in concentration of t-butyl chloride is observed. (c) When methyl bromide reacts with NaCN, major product is CH3CN, whereas when CH3Br reacts with AgCN, major product is CH3NC. PROBLEM 855 Explain the mechanism of following reactions: (a)

OH + HCl

(b)

OH + HCl

Cl Cl +

Cl (only product)

PROBLEM 856 Account for the following observations: (a) t-BuF is solvolysed only in very acidic solution. (b) t-BuCl is solvolysed more slowly than 2-chloro-2,3,3-trimethylbutane. (c) t-BuCl is solvolysed much faster than 2-chloro-1,1, 1-trifluoro-2-methy- lpropane. PROBLEM 857 Explain the following observations in S N 1 reactions: (a) AgNO3 increases the rate of solvolysis (b) the more acidic the nucleophilic solvent, the faster is the rate of solvolysis. PROBLEM 858 Complete the following reaction sequence, showing the major product you expect for each stage. Also indicate stereo-chemistry where possible. TsCl

NaN

3 (a) CH 3CHCH 3 → A → B Pyridine  OH

CH2CH2CH2OCH3 (b) H H3C

CH3SNa

C

C

Br

PROBLEM 859 Explain the relative rate of E-2 reaction of the following compounds: Cl

Cl CH3

H 3C

(I)

Cl CH3

H 3C

(II)

CH3

H 3C

(III)

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PROBLEM 860 Explain the followings: (a) When CH3Br is dissolved in ethanol, no reaction occurs at 25°C. When excess of C2H5ONa is added, a good yield of ethyl-methyl ether is obtained. (b) Explain the role of solvent in the reaction of potassium ethanoate with ethyl iodide. (c) Which nucleophile reacts more rapidly with CH 3I : Et3N or Et3P? PROBLEM 861 Predict product in the following reaction: Br

NaN3

(a) I

(b)

H 3C

BH3-H2O2

SO2Cl CH =CHCH ONa 2 2

NaOH

Br (CH3)3COK

(c)

(CH3)3COH

Br

Br

NH3

(d) MeO Br (e)

Br2 hν

PROBLEM 862 In attempting to synthesize 1-methylcyclohexene, you are trying reaction of trans-1-iodo-2-methyl cyclohexane with potassium tri-tertiary butoxide. The reaction fails to produce the desired product and the isolated product is 3-methylcyclohexene. Using a detailed mechanistic analysis, explain why this alternative product is formed? PROBLEM 863 The two tosylate shown below react differently under identical conditions. One will undergo a facile S N 2 reaction while other will easily eliminate via E-2 mechanism. Draw structures of major product for each reaction and provide mechanistic explanation for them: OTs OTs

CH3ONa

OTs and

CH3ONa

OTs

PROBLEM 864 When piperidine undergoes the indicated series of reactions, 1,4-pintadiene is obtained as product.

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Problems i) excess of CH3I ii) Ag2O/H2O iii) Heat

N

i) excess of CH3I

N

ii) Ag2O/H2O iii) Heat

H

When four different methyl substituted piperidine is subjected to same series of reactions, each forms a different dienes-1,5-hexadiene, 1,4-pentadiene, 2-methyl-1,4-pentadiene and 3-methyl-1,4-pentadiene. Which methyl substituted piperidine yields which diene?

PROBLEM 865 Give substitution product mentioning their stereochemistry: H H H2O HO– (a) H 3C (b) H 3C Br → Br → C4 H 9 C4 H 9 Cl H (c)

CH 3 CH3ONa

H

(d)

H H 3C

CH3

Cl H

CH ONa

3 →

C2 H 5

CH 3 (e)

H

Cl

H

CH 3

CH ONa

3 →

C2 H 5

PROBLEM 866 In the following pair of reactions, which will occur at faster rate-explain your choice:

Cl + (CH3)2CHS

or

(b)

Cl

+ HO –

or

(c)

S

Cl + CH3S

(a)

O

S

OH –

– Cl

O

OH

Cl + H2O

or

–

– Cl Cl + HO

+ Cl

Cl + H2O

OH

OH

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PROBLEM 867 Propose mechanism: OH (a)

Br + H2O

∆

OH + H2O

(b)

∆

H

Br

CH3 +

CH3

CH3

H

OH

CH3

PROBLEM 868 Explain the following observations: (a) Azide ion (N –3 ) react with 2-bromopentane thousand times faster than with neopentyl bromide in a S N 2 reaction though former is a secondary halide while latter is primary. (b) What will happen to the stereochemistry of product of the following reaction: Br H

– CH3 + N3

S N2

D

(c) (d) (e) (f)

What will happen to the rate if the concentration of alkyl bromide in (b) is doubled? What will happen to the rate if the concentration of azide ion in (b) is doubled? How the sign of optical rotation of reactant and product are related in (b) When allowed to stand in dilute H 2SO4, laevo-rotatory 2-butanol slowly loses optical activity.

PROBLEM 869 Assuming that the following reaction takes place by S N 2 mechanism, select the faster reaction in each pair and explain reason for your choice: (a) Reaction of cyanide ion with n-iodoheptane or n-chloroheptane. (b) Reaction of ethanol or C 2H5ONa with n-butyl bromide. (c) Reaction of azide ion (N 3–) with n-butyl tosylate or s-butyl tosylate. (d) Reaction of isopropoxide with ethyl bromide or reaction of ethoxide with 2-bromopropane. PROBLEM 870 Synthesize the following compounds from indicated starting material: OMe (a)

from 1,4-dihydroxy benzene

NH2

(b)

from benzyl alcohol

OMe S (c)

from phenol

(d)

from O

Cl

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Problems

PROBLEM 871 Explain why allyl bromide (CH2 ==CHCH2Br) undergoes rapid substitution under either S N 1 or S N 2 reaction condition. PROBLEM 872 Following reactions are not feasible as indicated. Briefly explain the reason for their failure: Br

KCN

(a)

H + O H

HO (c) H 3C

(b)

CN

–H+

+ CH2

+

Br

HO

OH

H 3C

CH 3

→

CH 3

PROBLEM 873 Provide mechanism: I (a) OH

∆

(b) HO

Br

KOH

O

O

PROBLEM 874 Arrange the following in the increasing order of their ability as a leaving group: (b) CF3SO 3– , CH 3SO 3– and CH 3COO – (a) CH 3S – , CH 3O – , CF3– and F – PROBLEM 875 An organic compound A has molecular formula C6H11Br. A decolorizes brown colour of bromine water but it can’t be resolved into enantiomers. A on treatment with HBr produces B(C6H12Br2) which is still non-resolvable. Also A on treatment with HBr in presence of a peroxide (R2O2) yields C-an isomer of B, but it is resolvable. B on heating with Zn-metal yields D(C6H12) which on treatment with ozone followed by hydrolysis of product in presence of Zn-powder yield acetone as the only organic compound. Also A on treatment with ethanolic KOH solution yields E(C6H10) as only possible product. E on treatment with excess of HBr/R2O2 in cold produces isomeric F and G with their molecular formula C6H12Br2, in which F is a meso compound while G is resolvable. Also F on hydrolysis in aqueous NaOH yield a diol C6H14O2 which O on refluxing with dilute H 2SO4 yields the adjoining compound. Deduce structures of A to G. PROBLEM 876 An organic compound is optically active and has molecular formula C14H23Br. A does not decolourise cold, dilute and alkaline solution of KMnO4. A on treatment with potassium tertiary butoxide in tertiary butanol yields B(C14H22) as the major product. B on treatment with acidified permanganate solution yields C(C14H22O4). C on heating with sodalime yields D(C12H22) which can also be formed by dissolving cyclohexyl bromide in etheral solution of Mg. Identify A to D. PROBLEM 877 An optically active organic compound A has molecular formula C7H16O. A neither changes the colour of bromine water nor evolve any gas on reacting with MeMgBr. A on refluxing with dilute H2SO4 produces B(C3H8O) and C(C4H10O). B is non-resolvable and gives yellow precipitate with

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NaOH/I2 while C is resolvable. C on heating with concentrated solution of H2SO4 yield stereo-isomeric compounds D and E with their molecular formula C4H8. D on treatment with Br2/CCl4 yields a meso dibromide while E on similar treatment yields racemic mixture of products. Identify A to E.

PROBLEM 878 An organic compound A(C5H11Cl) is optically active and on treatment with ethanolic KOH solution yields B(C5H10) as major product which does not show stereo-isomerism. Also A on treatment with (CH3)2Cu yields C(C6H14) which is optically inactive. Deduce structures of A to C. PROBLEM 879 An organic compound A has molecular formula C10H17Br and it is non-resolvable. A does not decolourise brown colour of bromine water solution. A on treatment with (CH3)3COK/(CH3)3COH yields B(C10H16) as major product. B on treatment with H2/Pt yields C10H18 which on treatment with Cl2/hν yields three monochloro derivative. Also B on boiling with acidic permanganate solution yields C(C10H16O3). C on heating with soda-lime yields D(C9H16O). D on reducing with LiAlH4 followed by heating the product with concentrated H2SO4 yields E(C9H16) as major product. E on treatment with ozone followed by work-up with Zn-H2O yields 6-keto nonanal. Deduce structures of A to E. PROBLEM 880 An organic compound A has molecular formula C5H10O and it neither decolourise bromine water solution nor evolve any gas on heating with Na-metal. A on refluxing with aqueous HI yields B(C5H10I2) which is non-resolvable. B on treatment with NaCN in acetone yield C(C6H10IN) in significant amount while B on treatment with aqueous Ag2O yields D(C5H11IO) in significant amount. D does not change the orange colour of acidic dichromate solution. Deduce structures of A to D. PROBLEM 881 An optically active organic compound A has molecular formula C11H16O and it gives off a colourless gas on treatment with CH3MgBr. A on heating with concentrated H2SO4 yields B(C11H14) which does not show stereo-isomerism. B on treatment with HBr in absence of a peroxide yields C(C11H15Br) which is resolvable while addition of HBr on B in presence of a peroxide yields D, an isomer of C but it is non-resolvable. Also B on treatment with ozone followed by work-up with Zn-H2O yields C8H8O and C3H6O in which former can also be obtained by Friedal Craft’s acylation reaction of benzene. Both these ozonolysis products gives yellow precipitate on treatment with NaOH/I2. Deduce structures of A to D. PROBLEM 882 Bring about the following transformations: (a) C 6 H 5CH 2OH → C 6 H 5CH 2SCH 2CH 3 (b) n-pentanol → n-pentyl iodide (c) C 6 H 5SH → C 6 H 5SCH 2CH 2C 6 H 5 (d) n-butyl bromide → 2-heptyne CH3

CH3

(e) HO H5C2

C

H H5C2

H

O (f)

C NH2 O

O

Cl O

H N CHCH2OH CH=CH2

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Problems

PROBLEM 883 Both mesylates and triflates are good leaving groups. However, triflate displacements proceed about 5000 times faster than mesylate displacements. Explain why triflate is a better leaving group than mesylate. PROBLEM 884 Upon treatment with Ag + ions, on α-halo-ether undergo substitution via SN-1 mechanism. Consider the following two reactions and explain why 1 produces a mixture of two possible stereoisomers while 2 provides only one. O

O O

O Ph

Br

Br O

Ph

(1)

(2)

PROBLEM 885 For each of the following of substrate, indicate with reasoning, which will react at faster rate under the given reaction conditions. In addition, indicate whether you expect each reaction to follow SN-1 or SN-2 pathways. Br

(i)

(iii)

—SNa.

Br

Br

(ii)

Br with

or

or

with CH3COOH.

or

Br

Br

with CH3COOH.

PROBLEM 886 Write products in the following reactions explaining mechanism of their formations: H I Et n-Bu NaOH NaCl (a) → (b) EtOH, ∆ H Ph CH3COOH CH3—CH—CH2— OTs i-Pr (c)

OMs CH3

NaOH HCOOH, ∆

(d) OTs

CH3 H

(e)

Me CH(CH3)2

CH3

H 2O

Br

∆

H

(f) Et

Et

KOH EtOH, ∆

COO–

CH3I

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(g)

Et

H

H

Cl

PROBLEM 887

(a)

KOH EtOH

Write products in the following reactions indicating their stereochemistry: I Et Me   H 2O EtONa E2 (b) CH 3CH 2  C  C  CH 3 → S N 1 ∆ EtOH/∆   CH 3 H OMs CH3

(c) CH3

(CH3)3COK (CH3)3COH

OTs E2 (d)

(CH3)2CHSH NaH

SN2

CH3

Br

CH 3 Ph NaCl Acetone

Et → S N 2

(e) ICH 2 H

(f)

CH 3

OTs

Ph

CH 3

NaOH ∆

→ E–1

CH 3 Cl CH 3   KOH (g) CH 2 ==CH  CH 2  CH CH  CH 3 → E–2 EtOH

PROBLEM 888

Write mechanism for the following reaction:

CH 3 Cl CH 3    heat CH 2 == C  CH 2  CH  CH 3 + CH 3OH → CH 2 == C  CH  CH 2 CH 3  OCH 3 CH 3  + CH 3OCH 2  C== CHCH 2CH 3

PROBLEM 889

Bring about the following transformations: I

CN

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Problems

PROBLEM 890

Write major products in the following reaction and discuss their stereochemistry: F PhNH2 Et2O

H3C

OTs

PROBLEM 891 5-Bromononane on treatment with potassium tertiary butoxide in ethanol produces a mixture of cis-4-nonene (23%) and trans-4-nonene (77%). Draw Newman’s projections of 5-bromononane, looking down the C5–C6 bond showing the conformational forms that lead to cis-4-nonene and trans-4-nonene respectively. Based on conformational analysis, suggest reason why transform of product predominate. PROBLEM 892

Give stereochemical structures of product formed in the following reactions: CH2CH3

(a)

H H3C

+ CH3CH2CH2ONa

SN2

CH3 CH3ONa

(b)

SN2

Br

Cl

CH3 CH3OH

(c)

∆/SN1

Cl

PROBLEM 893 An alkyl bromide A has molecular formula C 8 H17 Br and four different structures can be drawn for it, all optically active. A on refluxing with ethanolic KOH solution yields only one elimination product B (C 8 H16 ) which is still enantiomeric. B on treatment with H 2 /Pt yields C (C 8 H18 ) which does not rotate the plane polarized light. B on ozonolysis followed by work-up with H 2O 2 yields D(C 7 H14O) as one product which is still resolvable. Deduce structures of A to D. PROLBEM 894 Provide missing products, starting material or reagents/conditions, for each of the following reactions: Br

SOCl2

(a)

OH

H

NaCN

(b)

Pyridine

OMe (c)

(e)

OH

OH

Ph

(d)

(ii) EtBr

TsCl Pyridine

OH

H

(i) NaH

NaCN

(f)

H2SO4 MeOH/∆

CH3Na

OH

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PROBLEM 895

Propose mechanism of the following reaction: Br +

+ N

CH3

N

N

H

PROBLEM 896 SN2 reaction of simple carboxylate ions with haloalkanes in aqueous medium generally do not gives good yields of esters, explain. Reactions of 1-iodobutane with sodium acetate gives an excellent yield of ester if carried out in acetic acid medium. Why acetic acid is a better solvent for this process than water? PROBLEM 897

Provide major substitution product in the following reaction: heat

heat

(a) 2-butanol + HBr →

(b) 3,3-dimethyl-2-butanol + HBr →

OH Heat

+ HCl

(c)

OH

(d)

+ HCl

Heat

PROBLEM 898 Provide structure of major product in the following reaction indicating stereochemistry where appropriate: D

D (a)

C CH3

OH

TsCl

NaCN

(b)

H

HBr Heat

OH

C

NaCN

H

CH3

PROBLEM 899 How 1-butanol can be converted efficiently into: (a) 1-iodobutane (b) N-ethylethanamine (c) Butylpropanoate (d) CH 3CH 2CH 2CH 2SH (e) CH 3CH 2CH 2CH 2OCH 3 PROBLEM 900 Which of the following alkyl halide could be successfully used to synthesize Grignard reagent and why other fail? Br

N

Br

HO

II

I

O Br

OH III

H2 N

Br IV

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Problems

Propose mechanism of the following reactions:

PROBLEM 901 Cl

O

(a)

O

CH3OH

+ CH3O–

O

O

O

(b)

O

CH3O–

Br O + HO–

(c)

O

HO

OH

O

Explain mechanism of formation of products in the following reactions:

PROBLEM 902 (a)

Mg

—Br

—MgBr + Br

(b)

—OH

Discuss the relative reactivity of reactants in the following pairs:

PROBLEM 903 (a)

HBr

(CH3)3CCl + HO

–

–

and (CH3)3CI + HO

H

H

H

CH3

–

(b)

CH3 H H3C I

CH3O

CH3OH

H3C

Br H

H

H –

CH3 Br

and H3C II

H

CH3O

CH3OH

H3C

CH3

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PROBLEM 904 (a) Give structures of products obtained from the reaction of both enantiomers of cis-1-chloro-2-isopropyl cyclopentane with CH 3ONa. (b) Are all the products obtained chiral? (c) How would products differ if the starting material were trans isomer? (d) Which pair of structure (enantiomers) form substitution product more rapidly-cis or trans and why? (e) Which pair of products form elimination product more rapidly cis or trans? Write elimination products in the following reaction and compare their relative

PROBLEM 905 reactivity:

CH3

CH3

PROBLEM 906 (a)

(CH3)3COK

(CH3)3COK

(CH3)3COH

(CH3)3COH

Cl

Cl

I

II

Compare the following compound for their reactivity in SN–1 reaction: OCH3 OCH3 Cl Cl

and

and

(b)

Cl

Cl Cl

(c)

and Cl Cl

(d)

and

N (e)

N and

Cl

Cl

Cl

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Problems

PROBLEM 907

Propose mechanism of the following reaction:

I

NaCl CH3OH

Cl

+

I

II

+

Cl

OCH3 + III

IV

OCH3

ALCOHOLS AND ETHERS PROBLEM 908 Why does ethylene oxide react readily with nucleophiles such as ammonia, whereas THF is inert to nucleophilic attack by ammonia? PROBLEM 909 Give the products and mechanism of each of the reaction shown below: O CH3ONa/CH3OH

H2SO4/CH3OH

PROBLEM 910 Bring about the following conversion from the indicated starting materials: O (a)

CH2OH 1,6-hexan diol

and

(b) O

O

PROBLEM 911 An optically inactive organic compound A(C7H11Br) is treated with Mg in ether to give B(C7H11MgBr), which react violently with D2O to give methylcyclohexene with deuterium atom on the methyl group (C). Reaction of B with acetone followed by hydrolysis gives D(C10H18O). Heating D with concentrated sulphuric acid gives E(C10H16). Determine structure of A to E. PROBLEM 912 When ethanol is heated with conc. H2SO4 at 140oC, diethyl ether is obtained, whereas at 180oC ethylene is the major product. Suggest reason. PROBLEM 913 Compound A is an optically active alcohol. Treatment with chromic acid convert A into a ketone B. In a separate reaction, A is treated with PBr3, converting it into C. C on reaction with Mg is added to B to yield D, which after hydrolysis gives 3,4-dimethyl-3-hexanol. Identify A to D.

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PROBLEM 914 A compound ‘A’ C10H14O exist in diastereomeric form, liberates a gas on reaction with CH3MgBr. Treatment of A with aqueous HBr gives B(C10H13Br), which exist in enantiomeric forms. B on treatment with alcoholic solution of KOH gives a compound C. Ozonolysis of C followed by work-up with H2O2 gives a compound D and acetophenone. Identify A to D. PROBLEM 915 Synthesize the followings from the indicated starting materials: O from

(a)

O

CH3 from

(b)

O

CN

H

from

+ Br

(c) O

O

O

O (i) CH3MgI

A

(ii) H3O+

H2/Pt

H2SO4

B

C

H2SO4

KMnO4

C6H12O (E)

C10H16 (H)

CH3MgBr H 3O +

H2SO4

C10H18O (G)

(i) Mg/ether (ii) (iii) H3O+

C5H9Br (F) O

BrH

PhCO3H

C5H8O (D)

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Problems

PROBLEM 916 Determine the structure of compound A through H OH

(a)

H2SO4

OH

(b)

H2SO4

O OH

(c)

H2SO4

+

OMe

(d)

+

O H2SO4

PROBLEM 917 Propose mechanism of the following reactions: PROBLEM 918 Convert: (a) 2-methyl-2-butene to (CH 3)2C(OH)CHDCH3 (b) Synthesize 2,5-dimethyl-3-hexanone using alcohol containing not more than four carbon. O (d)

O OH OH

(c) C6H5Br to (C6H5)2C(OH)CH3 using any additional compound containing not more than three carbon. OH

Cl

PROBLEM 919 (a) Convert: (b) Cyclohexyl bromide when treated with potassium ethoxide, the major product is a cyclohexene whereas when ethyl bromide is treated with potassium salt of hexanol, the major product is ethyl-cyclohexyl ether. Explain the difference. OH (a)

cold, dil, alkaline KMnO4

A

HIO4

B

(b)

+

H

A

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PROBLEM 920 Predict major product: (c) OH

OH

HIO4

OCOCH3 (i)

(d)

A

(ii)

LiAlH4 NaBH4

O

PROBLEM 921 An optically active compound A(C10H17Br) when treated with alcoholic solution of KOH yield two compound B and C of molecular formula C10H16 of which ‘B’ is optically active. Ozonolysis of A followed by treatment with Zn-H2O gives acetone as one of the product. Hydrogenation of either B or C yield 4-isopropyl-1-methyl cyclohexane, identify A,B and C. PROBLEM 922 When cis-2,3-dimethyl oxirane is treated with water containing a trace of HClO4, a racemic mixture of 2,3-butandiol is formed, trans-2,3-dimethyl oxirane give meso-2,3-butandiol under identical conditions. Write mechanism for these reactions. PROBLEM 923 When 3-methyl-2-pentanol is treated with ZnCl2 in concentrated solution of HCl, a mixture of chloroalkanes forms, including chiefly 2-chloro-3-methylpentane, and 3-chloro-3-methyl pentane. When 3-methyl-2-pentanol is treated with thionyl chloride in pyridine, only 2-chloro-3-methyl pentane is formed. Write detailed mechanism that accounts for the observation. PROBLEM 924 Synthesize the following by Williamson’s method, choosing alkoxide anion and alkyl halide that will give the best yield: (a)

(c)

(b)

O

(d)

O

O

O

PROBLEM 925 Treatment of 3-methyl-2-butanol with HCl, only a trace of 2-chloro-3-methyl butane is formed. An isomeric product was isolated in 97% yield. Suggest a reasonable structure of this product. PROBLEM 926 Suggest reasonable explanation for the following observations: (a) The first order rate constant for the solvolysis of (CH3)2C==CHCH2Cl in ethanol is 600 times greater than that of allylchloride. (b) After a solution of 3-buten-2-ol in aqueous H2SO4 be allowed to stand for a week, it was found to contain both 3-bute-2-ol and 2-buten-1-ol. PROBLEM 927 Propose mechanism of the following reactions: Ph O + H Ph (a) Ph OH OH Ph OH (b)

H

+

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Problems

OH

O

OH H

(c)

+

H

(d)

+

HO

O

PROBLEM 928 A compound X(C14H14O) on mild oxidation yields C14H12O(Y). If X is treated with a dehydrating agent, it loses a molecule of water and resulting product on vigorous oxidation yields two molecules of benzoic acid. Give structure of X and Y. PROBLEM 929 Convert: R

R

(a)

(b) Propose mechanism: H+

OH

O

H 2O

O

(c) Write products: (i)

OsO4

A

NaIO4

B

(ii)

OMe (iii)

HI(excess)

D

(iv)

O

PhSNa H 2O

C

BH3/H2O2/NaOH

E

OMe

PROBLEM 930 Two optically active compounds A and B have same molecular formula C9H12O and do not decolourise bromine water solution. Both A and B are oxidised by KMnO4 to give benzoic acid and react with Na to give a colourless, odourless gas. Compound A gives yellow precipitate with I 2 /OH – whereas B does not. Also A loses optical activity on treatment with PCC while B retain optical activity on similar treatment. Deduce the structure of A and B.

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PROBLEM 931 Compounds A, B and C are isomeric alcohols with formula C5H12O. A and B reacts with chromic acid solution, B giving an acid D. The three isomeric alcohol reacts with HBr with decreasing relative rates C > A >> B, all giving same C5H11Br(E). in varying yields. A alone can be oxidised by I 2 /OH – to F. Write structure of A to F. PROBLEM 932 C7H14 (A) decolourises Br2 in CCl4 and reacts with Hg(OAc)2 in THF-H2O followed by reduction with NaBH4 to produce a resolvable compound B. A undergoes reductive ozonolysis to give the same compound C as obtained by oxidation of 3-hexanol with KMnO 4. Identify A, B and C. PROBLEM 933 An organic compound A (C7H14) reacts with BH3-THF and then with H2O2/OH – to give a chiral B. Oxidation of B with KMnO4 affords a chiral carboxylic acid C. Ozonolysis of A followed by work-up with (CH3)2S produces C6H12O(D) as one product. D on treatment with LiAlH4 produces another compound E (C 6 H14O) which is optically active. E on heating with conc.H2SO4 produces F(C6H12) as major product. F on ozonolysis followed by work-up with DMS produces acetone as one product. Identify A to F. PROBLEM 934 Treating 3,3-dimethyl-1-butene with dilute sulphuric acid is largely unsuccessful as a method of preparation of 3,3-dimethyl-2-butanol because an isomeric compound is major product. What is the isomeric product and how it is formed? PROBLEM 935 When C2H5ONa reacts with 1-(chloromethyl) oxirane, labelled with 14C as shown by astric in (*) I, the major product is an epoxide bearing the labelled carbon as shown in II. Provide mechanism to explain this fact. Cl *

*

C2H5ONa

O I

OC2H5

O II

PROBLEM 936 An unknown organic compound A(C4H10O2) reacts with sodium metal to liberate one mole of hydrogen gas per mole of A. Although A is inert towards periodic acid, it does reacts with CrO3 to form B(C4H6O3). Identify A and B. PROBLEM 937 Convert: (a)

OH

D

(b) CH2

CH2

O

O

PROBLEM 938 Synthesize the following compounds starting with isobutane: (a) ter-butyl bromide (b) 2-methyl propene (c) iso -butyl bromide (d) iso butyl methyl ether (e) (CH3)2CHCH2OCOCH3 (f) (CH3)2CHCH2CN PROBLEM 939 A neutral compound A has molecular formula C10H16O2. A does not decolourise aqueous solution of Br2 and evolve no gas on treatment with Na. A on acidic hydrolysis produces two compounds B(C4H8O) and C(C6H10O2). B on treatment with SOCl2 gives D(C4H7Cl) which on further treatment with aqueous solution of KCN followed by hydrolysis of product yielded E(C5H8O2). E on reducing with LiAlH4 gives F(C5H10O) which on heating with concentrated sulphuric acid solution

Problems

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yields G(C5H8). G on treatment with HCl yields H(C5H9Cl). C on heating with soda lime gives I(C5H10). I on monochlorination with Cl2/hν give H as sole product. Identify A to I describing mechanism of their formation in each step.

PROBLEM 940 An organic compound A has molecular formula C6H10O and known to decolourise aqueous solution of bromine. A is also resolvable and on catalytic hydrogenation with H2/Pt yields B(C6H12O) which is optically inactive. Also B does not change the orange colour of an acidic solution of dichromate ion. B on heating with concentrated sulphuric acid solution yields C(C6H10) which on reductive ozonolysis gives the following compound: O O   H  C  CH 2CH 2CH 2  C  CH 3 Deduce structures of A, B and C.

PROBLEM 941 An organic compound A(C5H8O2) does not decolourise aqueous solution of bromine and evolve no gas on treatment with CH3MgBr. A on reaction with concentrated solution of HI produced methylene iodide and an another compound B(C4H6I2). B on heating with sodium metal produced C(C4H8) which on monochlorination gave a single isomeric product C4H7Cl. Deduce structures of A, B and C. PROBLEM 942 Starting with 2-methyl propene and using any other needed reagents, synthesize the following compounds: (a) (CH3)2CHCH2OH (b) (CH3)2CDCH2T (c) (CH3)2CHCH2T (d) (CH3)2CHCH2OCH2CH3 PROBLEM 943 Propose mechanism: OH

+

H

PROBLEM 944 Coniferyl alcohol(X) is not soluble in water or NaHCO3, has molecular formula C10H12O3. A solution of Br2 in CCl4 is decolourised by X forming C10H12O3Br2(A). Upon reductive ozonolysis of X, 4-hydroxy-3-methoxy benzaldehyde and B(C2H4O2) are produced. X reacts with benzoylchloride C6H5COCl in presence of a base to form C(C24H20O5). This product rapidly decolourises aqueous solution of KMnO4 and is insoluble in NaOH. X reacts with cold HBr to form D(C10H11O2Br). X reacts with HI to produce E(C9H9O2I) and CH3I. In aqueous base, CH3I and X forms F(C11H14O3), which is not soluble in strong base but decolourises aqueous solution of Br2. Deduce structures of X and A to F. PROBLEM 945 An optically active organic compound A has molecular formula C7H12O2 and does not decolourise cold, dilute and alkaline solution of potassium permanganate. A on treatment with concentrated HCl gives CH2Cl2 and B(C6H10Cl2). B on treatment with aqueous solution of KI gives C(C6H10) which decolourise bromine water. C on treatment with acidic solution of KMnO4 gives 2-methyl pentan-1,5-dioic acid. Deduce structures of A to C. PROBLEM 946 An optically active organic compound A has molecular formula C9H18O which neither change colour of Bayer’s reagent nor of acidic dichromate solution, but evolve a colourless gas on heating with Na-metal. A on dehydration with conc.H2SO4 produces B(C9H16) which exist in two stereomeric forms. Ozonolysis of B followed by work-up with Zn-H2O produced ethanal and

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C(C7H12O). C on treatment with LiAlH4 produced D(C7H14O) which is enantiomeric and change the colour of acidic dichromate solution from orange to blue-green. D on dehydration produced E(C7H12) which on ozonolysis followed by hydrolysis of product in presence of Zn produced cyclopentanone. Deduce structures of A to E showing stereochemical structures of B.

PROBLEM 947 An organic compound A(C11H22) when ozonolyzed yields B(C6H12O) and C(C5H10O). Mild oxidation of C yields D(C5H10O2). Compound B reacts with NaOCl to yield chloroform and sodium salt of D. Reduction of B with LiAlH4 yields E(C6H14O). Dehydration of E gives F(C6H12). Ozonolysis of F yields G(C2H4O) and H(C4H8O). Compound H does not react with Tollen’s reagent or Fehling solution but does form an oxime and semicarbazide. Write structures of A to H describing the reaction involved. PROBLEM 948 An optically active organic compound A(C8H16O) does not decolourise the brown colour of bromine-water but turns orange colour of acidic dichromate solution to blue green converting itself into another optically active compound B. A on heating with conc.H2SO4 produced C(C8H14) as major product which does not show stereomerism. C on reductive ozonolysis produced D(C5H8O) as one of the product which does not change the colour of acidic dichromate solution. D on reduction with LiAlH4 produced E(C5H10O) which on dehydration followed by reductive ozonolysis of product yields a non-resolvable F(C5H8O2). F gives positive Fehling’s solution test but negative haloform test. Deduce structural formula of A to F. PROLBEM 949

Predict products in the following reactions showing stereochemistry where applicable: O 1. BH3

(a)

1. NaBH4

(b)

2. H3O+

2. H2O2-NaOH

OCH3 OH OH

(c)

Ac2O DMSO

1. TsCl 2. NaI, acetone

(d) O

(e) SH

1. NaOH 2. CH3I

1.CH3MgBr (excess)

(f) OCH2CH3

2. H3O+

OH O (g) H

1. Na2Cr2O7, H2SO4 2. CH3CH2MgBr 3. H3O+

(h)

O

CHO

1.LiAlH4 2. H3O+

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Problems

OH

(i)

1.NaH 2. CH3CH2I

OH

(j)

1. HBr 2. Mg/ether 3. CH3CHO 3. H3O+

PROBLEM 950 Provide reagents that will effect the following transformations (more than one steps may be required): CHO OH (a) OH

O

O

O (b) OCH3

Enantiomers

(c)

O

O Br

Br

(d)

OH

OCH3

(e)

(f) OH O

OH HO

O

OH

(g) CHO

CHO

OCH3

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O OH (h)

PROBLEM 951 Devise synthesis of the following compound starting from the indicated starting materials and any other reagents you need. O

O Br from

PROBLEM 952 applicable:

OH

OCH3

CH3CH2Br

NaH

O

(b)

HBr Excess

S CH3CH2Br

(c) S

H2O2 (1 eq)

(d)

CH3COOH

Br NaOH

(e)

H2O

OH OH (f)

O

+

H

Predict all products of the following reactions and show stereochemistry where

(a)

(g)

+

SBr –

BuLi

H2SO4 CH3OH Cyclohexanone

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Problems

(h)

(i) CH3CH2SNa

O

(i)

—S—

(j)

O

(2) H3O+ mcpba excess (1) CH3MgBr (2) H3O+

Silicon-oxygen bonds are easily cleaved by fluoride ions as:

PROBLEM 953

H3C

Si

O

CH3

l l

(CH3)3SiF +

O

H3C

–

F–

Based on the above information, provide a structure of A(C 6 H10O 2 ). Provide mechanism of formation of A. CH3 —O—Si—CH3

CF3CO3H

KF

H 3O +

A (C6H10O2)

CH3

Provide reagents that will effect the following transformations:

PROBLEM 954

CH3O

OCH2CH3 B

O

OH A

C E

D

S OCH3

HO

mCPBA (excess)

NaH CH3I

F

PROBLEM 955 The reaction of a dialkyl ether with excess of HI provides two alkyl iodides. However, under the similar reaction conditions, aryl-alkyl ethers affords an alkyl iodide and a phenol. Explain the observations with the help of following reactions.

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O

CH3

(1)

+ CH3I OH

O

CH3

(2)

PROBLEM 956

I HI excess

HI excess

+ CH3I

Bring about the following transformations: OH OH

(a)

O N

OH

(b)

H OH

(c)

O OH (d)

PROBLEM 957 Predict products of the following reactions. Show stereochemistry where important. In case where more than one stereoisomers are formed, draw both the products and indicate which one will be the major product. (a)

PCl3

—OH

Mg Ether

OH Ac2O

(b) NaBH4

(c) O

D 2O

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Problems

NaSCH3

TsCl

(d) OH

H

PCC

(e)

CH3CH2MgBr H 3O +

OH CH3 O

O

LiAlH4

(f)

H 3O +

O

O OH

(g)

TsCl

NaI

(1 eq)

OH BH3/H2O2

(h)

NaH

CH3CH2I

NaOH

PROBLEM 958

Provide reagents that will bring about the following transformations: OH

(a)

OH

OH (b)

(c) OH

OH

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H

H (d) OH

OH CH3

CH3

PROBLEM 959 Reaction A proceeds more quickly than B. Explain why this is so. On the basis of your explanation of first part, what product would you expect from reaction C? O A:

B:

H

CH3

Na2Cr2O7

OH

H2SO4

D

Na2Cr2O7

C:

H

D

O

H2SO4

OH

PCC CH2Cl2

OH

PROBLEM 960 When benzyl alcohol (A) is treated with TsCl and pyridine at room temperature, the tosylate (B) is formed. However, when the same reaction is carried out at 60°C, the chloride (C) is produced. Draw a mechanism of formation of C from B and explain the effect of temperature on reaction. OH

OTs

TsCl

Cl

60°C

Pyridine A

B

PROBLEM 961 Predict products in the following reactions : where appropriate):

mCPBA excess

(a)

(Provide structures of stereoisomers

NaOH

(b)

SH CH3CH2I

S

(c)

O

H2SO4 t-Butanol

NaH

CH3I

mCPBA

(d) O

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Problems Br Br NaHCO3

(e)

excess

(f)

O

(h)

O

(1) NaN3 (2) H3O+

Br OH

(g)

NaH

(1) PhONa (2) H3O+

Br

OH H3

(i)

O+

I NaH

(j)

HO OH CH2N2

(k) O

PROBLEM 962 HO

Provide the reagents that will effect the following transformations: Br

CH3O

N3

(a)

HO

Br

CH3O

HO

(b)

PROBLEM 963

OCH2CH3

Bring about the following transformations: O

HO

OCH2CH3

H O

I

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PROBLEM 964

Hydrolysis of ether proceed via carbocation intermediate as shown below: H OCH3

O

O

O—CH3

O

+

H+

+

+ CH3OH

Rate determining step

Based on the above information, rationalize the relative rate of hydrolysis of compounds A to D below: O

O

OCH3

I

PROBLEM 965 appropriate:

II

OCH3

O

OCH2CH3

III

OCH3

F

O

IV

OCH3

NH2

Write products in the following reactions. Provide structures of stereo-isomers where

Br

(a)

CH3CH2ONa

m-CPBA

OH PBr3

(b)

Li

O

CH2

Br Mg ether

(c)

D 2O

HBr excess

OCH3 O

OCH3 CH3MgBr

(d)

excess

Ph

H 2O H+

H2SO4 Heat

OH

(e) H CH2CH3

PCC

(i) CH3MgBr (ii) H3O+

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Problems

PROBLEM 966

Provide products in the following reactions: O

OH

(1) NaH

(a)

(2) CH3CH2I

BH3–THF

CH3Li

(b)

H2O2–NaOH

H3O+

OH

O

NaBH4

(c)

K2Cr2O7, H2O

(d)

H2SO4

OCH3 OCH2CH3 (e)

PROBLEM 967

Bring about the following transformations: OCH3

OH (a)

OH

(b)

O (c)

O

O OCH3

O (d)

OH

S O

CH2N2

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PROBLEM 968 The conversion A into B by the reaction sequence below does not work well. Explain why this is the case and provide a better way of doing this. OH

OTs LiAlH4

TsCl

PROBLEM 969

Bring about the following transformations: OH CN

(a) O Br

(b)

OCH3 H N

OH (c)

PROBLEM 970 (a)

O

Provide products in the following reactions: —OH

SOCl2 Pyridine

mCPBA

(c)

CH3 (e)

Hg(OAc)2

(b)

(d)

CH3OH

O

POCl3

OH

Bring about the following transformations: OH  (a) CH 3 (CH 2 ) 4 CH 2CH 2 Br → CH 3 (CH 2 ) 4 CH  CH 2CN

PROBLEM 971

NaBH4

HI

(CH3)3COK excess

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Problems (b)

—OH

PROBLEM 972

Complete the following reactions: CH3 SOCl2

(a)

(b)

O

H3 C

CH3MgBr H 2O

H3 C

OH Ph (d)

HI

O

PROBLEM 973

O

OsO4

(d)

NaHSO3

Propose mechanism of the following reaction: O

PhSNa H2O

OH

SPh

O

PROBLEM 974

OH

O

O

OH

Explain the following observation: CH3

CH3

OH O

CH3MgBr

H2O

H

H

CH3

CH3

CH3

CH3 but

O

CH3MgBr

H2O

OH H

PROBLEM 975

H

Bring about the following transformations:

(a) PhCH 2CH 2CH 2 Br →

OH  C  CH 2 Ph  CH 3

H 3C C==C H

H

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O +

(b)

—C—

—Cl

O (c)

CH3CH2—C—CH2CN

OH

PROBLEM 976 Starting from propene, propanal methylbromide as the only organic reagents, and any other inorganic reagent, synthesize the following compound: OCH 3  CH 2 ==CH  CH 2  CH  CH 2CH 3 PROBLEM 977

Bring about the following transformations starting from indicated materials:

(a)

OH H

+ PhOH

CH3

CH3 Ph

OH

O

O (b)

C—C2H5

CH3CH2—C—CH2—C

PROBLEM 978

Propose mechanism: CH2CH3 + HBr

HOCH2CH2—C—CH2CH3

O

Br

Bring about the following transformations: OH  (a) CH 3CH 2CH 2CH 2 Br → CH 3CH 2  C  CH 3  CH 2CH 3

PROBLEM 979

(b) PhCH 2CH 3

Br  → Ph  CH  CH 2OH CH3

CH3 (c) CH3—C—CH CH3

CH2

PhI

O

CH3—C—CH2—C— CH3

SO3H —I

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Problems

OH  (d) CH 3CH 2OH + CH 3  → CH 3  CH  C ≡≡C  CH 3 O

CH 3 O  (e) CH 3  CH  CH 2 Br + 

PROBLEM 980 appropriate:

CH 3 OH   → CH 3  CH  CH 2  CH  CH 2CN

Complete the following reactions and indicate stereochemistry of product where

CH 3  HBr (a) CH 3  C  CH  CH 2CH 3 →   CH 3OEt + CH3C

(b) HO

CMgBr

O O KOH EtSH

(c)

H CH 3   PhNH2 TsCl (d) CH 3  C  C  CH 3 → →   OH CH 3 S H2O2

(e)

H+

OEt PCl3

(f) HO

PROBLEM 981

(a)

OMe

NaBr Et2O

Synthesize the products from indicated reagents and other inorganic reagents required. OH Br O O + + OH HO

(b) Cl

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PROBLEM 982

Bring about the following transformations: Cl (a)

OH CH3 Cl (b)

CH3

CH3

PROBLEM 983 Synthesize the products from indicated starting materials and any other inorganic reagents needed: Cl (a)

+

O

CH2CH2OH + CH3OH

CH3 (b)

CH2OCH3

+ PhSNa SPh Cl

(c)

CHO + OH

NC CH3 (d)

O + CH3CH2OH

CH3

OH

PROBLEM 984

OCH2CH3

Show how the following compounds can be synthesized from cyclopentanol? OH

(a)

OH C6H5

O (b)

C6H5

(c)

Ph OH

PROBLEM 985 When enantiomerically pure (+)-2-phenyl-2-butanol is allowed to stand in methanol containing a few drops of H 2SO 4 , racemic 2-methoxy-2-phenyl butane is formed. Suggest a plausible mechanism.

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Problems

PROBLEM 986

Deduce structures of missing species in the following sequence of reaction: (i) Mg/Et 2O

Br2

(ii) CH2O (iii) H3O+

CCl 4

CH 2 ==CH  CH 2 Br → A → B(C 4 H 8 Br 2O)

KOH 25°C

→ C(C 4 H 7 BrO) Chiral

KOH heat

D(C 4 H 6O) ←

PROBLEM 987

NaBH4

→ E(C 4 H 8O) achiral

Bring about the following transformation: H

OH OH

PROBLEM 988

H

Bring about the folllowing transformations: OH Br OH

PROBLEM 989

Convert: CN

(a)

(b) H (c) CH3CH2C

CH H

CH2CH2OH

PROBLEM 990 An organic compound A has molecular formula C 8 H16O 2 and it can be resolved into enantiomers. A on treatment with acidified dichromate solution yields B (C 8 H14O 4 ) which is achiral. A on treatment with PCC/CH 2Cl 2 yields another achiral compound C (C 8 H14O 2 ). C on refluxing with dilute Na 2CO 3 solution yields D(C 8 H14O 2 ) which is distereomeric. D on heating in presence of H 2SO 4 yields E (C 8 H12O) which is still chiral. E on heating strongly in stream of H 2 /Pt yields 1,3-dimethylcyclohexane. Identify A to E.

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PROBLEM 991

Provide a detailed, step-by-step mechanism for the following: OH TsOH Heat

OH O

PROBLEM 992 Provide an ordered sequence of reagents that will convert the single enantiomer (indicated by *) of a starting material into the desired single enantiomer of product. * (a)

H

H S

OH

H

(b)

CN

OH

PROBLEM 993

H

Bring about the following transformations:

CH3

(a)

OH

+

N

CH3 CH3

(b)

C

PROBLEM 994 (a)

C

Devise synthesis of the following ethers from an alkyl halide and an alcohols: (b)

O

O

O (c)

PROBLEM 995

Beginning with alcohols containing not more than three carbons, synthesize: OH

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Problems

PROBLEM 996

Starting with alcohol containing not more than six carbon synthesize:

OH

PROBLEM 997

(a)

O

(b)

NH3

O

(g)

O

NaNH2

(c)

(e)

Predict major product in the following reactions:

CH3ONa

(d)

CH3OH

O

(i) LiAlD4

(h)

(ii) H3O+

PROBLEM 998

O

Dil. HCl

(i)

—Li (ii) H3O+

Predict product in the following reactions: O

O

(a)

(i) CH3MgBr (ii) H3O+

(f)

(ii) H3O+

O

O O

(i) CH3CH2SNa

HBr excess

OCH3

NaN3

HBr excess

(b)

O O Br (i) NaBH4

(c)

(d)

(ii) NaOH (aq)

OH NaH

(e) O

C6H5CH2Br

(f) O

H2SO4

O

EtOH

O

LiAlD4 NH4Cl

PROBLEM 999 Accounts for the following observations : 2,2-dimethyl oxirane is hydrolyzed at faster rate than oxirane in acid medium but reverse is true in alkaline medium.

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PROBLEM 1000

Propose mechanism: O HO

PROBLEM 1001

O

CH2N2

Propose mechanism: OH

O

O O * OH

HCl

O* +

O

Ph

Ph

Ph

PROBLEM 1002

OH

NaOH/H2O

Identify the missing species: TsCl

(a)

KCN

A

B

OH

CH3Li H3

O+

C

PPh3, CH3Br BuLi

CN O H+

(b)

EG

Pd/BaSO4

A

PbO

B

N 2H 4 HO–

C

H 3O + H 2O

D

H

PROBLEM 1003

Provide reagents necessary for the following transformation:

(a)

OCH3 OH N3

(b)

OH

(c) N

N

D

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Problems

Propose mechanism:

PROBLEM 1004

N 2H 4 H+/Heat

O

O

O

Synthesize the indicated product from the supplied starting materials:

PROBLEM 1005

(a) OH H

H

(b) HOOC

Cl

O CN

(c)

C O

and C

CN

O

O

PROBLEM 1006

Suggest mechanism of the following reaction: O

OH

CH3COOH ∆

N—NH2

PROBLEM 1007

Propose mechanism of the following reaction: O H+

OH

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Write product of the following reactions and indicate stereochemistry product where

PROBLEM 1008 appropriate: O (a) H3C

CH3 O

H+ CH3COOH

CH3

CH3

H+ H2O

O

(i) LiAlD4

(c)

O

(b) Ph Ph

HBr

(d)

(ii) H2O

OH NaH

(e)

CH3I

CH3

Bring about the following transformations:

PROBLEM 1009 (a)

—OH O

(b)

from alcohols containing not more than three carbons.

PROBLEM 1010 The following series of reactions was performed during synthesis of a target molecule. Give structures of each of the indicated compound: O OH

Br

PCC CH2Cl2

HO

A

OH H 3O +

PROBLEM 1011 reactions:

Mg

B

H+ D

PCC CH2Cl2

H

C

Et2O

E

NaOMe MeOH/∆

Predict structures including stereochemistry, of products formed in the following

O O (a)

F(C8H12O)

LiAlH4

(b)

Ph H

(i) PhLi

CH3

(ii) H3O+

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Problems

CH3 OH (c)

H CH3

CH2I2 (1.0 equivalent)

(d)

mCPBA

CH3

PROBLEM 1012

OH

Zn-Cu

Propose mechanism of the following reactions:

OCH3 OH O KH

(a)

∆

CH3

H OMe

H3C (b)

CH3

CH3

OH +

CH3

CH3

TsOH ∆

O

ALDEHYDES AND KETONES PROBLEM 1013 Synthesize 3-pentanone using alkyl bromide containing not more than two carbons. O CH3 (a) Br(CH2)5Br

(b) Br

O

PROBLEM 1014 Convert: PROBLEM 1015 A hydrocarbon A(C7H14) exist in two stereomeric forms. A on reductive ozonolysis yields B(C3H6O) and C(C4H8O). B on catalytic hydrogenation yields D(C3H8O) that can’t be resolved into enantiomers whereas C on catalytic hydrogenation yields E(C4H10O), which can be resolved into enantiomers. B on treatment with HCN followed by hydrolysis of product yields F(C4H8O3) which can be resolved into enantiomers. C on treatment with HCN followed by hydrolysis of product yields G(C5H10O3), which is also resolvable. Identify A to G and show the stereomeric structures of A. PROBLEM 1016 Which of the following pairs has larger equilibrium constant for addition of water? O  (a) Cl 3CCHO or Cl 3C  C  CH 3

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O O   (b) Cl 3CCH 2 C  CH 3 or CH 3  CCl 2  C  CH 3 O

O (c)

H or

C

O (d)

O or O N 2

C

NO2 (e)

H

C

MeO

C

O

NO2

NO2

O or

C O2N

C O2N

PROBLEM 1017 An organic compound A(C6H10) on catalytic hydrogenation with H2/Pt yields C6H12. A on ozonolysis followed by work-up with Zn-H2O yields B(C6H10O2). B on refluxing with dilute NaOH solution gives C (C6H8O). C on catalytic reduction yields D(C6H12O). D on treatment with chromic acid yields E (C6H10O2) which on heating with soda lime yields F(C5H10). F on monochlorination yields a single isomer. Identify A to F. PROBLEM 1018 Predict products in the following reactions:

(a)

A

+ cold,dil, alk KMnO4

HIO4

–

B

OH

CHO (b) A + OsO4

(c)

H2O2

B

HIO4

O3 Zn-H2O

–

C

–

A

OH

OH

B

C

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Problems

PROBLEM 1019 Predict products in the following Cannizzaro reactions: O  − (a) CH 3  C  CHO + OH →

PROBLEM 1020 Write products in the following reactions: (b) H N 2

(c)

CHO + HOC

Zn-H2O

+ O3

–

A

N O2

OH

–

OH

Only cross product

B

PPh

3 (a) Ph  C  CH 2  CH 2  CH 2CH 2Cl → BuLi  O

PPh3

(b)

BuLi

OHC

(i) O3/Zn-H2O

A

(ii) 50% aq NaOH

B

CH2Cl

PROBLEM 1021 Predict product in the following reactions O  LiAlH4 H+ H+ (a) CH 3  C  CH 2CH 2CHO + CH 3OH → A → B → C Reflux

Excess

O (b)

HOCH2

CH2OH

+

H

PROBLEM 1022 Predict products in the following reactions: (a)

O + H2NOH

A

H

+

B

O  – H+ NH2OH OH (b) C 6 H 5  C  CH 3 → C + D → G → H + I + ↓H OH– E → C 6 H 5 NH 2 + F

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164 NH2OH

(c) N

+

H

J + K

CHO

OH

N

–

O + P

+

H

L

OH

–

+ M N

NH2

PROBLEM 1023 An organic compound A has molecular formula C11H14O. A on treatment with hydroxyl amine yields two stereomers B and C with their molecular formula C11H15NO. B on treatment with concentrated sulphuric acid solution yields D, an isomer of B where C on treatment with concentrated sulphuric acid solution yields E, another isomer of C. D on alkaline hydrolysis yields a white crystalline solid and an optically active amine F(C4H11N). E on alkaline hydrolysis produces aniline as one product. Identify A to F. PROBLEM 1024 When 2-chloro-6-methyl cyclohexanone is treated with NaOH, ring contraction occurs and compound (i) is formed not (ii) explain through mechanism. COONa COONa

not CH4 CH3 (ii)

(i)

PROBLEM 1025 Synthesize (a) 5-nonanone from 1-butanol as only starting organic compound. (b) 2-butanone from ethyl bromide. PROBLEM 1026 Outlined below is a synthesis of a compound used in perfumes. Provide all missing structures. CH2OH

PCC

A (C11H14O)

Propanal NaOH H2/Pd-C

B(C14H18O) C (C14H20O)

PROBLEM 1027 Show mechanism of the following reaction O O Br

Br +

O EtONa

O

O O

PROBLEM 1028 An organic compound A(C12H20) decolourise bromine water, and on reductive ozonolysis produces two molecules of B(C6H10O). B on aldol condensation followed by dehydration of

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Problems

product gives C(C12 H18O) which on reduction with N2H4/OH – regenerate A. B on reduction with Zn(Hg)/C⋅HCl produces C6H12(D) which on monochlorination gives C6H11Cl(E) as a sole product. Identify A to E.

PROBLEM 1029 Complete the following reactions: dil. NaOH (b) C 6 H 5CHO + C 6 H 5COCH 3 → A+B O

dil NaOH

(a) C6H5CHO +

A+B

dil. NaOH (c) C 6 H 5CHO + CH 3 NO 2 →

O (d)

CHO

dil NaOH

(e) ?

(f) ?

dil NaOH

O

dil NaOH

O

PROBLEM 1030 Carry out the following transformations: O (a)

CH=CHCOC(CH3)3 O

(b) C6H5CHO

C6H5CH

NH 2  (c) C 6 H 5CHO → C 6 H 5CH 2 CHCH 3 O OH

(d) O

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166 O

(e)

PROBLEM 1031 An alkene (A) on reductive ozonolysis yields acetone and an aldehyde. The aldehyde is easily oxidised to an acid B. B when treated with Br2/P yields a compound C which on hydrolysis gives D. D can also be obtained from acetone by reaction with HCN/H 2O. Identify A to D. PROBLEM 1032 A hydrocarbon A containing C = 88.24% (MW=68) when reacted with Na followed by n-propylbromide, produces a compound B(C8H14). It gave a ketone C5H16O(C) with Hg2+/H2SO4. B on refluxing with H+/KMnO4 gave two isomeric acid D and E. Identify A to E. PROBLEM 1033 A compound having molecular formula C6H12 is treated with Cl2 in CCl4 to form a compound B(C6H12Cl2). B is treated with alcoholic KOH followed by NaNH2 resulting in the formation of C (C6H10). C when treated with Pt/H2 gave 2-methyl pentane. Compound C does not react with NaNH2 and with NH4OH/AgNO3. On ozonolysis A gave two aldehydes D and ethanal. Deduce the structure of A to D. PROBLEM 1034 Compound A(C7H16O) can be resolved into enantiomers and it reacts with Na2Cr2O7/H2SO4 to give B(C7H14O). When B is treated with NaOD at 25°C for several hours and then analyzed by mass spectroscopy, it is found to have MW = 116. Identify A and B. PROBLEM 1035 An organic compound A(C7H6O) on treatment with hydroxyl amine produces two isomeric compounds B and C with molecular formula C7H7NO. B and C on acidification rearranges to D and E respectively. D on alkaline hydrolysis produces a white crystalline solid F(C7H6O2) and ammonia gas whereas alkaline hydrolysis of E produces G and formic acid. Identify A to G describing mechanism of their formation. PROBLEM 1036 The following road-map problem centers on the structure and properties of A, a key intermediate in these reaction. Give structures for the compounds A through J: OH

H2CrO4

OH PCC

A +

H

H

NaBH4

G

warm H+

I

CH3MgI H3O +

Tollen's reagent

(ii) H3O +

D

Ethylene glycol

B

(i) PhNHNH2 (ii) H2SO4

OH (i) J

E

C

Zn(Hg) HCl

F

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Problems

PROBLEM 1037 An optically active alcohol A(C5H12O) on oxidation with PCC gives an optically active aldehyde B(C5H10O). B on treatment with dilute OH – gives C(C10H20O2). C on heating with H2SO4 gives D(C10H18O) which after reductive ozonolysis gives E(C6H10O2) and F(C4H8O). F on treatment with I 2 /OH – gives an yellow precipitate. Identify A to F. PROBLEM 1038 Provide reagents in the following transformations CHO

?

(a)

?

CHO CHO O ?

?

(b) O OH

X

(c)

O Y

OH CHO

CHO

Z

+

CHO

CHO O MeO

(d)

X

CH3

Y

CH3OCH=PPh3

H 3O +

Z

PROBLEM 1039 Propose mechanism of the following reactions: OH ROH

(a)

O

OH

O O +

H

(b)

+

H

O

O

OR

PROBLEM 1040 Synthesize the compounds from indicated starting material: O

Ph

(a)

COOH O

(c)

O

(b) PhCHO Ph

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PROBLEM 1041 Predict mechanism of the following reaction: OH

O (CH2)10 O

dilute NaOH

(CH2)9 O

(CH2)9

O OH

PROBLEM 1042 Give the structures for the indicated compounds in the following reactions:

O

(a) H

CH2CH2OH

CH3CH2MgBr

B

A

+

H

CH3(CH2)5CHO

C

+

H

Pd-CaCO3 H2

D

O

(b)

Br

Mg Ether

O (c)

E

H3

O+

F

HOCH2CH2OH +

H

PCC

G

BH3/H2O2/KOH

H

+

I

H

J

PROBLEM 1043 Complete the following equations. O (a)

H

O

+

NaOH

H 3O +

NaOH

H 3O +

O O (b)

+ CHO 2 equivalent

1 equivalent

(c) 2

CHO

NaOH

PROBLEM 1044 Plan a synthesis for each of the following compounds through shortest route. You have been supplied bromobenzene and any organic reagents containing three or fewer carbon atoms and any inorganic reagents you need. O (a)

CH3CH2 C

(b) H

C

CH2CH2OH H

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Problems

O Br

(c)

(d)

H O (e)

(f)

H

PROBLEM 1045 A neutral, resolvable organic compound A has molecular formula (C8H16O2). A on treatment with LiAlH4 gives isomeric B and C (C4H10O) of which only B is optically active. B on treatment with acidified dichromate solution gives C4H8O(D) which on refluxing with dilute NaOH followed by acidification of product gave E(C8H14O). E on heating with N2H4 in alkaline medium affords F(C8H16). F on treatment with B2 H 6 / H 2O 2 /OH – produced a resolvable G(C8H18O). G on treatment with acidified dichromate solution produced H(C8H16O) which on treatment with “m-chloroperbenzoic acid” produces A. Deduce structures of A to H. PROBLEM 1046 An organic compound A has molecular formula (C7H12O) and does not decolourise Bayer’s reagent and evolves no gas with Na-metal. A also does not react with either ammonical silver nitrate solution or phenyl hydrazine solution. A on refluxing with methanol in presence of trace amount of H2SO4 produces a non-resolvable B(C8H16O2). B on treatment with CrO3/pyridine produced C(C8H14O2) which is non-resolvable. C on refluxing in HI solution produced another non-resolvable compound D(C7H11OI). D on treatment with alcoholic KOH followed by oxidative ozonolysis of product produced E(C7H10O4) which does not show optical activity. E on simple heating produced C6H10O2 which on refluxing with dilute NaOH gave: O

Deduce structures of A to E.

PROBLEM 1047 Predict the structure of the products in the following reactions and show stereochemistry when it is known:

O (a)

C

CH2CH2CH2Cl

(i) HOCH2CH2OH (ii) TsOH/Benzene

CH2 NH A

2

H3O

B +

C

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170 O NaBH4/H2O

(b)

BH3

(c)

A

NaOH

A

+

Na2Cr2O7

CrO3 H

B

H

(e)

HC

NaNH2

CH

D

KOH/Ethanol

B

+

PhLi

CHO

TsCl Pyridine

H2O2

THF

(d)

A

C6H5CH2Br

PhNHNH2

CH3CH2MgBr

E

O (f)

H2NNH2

X

KOH ethylene glycol

PROBLEM 1048 Bring about the following conversions: OH

O (a)

CH3 OMe

CH3

O (b)

CHO OMe Br

O OMe

(c)

CH3

CH3

PROBLEM 1049 (a) Bring about the following transformation: CHO

OH CH2Br

B

O

C

F

C

mCPBA

D

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Problems (b) Propose mechanism Et O

O

+

OH

H

O OH

PROBLEM 1050 Propose mechanism: OH

+

H

(a)

OH O

(b)

+

H

O

O (c)

O

+

CH2OH

H

O

PROBLEM 1051 Propose mechanism: O (a) MeOOC

COOMe +

H

H

Heat

O

(b) EtOOC

O

O

O

COOEt

+

O

t-BuOK

O

O (c)

NaOMe

MeNHOH

H

O N Me

HO

COOEt

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PROBLEM 1052 Bring about the following conversions: O (a)

H O H

H

H H

(b)

(c)

OH

OH

Cl (d)

(e)

+

PROBLEM 1053 An optically active compound A has molecular formula C10H12O. A decolourises Bayer’s reagent as well as it forms salt with NaOH. A on treatment with HBr forms B(C10H13OBr) as major product which is still enantiomeric. B on treatment with alcoholic solution of KOH gives a stereomeric C that is also an isomer of A. A on treatment with acidic solution of KMnO4 gives ortho hydroxy benzoic acid as one of the product. A on treatment with B2H6/H2O2/NaOH produces D(C10H14O2) which on refluxing with dilute sulphuric acid produced E(C10H12O), E neither decolourises Br2-H2O solution nor forms any salt with NaOH. Deduce structures of A to E. PROBLEM 1054 An organic compound has molecular formula C10H6(A). A on treatment with HgSO4/H2SO4 gives B(C10H10O2) which forms an insoluble precipitate with 2,4-DNP but fail to react with ammonical solution of AgNO3. B on refluxing with dilute NaOH solution produced C(C10H8O). C on further treatment with alkaline solution of N2H4 gives D(C10H10) which decolourises Br2-H2O solution. D on treatment with H2/Pt produced the adjoining compound X. Deduce structures of A to D.

X

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Problems

PROBLEM 1055 Complete the followings: O

?

NH

(a)

NH CH3

H N

N ?

(b) O

I PPh3

(c)

?

BuLi ∆

O

NH2 H

(d)

COOH

?

O

O O

(e)

Me2CuLi +

H

(Propose mechanism)

PROBLEM 1056 Provide products in the following reactions: O (a)

LDA

O

(CH3)2CHI

OH–

(b)

OH2

Ph O

PROBLEM 1057 Propose mechanism: O CHO

BF3

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PROBLEM 1058 Convert: Ph

OH

O H3C

H H

H

Ph

PROBLEM 1059 Propose mechanism: O CH2

(a)

OMe

O

(b)

H3O+

X (C4H8O2)

NH2OH +

H

PROBLEM 1060 Starting from propanone, suggest synthesis of the following compounds: O

O (a)

COOEt

Ph

(b)

(c)

CN O

O (d)

(e)

(f) Ph

O

PROBLEM 1061 Offer mechanistic explanation for the formation of products in the following reaction: O O MgCl

OH +

OH +

PROBLEM 1062 Propose mechanism of formation of final products in the following reactions and identify the labelled product: COCl O CH2N2(excess) Ag2O/MeOH (a) A(C8H12N2O) O

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Problems

(b) Me2CO

750oC

MeOH

B(C2H2O Cool

O

NaN3

O

H2SO4

O

MeOH

C(C4H4O2)

(c)

CH3COOMe

O

D(C6H11NO)

LiAlH4

Me NH

H2SO4

(d) O

HO

O (e)

(f) O

H2SO4

CH2OH

H2SO4

HO

O

PROBLEM 1063 Supply a suitable technique for the synthesis of following compound from indicated starting material: O OCH3 from

CH3

OCH2CH3

PROBLEM 1064 A sweeting smelling organic compound A has molecular formula “C 9 H12O 3 ” and it does not change colour of moistened litmus paper. A on acidic hydrolysis yields an optically active compound B with their molecular formula “C 9 H14O 4 ”. B does not decarboxylate on simple heating. B on heating with soda-lime (NaOH/CaO) yields an optically inactive compound C “C 8 H14O 2 ”. C does not reduce Tollen’s reagent but yields a bright orange crystalline substance on treatment with 2,4-dinitrophenylhydrazine. C on oxidation with acidified K 2Cr 2O 7 followed by heating of product with soda-lime yielded D “C 7 H12O” which is still non-resolvable. D on heating with alkaline solution of hydrazine yielded methyl cyclohexane. Identify A to D. PROBLEM 1065 An organic compound A has molecular formula “C 9 H12 ” and decolourises bromine-water solution. A on treatment with H 2 /Pt gave an optically active hydrocarbon. A on treatment

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with acidified solution of KMnO 4 yields B “C 9 H12O 7 ” which can be resolved into enantiomers as well as gives bright orange crystalline substance with 2,4-dinitrophenyl hydrazine. B on simple heating gives an optically inactive C(C 7 H12O 3 ) which evolves a gas with aqueous NaHCO 3 and gives a yellow precipitate with NaOH/I 2 . C on treatment with NaBH 4 yields a resolvable D“C 7 H14O 3 ”. D on heating with H 2SO 4 yields E “C 7 H12O 2 ” which does not change the colour of moistened litmus paper. Identify A to E.

PROBLEM 1066 An organic compound A has molecular formula “C 9 H16O” and it gives an yellow precipitate on treatment with alkaline solution of iodine. A on treatment with hydroxyl amine (H 2 NOH) yields two stero-isomeric compound B and C with their molecular formula “C 9 H17 NO”. B on treatment with concentrated sulphuric acid undergo rearrangement to yield a stable product D which is an isomer of B. C on similar treatment undergo similar type of rearrangement to yield a stable product E which is an isomer of C. Hydrolysis of E in acid medium yields F(C 7 H15 N) as one of the product. F on treatment with nitrous acid followed by oxidation with acidified dichromate solution yields G(C 7 H12O 2 ). G on heating with sodalime yields a product C 6 H12 which on heating with nitric acid affords a single mono-nitro derivative. Identify A to G. PROBLEM 1067 important.

Predict all products of the following reactions and show stereochemistry where CHO

(a)

+ CH3NH2

HCl

O CH3OH (excess)

(b)

HCl

H H

HO

(c)

CH2

OH

PPh3

HCl H 2O

HCl

O O (d)

Ph3P/BuLi

Br

(e)

OCH3

HCl

OCH3

H 2O

O

(f) O

O

Acetone

CH3NH2

N 2H 4

X

Pd/C H2

KOH Heat

Y

CH3COCl

HCl H 2O

Z

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Problems

Provide reagents for the following transformations :

PROBLEM 1068

OCH3 (a)

OCH3

CH2

(b)

OCH3

NHCH3

OCH3

O

O OCH3 (c)

NOCH3

OCH3

OCH3 (d)

CH2

OCH3 OH

O

O

PROBLEM 1069 Arrange the following acetals/ketals in increasing ease of their hydrolysis in acidic medium—(provide mechanistic reasoning). O

O

O

O II

I

O

O III

PROBLEM 1070 The reaction of aldehyde A with HCN gives two cyanohydrin products B and C. Explain why C in produced in greater amount than B. OH

OH CHO

HCN

OH A

PROBLEM 1071 for your choice.

OH

OH CN

+

Write the structures of major and minor stereo-isomers of product explaining reasons

PCC

(a)

CH3MgBr

OH CH3 (b) H

CN OH C

OH B

H

Ph

OH

O H

CH3MgBr

H3 O +

H3 O+

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PROBLEM 1072

Bring about the following transformations: H

H

I

(a)

H

+ H

H OH

(b)

O O O

O

(c)

OH

H OH

OH

OH O

(d) O

H

O

PROBLEM 1073 Draw the product of the reaction A and B and predict which one proceeds more efficiently and why? O

A:

Cl

H H

B:

CN

HCN

H

HCN

H CN

O

PROBLEM 1074

OH

Cl

OH

Bring about the following transformations: O

O

(a) +

N Cl– H CH3 CH3 O (b)

N—NH2

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Problems O (c)

H

OCH3

O

O O

Provide a mechanistic explanation for the reaction:

PROBLEM 1075

O

O

O

HCl

+ NH2OH

N

Predict products in the following reaction. Provide stereoisomers where appropriate.

PROBLEM 1076

O CH3Li

(a)

NaH CH3I

H 3O +

CH3

O (CH3)2S

(b)

CH2

NaN3 H 2O

H OH

(c)

O (d)

K2Cr2O7

CH3CH2MgBr

H+/H2O

ether

HOCH2CH2OH

Ac2O

HCl

∆

OH

PROBLEM 1077

H 3O +

CH3CH

PPh3

H 2O HCl

Bring about the following transformations: O

O

O N

PROBLEM 1078

Propose mechanism for the reaction: O CH3O

OPh

O

H+ CH3OH

O

CH3O HO

OCH3

OPh

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Bring about the following transformations:

PROBLEM 1079 O

HO

(a) F NH

OH (b)

O CHO O

(c)

COOH H2N

H

PROBLEM 1080 The equilibrium established when cyclohexanone is treated with methanol and HCl lies in the favour of ketone. In contrast, the same reaction with cyclopropanone lies heavily in favour of hemiacetal. Explain. O

OCH3

CH3OH/H+

H

:

O

H

CH3OH/H+

OCH3

PROBLEM 1081 Within each set, select the compound which is more reactive in nucleophilic addition at carbonyl carbon: (b) CH 3CHO or CH 3COCH 3 (a) CH 3COCH 2CH 2 Br or CH 3COCH 2 Br (d) PhCHO or CH 3CHO (c) CH 3COCH 3 or CH 3  CO  CF3 PROBLEM 1082

Provide reagents for accomplishing the following transformations: OCH3 C Ph

OCH3

CH3 D

OH

CH3

A

C

C Ph

OH

O

CH2 CH

CH2

Ph

CH3 B

CH3

Ph

N

CH3

CH2

C Ph

CH3

OCH3

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Problems

PROBLEM 1083

Bring about the following transformations: CH3

O

O

O

O (a)

(a)

CHO

OH

CH3

PROBLEM 1084 Treatment of trans 2-chlorocyclohexanol with NaOH yields 1,2-epoxycyclohexane while reaction of cis isomer under the same condition yields cyclohexanone. Propose mechanistic explanation. PROBLEM 1085 When 4-hydroxybutanal is treated with methanol in the presence of acid catalyst, 2-methoxy tetrahedrofuran is formed. Provide mechanism. PROBLEM 1086

Give products of the following reactions (major): O NH3

(1) LDA

(a)

(b)

(2) CH3Br

HCN/H2O

H

O O

O (c)

H

O OH–

+

Heat

O

(d)

Ph +

H+

N H O

(e)

Br2/CH3COOH

KBuO+

PROBLEM 1087 When compound A is treated with CH 3ONa in CH 3OH, isomerization to compound B occurs. Provide mechanism and explain why equilibrium favour product B? H

O

H CH3ONa CH3OH

H A

H B

O

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PROBLEM 1088

Bring about the following transformations:

O OH (a)

C

CH3

C

H

H OH

H

(b) CH3 CH3

(c) H3C

OH CN

O O

CH3 O

(d)

PROBLEM 1089

Provide synthesis of the compounds from indicated starting materials:

(a)

OCH3

O

(b)

N H

PROBLEM 1090 An optically active organic compound A(C 4 H 8O 2 ) gives an orange precipitate with 2,4-dinitrophenylhydrazine and a silver mirror with ammonical silver nitrate solution. A is converted into B(C 4 H 6O) on treatment with concentrated sulphuric acid. B may exist as cis or trans isomers. B reacts with HCl to give C(C 4 H 7OCl) as major product and with chlorine to give D(C 4 H 6OCl 2 ). Reduction of C with Sn/HCl gives E(C 4 H 9OCl) which on hydrolysing with aqueous alkali yields F(C 4 H10O 2 ). F gives a yellow precipitate with I 2 and NaOH solution but A does not. Identify A to F. PROBLEM 1091

Synthesize products from indicated reactants and other inorganic reagents needed:

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Problems

(a) CH 3CH 2 —C ==CH 2 + CH 3  CH  CH 3   OH Br

CH 3  → CH 3CH 2  C ≡≡C  C== CH 2

CH 3 CH 3   (b) CH 3  C  OH → CH 3  C  COOH   Br CH 3 CH 3 CH 3 CH 3    (c) CH 3CH 2  CH  CH==CH 2 + CH 2 == CH 2 → CH 3  CH 2  CH C C ≡≡CH  OH CH2CH3 (d) CH3CH2—C

O

OH

CH2 +

CH3CH2—CH—C— CH2CH3

PPROBLEM 1092

Cl

Propose mechanism: C6 H 5

O O   CH3OK C 6 H 5CH 2 —C—CH 2C 6 H 5 + CH 2 ==CH—C —CH 3 → H3C— CH3OH

PROBLEM 1093

(a) H3C—

3

C6H5—CHO NaOH/H2O

PROBLEM 1094

Propose mechanism:

O

O Br

(a)

O —C6H5

Predict product in each of the following showing stereochemistry where appropriate: O O CH3 C6H5CH2SH NaOH (b) —CHO + CH3—C—CH3 NaOH/H2O H2O O CH O

(c)

—OH

(CH3)3COK

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O O O

O

(b) Ph—C—C—Ph + Ph—CH2—C—CH2—Ph

Ph

Ph

KOH C2H5OH

Ph

Ph

PROBLEM 1095 The acid catalyzed hydrolysis of two acetals shown below occurs at much different rates. Provide a brief explanations with structures of key intermediate to illustrate why these two acetals undergo hydrolysis at such different rates. H O

H

H 3O +

O

O

H O

Reaction is complete in 1.0 hour at 20°C

H 3O +

O

H Reaction is complete in 24 hour at 60°C O

PROBLEM 1096

Provide a stepwise mechanism for the following acid catalyzed reaction: CH3 O

H+

2CH3CHO

O CH3

PROBLEM 1097

Rank the following compounds in increasing order of electrophilicity: O  F3C

PROBLEM 1098

CH 2  CH 3

H I

NH  H

CH 3

II

III

CH 3

Provide a stepwise mechanism for the following acid catalyzed reaction: CHO H3O+ O

H2O

O

HO

OH

O O

PROBLEM 1099 H

Br

OH

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Problems

PROBLEM 1100

Identify A in the following reaction and propose mechanism of its formation: O NaOCH3

I

H

C5H10O2 (A)

CH3OH

H 3O +

HO

H 2O

H

O

NaOH — No reaction.

PROBLEM 1101 The enantiomerically pure cyclic ketone, shown below, loses optical activity upon standing in a solution of EtONa/EtOH. Provide mechanistic explanation. EtOH EtONa

O CH3

H

PROBLEM 1102

Racemic mixture

Bring about the following transformations: O CHO CH3

(a)

CH3 O

(b) OH

OMe OH

OH

(c)

OH

Br OH

PROBLEM 1103 An organic compound A has molecular formula C 6 H12O. A does not decolourize bromine water solution but evolves a gas on heating with NaH. A can be resolved into distereomers and it also change colour of acidified dichromate solution from orange to blue green. A on heating with concentrated sulphuric acid yields B(C 6 H10 ) as a major product which can’t be resolved into enantiomers. B on oxidation with ozone followed by work-up with Zn  H 2O yields C(C 6 H10O 2 ). C on refluxing with dilute solution of NaOH gives D–an isomer of C. D also evolves a gas wtih NaH. D on heating dehydrate to yield E(C 6 H 8O). E on heating with H 2 gas in presence of platinum catalyst at high temperature and pressure yields F(C 6 H12 ) which gives single monochloro derivative on free radical chlorination. Identify A to F. PROBLEM 1104 An organic compound A has molecular formula C 7 H14O 2 and it can be resolved into enantiomers. A on treatment with H 2CrO 4 yields B(C 7 H12O 2 ) which can’t be resolved into enantiomers.

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Refluxing B with dilute alkali solution yields “C” which is an isomer of B. C evolves a gas with NaH but does not change colour of acidified dichromate solution. C on heating with concentrated sulphuric acid yields D(C 7 H10O) which on heating with Pt at high pressure yields methyl cyclohexane. Identify A to D.

PROBLEM 1105 An organic compound a has molecular formula C11 H 22 . A on treatment with concentrated solution of KMnO 4 in acidic medium yields B(C 3 H 6O) and C(C 8 H16O), both form orange colour precipitate with 2,4-dinitrophenyl hydrazine but neither gives positive Tollen’s test. C on treatment with LiAlH 4 yields D(C 8 H18O) which on heating with concentrated sulphuric acid solution gives E(C 8 H16 ). E on further treatment with acidified permanganate solution yielded B and 2,2-dimethyl propanoic acid. Identify A to E. PROBLEM 1106

Identify the compounds labeled A to F: NBS

(CH3 )3 COK

O3

hν/CCl 4

(CH3 )3 COH

Zn  H2O

MeMgBr

A(C 8 H10 ) → B(C 8 H 9 Br) → C → D(C 7 H 6O) → H 3O +

PCC CH2Cl 2

N 2H 4

A → E → F → – H 2O

PROBLEM 1107

OH /∆

Arrange the following hydrates in order of increasing K eq : O R

A:

C

H 2O

A

HO

C

R

OH A

H, (CH 3 ) 2 CH—, ClCH 2 — Ph—.

PROBLEM 1108 Propose a reasonable synthesis of the product in the following reactions starting from the indicated materials. CHO

(a)

CHO

O N

(b) Compound containing 5 or less carbons

PROBLEM 1109

Give structures of all the products in the following reactions:

O (a)

NaBH4

B2H6

(b)

H2O2/NaOH

NH4Cl

O

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Problems O

OH Ph

(c) H

(i) (CH3)2CHMgBr Et2O

(ii) H+

H

(d)

(iii) H+/H2O

H 2O

H

CH3 OH NaBH4

(e)

(f) Ph—CH—CH—CH3

NH4Cl

O

(i) CrO3/H2SO4 (ii) PhCH2Li

(i) PCC (ii) MeMgBr (iii) H+/H2O

CH3

O O (i) LiAlH4

(g)

(ii) NH4Cl

PROBLEM 1110

Complete the following reactions:

O

O NH2(excess)

H

(a) H O

H2/Pt

C—NH2

(b)

CHO

CH3COCl

O

O Ph ?

(c)

Br2 NaOH

(i) N2H4/OH–

(d)

(ii) O3/Zn–H2O (iii) EG/H+

NH2

(e)

H+

O

PROBLEM 1111

Bring about the following transformations: OH

( a)

O ( b)

O

O NO2

I

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(c)

NH2

PROBLEM 1112

N

Propose mechanism: O

OH

PROBLEM 1113 (a) Draw all the enol forms possible 4-methyl-1,3-cyclopentandione. (b) Give product of the reaction of above diketo compound with: (i) Excess of D 2O/H + , (ii) One mole of Br 2 /H + , (iii) LDA/THF then CH 3I. PROBLEM 1114

Complete the following reactions: O KMnO4

(a)

OMe

(b)

H+

OH–

OH

OH

(c) 2PhCH2CHO

O

HO– Warm

NaBH4

(d)

CH3OH

O

(e)

CH3NH2

CHO

LDA

MeI

CHO

Br2

(f)

H+

O

PROBLEM 1115

Consider the compound shown below: O O H

(a) Give the product of its reaction with excess of D 2O. (b) Give product of reaction with one mole of Br 2 in dilute acidic medium.

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Problems

(c) Give product of its reaction with one mole of C 6 H 5CHO in presence of strong base EtONa. (d) Write structure of its most stable enol form.

PROBLEM 1116 Of the two compounds A and B shown below, A losses its optical activity upon standing an enantiomerically pure form of it in a solution of NaOEt but no such loss of optical activity is observed on similar treatment to B. Explain:

A

B O CH3

O

Give a complete stepwise mechanism for the reaction:

PROBLEM 1117

O O

O

NaOH ∆

Complete the following reactions:

PROBLEM 1118

O (a)

+ LDA

MeI

H3O+

KCN

(b)

H2O

CHO O

O

O OEt

(c)

H3O+

O

heat

Br2

(d)

H+/H

2O

O H2CO

(e)

(f)

Li

HO–/∆

CH2O HO–/∆

O O (g) Cl

Cl

H2O H+

(h)

H+

+ N H

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PROBLEM 1119

Synthesize the following compound starting from alcohols of three or less carbons:

PROBLEM 1120

Write product in the following reactions: OH

(a) O

O

H 3O +

PhMgBr

PCC CH2CH2

PCC CH2Cl2

O

(b) H

HO–

D 2O

H

excess

O (c)

H

HO– warm

KCN

(d)

H 2O

O COOH (e)

CH3Li

H 2O

excess

H+

O (f)

H2

H+

H + H—N

Ni

H N

(g)

H+

O

PROBLEM 1121

Give a complete, stepwise mechanism of the following reactions: O

(a) HO

H+

H

CH3OH

O

OCH3

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Problems

O

(b)

O

CH3

O CH3

H2O HO–

A

H3O+ heat

CH3 O–

CN

NO2 —CHO + H2N—O—

(c)

—NO2

H+

[A]

CH3O–

+

CH3OH

NO2

PROBLEM 1122 O

NO2

Rationalize the difference in enol contents shown below: O

OH

O

OH H

I

PROBLEM 1123

H

II

0.001%

O

92%

Propose mechanism of the following reaction:

O N

NaOH

OH + NC—N OMe

O

H

N

OMe

H

PROBLEM 1124

Bring about the following transformations: OH CHO (b)

(a)

O

O OH (c)

COOMe

O

O

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Explain, with the aid of mechanism, the differing outcome in the following pair of

PROBLEM 1125 reactions:

OH

O HNO2

( a)

H2SO4

NH2 OH HNO2

but

H2SO4

NH2

CHO

O

O ( b)

CH3

CH3

NaOMe

—P h

O Ph

OH O

O NaOMe

but

O

OH

PROBLEM 1126

Propose mechanism:

TsOH

(a)

Ph

Ph

O

O O

OH

MeMgBr

(b) O

HClO4

NaOH

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Problems

PROBLEM 1127

Explain the following observation: Br

O

LDA Et2O

O

O Br LDA Et2O

but

O

O

PROBLEM 1128 An organic compound A has molecular formula C 9 H18O and it can be resolved into enantiomers. A evolves a colourless gas on heating with Na-metal, but does not change the colour of acidified dichromate solution or bromine water solution. A on heating with concentrated sulphuric acid undergo dehydration to yield three isomeric products B, C and D all having molecular formula C 9 H16 . B shows cis-trans isomerism and on ozonolysis followed by work-up with Zn  H 2O produced E (C 7 H12O) as one product. E formed an yellow precipitate on treatment with alkaline solution of iodine. Ozonolysis of C followed by work-up with Zn  H 2O produced F (C 5 H 8O) which did not produce yellow precipitate with alkaline NaOH but produced a single isomer G(C 5 H 9 NO) on treatment with H 2 NOH. Also D on ozonolysis followed by work-up with Zn  H 2O produced nonan-2,7-dione. Identify A to G. PROBLEM 1129 An organic compound A has molecular formula C 8 H16 . A on treatment with aqueous (CH 3COO) 2 Hg followed by reduction with NaBH 4 yields yields B (C 8 H18O) which cannot be resolved into enantiomers. A on treatment with B2 H 6 /H 2O 2 in alkaline medium yields C, an isomer of B, but it can be resolved into enantiomers. Ozonolysis of A, followed by work-up with dimethylsulphide, yields D(C 6 H12O) as one product which is non-resolvable. On catalytic reduction with LiAlH 4 , D yields E (C 6 H14O) which can be resolved into enantiomers. E on heating with concentrated H 2SO 4 yields two isomeric alkenes F and G, of which F is stereo-isomeric while G is not. Deduce structures of A to G.

CARBOXYLIC ACID AND ITS DERIVATIVES PROBLEM 1130 How could you use an acid base extraction technique to separate a mixture of carboxylic acid and phenol into its pure components? PROBLEM 1131 A carboxylic acid (A) of unknown structure was found to contain only C,H and O. 150 mg of A required 11.9 mL of 0.22 N NaOH to reach the equivalent point. Gentle heating of A evolves CO2 and a new carboxylic acid (B) was formed whose equivalent weight was found to be 74.Identify A and B. PROBLEM 1132 Propose mechanism: OH (a)

Br2/CCl4

O

O

O Br

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(b)

H2SO4

OH O

C—CH3 O COCH3

+

OH

(c)

N

H

NHCOCH3

PROBLEM 1133 Predict major product in the following reaction: (a)

COOH

SOCl2

CH3NH2

COOH

(b)

(CH3CO)2O Heat

COOH

OH

O

OH (c)

+ COCl 2

CH3OH

O

(d)

O

OH

ClH

PROBLEM 1134 Bring about the following transformations: (a)

COOH

(b)

COOH O COOH

(c)

PROBLEM 1135 Complete the following reactions: O

O (a) MeO

OMe

NaOH(aq)

(b)

O

CH3CH2OH

O (c) PhCH2COOEt

EtONa PhCOOEt

(d) EtO2CCH2COOEt

EtONa CH3CH2I

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Problems

H3O +

(e) H3C

(i)

LiAlH4

CN

(ii)

PhMgBr

(iii)

+

H

PROBLEM 1136 Convert: O COOEt COOH

(a) HOOC

COOEt COOEt

EtOOC

(b) CH3COOH

PROBLEM 1137 Bring about the following conversions: (a) PhCH 2OH → PhCH 2CN (b) PhCOOH → PhC(CH 3 ) 2 OH

(c) EtOOC

COOEt

COOEt

PROBLEM 1138 Show how would you accomplish the following synthesis in good yield? O

O

O

(a)

O

(b)

O

O O

O (c)

COOH

O

(d)

O

COOH

COOH

O

PROBLEM 1139 Show how would you accomplish the following multisteps synthesis? (a) 6-hepten-1-ol → caprolactum CH2Br

NH2

(b)

CHO (c)

CN

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PROBLEM 1140 Synthesize the following compounds from indicated starting materials: O (a)

O

from

COOH

O O (b)

from EtOOC

COOEt

PROBLEM 1141 Show how would you accomplish the following synthesis: (a) Isobutyl amine → N-isobutyl formamide (b) Ethylacetate → 3-methyl-3-pentanol. (c) Cyclohexyl amine → N-cyclohexylacetamide (d) Bromocyclohexane → dicyclohexylmethanol PROBLEM 1142 A neutral organic compound A has molecular formula C5H8O2 and does not decolourise Bayer’s reagent. A on acidic hydrolysis produced B(C5H10O3) which is resolvable. B on heating with concentrated H2SO4 undergo dehydration to produce C, an isomer of A. C decolourises Bayer’s reagent and show stereoisomerism. Also B changes colour of acidic dichromate solution from orange to blue-green converting itself into D(C5H8O3) which is non-resolvable and forms an yellow precipitate on treatment with NaOI solution. Deduce structures of A to D. PROBLEM 1143 A neutral organic compound A(C5H8O2) does not decolourise Bayer’s reagent and on hydrolysis with dilute H2SO4 produces B(C5H10O3) which is diastereomeric. B on heating with concentrated H2SO4 undergoes dehydration producing C(C5H8O2) which shows geometrical isomerism. Also B on treatment with acidic dichromate solution produced D(C5H8O3) which is enantiomeric and gives an yellow precipitate with NaOI. D on gentle heating produces E(C4H8O) which is nonresolvable. Deduce structures of A to E. PROBLEM 1144 An organic acid A(C5H10O2) reacts with Br2 in presence of phosphorus to produce a resolvable B. B on dehydrobromination yields C. C does not show geometrical isomerism and on decarboxylation gives an alkene D. D on ozonolysis gives E and F. Compound E gives positive Schiff’s test but F does not. Give structures of A to F. PROBLEM 1145 A monobasic acid A has neutralization equivalent 116 and does not decolourise Bayer’s reagent. Also A is enantiomeric. A on treatment with Br2/red phosphorus produces B which is still resolvable. B on dehydrobromination produced C which show stereo isomerism. C on decarboxylation produced D which does not show stereoisomerism. Deduce structures of A to D. PROBLEM 1146 An organic lactum A on acid hydrolysis produced B, an amino acid. B on treatment with nitrous acid gives C. C on heating with concentrated H2SO4 produces a lactone D. A can also be synthesized by the reaction of cyclopentanone with hydroxylamine followed by treatment of product with concentrated H2SO4. Deduce structures of A to D. PROBLEM 1147 A neutral organic compound A has formula C9H16O3 and rotates plane polarized light. A on acid hydrolysis produces B and C(C3H8O). C on partial oxidation with PCC gives D(C3H6O) which

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Problems

does not gives iodoform test. D on treatment with dilute solution of NaOH gives E(C6H12O2) which is diastereomeric. E on treatment with acidic solution of KMnO4 produces B. Deduce structures of A to E.

PROBLEM 1148 An organic compound A(C9H6O3) does not reacts with aqueous solution of NaHCO3 and does not change the colour of litmus paper. A on acid hydrolysis gives B(C9H8O4) whereas A on treatment with methanol in acid medium gives C(C10H14O4) as the major alcoholysis product. B on heating with excess of soda-lime produces toluene. Deduce structures of A to C. PROBLEM 1149 Diethyl malonate on treatment with one mole of C2H5ONa followed by reaction with one mole of 5-bromoethyl pentanoate produces A(C14H24O6). A on heating with dilute H2SO4 yields B(C7H12O4). B on treatment with excess of C2H5OH in presence of catalytic amount of H2SO4 yields C(C11H20O4). C on treatment with one mole of EtONa followed by work-up with dilute HCl yields D(C9H14O3). D on further treatment with one mole EtONa followed by reaction with CH3I yields E(C10H16O3). E on heating with HCl yields 2-methylcyclohexanone. Deduce structures of A to E. PROBLEM 1150 An organic compound contain C,H and oxygen and its aqueous solution is neutral. 1.44 g of A on refluxing with dilute H2SO4 produces two organic compound B and C of which only B is resolvable and gives effervescence with NaHCO3. Also neutralization of entire B require 0.4 g of NaOH. B on reduction with HI/red phosphorus produces a hydrocarbon D which is non-resolvable. Also D on monochlorination yields three alkyl halide in which only one is resolvable. Deduce structures of A to D. PROBLEM 1151 Identify the labelled product and explain their formations: O

O

Heat

O

A (C8H14O)

NaBH4

B (C8H16O)

PROBLEM 1152 Give structural formula of all possible products in the following reactions: (a)

MeO

(b)

NH2OH

COOEt

CH2COOH

(c)

(CH3CO)2O

NH2

(CH3CO)2O heat

COOH (d)

CHO

+

Ag2O/H2O

X

O

(e) O

Li

X

H

Y

(CH3CO)2O

Y

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PROBLEM 1153 Complete the following reactions: O (a)

NaBH4

OMe

SOCl2

A

B

O O

NaCN

(b)

C

(CH3)2SO4

+

D

CF3

H heat

E

PROBLEM 1154 Complete the following sequence of reaction: Li

TsOH

HOCH2

A

BrCH2CH2Cl

B

O

C

NaCN +

E

H

NaOH OH2

D

PROBLEM 1155 Complete the following sequence of reaction explaining the formations of products in the second step: O mcpba

A

O

CH3COCl/LiCl H C 3

O Cl

+ H3C

O Cl

PROBLEM 1156 Bring about the following transformations: (a)

H 3C H

(b)

H 3C H

H 3C

H CH2Cl

H

H CH2CH2Br

H 3C

H COOC2H5

H

H

CHO

O (c)

CH2OH O HO

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O

O (d)

O

CH2

O O

NH2

CN

O

(e) HO COOC2H5

Et

O

O

PROBLEM 1157 Bring about the following transformations: CH2OCOPh COOH O CD3

O

O

CD3 CD2OCOPh CD3

PROBLEM 1158 An organic compound exist in three isomeric forms A, B and C with their molecular formula C10H12O2 and all are resolvable. Decarboxylation of either B or C gives the same D(C9H12) whereas decarboxylation of A gives E an isomer of D. Both D and E are non-resolvable. Treatment of either D or E with KMnO4 in acidic medium produces benzoic acid as one product. Monobromination of E in presence of sunlight yields only two isomeric products whereas D on monobromination yields three isomers. Also B is more reactive than C towards decarboxylation reaction. Deduce structures of A to E. PROBLEM 1159 An organic compound A(C6H10) decolourises Bayer’s reagent and it is nonresolvable. A on treatment with acidic permanganate solution gave B(C6H10O4) which is still nonresolvable. B on treatment with excess of ethanol in presence of acid catalyst gave C(C10H18O4) which is a neutral sweet smelling liquid. C on refluxing with ethanolic solution of C2H5ONa gave D(C8H12O3) which is resolvable. D on treatment with acidic solution of ethylene glycol yields E(C10H16O3). E on treatment with one equivalent of Grignard reagent followed by reduction of product with Zn(Hg)—HCl and finally hydrolyzing product yields 2-ethylcyclopentanoate. Deduce structures of A to E. PROBLEM 1160 Provide products: OMe (a) O O (b)

CH3ONa CH3OH

O OMe

CH3ONa Br

+

H OH2

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PROBLEM 1161 Predict products and write mechanism of their formations: O 18

(a)

H2SO4

O

H N

(b) H2N

heat

KOH

O

O NH3/H2O

O

(c)

COOH

O

PROBLEM 1162 Provide organic products in the following reactions: O COOCH3 MeONa

(a)

CH3CH2CH2Br

O

Y

H3O+ Heat

Z

i) MeONa ii) CH3(CH2)8CH2I

O

(b) MeO

NaOH Heat

X

OMe

iii) NaOH/H2O iv) H+/heat

PROBLEM 1163 Indicate how each of the following compounds can be synthesized from the given starting material? O

O

O

(a)

H O

O

O

(b)

CH3

O O

C

CH2 O

CH2CH3

(c)

CH2CH2CH2COOH O

O

O (d)

OMe O

O

O OC2H5

(e) C2H5O O

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PROBLEM 1164 Bring about the following transformations: Ph

COOH (a)

H

H

O (b) Ph

Ph

OMe

(c) H2C(COOCH3)2

COOCH3

Ph

COOCH3

PROBLEM 1165 Propose mechanism: O

OH H

(a)

+

H

O

OH

O

O

O

O

(b)

COOEt

O

HCl

PROBLEM 1166 Provide mechanism: O

O

(i) EtONa

H3O+

O

heat

OEt

EtO

(ii)

PROBLEM 1167 Starting from ethyl propanoate synthesize: O

O

(a)

(b) O

O

Ph (e)

(d) HOOC

Ph

(c)

N

N O

O

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PROBLEM 1168 Bring about the following conversions: O

O (a)

O OEt

OEt O

O

O

(b)

OEt O

O

(c)

O OEt O

PROBLEM 1169 Propose mechanism of formation of products in the following reactions: O

O (a) MeOOC

COOMe + H

O O

O

O OEt +

(b) EtO

H

H CH3ONa heat

H COOEt

t-BuOK heat

O

O

HO

O O

O

(c)

COOEt

PROBLEM 1170

EtONa

COOEt

Propose mechanism of the following transformations:

OH

K2Cr2O7

CH3OH

H2SO4/H2O

H2SO4

COOH

PROBLEM 1171 materials:

Devise synthesis of the compounds shown below from the indicated starting O

O (a)

COOCH3

Br

N

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NH2 CHO OH

(b)

NH2

NH2

PROBLEM 1172 Propose an efficient synthesis of the compound shown below from indicated starting material and methanol. O

COOH

O HO

OH O

PPROBLEM 1173

Propose mechanism of the following reactions: O

O OH

(a)

CH3

CH3Li Excess

SOCl 2

H 2O

(b) R  CN → RCOOH → R  COCl HCl

O  H 2O (c) R  C  N 3 → R  NH 2 + CO 2 + N 2 heat

PROBLEM 1174

Provide a mechanism for the following reaction: H CH3

N

CH3O

O

POCl3

N CH3O CH3

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PROBLEM 1175 (a) CH3O

Bring about the following transformations: O CH3O

OCH2CH3

O

OCH2CH3

COOH COOH

(b)

PROBLEM 1176 Arrange the followings in increasing order of reactivity towards nucleophile: O O O O     (a) H 3C  C  NH 2 (I), CH 3  C  Cl(II), CH 3  C  OAc(III), H 3C  C  OCH 3 (IV) O O O O     (b) CH 3  C  Cl(I), CH 3CH 2  C  Cl(II), (CH 3 ) 2 CH  C  Cl(III), (CH 3 ) 3 C  C  Cl(IV) O O O    (c) CH 3  C  OCH 3 (I), CH 3  C  OCH 2CH 3 (II), H 3C  C  OCH(CF3 ) 2 (III)

PROBLEM 1177 Convert (a) CH 3CH 2CH 2COOH → CH 3CH 2CH 2CH 2COOCH 3 O

(b)

O O

PROBLEM 1178

O

Consider the following acid derivative: H O 1

4

O

O

2

O B

3

H

CH3

A

6 5

C

N

O

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Problems

(a) (b) (c) (d)

Which of the groups in square will hydrolyze first in base catalyzed medium? How the rate of hydrolysis be affected if a nitrogroup is substituted at C–4? How the rate of hydrolysis be affected if an amino group is substituted at C–4? How the rate of hydrolysis be affected if a methoxy group is substituted at C–3?

PROBLEM 1179 Propose mechanism of the following reaction and calculate enthalpy of reaction from the given bond-energies data: O

O

O

Ph OCH3

(1) CH3ONa

+ Ph

OCH3 (2) H3

O

Ph

OCH3 + CH3OH

O+

Ph

BE : C  C = 85 kcal/mol; C  O = 91 kcal/mol; O  H = 102 kcal/mol; C  H = 99 kcal/mol.

PROBLEM 1180

Predict major product in the following giving mechanistic reasoning: O

O

CH3OH

H3 C

* O

CF3

PROBLEM 1181 An optically active organic compound A has molecular formula (C 7 H12O 3 ) and found to produce an yellow precipitate with alkaline solution of Iodine. Also A neither decolourised brown colour of bromine water nor evolved any gas with aqueous solution of NaHCO 3 but evolved a gas on heating with sodium metal. On hydrolysing in acidic medium, A produced antoher optically active compound B(C 7 H14O 4 ) which also produced yellow ppt with NaOH/I 2 . Also B evolved a colourless, acidic gas on treatment with aqueous NaHCO 3 solution. A on treating with CrO 3 /HCl/Pyridine in CH 2Cl 2 produced another optically active compound C(C 7 H10O 4 ). C on refluxing with aqueous Ag 2O produced an optically inactive compound D(C 7 H10O 5 ). D does not decarboxylate (does not evolve CO 2 ) on simple heating. However, heating D with sodalime gave E(C 5 H10O). In a separate analysis, 3-oxo ethylbutanoate was heated with excess of sodium metal and then with CH 3I to produce X (C 8 H14O 3 ). X on hydrolysis followed by simple heating of product yielded E. Identify A to E and X. PROBLEM 1182 An oily liquid A is insoluble in water but on heating with aqueous NaOH for half an hour it dissolves. From the reaction mixture a liquid B can be distilled, which gave an yellow precipitate with iodine and NaOH. On careful oxidation, B gives an aldehyde, C, which also gives an yellow precipitate with iodine and NaOH solution. If sulphuric acid is added to the solution obtained from heating A with NaOH solution, a white precipitate D is obtained. D liberates a gas on treatment with aqueous NaHCO 3 . Heating D with sodalime produced benzene. Identify A–D. PROBLEM 1183 An organic compound A has molecular formula C 9 H 8O 2 and if exist in stereoisomeric forms. A effervesces with NaHCO 3 . A decolourises brown colour of bromine water forming B(C 9 H 8O 2 Br 2 ) which is distereomeric. B on hydrolysing with aqueous NaOH followed by refluxing of product with acid gave C(C 9 H10O 4 ). C on oxidising with acidified dichromate solution gave D(C 9 H 6O 4 ) which gives orange precipitate with 2,4-dinitrophenylhydrazine but did not react with ammonical AgNO 3 solution. D on reacting with PCl 5 yielded a steamy fumes of E(C 9 H 5O 3Cl). E on treatment with aqueous ammonia yielded F(C 9 H 7O 3 N). Both A and F are oxidized by hot aqueous KMnO 4 solution to yield a white crystalline substance which on heating with sodalime yielded benzene. Identify A to F.

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PROBLEM 1184 O

Propose mechanism of the following reactions: O O (i) NaOH/H2O

(a)

COOH

(ii) H3O O NC

(i) TsCl

(b)

COOH

(ii) NaOH (iii) H3O+

HO

N

H O

(c)

O

BuLi

N—OTs +

N O

O

PROBLEM 1185 O

H

Propose mechanism of the following reactions: O O O K2CO3

(a)

BrCH2CH2Br

OEt

OEt O COOEt

Na/excess

(b) EtOOC

COOEt EtOOC O

PROBLEM 1186 EtOOC

R

Propose mechanism: O

N

(i) t-BuOK

CH3

O

OAc

(i) MeMgBr

(iii) Ac2O

N R

(ii) aq NaOH (ii) Heat

CH3

O

N R

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Problems

(a) Indicate a mechanism of the following reaction:

PROBLEM 1187

COOEt

(i) EtONa/EtOH/heat COOEt

(ii) CH3COOH/H2O O

(b) Devise a sequence by which product of the reaction in part “a” converted into the following comopunds: O (i)

(ii) O

O

O R

COOH (iv)

N

(v)

O

PROBLEM 1189

N O

Bring about the following transformations:

PROBLEM 1188

O

COOCH3

CH3

Bring about the following transformations: Br —Br

(a)

COOH

(b) CH 3CH  CH 2CN → CH 3CHCH 2 CH 2OH   C6 H 5 C6 H 5

PROBLEM 1190

Propose mechanism: O

O

O

H+/H2O COOH

Ph

(iii)

O OH

O

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PROBLEM 1191

Propose mechanism of the following reactions: MgBr

+

(a)

H3O+

O

O

OH HO MgBr heat

(b)

SH

S

O

N

O

NH2

PROBLEM 1192

Provide missing reagents/products in the following reactions: O Br2/NaOH

NH2

(a)

H2O NH2

(b) O

?

O

O ?

(c)

(d)

HO

OCH3

Br

OCH3 ?

O

PROBLEM 1193

Bring about the following transformations:

PhBr

PhCH2—N

PROBLEM 1194 A neutral organic compound A(C 4 H 6 ) reacts with Br 2 /CCl 4 solution to form a compound B(C 4 H 6 Br 2 ). A on treatment with acidified permanganate solution yields C(C 4 H 6O 3 ). C gives an orange precipitate with 2,4-dinitrophenylhydrazine. C on refluxing with NaHCO 3 evolves a

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gas. Treating C with NaBH 4 gave a new compound D(C 4 H 8O 3 ) which also produces a gas with NaHCO 3 solution. Deduce a structure of A  D.

PROBLEM 1195 2.81 g of an optically active diester A, containing only C, H and oxygen was saponified using 30 mL 1.0 M NaOH solution. Following saponification, the solution required 6 mL 1.0 M HCl solution to titrate unreacted base. The saponification products were an optically inactive diacid B, methanol and an optically active alcohol C. Alcohol C reacted with alkaline solution producing an yellow precipitate and C 6 H 5COONa. Also, the diacid B reacted with Br 2 in CCl 4 to give a single, optically inactive product D. Ozonolysis of B gave only one product. Also, the diester A reacts with Br 2 /CCl 4 to give a mixture of E and F, both optically active. Identify A to F. PROBLEM 1196 The following three reactions occurs by a common mechanism. Write detailed mechanism for the reaction A. Then indicate, which reaction will be the fastest and which reaction will be the slowest and explain briefly, why? Br

O NH—C—CH3

CH3CN

A:

H2O/∆

H3CO

H3CO

Br

O NH—C—CH3

CH3CN

B:

H2O/∆

O2N

O2N

Br

O NH—C—CH3

CH3CN

C:

H2O/∆

NO2

NO2

PROBLEM 1197

Propose mechanism of the reaction: OH

O O

LiAlH4 NH4Cl

OEt

TsCl

NaOH

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PROBLEM 1198

Complete the following reactions: COOCH3 COOCH3 ∆

(i) NaOH

(a)

(ii)

H3O+

COOCH3 COOCH3 O

O (i) [(CH ) CH] NH 3 2 2

(b)

(ii) H2O

O

O

(i) Br—(CH2)4—Br

(c)

H2O

(ii) Mg/Et2O

I2/H2O

(d) COOH

NH2

(e)

(i) KMnO4/NaOH (ii) SOCl2 Cl

(f) O

PROBLEM 1199

(i) Me2NH

(i) H3O+

(ii) LiAlH4

(ii) NaOH

Propose mechanism of the reaction: O OH NH2

Ph N

(i) NaOH, PhCOCl O

O

Ph

(ii) O

O O

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Problems

Provide products of the following reactions: O

PROBLEM 1200

CH3CH2SNa

(a)

OCH3

H 3O + H 2O

Et2O

O (b)

PCl5

—C—OMe

∆

CCl4

(c)

OEt

Peroxide

(d)

H 3O +

COOEt

∆

COOEt O

O

(e)

OEt

(f) PhCH2Li + HO—

PROBLEM 1201

EtONa excess

Br(CH2)5Br

—CHO

Provide selective reagents that will bring about the following transformations:

OH

OH

(a) CHO

COOH O

O (b)

O

O

(c)

H

OCH3

Cl

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MeO

MeO

OH

H O

O

PROBLEM 1202

Propose mechanism of the following reactions:

O OEt

(a)

CH3MgI

CN

O

Et2O

O

(b)

O

H 3O +

O CH3Li

H 3O +

Et2O

PROBLEM 1203 Synthesize the following compound starting from diethylmalonate and other necessary reagents: CONH2

Write product of the following reactions:

PROBLEM 1204

O

O (a)

ONa

Cl O LDA

(b)

Br

OEt O

(c)

O

N—

Cl

H 3O + heat

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O (d)

OEt

H3O+

EtO– EtOH

PROBLEM 1205 An organic compound A reacts with I 2 in basic medium to give a yellow precipitate. A on treatment with Br 2 in acetic acid medium gives B which readily undergoes reaction with cyanide ion to give C. C on reduction with NaBH 4 forms D–an optically active substance. Acid hydrolysis of D followed by treatment with acidic ethanol gives E. Gentle oxidation of E with pyridinium chlorochromate gives F, which on treatment with potassium ethoxide in ethanol followed by reaction with ethyl iodide and then heating the final product yields 1-phenyl-1-butanone. Identify A to F.

AMINES PROBLEM 1206 Provide appropriate reagents for the following transformations: (a) R —OH → RCH 2 NH 2 (b)

OH

NH2

(c) C 6 H 5 NH(CH 3 ) → C 6 H 5 —N—CH 2CH 3  CH 3 (d) C 6 H 5 NH 2 → C 6 H 5CH 2 NHCH 2 NH 3

PROBLEM 1207 An organic compound A has molecular formula C 5 H13 N and it reacts with benzene sulphonyl chloride in aqueous potassium hydroxide to give a clear solution, acidification of solution gives a precipitate. Also A can be resolved into enantiomers. A on refluxing with nitrous acid gives B (C 5 H12O) which is still resolvable. B on treatment with CrO 3 in pyridine solution of HCl yields another resolvable compound C (C 5 H10O) which on further treatment with Zn(Hg) in concentrated HCl solution yields a non-resolvable D(C 5 H12 ). Identify A to D. PROBLEM 1208 An organic compound A has molecular formula C 7 H 9 N and it forms a clear solution when dissolved in aqueous KOH solution of benzene sulphonyl chloride. A on treatment with NaNO 2 and HCl at 0°C, and then with 2-naphthol forms an intensity coloured compound. Also A on treatment with CH 3COCl followed by electrophilic substitution yield a single substitution product. Identify A. PROBLEM 1209 An organic compound A has molecular formula C 9 H13 NO and it can be resolved into enantiomers. A does not decolourise bromine water solution. A on refluxing with dilute H 2SO 4 solution yields another resolvable compound B (C 9 H14O 3 ) which gives effervescence with NaHCO 3 . B on treatment with NaBH 4 yields C (C 9 H16O 3 ). C on heating with concentrated H 2SO 4 solution yields a

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sweet smelling neutral compound D(C 9 H14O 2 ). Also A on reduction with LiAlH 4 yields E (C 9 H19O) which on treatment with H 2SO 4 yields the following compound : H

N

Identify A to E.

PROBLEM 1210 Provide structures of A to F in the following sequence of reaction: Ag2O

Heat

+ CH 3I → A (C 7 H16 NI) → B (C 7 H17 NO) → C (C 7 H15 N) H 2O

N CH3

Ag O

CH I

Heat

2 3 C –→ D (C 8 H18 NI) → E (C 8 H19 NO) → F (C 5 H 8 ) + H 2O + N(CH 3 ) 3

H 2O

PROBLEM 1211 Identify the labelled product in the following sequence of reaction: Br CH3NHCH3

(a)

2 equivalent

A

BaO

B

CH3I 2 equivalent

C

Ag2O H2O

D

Heat

E + 2 (CH3)3N

Br

O Ph (b)

Ph

NH2OH

A

LiAlH4

B

CH3I Excess

C

Ag2O

Heat

D + (CH3)3N

H2O

PROBLEM 1212 In the following sequence of reaction: Br

+ CH3NHCH3

A

KOH Heat

Br

Deduce of A and B and explain formation of C.

B

CH3I

KOH Heat

+ (CH3)3N C

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Problems

PROBLEM 1213 Propose mechanism of following reaction: O N3

Pd/CaCO3 H2, quinoline

NH O

PROBLEM 1214 How butylamine and dibutyl amine can be separated? PROBLEM 1215 An organic compound A has molecular formula C 5 H10 N 2 and it can be resolved into enantiomers. A on hydrogenation yields optically inactive B (C 5 H12 N 2 ). B on treatment with nitrous acid forms an oily compound C (C 5 H10 N 2O 2 ) which is yellow coloured. C when warmed with a crystal of phenol and few drops of concentrated sulphuric acid, turns green which, when made alkaline, turns deep blue. Also A on treatment with excess of iodomethane forms D(C10 H 22 N 2I 2 ). D on treatment with aqueous Ag 2O followed by heating yields pyridine. Identify A to D. PROBLEM 1216 Complete the following reactions: H O (a)

+

N

Heat

A

CH2==CHCOOEt

B

hν (b) HN 3 → D C2H4

Na C2H5OH

KCN (c) BrCH 2CH 2 Br → → E

PROBLEM 1217 Bring about the following transformation: (a) R —CN → R —NC (b) R —NC → RCN PROBLEM 1218 Complete the following reactions: H SO

2 4 (i) CH 2 == CH—CN + PhCH 2OH → ?

Et O

2 (ii) PhCH 2COCHN 2 + HCl( g ) → ?

(iii) MeNH 2 + ? → MeNHCOOEt HBr Heat

(iv) HOCH 2CH 2CN → ?

PROBLEM 1219 Complete the following reactions: B H

ClNH

2 6 2 (a) C 2 H 5CH==CH 2 → A → B

H3O

+

C

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– CO

PhNCO 2 2 (b) PhNCO → C → D → E LiAlH

KOH 4 (c) C 2 H 5 NH 2 + KCN + Br 2 → F → G

PROBLEM 1220 MeCOCH 2Cl + CH 2 N 2 → A (C 4 H 7ClO) + B (C 4 H 7ClO) Ketone

Epoxide

Suggest structures of A and B and mechanism of their formations.

+

−

PROBLEM 1221 An amine A(C 6 H15 N), on treatment with CH 3I and then KOH gives B (C 8 H 20 NOH) B on heating produces C (C 4 H 8 ) and D(C 4 H11 N). C on treatment with aqueous (CH 3COO) 2 Hg follows by reduction with NaBH 4 yields E (C 4 H10O) which does not change the colour of chromic acid solution. Deduce structures of A to E. PROBLEM 1222 Complete the following reactions: KHCO3

(a) EtNO 2 + 2CH 2O → ? (b) RCH 2 NH 2 + 2CH 2O + 2HCOOH → ? H C==C==O

2 (c) CH 2 ==C==O + HCN → ? → ?

PROBLEM 1223 Provide mechanism of reaction NaOH

(CH 3 ) 2 CHCONH 2  –→ (CH 3 ) 2 CHNH 2 Br2

PROBLEM 1224 Prepare PhCH 2 NH 2 by Gabriel synthesis and nitrile reduction starting from PhCH 2 Br in both cases. PROBLEM 1225 Synthesize following compounds starting from cyclohexanone and any amines needed. H

H

N (a)

N (b)

N

N

H

H

AROMATIC COMPOUNDS PROBLEM 1226 A resolvable organic compound A has molecular formula C10H14O. A does not forms any salt with NaOH but evolve a colourless gas on heating with sodium metal. A gives an yellow precipitate with NaOH/I2 but does not decolourise Br2-water solution. A on treatment with CrO3/HCl/pyridine produces B(C10H12O) which is non-resolvable but gives iodoform test. B on

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Problems

treatment with Cl2/AlCl3 gives C(C10H11OCl) as the only possible isomer. Also A on heating with alkaline KMnO4 solution yields C8H6O4(D) as one of the product which evolves a gas with NaHCO3. D on heating with ethan-1,2-diol in presence of an acid catalyst forms a polyester E. Deduce structures of A to D and write the formula of repeat unit of E.

PROBLEM 1227 Predict products/reagents: COOH +

(a) O

O

Cl

NaOH

CHO

HNO3

(b)

Heat

COOH

H2SO4

COOH (c)

OH

Br2/CHCl3

Br2/AlBr3

(d) S

CH3 O (e)

OMe CH3

HNO3

(f)

H2SO4

+

H2SO4

CH3

CH3 O 2N

O (g)

X

C

CH2

CF3 Y

O 2N

CF3 CH2

CF3

PROBLEM 1228 Propose mechanism: Ph

+

H

HO

PROBLEM 1229 Bring about the following transformation: Ph

PROBLEM 1230 An organic compound A(C11H14O) exist in two stereo-isomeric forms, none of which is resolvable. A does not evolve any gas with sodium metal but decolourise bromine water solution. A on

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treatment with ozone followed by work-up with (CH3)2S yields B(C9H10O2) as one of the product. B neither forms any salt with NaOH nor does it evolve any gas with sodium metal. Also B does not react with Tollen’s reagent. B on treatment with concentrated solution of HI forms C(C8H8O2) as one of the product which forms salt with NaOH but does not evolve any gas with NaHCO3. B on treatment with AlCl3/Cl2 gives D(C9H9O2Cl) as the only possible product. C can also be obtained by treatment of an ester E(C8H8O2) with AlCl3. Deduce structures of A to E and explain the formation of C from E.

PROBLEM 1231 Bring about the following transformation: O PhCH2CH2

CHO

(a)

(b) Cl COOH

(c)

(d) SO3H

NO2

PROBLEM 1232 An optically active organic compound A has molecular formula C11H16O2 and it does not forms salt with NaOH as well as it does not decolourise Br2—H2O solution, but evolve a gas on treatment with MeMgBr. A on treatment with HI gives another optically active compound B(C9H10O2). B on treatment with aqueous KOH yields another optically active compound C(C9H12O2). C on treatment with PCC yields D(C9H8O2) which gives positive iodoform test as well as positive Tollen’s test. D is optically inactive and on treatment with Cl2/AlCl3 gives one and only one possible monochloro derivative E(C9H7O2Cl). Deduce structures of A to E. PROBLEM 1233 Synthesize: OH (a) Ph—CH —C 2

C==C

CH3

CH3

from Ph—CH2CH2CH2Br

H H Cis

Br (b) O2N

—CH—CH2OH O

(c)

from Cl

from benzene, chloroethane and any inorganic reagent

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Problems

(d) C6H5CH2COC6H5 from benzene, oxirane and inorganic reagent O I from iodo benzene and

(e) (CH3)3CCH2

SO3H

PROBLEM 1234 Complete the following reaction: O

NO2

(a) Ph

TiCl4 DMG

X

OH –

TsOH

Y

Z

O LDA

(b) Ph O (c)

X

CuCl2

Y

KCN/AcOH

DIBL-H

X

Ph

H3PO4

Z

Y

TsOH

Z

PROBLEM 1235 Bring about the following transformation: (a) Br

O

COOMe

O

O (b) Ph

Ph

O

O 2N (c) S

S OH

PROBLEM 1236 (a) Synthesize the following compounds starting from benzene: F

CH3

COOH

NO2

F

F

F

(b) Write mechanism of the reaction of p-nitrofluoro benzene with (C2H5)2NH. Explain why m-nitro fluoro benzene does not react with diethyl amine under identical conditions.

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(c) Synthesize the following compounds starting from benzene and any other reagents needed: O

O

O R

OMe

(i)

(ii) R'

O COOCH2CH3

(iv) H2N

Ph

(iii)

PROBLEM 1237 Synthesize the following compounds starting from benzene: OCOPh

OCOPh (b)

(a) O 2N

Br

O OPh

(c)

NO2

(d) O2N

O2N

OH

OH (e)

Br

Br

(f) NC

PROBLEM 1238 Synthesize the following compounds starting from benzene: I I

I (b)

(a)

(c)

Cl

Cl

Cl

Cl

PROBLEM 1239 Bring about the following transformation: Br NO2

Ph (a)

(b) NO2

Br

Br

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Problems

COOH (c)

O

O

NH2

H2N

O

NH2

PROBLEM 1240 Propose mechanism: O

H

MeO

N

COOMe

NHNH2

MeO COOMe

PROBLEM 1241 Bring about the following transformation: CHO Br

CH3

CHO (b)

(a)

OH

OH CHO

(c)

OH

OH

(d) CH3 O

OH

(f)

(e)

Ph

O (g)

H N (h) NHNH2

OH

CHO

O

COOH Ph

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PROBLEM 1242 Bring about the following conversions: CH3

SO3H

CH3

OMe

(b)

(a)

NH2

CHO OH OH

OH

Br

(c)

OH

(d) Br

Br CH2OH OMe (e)

PROBLEM 1243 Predict product of the following reaction explaining mechanism of their formation: —CH3 + CBrCl3

hν

PROBLEM 1244 Prepare the following compounds starting with either benzene or toluene: OH CH3

(a)

CH3

HO (b)

HO (c) CH3

CN

Cl

OH (f)

(e)

(d)

CN

Cl

I CN (h)

(g)

NH2

CH3 Br

O 2N

(i) Br

Br

Br

Br

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Problems OH

CN

COOH Br

Br

(j)

(k) Br

Br

(l) Br

Br

Br

Br CH3

Br

OH NO2

I (m)

(n)

(o)

Br

NO2 Br

CH3

CH3 CN (q) H3C

(p)

N N

OH

Br

CH3 (r) H C 3

N N HO

PROBLEM 1245 Bring about the following transformation: NO2

NO2

CHO OHC

(a)

(b) NO2

Cl NO2

NO2

Cl

NO2

CN Cl

(c)

COOH

(d)

NO2

NH2

NO2

CH3

NH2

(e)

CH3

(f) Br

Br CH3

CH3

NH2

Br

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NO2

CH2NH2 CHO

NO2

CHO

(g)

NO2

(h) HOOC CH3

NH2

CH3

Br

Br (i)

(j) NO2 NO2

Br

PROBLEM 1246 Bring about the following transformation: NO2

CONH2 NO2 (a)

Br

Br (b) Br

NO2

NO2

NH2

Cl

(c)

(d) Cl

Cl Br

PROBLEM 1247 An organic compound A(C8H10) on treatment with fuming sulphuric acid yields two isomeric products B and C. B on fusion with KCN gives D(C9H9N) which on hydrolysis yields E(C9H10O2). E on treatment with hot alkaline KMnO4 yields F(C8H6O4). F on heating with P2O5 undergo intramolecular dehydration to yield G(C8H4O3). Identify A to G. PROBLEM 1248 Bring about the following transformation: CH3

OMe

CH3

OMe Br

(b)

(a)

NH2 Cl

CH2OH CH3

NH2 NO2 (c)

(d) Cl

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Problems

CH3

Br

NO2

CH3

Br

Br (e)

(f) NO2 NO2

NO2

NO2

CH3

OH (g)

OH

PROBLEM 1249 Identify the aromatic product formed in the following reaction: CN (a)

C2H5MgBr

HCl/H2O

COOH

COOH (c)

(d)

HCl/H2O

EtLi

(b)

CHO

(CH3)2Cd

SOCl2

Br

COOEt Zn/HCl

PROBLEM 1250 Synthesize the following compounds starting from benzene:

(a)

(b)

(c)

(d)

OH OH (e)

(f)

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PROBLEM 1251 Predict major substitution product in the following reactions: OMe

N Br

HNO3

(a)

Na2CO3

NHCOCH3

OMe CH3COCl

(c)

Cl2

(b)

H2SO4

Br2

(d)

AlCl3

FeBr3

Br

COOCH3

PROBLEM 1252 Starting from benzene synthesize the following compounds as major product: CH3

NO2

NH2 NO2 (c)

(b)

(a)

Br COCH3 NH2

NO2

Cl

(e)

(d)

(f) NO2

PROBLEM 1253 How would you rationalize the following experimental result: N

N HNO3

NH3

H2SO4

OH2

NO2

PROBLEM 1254 Propose synthesis of the following compounds starting from benzene: NO (a)

O

COOH (c)

(b) NO2

COOH

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Problems

H2N

OH (d)

(e)

Br

CH3 Cl

NO2 (g)

(f) Br

O

(h)

(i) Br

SO3H

NO2 N (j) CH3

PROBLEM 1255 Identify the starting material in the following sequence of reaction: O

(a) X

CH3COCl AlCl3

N

(b) X

(c) X

N2H4

Br2

NaOH

AlBr3

Br2

Br

Sn/HCl

AlBr3 Br

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OMe (d) X

(e) X

Br2

NaOEt

CH3Br

Zn(Hg)

HNO3

Sn/HCl

ClH

H2SO4

Br

H 2N

PROBLEM 1256 Predict the major product in the following reaction: CH3 (a)

CH3COCl

HNO3

(b)

AlCl3

H2SO4 NO2

O Cl2

(c)

(d)

AlCl3

CH3 (e) Cl

OPh

Br2 AlBr3

Br2 AlBr3

PROBLEM 1257 p-nitro phenol is shaken with one equivalent of D2O in the presence of a strong acid (perchloric acid) at 100°C for a long time. The product has two deuterium atom in it. One of these is lost instantly when the product is treated with ordinary water. Propose a structure for the original product and explain the difference in the ease with which the two deuterium atoms can be removed from the molecule. PROBLEM 1258 Bring about the following transformation: CH3

COOH (a)

(b) NO2

O2N

COOH HO

CH3 NO2

(c)

(d) NO2

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Problems O NH2

(e)

(f)

NO2

PROBLEM 1259 Complete the following reactions: O OH

O O AlCl3

(a)

Cl +

(b)

excess

Cl

O

O

OMe

O H 3C

(c)

AlCl3

O

+

AlCl3

H 3C O

PROBLEM 1260 Synthesize the following compounds starting from benzene: H

O

H

H3C C2H5

(a)

(c)

(b)

Ph OH

PROBLEM 1261 Predict the major substitution product in the following sequence of reaction: OMe

OH D H3PO4

(a)

A

SO3

(b)

H2SO4

Cl O (c)

+

AlCl3

O

NH2 CH3 (d)

O

(CH3CO)2O

HNO3 H2SO4

H3O

+

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PROBLEM 1262 Predict synthesis of the following products starting from benzene: Br CH3

(a)

O

C H

(b)

CH3

O NO2 (c) H3C

OH

(d)

COOH

(f)

(e)

(g)

PhCH2

O

CH2Ph

H

H

C 2H 5

PROBLEM 1263 1,3,5-trimethyl benzene undergoes electrophilic aromatic substitution with iodine monochloride (ICl). Write mechanism for this reaction showing major product. PROBLEM 1264 When allyl alcohol is treated with HF in presence of benzene, two products are formed: 3-phenyl-1-propene and 1,2-diphenyl propane. Write equations showing the mechanism for the formation for these products. PROBLEM 1265 When 2-hydroxy benzoic acid is heated with isobutyl alcohol in presence of sulphuric acid, compound A is formed. The same product is formed if tertiary butyl alcohol and sulphuric acid is used. What is the structure of compound A? Write mechanism of formation of this product. PROBLEM 1266 Bring about the following synthesis starting from benzene: (*direct F.C. alkylation of nitro benzene does not succeed) CH3 CN *(b)

(a) Cl

NO2 Br (d)

(c)

COOH

HO

Cl

F (e) H3C

NO2 COOH

(f) NO2

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Problems OMe (g)

(h) HO

I

NH2

PROBLEM 1267 Complete the following reactions: NO2

MeOH

+

(a)

H

N

+

C6H5SNa

Br NO2 NO2 (b)

MeOH

F NO2 NO2 OTs +

(c)

NH2

NO2

O 2N

H 2N

NO2

N

Cl +

(d) O 2N

Heat

CHCl3

NO2 CH3

(e) O2N

+

F + H3N—CH—COO

–

H 2O

NO2

PROBLEM 1268 Give structural formula for all reagents, intermediates and products designated by letters in the following equations: CHO OMe (a)

OMe

NaBH4

A

H3O

+

B

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COOH

HOOC

H3PO4

C + D

(b) CH3

OH

CH3I

NaH

(c)

B2H6

E

CrO3

F

G

H2SO4

H2O2/NaOH

H3PO4

H

LiAlH4/H2O

I

CHO (d)

OH

NaBH4/H2O

J

NaH

K

CH3I

L

OMe NO2 Cl

Br

Na

(e) H-CH(COOEt)2

NO2

M

N

COOH

HO (f)

CH2OH

MeO

COOCH3

MeO

P

O OH

OMe

OMe

Br CH2OH

MeO Q OMe

PROBLEM 1269 Identify the missing reagents/reactants/products in the following sequence of reactions: (a) A

(b) O2N

HNO3

B

Sn/HCl

C

O NO2

NaNO2 HCl/0oC

D

CuCN

NO2 + H N

OMe

NC

OH2

E+ F

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Problems NO2

(c) H2N

(d) CH3CH2O

NH2

G

KI(aq)

H

HCl/0oC NaNO2

Br2

NH2

(e) HO

NaNO2

ICl excess

I

NaNO2

J

HCl/0oC

H3PO2

K

KI(aq)

L

HCl/0oC

M

CH3 (f) O2N

CH3

H2/Ni

N

NaNO2

CuCN/heat

O

HCl/0oC

P

H3O +

Q

PROBLEM 1270 Write structural formula for all the products in the following sequence of reaction: CHO

HNO3

(a)

A

H2SO4

Sn/HCl

NaNO2

B

C

HCl/0oC

CuCl/HCl

NH2 NaNO2

(b) MeO

H2SO4 , 0oC

OMe

H2O/H2SO4

E

F

CH3 (c)

NO2

O2 N

H2/Ni

NaNO2

G

HCl/0oC

COOH NH2

(e)

C6H5

N CH3I

(f) N H

C6H5

M

NaNO2 HCl/0oC KOH

K

J

HCl/0oC

CH3

H2O/H2SO4

N

NaNO2

(d)

H

N

L CH3I

O

KOH

D

P

+

Q

I

D

234

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PROBLEM 1271 An organic compound A(C8H10) does not decolourise aqueous solution of Br2 and on treatment with Br2/FeBr3 in dark yields two product in principle, B and C but due to steric reason, B predominates. A on refluxing with alkaline solution of KMnO4 yields D(C8H6O4) which gives off gas on treatment with NaHCO3. D on treatment with Br2/FeBr3 reacts very slowly to produce E(C8H5O4Br) as the only isomer. Also D on treatment with excess of SOCl2 followed by work-up with benzene solution of AlCl3 produces F(C14H8O2) which does not yields any gas with NaHCO3. F on heating with N2H4 produces G(C14H8N2). Deduce structures of A to G. PROBLEM 1272 An organic compound A (C9H10O2) does not change colour of Br2-H2O and produces no gas with NaH. A on treatment with Br2 in presence of FeBr3 yields B(C9H9O2Br) as only isomer. Also A gives orange precipitate with 2,4-dinitro phenyl hydrazine and produces a resolvable compound C(C9H12O2) on reduction with NaBH 4. Deduce structures of A, B and C. PROBLEM 1273 An organic compound A(C9H8O2) does not decolourise bromine water solution and evolves no gas with CH3MgBr but gives orange precipitate with 2,4-dinitro phenyl hydrazine. A on refluxing with dilute H2SO4 produces B(C9H10O3) which forms salt with NaOH and on treatment with excess of CH3COCl yields C13H14O5. B is a non-resolvable compound which on heating with N2H4/NaOH yields C(C9H12O2). C on dehydrating with concentrated H3PO4 yields D(C9H10O) as major product. D on ozonolysis followed by work-up with (CH3)2S yields E(C7H6O2) which can also be obtained by the action of phenol with alkaline solution of chloroform followed by acidification of product. Identify A to E. PROBLEM 1274 An oily liquid A is insoluble in water, but on heating with aqueous solution of sodium hydroxide for one hour, it dissolves. From the reaction mixture, a liquid B can be distilled, which gives a yellow precipitate with NaOH/I2 as well as it is resolvable. B on treatment with acidic dichromate solution yields C which also gives positive iodoform test. If sulphuric acid is added to solution obtained on heating with NaOH, a white precipitate D is obtained. D gives effervescence with NaHCO3 and heating D with soda lime converts it into toluene. Also A on treatment with Br2/FeCl3 in dark yield single mono bromo derivative as a substitution product. Compound B on heating with concentrated H2SO4 yields stereomeric alkene, one of which on treatment with cold, dilute and alkaline KMnO4 yields a meso diol. Deduce structures of A to D. PROBLEM 1275 An organic compound A has molecular formula C10H14 and it does not decolourize bromine water solution. A on treatment with Br2/Fe yields three products in principle but in actual practice, only two of these are produced as mono-bromo derivative. A on heating with Br2 yields B (C10H13Br) as the mono-bromo derivative which is optically inactive. A on heating with alkaline KMnO4 yields C(C8H6O4) which does not forms anhydride on heating. Deduce structures of A to C. PROBLEM 1276 An organic compound A(C8H8O3) evolves a gas with NaHCO3 as well as forms salt with NaOH. A on treatment with HI yields B(C7H6O3) which when treated with excess of acetyl chloride (CH3COCl) yields C(C11H10O5). B on treatment with Br2/FeBr3 in CCl4 should yield three products D, E and F in principle but only two products E and F are produced in reality and E is the major product. Deduce structures of A to F. PROBLEM 1277 An oily liquid is insoluble in water, but on heating with aqueous solution of sodium hydroxide for one hour, it dissolves. From the reaction mixture, a liquid B can be distilled, which gives a yellow precipitate with iodine and NaOH solution. On careful oxidation, B gives an aldehyde C, which also gives positive iodoform test. If sulphuric acid solution is added to the solution obtained from heating

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Problems

A with NaOH, a white crystal D is obtained. D liberates a gas with NaHCO3. Heating D with NaOH/CaO, converts it into C 6H6. Identify A to D.

PROBLEM 1278 Identify the followings as aromatic, antiaromatic or non-aromatic, supporting your answer with brief explanations. OH

(a)

(b)

s

+

(c)

(d)

PROBLEM 1279 The pK a of benzylic hydrogen in A is ~ 16, whereas the pK a of benzylic H in B is ~ 25. Explain the difference. H

CH3

B

A

PROBLEM 1280 appropriate:

CH3

H

Predict products in the following reactions and show stereochemistry where

OCH3 Br2

(a)

NaNH2

(b)

NH3

FeBr3

Br

O3

Na

(c)

liq. NH3

AlCl3

(d)

CH3COCl

(CH3)2S

NO2 CH3ONa

(e)

(f)

CH3OH

Br O

HNO3 H2SO4

Cl2

Zn(Hg)

AlCl3

HCl

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NO2

O

O

C HNO3

(g)

H2SO4

AlCl3

(i)

(j)

OH

CH3Li

H3O+

excess

O3 (CH3)2S

O O

(h)

O

I2/HNO3

t-BuCl AlCl3

I

PROBLEM 1281 Provide a mechanism for the nitration of benzene using concentrated nitric acid-sulphuric acid mixture. Draw an energy diagram for the reaction. Will the rate of reaction change if benzene is replaced by hexadeutrobenzene? Why? PROBLEM 1282 A by product of sulphonation of benzene is diphenyl sulphone. Propose a mechanism by which diphenyl sulphone is produced during sulphonation reaction. PROBLEM 1283

Synthesize the following compounds starting from benzene: O O

O H

H

O

and

H

H O

PROBLEM 1284

Provide reagents that will accomplish the following transformations: O

(b)

(a)

Br

OH

O

Br

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Problems OH

OCH3

O

OCH3 NO2

H (d)

(c) OH

O H (e)

(f) O

Br (g) O2N

PROBLEM 1285 Reaction of acetophenone with Br 2 /FeBr 3 yield meta-bromo acetophenone. Does this reaction occur at faster rate or slower rate than the treatment of benzene with bromine under similar conditions? Why? PROBLEM 1286 Chloropyridines undergo nucleophilic substitution with sodium methoxide to give methoxy pyridine as: CH3ONa CH3OH

N

Cl

N

OCH3

Order the following derivatives from high to low in term of their reactivity towards nucleophilic aromatic substitution. Cl Cl

N A

N+ Cl–

N C

B

CH3

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Propose a mechanism for the following reaction:

PROBLEM 1287

+

O

H2SO4

Devise synthesis of indicated compound starting from benzene:

PROBLEM 1288

N

O

H

O

Provide a mechanism for the following reaction:

PROBLEM 1289 CH3O

NH2 CH3COCl

CH3O

HCl

CH3O

NH

CH3O

CH3

Identify the followings as aromatic, antiaromatic or nonaromatic:

PROBLEM 1290 (a)

B—

(b)

(c)

l

(d)

PROBLEM 1291

O

+

—OH

Predict major products in the following reactions:

Cl (a)

HNO3 H2SO4

(b)

(1) NBS hν (2) NaOCH3

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Problems

O

SO3

(c)

AlCl3

NO2 OCH3

O2N O

O

Na/NH3

(e)

(f)

EtOH

PROBLEM 1292

CH3COCl

(d)

H2SO4

Zn(Hg) HCl

Predict major products in the following reactions:

KMnO4

(a)

NaOH/∆

H 3O +

(b)

(1) C6H5ONa (2) Br2/FeBr3

O2N

O

Br2 FeBr3

(c)

—NO2

Br

(d)

—NO2

(f)

—OCH3

(1) Br2/FeBr3 (2) NaNH2/NH3

OCH3 SO3H (e)

NO2 HNO3 H2SO4

(1) C6H5COCl/AlCl3 (2) Fe(Hg)/HCl (3) H2/Rh/Heat

PROBLEM 1293 Nitration of toluene gives exclusively the ortho and para substitution products. Using mechanism, explain the reason for regioselectivity. PROBLEM 1294 Provide reagents that will effect the following transformations. More than one steps may be required: (a) OCH3

(b)

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COOH (c) HO3S (d)

CH3O

PROBLEM 1295 Explain.

Will the following reaction give mostly ortho/para products or mostly meta product.

—B

PROBLEM 1296

OH

HNO3

OH

H2SO4

Devise synthesis of the following compounds from the indicated starting materials: N3

(a) Cl

O

SO3H (b) Br

PROBLEM 1297

Propose mechanism for the following reaction:

—CH2Br

DMSO heat

NaOH

—CHO

PROBLEM 1298 When treated with H 2SO 4 , α-methyl styrene gives the indicated dimeric product. Propose mechanism of its formation:

H2SO4 α-methyl styrene

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Problems

PROBLEM 1299

Propose a mechanism of the following reaction: CH3

CH3 D

D2SO4

D

D2O

D

PROBLEM 1300 Would you expect Birch reduction (Na/liq. NH 3 /EtOH) of benzoic acid or anisole to proceed more quickly, explain. PROBLEM 1301

Devise a synthesis of the indicated compound starting from benzene:

N—OH

Br

PROBLEM 1302

Write products in the following reactions:

Br

HNO3

(a)

NaNO2

Fe/HCl

CuCN

HCl/0°C

H2SO4

Br

(b)

CH3COCl

N2H4

AlCl3

HCl

Fe/Br2

(c)

H3O+

NaBH4

O (d) COOH

Na NH3

EtOH

SOCl2

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PROBLEM 1303

Bring about the following transformation: O

(a) Br OH O

O

(b) O2N

PROBLEM 1304 Provide structure of one example of the following species: (a) An antiaromatic, (b) An aromatic cation, (c) A cyclic non-aromatic compound containing 6 π electrons, (d) A neutral, aromatic heterocyclic compound. PROBLEM 1305 (a)

Predict products in the following reactions: HNO3

Br2

H2SO4

FeBr3

NO2

CH3O

NO2

(b)

Cl2/AlCl3

KMnO4

(c)

H 3O +

NaOH/heat

Na/NH3 EtOH

O

(d)

OH

(1) CH3Li (excess) (2) H3O+

PPh3

H+ H2SO4

O

O2N

CH2

Zn(Hg) HCl

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Problems

(1) NaBH4

CH3COCl

(e)

AlCl3

PROBLEM 1306

(2) H3O+

Predict products of the following reactions:

(a)

H3O+

PhLi excess

CN

CH3MgBr

HO

D2, Ph

(b)

heat, pressure

(c)

Cl

Br2

NaNH2

FeBr3

NH3

O

F3C NaOH

(d)

H 3O +

CH3Li excess

N CH3O Na

(e)

NH3/EtOH

O3

HCl H 2O

(CH3)2S

O

(f)

(1) NBS/hν (2) NaOCH3

LiAlH4 H 3O +

H3O+

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PROBLEM 1307

Provide reagents that can be used to carryout the following transformations: OCH3

(a) H3CO (b)

H (c) O

O2N

PROBLEM 1308

Identify A, B and C in the following sequence of reaction: Ph 3P

C H CHO

HCl H 2O

6 5 CH 3OCH 2I → A → B → C

BuLi

PROBLEM 1309 Nitration of A can give rise to seven possible products in principle (B H). However, only three are produced. Indicate, which are formed and explain with the of aid of mechanism. NO2 NO2 NO2 O2N

HNO3 H2SO4 O2N

A NO2

NO2

C

B NO2

NO2

NO2

D

F

E NO2

NO2

G

NO2

NO2

H NO2

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Problems

PROBLEM 1310

Provide reagents that will bring about the following transformations: O

O

(a)

C

H

O CH3

H

N

NH2

CH3 N

N

(b)

CH3

CF3

PROBLEM 1311

CF3

Propose mechanism of the following reaction: OH

OH CH3

H3 C

+ Br2

CH3

H3 C

FeBr3

Br

SO3H

PROBLEM 1312

In each of the following pair, select the more stable one: –

+

or

(a) I

(b)

+

or

–

II II

I –

(c)

+

(d)

or I

II

or I

II

PROBLEM 1313 Rank the compounds in each group in increasing order of their reactivity towards electrophilic aromatic substitution reaction: (a) Chlorobenzene (I), fluorobenzene (II), cyanobenzene (III), (b) o-dichlorobenzene (I), benzaldehyde (II), methoxybenzene (III), (c) methoxybenzene (I), o-xylene (II), benzene (III). PROBLEM 1314 Discuss the orienting effect of nitroso group (  NO) in aromatic electrophilic substitution reaction.

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PROBLEM 1315

Provide synthesis of the following compounds starting from benzene:

(a)

(b) O2N COOH NO2

(c)

Cl

PROBLEM 1316

Write products in the following reaction: CHO HCN KCN

(a)

Br

(b)

Mg

CO2 H3O+

O OCH3 CH3MgBr

(c)

excess

OH (e)

PROBLEM 1317

H3O+

OCH3 (d)

Na NH3/EtOH

+

C6H5N2Cl–

Starting from benzene synthesize the following compounds: NH2

CH3 OH (b)

(a)

Br CH3 (c)

CH3 Br

PROBLEM 1318 Furan undergo electrophilic aromatic substitution. With the help of mechanism, explain the most probable site for the attack of electrophile on furan ring.

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Problems

PROBLEM 1319

Bring about the following transformations:

NO2

OH OCH3

(a) (b) CH3

N (c)

PROBLEM 1320

Which of the following pairs is more stable:

+

(a)

●●

N I

or

(b)

N II

or I

●●

II

PROBLEM 1321 Arrange the following compounds in increasing order of reactivity towards electrophilic aromatic substitution reaction: (a) Chlorobenzene (I), benzene (II), nitrobenzene (III), (b) m-chloroanisole (I), p-chloroanisole (II), anisole (III), (c) Cyanobenzene (I), p-cyanotoluene (II), p-cyanoanisole (III). PROBLEM 1322

Bring about the following transformations: (b)

(a)

PROBLEM 1323

Starting from toluene, synthesize: CH3

N

CH3

H2N—

—

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PROBLEM 1324

Predict the direction of electrophilic attack on the following compound: CN

PROBLEM 1325

Provide structures of A and B and suggest mechanism of their formation: Cl Cl

(a)

O

C10H14(A)

AlCl3

AlCl3

NO2

P(OMe)3

(b)

heat

H2N C12H9N B

HCl/NaNO2 0°C, then heat

N H

PROBLEM 1326

Propose mechanism: O

OMe

OMe

COOMe

COOMe

NHNH2

then HCl

N H

PROBLEM 1327

Bring about the following transformations: Br

CH3

CHO

(a)

CHO

(b)

OH

OH

OH

(c)

OH

(d) CHO

C O

CH3

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Problems

O

OCH3 (e)

OH

CHO (f)

Ph

COOH Ph

O

(g)

(h) N

NHNH2

H

PROBLEM 1328

Provide missing species in the following reactions: Na

(a)

KMnO4 HO–/heat

(c)

HBr Peroxide

(b)

liq. NH3

NBS hν

(d)

OH

Br NaCN

(e)

(f)

HF

+

Br O

Cl

(g)

AlCl3

(h)

H2SO4 SO3

PROBLEM 1329

Propose synthesis of the target compound starting with the substrate provided: OH

(b)

(a) O2 N

Cl

Br

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PROBLEM 1330

Complete the following reaction: +

NH3 (a)

H2SO4

Br2

(b)

HNO3

FeBr3

Br2

(c)

heat

PROBLEM 1331

Starting from benzene, synthesize: COOH —N

O2N—

PROBLEM 1332

N—

—OH

Bring about the following transformations:

(a) MeO

MeO Br (only product) O

(b)

O

O

NH2

(c)

(d) (cis/trans)

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Problems

PROBLEM 1333

Classify each of the followings as aromatic, antiaromatic or nonaromatic:

l

(a)

(b)

(c)

(d)

N (f)

(e)

PROBLEM 1334 Ordinarily barrier to rotation about carbon-carbon double bond is quite high (≈ 40 kcal/mol) but the compound shown below was observed to have a rotational barrier of about only 20 kcal/mol, explain.

PROBLEM 1335 Specify whether you expect the benzene ring in the following compounds to be activated/deactivated for electrophilic aromatic substitution reaction: O NO2 (a)

(b)

NO2 O (d) H2N—

(c)

—CH3

O SO3H (f)

(e) CH3

CD3

PROBLEM 1336 from benzene:

Propose a reasonable synthesis of each of the following multiply substituted arenes

Br (a) Cl—

—NO2

(b) Br—

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(c) SO3H

Predict products of the following reactions:

PROBLEM 1337

O Br2

(a)

CH3COCl

(b)

FeBr3

AlCl3

O

O HNO3

(c)

HNO3

(d)

H2SO4

H2SO4 CH3

O COOH HNO3

(e)

PROBLEM 1338

CO, HCl

(f)

H2SO4

AlCl3

Predict product of the following reactions: Cl

O

(a) NO2

O2N

NaNH2 NH3

Na/NH3 EtOH

(b)

O F

OH

(c)

NaOH Br2

NO2

O2N (d)

NO2

CH3CH2NH2

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Problems Cl

O LDA

(e)

OPh

(f)

Cl2 FeCl3

[(CH3)2CH]2NH

CF3 Cl OH (CH3)2CHCH2Cl

(g)

(h) H+

AlCl3

OCH3

PROBLEM 1339

Propose mechanism: NHCH3

O O2 N

NO2

O2 N

H 2O

NO2

PROBLEM 1340 aromatic.

(b)

⊕

NO2

NO2

Classify each of the following species as either antiaromatic, aromatic or non

(a)

(f)

OH

N

⊕

(g)

(c)

(d)

s

(e)

s

(h)

PROBLEM 1341 Discuss the reasons for low energy barrier to rotation about the double bond connecting the two rings, as compared to normal double bond.

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Suggest a method of synthesis of the following compounds starting from benzene: NO2 COOH

PROBLEM 1342

(b)

(a)

NO2 COOH

F

OH

NH2 OH (d)

(c) CH3

OH

SO3H (f)

(e)

CH3

NO2

PROBLEM 1343

Predict product of the following reactions:

SO3H

N Br2

(a)

(b)

FeBr3

O

NO2 HNO3

(c) O

CH3 COOH

SO3 H2SO4

CH3

FeCl3

H2SO4

Cl

(e)

Cl2

(d)

(f)

HNO3 H2SO4

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Problems

COCH3

H CH3

N (g)

Br2

Na

(h)

NH3/EtOH

FeBr3

O

COCH3

CH3

PROBLEM 1344

Bring about the following transformations:

O

PROBLEM 1345

Synthesize following compounds starting from benzene: OH

(b) H2N—

(a)

PROBLEM 1346

—COOH

Identify A to F in the following reaction: O Cl

(a)

A AlCl3 Cl

(b)

PROBLEM 1347

E AlCl3

HNO3 H2SO4

HNO3 H2SO4

B

F

N2H4 NaOH/heat

C

Co(OAc)2 AcOH/HBr/O2

Co(OAc)2 AcOH/HBr/O2

D

No reaction.

Bring about the following transformations: (b)

(a)

from benzene

O CH2NH2

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NO2 Br (d)

(c)

NH2

On the basis of Huckel rule, discuss the aromaticity of the followings:

PROBLEM 1348

l

(b)

(a)

l

l

l

N

l

l

l l

l

O

l l

l l

l

Give formula of the major organic product in each of the followings:

PROBLEM 1349 OCH3

F (a)

O

O O

(b) CH3O—

AlCl3

CH2Cl

NO2 HNO3

(c)

(d)

F +

H2SO4

AlCl3

Br2

—OH

(e)

H2SO4

—CH3 +

CHCl3

PROBLEM 1350 The following compounds undergo intramolecular Friedel-Craft reaction in presence of a Lewis acid catalyst. Give structures of most likely product in each case. O Cl

(a)

(b)

OMe (c)

OMe

O

Cl

O Cl

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Problems

PROBLEM 1351 Synthesize the following compounds starting from benzene: (a) 2-bromo-4-nitro toluene, (b) p-bromoanisole, (c) p-nitrobenzene sulphonic acid. PROBLEM 1352

Bring about the following transformations: COOH COOH

(b)

(a)

COOH

SO3H OCH3

COCH3

OCH3 NO2

(c)

(d) OCH3

Cl

Cl

CH3O F

(e)

(f) NO2

COOCH3 Br OH

NO2

(g)

(h) Cl Br OMe

(i)

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PROBLEM 1353

Starting from benzene, synthesize the following: Br Br

(a)

(b) Br CN D

(c)

(d)

NO2 OMe

(e) F—

(f)

—COCH2CH3

CH2OH CH3

NO2 Br

Cl (g)

(h)

Br

PROBLEM 1354

Complete the following reactions: NH2 NaNO2

(a)

HCl/0°C

Br NO2 CN

(b)

HNO3 H2SO4

CuCl HCl

Br

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Problems

O NaBH4

(c)

NaOH

KMnO4

(d)

CaO/heat

H2O/NaOH

Cl

H+ H 2O

H N

(e)

AlCl3 H 3O +

O CHO

(f)

H+ C6H 6

EtOH 1 mol

PROBLEM 1355 A sequence of reaction has been carried out on benzene as described below. Provide reagents which are needed for each observed transformations and give a structure of product E which results. O OH

H A

O

O

B

OH

O

H C

O

OH CHO

COOH

D

H+

E

PROBLEM 1356 Each of the following reaction is reported to yield single predominant product. Write structural formula for one of them. (a)

+

H2SO4

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NHOAc

O

CH2CH3 Cl

(b)

AlCl3

CH3 OH Br2

(c) CH3

CHCl3

C(CH3)3 HNO3

(d) CH(CH3)2

Cl2/AlCl3

(e) O2N—

PROBLEM 1357 them?

As written, the following synthesis has certain flaws. What is wrong with each of

CH3

COOH (i) Br2/Te (ii) K2Cr2O7/H+

(a)

(iii) Li/Et2O (iv) CO2/H2O

COOH CH3

CH3 (i) HNO3/H2SO4

(b)

(ii) NaNO2/HCl/0°C (iii) CuCN

CN Cl

(c)

Cl (i) HNO3/H2SO4/heat (ii) CH3Cl/AlCl3

CH3

(iii) Sn/HCl (iv) NaOH/H2O

NH2

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Problems

Cl (i) Li/Et2O (ii) HCHO/H2O

(d)

COOH

(iii) Li/Et2O (iv) CO2/H2O

PROBLEM 1358 Cl

Bring about the following transformations: COOH Cl

COOH

(b)

(a)

COOH

COOH Cl

F

OCH3

OCH3 (d)

(c)

HOOC OH

PROBLEM 1359 If we do the electrophilic bromination on the compound below, only one product (major) will be formed. Predict structure of the predominant product in each case: O (a)

Br2

—OCH3

FeBr3

O (b)

—NO2

Br2 Fe

CH3O (c)

—OCH3

Br2 FeBr3

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PROBLEM 1360

Write product (major) in the following reactions: O

(a)

NO2

Cl

NH2

Br2

H2SO4

(b)

SO3/100°C

NaOH

MeO OMe

OMe

NH2

OMe NaNO2

(c)

HPF6

Heat

(d)

CuCl2

H2SO4/0°C

Br NO2 O (e)

I2

Cl +

AlCl3

PROBLEM 1361 In the following sequence of reaction, every steps has certain flaws. Correct them by providing appropriate reagents: Br

Br

NO2

HNO3/H2SO4

NH2

CF3CO3H (B)

(A)

Br

NHCOCH3

Ethanal H+ (C)

Br NH2

NaNO3

H2O

NHCOCH3

LiAlH4

H2SO4/0°C (F)

CH3CH2CH2Cl AlCl3(D)

(E)

Br CuBr

Br

Br

COOH Br

(i) Li/Et2O (ii) CO2/H2O (G)

COOH Br

Br

F2 Fe (H)

F

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Problems

PROBLEM 1362

H

On the basis of Huckel rule, decide whether the following species are aromatic or not:

N

(a)

(c)

(b)

(d)

●●

PROBLEM 1363 The following series of reaction was performed during synthesis of E. Provide appropriate reagents needed to perform the indicated transformations: CH3

COOH A

O

COCl B

O

C

NEt2

O2N NO2

NO2

NO2

O NEt2

O

D

H2N

E

PROBLEM 1364 Two, isomeric, aromatic lactones A and B have molecular formula C 9 H 8O 4 . Both A and B dissolves in dilute NaOH solution but none gives gas with NaHCO 3 . Both A and B gave violet colouration with FeCl 3 . Reaction of A with NaOH followed by work-up with CH 3I yields C (C10 H10O 4 ). Selective demethylation of C with BCl 3 followed by aqueous work-up yields D which is also an isomer of A. D showed presence of an intramolecularly hydrogen bonded hydroxyl group. Complete hydrolysis of either A or C with dilute H 2SO 4 produced the same compound E (C 8 H 8O 5 ). Oxidising E with alkaline permanganate followed by acidification of product yields F (C 8 H 6O 6 ). Heating F with sodalime followed by acidification of product yields 1,3-dihydroxy benzene. Identify A to F. PROBLEM 1365 (a) Cl—

Write the major product in the following reactions: HNO3 H2SO4

(b) O2N—

Cl2 Fe

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(c)

CH3Cl

H3CO—

AlCl3

Cl2/Fe

(d) O2N— NO2

PROBLEM 1366

Bring about the following transformation in good yield. Cl

NH2

CARBOHYDRATES, AMINO ACIDS AND POLYMERS PROBLEM 1367 (a) What do you understand from reducing and non-reducing sugar? (b) How might you distinguish between a α-glucopyranose and methyl α-D-glucopyranoside? PROBLEM 1368 Name two disaccharides and write their hydrolysis reaction. PROBLEM 1369 Name two common polysaccharides and write their hydrolysis products. PROBLEM 1370 What type of saccharide is lactose? Is it a reducing sugar? Write hydrolysis reaction of lactose. PROBLEM 1371 With the help of a chemical reaction show that D-glucose and D-mannose are epimers. H HO H H

CHO OH H OH OH CH2OH D-glucose

HO HO H H

CHO H H OH OH CH2OH D-mannose

PROBLEM 1372 A, B and C are all aldohexose. Compound A and B yield the same optically active alditol on reduction with H 2 / Ni but they gives different phenyl hydrazones when treated with C 6 H 5 NHNH 2 . On the other hand B and C gives same phenyl hydrazones on reaction with C 6 H 5 NHNH 2 , but different alditols on reduction. Assuming that all three are D-sugars, deduce structures of A, B and C. PROBLEM 1373 Why an aminoacid is usually solid at room temperature? PROBLEM 1374 Write the structures of predominant form of 2-amino-3-methyl butanoic acid (valine) in its aqueous solution at pH (a) 2 (b) 12.

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Problems

NH 2  PROBLEM 1375 What form of glutamic acid (HOOC—CH 2 —CH 2 —CH—COOH) predominate in: (a) Strongly acidic solution? (b) Strongly basic solution? (c) At its isoelectric point?

PROBLEM 1376 Glutamic acid has lower isoelectric point O  (H 2 N—C—CH 2CH 2 —CH—COOH) has higher isoelectric pH. Explain?  NH 2

while

glutamine

PROBLEM 1377 An α-amino acid A has molecular formula C 7 H 7 NO 3 . A on treatment with methanol in presence of HCl yields B (C 4 H10 NO 3Cl). B on further treatment with PCl 5 yields C (C 4 H 9 NO 2Cl 2 ) which on acidic hydrolysis yields D(C 3 H 6 NO 2Cl). D on reduction with Na(Hg) in dilute acidic medium yields alanine. Deduce structures of A to D. PROBLEM 1378 Synthesize alanine from acetaldehyde and any other inorganic reagent needed. PROBLEM 1379 (a) What do you understand from step growth polymerization? (b) How decaron is synthesized by this method? PROBLEM 1380 How teflon is synthesized by radical chain polymerization? Device synthesis using benzoyl peroxide initiator and mention few of its common applications. PROBLEM 1381 (a) How nylon-6,6 is synthesized? (b) Why nylon has very high melting point compared to other polymers.

MISCELLANEOUS Complete the following reactions:

PROBLEM 1382

—NH2

O+

(a)

Major

O HO

(b) H (c)

CN

O

O

OH H+ H3O+

LiAlH4

H3O+ H2O

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O (C2H5)2CuLi

(d) O

(C2H5)2AlCN

(e)

(CH3COO)2Hg

(f)

H+

Write major products of the following reactions:

PROBLEM 1383

NH2 (1) NaNO2/H+/0°C

(a)

(2) HCl, Hg2Cl2

O

HgO, Br2

(b)

CCl4/∆

OH

(c)

O

O

O

H+/H2O

(d)

Heat

HO

CH3

CH3O

(1) CH3O– (2) H+/H2O

O OH SOCl 2

(e)

O

O

N—H

—C—O—CH2—

(f)

H+ H 2O

CH3 Br (f)

—N

+

N+

—OH

(h)

NaN3

H

LiAlH4

H

PROBLEM 1384 Aminomethylcyclohexane can be prepared from simple starting materials using the four methods shown below. Provide a suitable starting material for each of the reaction shown: A

N3– then LiAlH4 LiAlH4

B

O NH2 N– , H2 O O

C D

LiAlH4

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Problems

PROBLEM 1385

Starting with acetophenone, suggest a synthesis of each of the following compounds. O O

(A)

(C)

(B)

N

PROBLEM 1386 Cocaine (A) on alkaline hydrolysis produces benzoic acid, methanol and B (C 9 H15 NO 3 ). B on treatment with acidified solution of chromic acid produces “C” which on simple heating yields the following compound: H3 C

N

O

Deduce structures of A, B and C.

PROBLEM 1387 Starting with benzyl alcohol and any alkylhalides containing five or less carbons, propose a step-by-step synthesis of the following compound. OH

Ph

PROBLEM 1388 Propose synthesis of Dimestrol starting from p-methoxypropiophenone. Show mechanism of each step. OCH3

H3CO (Dimestrol)

PROBLEM 1389 Compound A(C 4 H 8O) gives positive iodoform test but does not decolourize bromine water solution or it does not evolve any gas on treatment with NaH. Reaction of A with excess of benzaldehyde in alkaline medium gives two products B and C. These products are unstable and readily dehydrate into stable products D and E. D reacts with I 2 in basic medium to give iodoform; E does not. Treatment of D with bromine in acetic acid medium yields F. F can be reduced by NaBH 4 to give G. Warming G in basic solution results in formation of H. Ozonolysis of H followed by reductive work-up, gives benzaldehyde and a ketone I (C 4 H 6O 2 ). Identify A–I.

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268

PROBLEM 1390

Beginning with benzyl bromide, suggest a synthesis of the following compounds: O

(a)

and

(b)

OCH3

PROBLEM 1391 appropriate:

Complete the following reaction indicating stereochemistry of product where O (i)

(a)

MgBr

CH2CH3 (b)

O

(ii) H+/H2O

O

H+/H2O

(i) CH3CH2O–

(c)

(ii) H+/H2O

PROBLEM 1392 Compound A, on reaction with CH 3Cl/AlCl 3 gives compound B. The mass spectrum of B is dominated by a peak at m/e = 91, which is identified as tropylium ion. B-undergoes free radical bromination to give C. C reacts with diethylmalonate in presence of sodium ethoxide to give D. Treating D with aqueous acid followed by gentle warming gives E and CO 2 . Identify A to E. PROBLEM 1393 Br

Draw structures of major organic product in each of the reaction below: O H

(a)

NaCN, Et2O–H2O then HCN

(b)

(i) NaBH4 (ii) H2O

O O OMe PhMgBr(excess)

(c)

Et2O

CH3 (d)

O

EtOH H2SO4

H2O

H+ Heat

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Problems CH3 Br2–H2O

(e)

NaOH

PROBLEM 1394 Suggest reaction sequences, including reagents and conditions, which could be used to effect the transformation of 1 to compound 2 and 3. You may use the indicated starting material and any reagents, which you deem necessary.

OH

HO

OH

OH

2

1

3

Bring about the following transformation:

PROBLEM 1395

O O

O OEt

O

PROBLEM 1396 Exposure of 1,3-diketone 1 to aqueous acid catalyzes the formation of enols 2 and 3. Justifying your answer, indicate, which is thermodynamically more stable enol. Draw a mechanism of formation of most stable tautomer of 1 in presence of an acid catalyst. O

OH

O H3O+

O

O

CH2

+

1

OH

3

2

PROBLEM 1397 For the following carbonyl compounds X, Y and Z, draw the structures of thermodynamically most stable enolate anion that could be formed in presence of a strong base: O

O X

O

Y

Z

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270

PROBLEM 1398 O

Place the following dicarboxylic acids in order of increasing acidity. O O O O

HO

OH

HO

OH HO

OH HO

A

O

B

O C

OH D

O

PROBLEM 1399 Suggest reaction sequence, including reagents and conditions, which could be used to bring about the following transformations: O

O O

A

O

O OCH3

Br

O

B

O N

C

PROBLEM 1400

Propose mechanism of the following reaction:

O

O CH3

O CH3O–Na+

+

KOH

MeOH

MeOH, Heat

O

PROBLEM 1401

O

Provide structures of missing species: O

A

Acetone H+

O

O Ph3P

O H

CH2

B

C

O

O

HO (CH3)2NH

OH HO N

CH3 CH3

E

D

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Problems

PROBLEM 1402 Unknown monosaccharide 1 was converted to tetramethylated aldohexose 4 through the sequence of reaction shown below: CHO

CH3OH HCl Cu2+

1

Cu2O precipitated

2 Cu2+

CH3I (excess) Ag2O, CH3OH

3

H+ H2O

No reaction

CH3O CH3O H H

H H OCH3 OH CH2OCH3 4

Using the information provided, draw the correct structures of 1, 2 and 3. Bring about the following transformations in good yield:

PROBLEM 1403

O

O Br

CN

PROBLEM 1404 When the diester (A) was treated with a solution of sodium ethoxide in ethanol and the reaction mixture then quenched with aqueous acid, only (B) was formed. The alternative product (C) was not formed under these reaction conditions. Explain the result. OEt

EtO O

A

O

O

O (1) NaOEt/EtOH

O OEt

OEt

(2) H3O+

O

C

B

PROBLEM 1405 Synthesize the following compounds starting from styrene: O  (a) Ph  C  CH 3 (b) Ph  CH 2  CHO (c) PhCH 2CH 2CH 2OH PROBLEM 1406

Bring about the following synthesis starting from phenyl acetylene:

Br (a)

HO —COCH3

(b)

—CH2CH2OH

(c)

QUALITATIVE ANALYSIS PROBLEM 1407 A is a white compound, on heating gives yellow residue B and colourless gas that turns lime water milky. If B is heated in air for several hours at 500°C, it is converted into a scarlet powder C. C on heating with dilute aqueous nitric acid gives a colourless solution D and a brown solid E. If sodium

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hydroxide is added to solution D, a white gelatinous precipitate is formed initially, which dissolves in excess of base. Compound E reacts with concentrated hydrochloric acid to give a white solid F and a green coloured gas. F is soluble in hot water but insoluble in cold, and forms a soluble complex with excess of chloride ion. With KI solution, D gives a bright yellow solid G which is insoluble in cold water, but in hot water a colourless solution, which on cooling gives a shimmering yellow plates like crystal. Identify A to G.

PROBLEM 1408 A metal reacts with 50% nitric acid solution to give a blue coloured solution ( A ) and a brown gas ( B ). If the solution A is cautiously treated with dilute NaOH solution, a gelatinous blue precipitate (C ) is formed, which if warmed, forms a black solid ( D ). Addition of concentrated ammonia solution to C, gives a deep blue solution that contains the ion ( E ) while addition of concentrated HCl solution to C gives a green solution of ion ( F ). If brown gas B is passed through water, a mixture of two monobasic acids are formed. Identify A to F and write reaction of B with water. PROBLEM 1409 A white powder turns yellow (B ) on heating and evolves a gas which turns lime water milky, as well as water vapour. The yellow residue turns white on cooling, but will turns yellow again when heated. B reacts with dilute sulphuric acid to give a colourless solution (C ). If dilute NaOH is added to C, a white precipitate ( D ) is formed initially, which dissolves on adding excess base. With dilute ammonia solution, C gives a white precipitate which dissolves in excess ammonia, giving a clear solution ( E ). Identify A to E. PROBLEM 1410 A metal ( A ) reacts moderately quickly with dilute sulphuric acid to give a pale green solution ( B ) and a colourless, neutral gas. If solution B is allowed to crystallize, a pale green solid is obtained which on strong heating gives a solid (C ) and two acidic, non-metallic gases. A reacts with steam to give another solid ( D ) and the same gas as obtained in the first experiment with dilute sulphuric acid. A reacts with dry Cl 2 ( g ) to produce a brown covalent solid ( E ), which sublime on heating. Aqueous solution of E reacts copper with metal, and for that reason are used to each printed circuit boards in electronics. Solution B gives a dirty green precipitate ( F ) if sodium hydroxide is added and remains insoluble in excess NaOH. If F is allowed to stand in air, it forms a foxy-red compound G, which can also be obtained by adding NaOH( aq ) to aqueous solution of E. Identify A to G. PROBLEM 1411 A compound of sulphur (one atom per molecule), oxygen and one or more halogen atoms, was examined. A small amount of the substance reacted with water, it was completely hydrolyzed without any oxidation or reduction and all reaction products dissolved. 0.1 M solution of a series of test reagents were added to separate, small portions of a dilute solution of the substance. (i) Addition of HNO 3 / AgNO 3 (ii) Addition of Ba(NO 3 ) 2 (iii) Adjustment to pH = 7 with NH 3 and addition of Ca(NO 3 ) 2 (iv) Addition of KMnO 4 followed by Ba(NO 3 ) 2 to an acid solution of the substance. (v) Addition of Cu(NO 3 ) 2 . The above tests gave the following results: (i) A yellowish precipitate (ii) No precipitate (iii) No visible reaction (iv) The main features were that purple colour disappeared and a white precipitate was formed on addition of Ba(NO 3 ) 2 . (v) No precipitate. Finally 7.2 g of the substance was dissolved in water and volume made to 250 mL. 25 mL of this solution on treatment with excess of AgNO 3 gave 1.425 g precipitate. Identify the original compound.

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PROBLEM 1412 A white salt ( A ) evolves a colourless gas on treatment with dilute HCl solution, which turned moistened litmus paper red. A small sample of A was moistened with concentrated HCl and placed on a platinum wire and introduced into a bunsen burner flame, a green colouration was observed. On strong heating, A decomposes to produce a white solid B, which turned red litmus blue. 1.54 g of B was dissolved in 250 mL H 2O and its 25 mL required 20.4 mL 0.0985 M HCl solution. Identify A and B. PROBLEM 1413 A metallic chloride A, when treated with NaOH/H 2O 2 , gives yellow coloured solution due to formation of B. The colour of this solution changes to orange when dilute H 2SO 4 is added. It is due to the formation of compound C. When a compound D is heated with C in presence of concentrated H 2SO 4 , a red volatile liquid E is formed. E, when absorbed in NaOH solution, gives a yellow coloured solution of B, which when treated with (CH 3COO) 2 Pb solution gives a yellow precipitate. Compound C when treated with NH 4Cl, forms a compound F, which decomposes on heating, giving a colourless gas, water and a green residue G. Also D gives golden yellow flame colouration. Identify A to G. PROBLEM 1414 A white powder A on heating gives a colourless gas B and a solid residue C. The compound C turns yellow on heating and changes white on cooling. It dissolves in dilute acid and the resulting solution gives a white precipitate ( X ), with K 4 [Fe(CN) 6 ] solution. Further A dissolves in dilute HCl with evolution of a gas B, which turns lime water milky. The solution thus obtained gives a white precipitate D with H 2S in slightly alkaline medium. Another portion of the solution gives initially a white precipitate E with NaOH, which dissolves on addition of excess base. Identify A to E. PROBLEM 1415 A white solid A on heating with excess of dilute HCl gave a pungent smelling gas B and a solution C. Solution C on treatment with aqueous ammonia did not give any precipitate, but on treatment with NaOH, precipitate D is obtained which dissolves in excess reagent. A on strong heating in air gave a pungent smelling gas E and a solid residue F. F dissolves completely in dilute HCl solution and the resulting solution produced a precipitate with BaCl 2 . Identify A to F. PROBLEM 1416 A white compound A on heating yields a basic oxide B. Compound A on treatment with dilute HCl evolves a gas which turns lime water milky. A is insoluble in water but dissolves in mineral acid. When HCl solution of A is made ammonical and a small amount of ammonium oxalate is added, a white precipitate C is obtained. Solid C decolourises acidic solution of KMnO 4 .Solution of C in dilute HCl evolves CO 2 with MnO 2 and the resulting solution gives a pink coloured precipitate with Na 2S solution. Also A gives brick red colouration in bunsen burner flame. Identify A to C. PROBLEM 1417 A mixture of salt ( A ) is yellow coloured and it did not lose any weight on simple heating. A dissolved in cold, dilute nitric acid solution giving an orange coloured solution. This solution on treatment with aqueous NaOH gives a white coloured precipitate ( B ) which dissolves in excess of reagent leaving some white coloured residue (C ). Filtrate of the above solution forms a brown precipitate ( D ) on treating with H 2O 2 . Compound D reacts with concentrated HCl solution evolving a green coloured gas and a white solid E is formed which turned black on passing H 2S gas. Also C dissolves in dilute acid giving, a clear solution, to which if KI is added slowly, a black precipitate ( F ) is formed initially which dissolves in excess KI forming an orange solution. Identify A to F. PROBLEM 1418 A green salt mixture ( A ) consists of halides of two metal and both salts are soluble in water. A on dissolving in dilute HCl and passing H 2S gas yields a black precipitate ( B ). B was removed by filtration and filtrate was made alkaline by adding NH 3 solution and finally treated with H 2O 2 giving a yellow coloured solution of ion (C ). On acidifying C with dilute H 2SO 4 followed by treatment with few drops of amyl alcohol and finally with excess of H 2O 2 , a blue colouration was observed due to formation of ( D ), which can be extracted into organic phase by gentle shaking. Compound C is insoluble

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in boiling, dilute H 2SO 4 , in NaOH as well as in aqueous Na 2S, but dissolves in concentrated solution of nitric acid, leaving behind a white precipitate. If the above solution is boiled for a long time, white precipitate dissolved and a clear blue solution was formed. Sodium extract of the original salt mixture gave a white precipitate with dilute AgNO 3 solution, which dissolved in excess of aqueous ammonia solution. Identify A to D.

PROBLEM 1419 A mixture consists of a two metal oxide A (green) and B (white). Mixture was dissolved in 20 mL 2 M NaOH solution containing some H 2O 2 to give a clear yellow solution leaving no residue. The above solution was separated into two part. One part was acidified with acetic acid and then treated with (CH 3COO) 2 Pb solution to give an yellow precipitate (C ). C dissolve in dilute nitric acid forming a clear orange solution. Other part of the solution was acidified with 2 M HCl and the 2 M NH 3 was added till solution became alkaline and finally boiled. A white gelatinous precipitate ( D ) was obtained. D was then dissolved in dilute HCl and some 6 M ammonium acetate was added. The solution was finally treated with few drops of “aluminion” reagent and made basic by adding ammonium carbonate. A red precipitate was obtained. Identify A to D. PROBLEM 1420 A solid mixture consists of a red metal oxide ( A ) and a white hydrated salt B. The mixture was dissolved in dilute HNO 3 and a portion of this solution was treated with NH 3 solution, just to make the solution neutral and then finally treated with NH 3 / NH 4Cl buffer solution when a reddish brown gelatinous precipitate (C ) was obtained. C was then dissolved in dilute HCl and few drops of NH 4SCN solution was added when a deep blood red colouration was observed. Filtrate obtained after removal of C was then treated with excess of (NH 4 ) 2 S solution when a pink coloured precipitate ( D ) was formed. D was then dissolved in dilute HNO 3 and treated with NaBiO 3 solution. A deep purple coloured solution was formed. In a separate experiment a pinch of the original salt mixture was mixed with solid K 2Cr 2O 7 and then dissolved in concentrated H 2SO 4 and finally boiled. A deep red fumes of ( E ) was obtained which made aqueous solution of NaOH yellow, when passed through it. Identify A to E. PROBLEM 1421 A solid substance is a mixture of a scarlet-red oxide ( A ) and a brown substance (B ). The solid dissolve in boiling solution of dilute nitric acid giving a brown precipitate (C ) and a clear orange solution. Solution is filtrated off and filtrate as well as precipitates were preserved for further analysis. Filtrate was further acidified by adding dilute HCl and H 2S gas was passed. A black precipitate ( D ) was obtained. D was then filtered off and filtrate was treated with excess H 2O 2 and finally boiled to decompose off any unreacted H 2O 2 . The solution was finally cooled and treated with K 4 [Fe(CN) 6 ] solution when a deep blue coloured precipitate ( E ) was formed. Compound ( D ) was dissolved in dilute HNO 3 and boiled, resulting in formation of a white precipitate. Brown solid C was analysed by dissolving in a concentrated HCl solution which resulted in formation of white precipitate ( F ) and a green coloured gas was evolved. Compound F was dissolved in boiling water and treated with aqueous sodium chromate solution, resulting in formation of a yellow precipitate. Identify A to F. PROBLEM 1422 A salt mixture consists of two salts, salt-A is scarlet coloured and insoluble in water, while salt-B is yellow coloured and soluble in water. The mixture was shaken with water in a test tube and filtrated off. The precipitate is pure A and filtrate is solution of B. A dissolves in concentrated solution of KI forming a colourless solution of C, which when treated with aqueous ammonia, forms a brown precipitate D. A portion of original filtrate containing ions of salt B was treated with aqueous NaOH solution resulting in formation of a white gelatinous precipitate ( E ), soluble in excess reagent. E was dissolved in

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dilute HCl solution and few drops of this solution was placed on a filter paper, previously moistened with an alcoholic solution of alizarin. Filter paper, on drying at 100°C, gave a red lake. In a separate experiment, 1 mL of the original filtrate was mixed with 1 mL starch solution and few crystals of KNO 2 were added, a deep blue colouration was observed. Identify A to E.

PROBLEM 1423 A mixture consists of a white oxide A and a white salt B in which latter is water soluble. Mixture was dissolved in water and filtered to separate precipitate A and filtrate containing B. A was dissolved in dilute hydrochloric acid solution and then treated with aqueous ammonia when a white gelatinous precipitate C was formed initially, which dissolved in excess reagent, forming a clear solution of D. A portion of original filtrate was treated with aqueous (NH 4 ) 2 CO 3 solution when a white precipitate E was formed. E gave a brick-red colouration in Bunsen burner flame. An other portion of original filtrate was mixed with same concentrated sulphuric acid in a test tube and a freshly prepared ferrous sulphate solution was add through side wall of the test tube. A brown ring appeared at the junction of two solution. Identify A to E. PROBLEM 1424 A white mixture contain two salts A and B, both water soluble. A pinch of this salt mixture was dissolved in water and treated with excess of ammonium oxalate solution, resulting in formation of a white precipitate. Precipitate was added to a boiling solution of dilute acetic acid, where a portion of it went into solution, leaving behind a white solid C. Solution was cooled, filtered and filtrate was divided into two part. One part of the filtrate was treated with aqueous K 2CrO 4 solution, giving a yellow precipitate ( D ), insoluble in dilute acetic acid. Other part of filtrate was treated with excess of (NH 4 ) 2 CO 3 solution giving a white precipitate ( E ). Solid compound C and E was subjected to flame test separately. C gave brick-red colouration while E gave apple-green colouration in the Bunsen burner flame. In a separate analysis, an aqueous solution of original salt mixture was prepared and divided into two parts. One part of the solution was mixed with concentrated sulphuric acid solution and then a freshly prepared ferrous sulphate solution was added slowly, giving a brown ring at the junction. Other part of the solution was treated with aqueous AgNO 3 solution when a white precipitate soluble in aqueous ammonia, was formed. Also salt B is hydrated one, and on heating lost 49.3% of its weight to become anhydrous. Identify A to E. PROBLEM 1425 A red coloured salt ( A ) is insoluble in water and dilute HCl but dissolved in dilute nitric acid, giving a clear orange solution. This solution when treated with a Na 2CO 3 solution, gave a white precipitate ( B ), leaving behind a clear orange solution B on strong heating gives off a colourless acidic gas leaving behind a brown solid (C ), soluble in aqueous ammonia as well as in dilute nitric acid. After removing precipitate B by filtration, filtrate was mixed with aqueous NaOH and turned into a clear yellow solution. Yellow solution on acidification with dilute sulphuric acid followed by treatment with excess of H 2O 2 gave a blue colouration due to formation of a new compound D. D can be extracted into organic phase by shaking the solution with ether. Identify A to D. PROBLEM 1426 A white metal ( A ) burns in nitrogen to produce an ionic compound B. B on treatment with lime water forms a white precipitate (C ), with evolution of a colourless gas ( D ). D when passed through an alkaline solution of K 2 [HgI 4 ], an yellow precipitate ( E ) was produced. Compound C was dissolved in dilute hydrochloric acid and some NH 4Cl crystals were added to this solution. Resulting solution was treated with oxine (8-hydroxy quinoline), when a yellow precipitate of complex salt was produced. Identify A to E. PROBLEM 1427 A white hydrated salt ( A ) loses 45% of its weight on heating and rendered anhydrous. A on strong heating yields a dirty brown solid ( B ) with evolution of two gases (C ) and ( D ). C turned

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orange colour of dichromate solution into blue-green. B dissolved in boiling HCl solution to produce an yellow solution of ( E ). Solution E was separated into two parts and one part was treated with NH 4SCN solution giving a blood-red solution. To the other part of solution, H 2S was passed after making it alkaline, where a white turbidity was observed initially and the solution became light green on standing for some time. An aqueous solution of A gave white precipitate when treated with aqueous barium chloride solution. Identify A to E.

PROBLEM 1428 A white salt ( A ) does not dissolve in water, but dissolve in dilute nitric acid solution. Passing H 2S gas through acidic solution of A, a black precipitate was produced which was insoluble in water, dilute HNO 3 and alkali but dissolved in concentrated solution of Na 2S as well as in aqua-regia. Adding SnCl 2 -dropwise to the solution of A, gave a white, silky precipitate ( B ), which turned into black precipitate (C ) on adding excess reagent. Addition of KI to the acidified solution of A gave a red coloured precipitate ( D ) in the beginning, which dissolved in excess reagent. Also acidified solution of A, when treated with concentrated sulphuric acid followed by K 2Cr 2O 7 and finally boiled, a dense brown fume was formed, which turned aqueous solution of NaOH, yellow coloured. Identify A to D. PROBLEM 1429 A black salt ( A ) is insoluble in dilute acid in cold but dissolve in boiling solution of concentrated hydrochloric acid, evolving a pungent smelling gas B. B turned a filter paper moistened with lead acetate, black. The above solution was cooled and separated into two parts. One part was treated with NaOH solution giving a white precipitate (C ), insoluble in excess of reagent. C, when added to a concentrated H 2O 2 solution, a clear, yellowish-brown solution of D was formed. Other part of the original solution was treated with KI solution, giving a black precipitate ( E ), which dissolved in excess of reagent forming a clear orange solution. Identify A to E. PROBLEM 1430 A white salt mixture contain two salts (anhydrous). A and B, both are water soluble. An aqueous solution of the salt mixture was prepared and treated with dilute NaOH when a white precipitate was obtained. Precipitate was shaken with concentrated NaOH solution and filtered again. The residue (C ) containing metals of salt B and filtrate were analyzed independently. Passing H 2S gas through filtrate yielded another white precipitate D. The residue C was dissolved in 5.0 mL diphenyl carbazide solution, when a violet red precipitate was formed. Also, a small portion of the original aqueous solution was mixed with some concentrated H 2SO 4 solution and boiled. An irritating gas ( E ) was formed, which gave a white cloud on the surface of a glass road moistened with ammonia solution. Also, the original aqueous solution did not give test of nitrate and sulphate ion. Identify A to E. PROBLEM 1431 A solid mixture consists of a red oxide ( A ) and a water soluble, hydrated salt (B ) which loses 51.2% of its weight on heating and becomes anhydrous. Solid mixture was dissolved in dilute HCl and the solution was used for further analysis. A portion of this solution was mixed with excess of (NH 4 ) 2 S, when a black precipitate (C ) was obtained. Solution was filtered off and filtrate was treated with aqueous NaOH when another white precipitate ( D ), insoluble in excess of reagent was produced. D was dissolved in small amount of dilute HCl and then excess of NaOH was added, followed by addition of a few drops of magneson-I [4-(4-nitrophenylazo) resorcinol], when a blue coloured precipitate formed. Solid C, when dissolved in dilute HCl, a white precipitate ( E ) was produced. Another portions of the original aqueous solution was treated with K 4 [Fe(CN) 6 ] and NH 4SCN separately, when deep blue, precipitate and deep blood-red colouration were observed respectively. Soda-extract of the original mixture gave a white precipitate ( F ) with BaCl 2 solution. F when fused with Na 2CO 3 on charcoal and the residue extracted with water and finally filtered into a freshly prepared solution of nitropruside {Na 2 [Fe(CN) 5 NO]}, transient purple colouration was observed. Identify A to F.

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PROBLEM 1432 A salt mixture contain a black salt ( A ) and a white salt (B ). Salt mixture dissolves in dilute sulphuric acid solution producing a clear blue solution. Passing H 2S gas through this solution gave black precipitate (C ). The solution was filtered off and colourless filtrate was treated with aqueous ammonia giving a white precipitate ( D ), that was readily soluble in aqueous ammonium chloride solution. Sodium extract of the salt mixture was prepared and divided into two part. One part of the extract was treated with BaCl 2 solution, giving a white precipitate ( E ), which dissolved into dilute HCl evolving a gas ( F ). F turned a filter paper, moistened with acidified K 2Cr 2O 7 solution, green. Other part of extract was mixed with excess of MnO 2 and concentrated sulphuric acid solution and finally boiled. A brown coloured vapour was formed which gave orange-red staining when passed through starch paper. Identify A to F. PROBLEM 1433 A mixture consists of a black oxide ( A ) and a yellow salt (B ). Mixture was shaken with dilute sulphuric acid solution resulting in formation of a yellow residue ( B ) and a clear solution on filtration. Filtrate was treated with excess of H 2O 2 and finally boiled to decompose-off unreacted H 2O 2 , cooled and divided into two parts. One part of the solution was treated with aqueous K 4 [Fe(CN) 6 ] solution giving a blue precipitate (C ), while other part of the solution gave a deep red colouration when treated with NH 4SCN solution. Residue B was insoluble in excess ammonia but dissolved in KCN solution as well as in Na 2S 2O 3 solution. Sodium extract of the original mixture was mixed with chlorine water when a brown colouration was observed. Brown solution when shaken with chloroform, organic phase turned violet. Identify A to C. PROBLEM 1434 A mixture contain a yellow salt ( A ) and a green salt (B ). Heating salt mixture gives-off a colourless gas which turned lime water milky. The residue thus obtained was shaken with dilute hydrochloric acid which gave a white precipitate (C ) and a clear blue solution. Solution was filtered off and H 2S gas was passed through filtrate when a black precipitate ( D ) was obtained. D was insoluble in boiling dilute H 2SO 4 as well as in NaOH solution but dissolved in a concentrated aqueous potassium cyanide solution forming a clear and colourless solution of E. Also C dissolved in excess of ammonia as well as in sodium thiosulphate solution. Sodium extract of the residue, obtained after heating of original salt mixture, when mixed with concentrated H 2SO 4 and some K 2Cr 2O 7 , a deep brown fume was produced on boiling, that turned NaOH solution yellow. Identify A to E. PROBLEM 1435 A mixture consists of an yellow salt ( A ) and a white salt (B ), both anhydrous. Salt mixture was dissolved in water and few drops of HCl was added to obtain a clear, yellow coloured solution. The solution was then treated with NH 3 / NH 4Cl solution when a reddish-brown precipitate (C ), insoluble in NaOH solution, was produced. Solution was filtered off and filtrate was treated with aqueous NaOH solution when a white precipitate ( D ), soluble in excess reagent, was formed. D was dissolved in dilute H 2SO 4 solution and few drops of copper (II) sulphate was added to it. To the resulting solution, 2.0 mL of ammonium tetrathiocyanato mercurate (II) [(NH 4 ) 2 [Hg(SCN) 4 ]] solution was added when a violet precipitate was formed. Also precipitate (C ), dissolved in dilute HCl and became blood-red coloured when NH 4SCN solution was added. Sodium extract of the original salt mixture gave the following results: (a) Extract solution in concentrated H 2SO 4 , when mixed with K 2Cr 2O 7 , gave a dense brown fumes on boiling, that turned aqueous NaOH solution yellow. (b) Extract solution when treated with Hg(NO 3 ) 2 , a yellow precipitate ( E ) was formed. Identify A to E.

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PROBLEM 1436 A white salt mixture contain two salts, salt A is water soluble and salt B is water insoluble. the salt mixture was leached with water so that soluble portion went into solution, leaving behind residue of B. Solution was filtered off and filtrate was preserved for further analysis. Precipitate B was insoluble in dilute HCl as well as in concentrated ammonium acetate solution but dissolved in boiling solution of concentrated sulphuric acid. The solution was cooled and treated with K 2CrO 4 giving a yellow precipitate (C ), that was insoluble in dilute acetic acid solution. Also the original filtrate gave a white gelatinous precipitate ( D ) on treatment with aqueous ammonia. D was dissolved in small volume of concentrated nitric acid and few drops of cobalt nitrate was added to the above solution. Now a filter paper was dipped into the solution and burned finally to give a blue fixed ash due to formation of ( E ). Sodium extract of the original salt mixture gave the following results: (a) Boiling with concentrated sulphuric acid solution gave off a colourless, pungent smelling gas which formed a white cloud on the surface of a glass rod moistened with ammonia solution. (b) Gave a white precipitate with lead acetate solution, that was insoluble in dilute acid but dissolved in concentrated solution of ammonium acetate. Identify A to E.

CO-ORDINATION COMPOUNDS PROBLEM 1437 Name the following compounds according to IUPAC convention: (i) [Co(NH 3 ) 6 ]Cl 3 (ii) [CoCl(NH 3 ) 5 ]2+ (iii) [CoSO 4 (NH 3 ) 4 ]NO 3 (iv) K 2 [OsCl 5 N] (v) Na 3 [Ag(S 2O 3 ) 2 ] (vi) K 2 [Cr(CN) 2 (O) 2 (O 2 )NH 3 ] (vii) [Ir(NH 3 ) 5 (ONO)]Cl 2 (viii) K 2 [PbCl 4 ] (ix) KCu[Co(CN) 6 ] (x) K 3 [Au(CN) 6 ] (xi) Li[AlH 4 ] (xii) Na[BH 4 ] (xiii) Na 3 [AlF6 ] (xiv) [CoCl 2 (H 2O) 4 ]Cl ⋅ 2H 2O

PROBLEM 1438 Name the following complexes according to IUPAC convention: (i) [Cr(NH 3 ) 6 ][Cr(NCS) 6 ] (ii) [Cu(NH 3 ) 4 ][PtCl 4 ] (iii) [Pt(NH 3 ) 4 ][PtCl 4 ] (iv) [Co(NH 3 ) 6 ]2 [Ni(CN) 4 ]3 (v) [Pt(NH 3 ) 4 Cl 2 ][PtCl 4 ] (vi) [Co(NH 3 ) 6 ]4 [Co(NO 2 ) 6 ]3 (vii) [Cr(NH 3 ) 5 NO 2 ]3 [Co(NO 2 ) 6 ]2 (viii) [Pt(Py) 4 ][PtCl 4 ] (ix) [Ni(NH 3 ) 6 ]3 [Co(NO 2 ) 6 ]2 (x) [Co(NH 3 ) 6 ][Cr(NH 3 ) 2 Cl 4 ] (xi) [Co(NH 3 ) 4 Br 2 ]2 [ZnCl 4 ] (xii) [Co(NH 3 ) 5 CO 3 ]2 [CuCl 4 ] (xiii) [Ag(NH 3 ) 2 ]4 [Fe(CN) 6 ] (xiv) [Cr(en) 2 Cl 2 ]2 [PdCl 4 ] PROBLEM 1439 Name the following complexes according to IUPAC convention: (i) [(NH 3 ) 5 CoNH 2Co(NH 3 ) 5 ](NO 3 ) 5 S O O (ii) [(H 3 N) 4 Co Co(NH 3 ) 4 ](NO 3 ) 3 NH 2

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CO (iii) [(CO) 3 Fe——CO——Fe(CO) 3 ] CO NH 2 (iv) [(H 3 N) 4 Co

Co(en) 2 ]Cl 4 OH NH 2

(v) [(en) 2 Co

Co(en) 2 ]Br 4 OH NO 2

(vi) [(en) 2 Co

Co(en) 2 ]Cl 4 NH 2

PROBLEM 1440 Write molecular formula of the following complexes: (i) pentaaminebromocobalt (III) sulphate (ii) dichlorobis(ethylendiamine) platinum (IV) sulphate. (iii) tetrapyridineplatinum (II) tetrachloroplatinate (II). (iv) potassium carbonylpentacyanoferrate (II). (v) cesium pentafluorotellurate (IV) (vi) triammine bromoplatinum (II) nitrate. (vii) dichlorobis (ethylenediamine) cobalt (III) chloride monohydrate. (viii) tetraquadibromochromium (III) chloride (ix) ammonium heptafluorozirconate (IV) (x) hexaammine nickel (II) hexanitrocobaltate (III). (xi) dichlorobis (ethylenediamine) chromium (III) tetrachloropalladate (II). (xii) aluminium tetrachloroacerate (III). (xiii) Iron (III) hexacyanoberrate (II). PROBLEM 1441 Two complexes of cobalt have same molecular formula Co(NH 3 ) 3 (H 2O) 2 ClBr 2 ⋅ 34.15 g of A loses 1.8 g of weight when heated to 200°C whereas no weight loss was observed on similar treatment with B. An aqueous solution of A has conductivity equivalent to that of a compound with two ions per formula unit while conductivity of B in aqueous solution is equivalent to a compound with three ions per formula unit. Also a 100 mL 0.1 M aqueous solution of A, on treatment with excess of AgNO 3 solution produced 1.88 g of precipitate while a 100 mL 0.1 M aqueous solution of B on treatment with excess of AgNO 3 solution produced 3.315 g precipitate. Identify A and B. (Atomic weight : Co = 59, Br = 80). PROBLEM 1442 A complex salt has molecular formula Co(NH 3 ) 5 SO 4 Br and exist in two isomeric forms A and B. An aqueous solution of A produced a white precipitate on treatment with BaCl 2 solution while B did not produce any precipitate on similar treatment. B on treatment with aqueous AgNO 3 produced a brown precipitate. Identify A and B and predict the type of isomerism being exhibited by them. PROBLEM 1443 An octahedral complex of platinum has molecular formula Pt(NH 3 ) 4 Cl 2 ⋅ Br 2 and exist in two isomeric forms A and B. An aqueous solution of A containing 49.4 g of salt give 37.6 g of

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precipitate on treatment with excess AgNO 3 solution whereas an aqueous solution of B containing same amount of salt gave 28.7 g of precipitate on treatment with excess of AgNO 3 solution. Identify A and B. [Atomic mass : Pt = 195, Ag =108, Br = 80].

PROBLEM 1444 A, B and C are three complexes of chromium (iii) with their formula H12O 6Cl 3Cr. Complex A does not react with concentrated H 2SO 4 solution whereas complexes B and C loses 6.75% and 13.5% of their weight, respectively, on treatment with concentrated H 2SO 4 solution. Deduce formula of A, B and C. PROBLEM 1445 An aqueous solution of titanium chloride, when subjected to magnetic measurement, measured zero magnetic moment. Deduce formula of complex assuming it to be octahedral in aqueous solution. PROBLEM 1446 A cobalt (ii) salt when dissolved in excess of concentrated HCl, it forms an octahedral complex whose magnetic moment was found to be 3.87 Bm. Predict shape of complex. PROBLEM 1447 How [Co(NH 3 ) 6 ][Cr(NO 2 ) 6 ] and [Co(NO 2 ) 6 ][Cr(NH 3 ) 6 ] can be distinguished by electrochemical method? PROBLEM 1448 How [Cr(NH 3 ) 6 ][Cr(NO 2 ) 6 ] and [Cr(NH 3 ) 4 (NO 2 ) 2 ][Cr(NH 3 ) 2 (NO 2 ) 4 ] can be distinguished from conductivity measurement in their aqueous solution? PROBLEM 1449 Two complexes of chromium have empirical formulae corresponding to Cr(NH 3 ) 3 (NO 2 ) 3 . In aqueous solution, one of these is non-conducting while other is electrolytic. What is the lowest possible formula weight of conducting reagent? What is the highest possible formula weight of non-conducting reagent? PROBLEM 1450 Write formula of all possible polymerization isomers of Pt(NH 3 ) 2 (NO 2 ) 2 . PROBLEM 1451 A metal complex having composition Cr(NH 3 ) 4 Cl 2 Br has been isolated in two forms A and B. A reacts with AgNO 3 producing a white precipitate which was soluble in dilute ammonia solution. B reacts with AgNO 3 producing a pale-yellow precipitate soluble in concentrated ammonia solution. Write formula of A and B and state hybridization of chromium in each. Also calculate their magnetic moment (spin only). PROBLEM 1452 A metal complex having composition Cr(NH 3 ) 4 ClBr 2 ⋅ H 2O has been isolated in three forms A, B and C. Heating 16.7 gram of either A or C loses 0.9 gram of its weight whereas no weight loss was observed on heating B. Adding 35.35 g of A to a concentrated solution of AgNO 3 resulted in 14.35 gram of precipitate whereas adding same amount of C to a concentrated AgNO 3 solution resulted in 18.8 g of precipitate. In another experiment, when 6.67 gram of complex B was added to a concentrated aqueous solution of AgNO 3 , when 7.52 gram precipitate was formed. Deduce formula of complexes A, B and C. PROBLEM 1453 An octahedral complex of cobalt has its composition Co(NH 3 ) 3 (H 2O) 3 (SO 4 )(NO 3 ) and it exists in four isomeric forms A, B, C and D. Simple heating of either A or B loses 5.6% of its original weight whereas similar treatment on C causes 11.2% loss in its original weight D did not lose any weight on simple heating. Either A or C did not result in any precipitation if treated with aqueous BaCl 2 solution, whereas B and D resulted a white precipitate when treated with aqueous BaCl 2 solution. Also C does not conduct electricity in its aqueous solution. Deduce formula of A, B, C and D. PROBLEM 1454 An octahedral complex of platinum has its composition Pt(NH 3 ) 4 Cl 2 Br 2 and exist in three isomeric forms A, B and C. A 4.94 gram of compound A on treatment with excess of aqueous

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AgNO 3 gave 3.76 g of precipitate. 2.47 gram of B on treatment with excess of AgNO 3 solution gave 1.66 gram of precipitate while 9.88 gram of C gave 5.74 gram of precipitate with aqueous AgNO 3 . Deduce structural formula of A, B and C. (Atomic weight Pt =195, Br = 80, Ag = 108)

PROBLEM 1455 A complex of cobalt has composition Co(NH 3 ) 3 (H 2O) 2 ClBr 2 and exist in two isomeric forms A and B. A on simple heating loses 5.3% of its weight whereas B does not lose any weight on simple heating. A on treatment with excess of AgNO 3 solution gives off precipitate that is 42% by mass of the complex. Similar treatment of B gives off another precipitate which is 110% by mass of the complex. Deduce structures of A and B. PROBLEM 1456 A solution was prepared by dissolving 0.5 g CrCl 3 ⋅ 6H 2O in sufficient water and volume made upto 100 mL. A 10.0 mL position of this solution when treated with excess of AgNO 3 , gave a precipitate, filtered dried and weighed to be 0.538 gram. Deduce structural formula of the complex. PROBLEM 1457 A complex [CoA 6 ]3+ is red coloured while [CoB6 ]3+ is green coloured. (a) Which ligand produces the lager crystal field splitting. (b) Which complex is expected to be more easily oxidised. PROBLEM 1458 In which complex [Cu(H 2O) 4 ]2+ or [Cu(NH 3 ) 4 ]2+ , are the bonds from acceptor to the donor atoms stronger? PROBLEM 1459 Which complex is more stable [Cr(NH 3 ) 6 ]3+ or [Cr(en) 3 ]3+ ? PROBLEM 1460 Why are complexes of Ti 4+ and Zn 2+ are typically white? PROBLEM 1461 Accounts for the followings : 2– is paramagnetic. (a) Ni(CN) 2− 4 is diamagnetic but [NiCl 4 ] (b) [Ni(CO) 4 ] is tetrahedral (c) [Ni(NH 3 ) 6 ]2+ is an outer orbital complex.

PROBLEM 1462 Why [CuCl 4 ]2– exist but [CuI 4 ]2− does not? 3+ PROBLEM 1463 Both Fe(CN) 3− 6 and Fe(H 2O) 6 are colourless in aqueous medium, explain.

PROBLEM 1464 Calculate the magnetic moment of K 3 [Mn(CN) 6 ] and K 2 [MnBr 4 ] and explain the geometries of these complexes. PROBLEM 1465 Select the complexes which will absorb in visible range : (a) [Ti(H 2O) 6 ]Cl 3 (b) [VF4 ] (c) [Cu(NH 3 ) 4 ]+ (d) [Cu(NH 3 ) 4 ]2+ (e) [Ni(CN) 4 ]2− (f) [Ni(CO) 4 ] PROBLEM 1466 CoF63− is paramagnetic while Co(CN) 3− 6 is diamagnetic, explain. PROBLEM 1467 Use an orbital diagram to show the electron distribution in 3d-orbitals of the central metal ion in each of the following complex ion. If more than one distribution seems possible, indicate whether you expect low-spin or high-spin state to be favoured and determine spin-only magnetic moment. (d) [Mn(CN) 6 ]3– (c) [CoCl 4 ]2– (b) [FeCl 6 ]3– (a) [Cr(H 2O) 6 ]3+

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PROBLEM 1468 Of the following two complexes: [Cr(H 2O) 6 ]Cl 3 and [Cr(NH 3 ) 6 ]Cl 3 , one is yellow coloured while other is violet. Identify them. PROBLEM 1469 Of the following two complexes : [Fe(H 2O) 6 ] (NO 3 ) 2 and K 4 [Fe(CN) 6 ] ⋅ 3H 2O, one is green coloured while other is yellow. Identify them. PROBLEM 1470 Explain why one of the cis-trans isomers of [CoCl 2 (en) 2 ] exhibit optical isomerism and other does not. PROBLEM 1471 Would you expect to find optical isomerism in either of the hypothetical tetrahedral complexes : [ZnA2 B 2 ]2+ and [ZnABCD ]2+ ? Explain. PROBLEM 1472 The complex [Pt(NH 3 ) 2 Cl 2 ] displays cis-trans isomerism but [ZnCl 2 (NH 3 ) 2 ] does not. Why do you suppose these two cases are different? PROBLEM 1473 The magnetic properties of octahedral complex ion [CrL 6 ]3+ are independent of identity of the ligand L. How do you account for this fact? PROBLEM 1474 Four structures are shown in the accompanying sketch. Indicate whether any of these structures are identical, geometrical isomers or optical isomers: en

Cl Co

Cl

3+

Co Cl

en

Cl

en 3+

Cl

en

3+

Co

en

Cl 3+

en

Co

Cl

en

en Cl

(a)

(b)

(c)

(d)

PROBLEM 1475 When ethylene diamine is added to a solution of cobalt chloride hexahydrate in concentrated HCl, a blue crystalline solid separate out. Analysis of this complex shows : N = 14%, C =12%, H = 5%, Cl = 53.25%. The magnetic moment of the solution was found to be 3.9 Bm. Determine structural formula and shape of this complex. PROBLEM 1476 Addition of TiCl 3 to an aqueous solution of urea followed by addition of KI gives a deep blue crystals of octahedral complex containing Ti, urea and iodine with this magnetic moment (spin-only) =1.76 Bm. Also 1.0 g of this complex on heating in excess of oxygen gives 0.12 g TiO 2 . Deduce structures of complex. PROBLEM 1477 (a) Explain why dissolution of a chromium (III) salt produces an acidic solution? (b) Explain why the slow addition of hydroxide ions to a solution of Cr(III) ions first produces a gelatinous precipitate that subsequently dissolves with further addition of hydroxide ion. PROBLEM 1478 Name the following complex ions: (c) [Pt(NH 3 )Cl 3 ]– (a) [Zr(OX ) 4 ]4– (b) [CuCl 4 (H 2O) 2 ]2–

(d) [Mo(CN) 4 (OH) 4 ]4–

PROBLEM 1479 By considering electronic configuration, suggest a reason why iron (III) compounds are readily prepared from iron (II), but conversions of Ni(II) and Co(II) to Ni(III) and Co(III) are much more difficult?

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PROBLEM 1480 Suggest a chemical test for distinguishing between: (a) [Ni(SO 4 )(en) 2 ]Cl 2 and [NiCl 2 (en) 2 ]SO 4 (c) [Ni(en) 2 I 2 ]Cl 2 and [NiCl 2 (en) 2 ]I 2 PROBLEM 1481 (a) Sketch the orbital energy-level diagrams for [MnCl 6 ]4– and [Mn(CN) 6 ]4– . (b) Which complex is expected to transmits the longer wavelengths of incident electromagnetic radiations? PROBLEM 1482 The complex [Co(CN) 6 ]3– is pale yellow. (a) Is short or long wavelength visible light absorbed? (b) How many unpaired electrons are available? (c) If ammonia molecules are substituted for cyanide ions as ligands, will the shift in absorbance of radiation be towards the blue or red region of electromagnetic radiation? PROBLEM 1483

Explain why high spin complex Mn 2+ are only faintly coloured.

PROBLEM 1484 The relative thermodynamic stability of two complexes can be predicted from a comparison of their standard potentials. Determine which complex of the following pair is the more stable? (i) (ii)

[Co(NH 3 ) 6 ]3+ + e → [Co(NH 3 ) 6 ]2+ [Co(H 2O) 6 ]

3+

+ e → [Co(H 2O) 6 ]

2+

E ° = 0.11 V E ° =1.81 V

PROBLEM 1485 (a) When excess of AgNO 3 is added to a 0.001mol Cr(III) chloride, 0.001 mol AgCl is formed. (b) When excess of AgNO 3 is added to a 0.001 mol Cr(III) chloride, 0.003 mol of AgCl is formed. Discuss the structure of complex based on above information. PROBLEM 1486 Identify the type of isomers represented by the following pairs : (a) [Cu(NH 3 ) 4 ][PtCl 4 ] and [Pt(NH 3 ) 4 ][CuCl 4 ] (b) [Cr(OH) 2 (NH 3 ) 4 ] Br and [CrBr(OH)(NH 3 ) 4 ]OH (c) [Co(NCS)(NH 3 ) 5 ]Cl 2 and [Co(SCN)(NH 3 ) 5 ]Cl 2 (d) [CrCl(H 2O) 5 ]Cl 2 ⋅ H 2O and [CrCl 2 (H 2O) 4 ]Cl ⋅ 2H 2O PROBLEM 1487 Following is a chelating ligand called nitrilotriacetic (NTA) acid. It act as a tetradentate ligand. Sketch the structures of [Co(NTA)(H 2O) 2 ]3+ ion and propose type of isomerism if present any : N(CH 2COOH) 3 : Nitrilotriacetic acid. PROBLEM 1488

The complex shown below is called diethylenetriamine (dien) : ••

••

••

H 2 N  CH 2CH 2  NH  CH 2CH 2  NH 2 (a) Sketch the structure of complex [Co(dien) 2 ]3+ (b) Which complex would be expected to be more stable in aqueous solution, [Co(dien) 2 ]3+ or [Co(NH 3 ) 6 ]3+ ?

PROBLEM 1489 Arrange the following complexes in order of increasing wavelength of visible light absorbed by them : (a) [Cr(H 2O) 6 ]3+ (b) [CrCl 6 ]3– (c) [Cr(en) 3 ]3+ (d) [Cr(CN) 6 ]3– (e) [Cr(NH 3 ) 6 ]3+

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PROBLEM 1490 Which complex should be expected to absorb light of highest frequency [Cr(H 2O) 6 ]3+ , [Cr(en) 3 ]3+ or [Cr(CN) 6 ]3– ? PROBLEM 1491 Which complex should absorb light at longer wavelength? (b) [Mn(CN) 6 ]3– or [Mn(CN) 6 ]4– (a) [Fe(OH 2 ) 6 ]2+ or [Fe(CN) 6 ]4– PROBLEM 1492 Sketch the structures of the following complexes indicating any cis-trans or optical isomerism: (b) [Pt(NH 3 )(Py)(Cl)(Br)] square planar, Py = Pyridine (a) [Pt(H 2O) 2 Br 2 ] square planar (c) [Zn(NH 3 ) 3 Cl]+ (d) [Ni(H 2O) 4 Cl 2 ] (e) [Co(C 2O 4 ) 2 Cl 2 ]3– PROBLEM 1493 Draw diagrams for any cis-trans and optical isomers that could exist for the following complexes: (c) [Cr(NH 3 ) 2 (H 2O) 2 Br 2 ]+ (b) [Co(en) 2 Cl 2 ]+ (a) [Co(en) 2 NO 2 (Cl)]+ 3+ (d) [Pt(NH 3 ) 2 Cl 4 ] (e) [Cr(en) 3 ] PROBLEM 1494 Determine the spin only magnetic moment: (c) [Mn(CN) 6 ]3– (b) [Co(en) 3 ]3+ (a) [CoF6 ]3–

(d) [MnCl 6 ]4–

PROBLEM 1495 If trimethylphosphene is added to solution of Ni(II) chloride in acetone, a blue compound of molar mass 280 is isolated. This compound on analysis gave the following composition : Ni = 21.5%, Cl = 26%, and (CH 3 ) 3 P = 52.5% This blue compound does not have any isomeric forms. What are the geometry and molecular formula of this blue complex? Also determine magnetic moment (spin-only) of the complex. PROBLEM 1496 The complex Na 2 [Ni(CN) 2 Br 2 ] has zero magnetic moment. Predict geometry and draw shape of the complex. PROBLEM 1497 Draw all geometrical isomers of the following complexes: (a) [Co(NH 3 ) 2 Cl 4 ]– (b) [Co(NH 3 ) 3 Cl 3 ] PROBLEM 1498 Draw structures of all the geometrical and optical isomers of each of the following complex: (c) [Co(C 2O 4 ) 3 ]3– (b) [Co(NH 3 ) 5 Cl]2+ (a) [Co(NH 3 ) 6 ]3+ PROBLEM 1499 The [Ni(CN) 4 ]2– ion is diamagnetic while [Ni(Cl) 4 ]2– ion is paramagnetic. Explain with the help of crystal field splitting diagram. PROBLEM 1500 Transition metal complexes containing cyanide (CN – ) ligands are often yellow in colour, whereas those containing H 2O ligands are often green or blue in colour. Explain. PROBLEM 1501 The absorption maximum for complex ion [Co(NH 3 ) 6 ]3+ occurs at 470 nm. Determine the crystal field splitting energy. PROBLEM 1502 For each of the following pair, choose the complex that absorbs light of longer wavelengths. (b) [FeF6 ]3– and [Fe(CN) 6 ]3– (a) [Co(NH 3 ) 6 ]3+ and [Co(H 2O) 6 ]3+ (c) [Cu(NH 3 ) 4 ]2+ and [CuCl 4 ]2–

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PROBLEM 1503 A solution made by dissolving 0.875 g of compound Co(NH 3 ) 4⋅Cl 3 in 25 g of water freezes at −0.56° C. Deduce structural formula of complex if cryoscopic constant of water is 1.86 K kg mol –1 . PROBLEM 1504 Oxalic acid H 2C 2O 4 is sometimes used to clean rust stains from bathtubs. Explain the chemistry underlying the cleaning action. PROBLEM 1505 When aqueous potassium cyanide is added to a solution of Cu(II) sulphate, a white precipitate is formed in the beginning which dissolves on adding excess of reagent. No precipitate is formed when H 2S( g ) is bubbled through this solution at the point. Explain. PROBLEM 1506 A concentrated aqueous copper (II) chloride solution is bright green in colour. When diluted with water, the solution becomes light blue. Explain. PROBLEM 1507 In dilute nitric acid solution, Fe 3+ forms a dark-red complex with thiocyanate ion (SCN – ) according to following reaction : [Fe(H 2O) 6 ]3+ + SCN –

H 2O + [Fe(H 2O) 5 NCS]2+

In one experiment,1.0 mL of 0.20 M Fe(NO 3 ) 3 was mixed with1.0 mL of a10 −3 M KSCN solution and 8 mL dilute nitric acid was added. Final concentration of [Fe(H 2O) 5 (SCN)]2+ was found to be 7.3 ×10 –5 M. Calculate formation constant of complex.

PROBLEM 1508 A student has prepared a cobalt complex that has one of the following three structures : [Co(NH 3 ) 6 ]Cl 3 , [Co(NH 3 ) 5 Cl]Cl 2 or [Co(NH 3 ) 4 Cl 2 ]Cl. Explain how the student would distinguish between these possibilities by an electrical conductance experiment. At the student’s disposal are three strong electrolytes—NaCl, MgCl 2 and FeCl 3 . PROBLEM 1509 Aqueous solution of CoCl 2 are generally either light pink or blue. Low concentrations favour the pink form while high concentration favours blue form. Adding HCl to a pink solution of CoCl 2 causes the solution to turn blue; the pink colour is restored by addition of HgCl 2 .Account for these observations. PROBLEM 1510 Suggest a method that would allow you to distinguish between cis-Pt(NH 3 ) 2 Cl 2 and trans-Pt(NH 3 ) 2 Cl 2 . PROBLEM 1511 You are given two solutions containing FeCl 2 and FeCl 3 at same concentrations. One solution is light yellow and the other is brown. Identify these solutions based on their colours.

REPRESENTATIVE ELEMENTS PROBLEM 1512 Write equations for the following process : (a) B2O 3 + Mg (b) The result of the addition of aqueous NaOH to a mixture of solid Al 2O 3 and Fe 2O 3 . (c) CO 2 + Na[Al(OH) 4 ]( aq ) →

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PROBLEM 1513 Explain how, during dimerization, each BH 3 molecule acts both a Lewis base and a Lewis acid. PROBLEM 1514 Suggest likely products for the following reactions: (a) BF3 + EtOH → (b) BCl 3 + PhNH 2 → (c) BF3 + KF → PROBLEM 1515 Comment on the following observations: (a) AlF3 is almost insoluble in anhydrous HF but dissolves if KF is present. Passage of BF3 through the resulting solution causes AlF3 to reprecipitate. (b) Borazine is similar to benzene in some ways but dissimilar in other ways. PROBLEM 1516 (a) Suggest why trimethylamine is pyramidal while trisilylamine is planar. (b) Suggest reasons why, at 300 K, CO 2 and SiO 2 are not isostructural. PROBLEM 1517 Write equations for : (a) the reaction of SiCl 4 with aqueous NaOH (b) the hydrolysis of SiH 3Cl, (c) the hydrolysis of SiF4 . PROBLEM 1518 What would you expect to form when: (a) Sn is heated with concentrated NaOH, (b) SO 2 is passed over PbO 2 , (c) SiH 2Cl 2 is hydrolysed, (d) 4ClCH 2SiCl 3 + 3Li[AlH 4 ] → PROBLEM 1519 Discuss the bonding in BeH 2 in its solid state. PROBLEM 1520 Write balanced equations for the following reactions: (a) NaH + H 2O → (b) NaN 3 + Heat → (c) KO 2 + H 2O → (d) NaF + BF3 → PROBLEM 1521 Write balanced equations for the following reactions: (a) Thermal decomposition of [NH 4 ]2 [BeF4 ] (b) The reaction between NaCl and BeCl 2 (c) The dissolution of BeF2 in water. PROBLEM 1522 (a) Suggest a likely structure for the dimer of BeCl 2 , present in the vapour phase. What hybridization scheme is appropriate for the Be centers? (b) Suggest structure of BeCl 2 in diethyl ether. PROBLEM 1523 How do anhydrous CaCl 2 and CaH 2 function as drying agents? PROBLEM 1524 Among NaF and NaCl, NaF has higher melting point instead of its lower molecular weight while reverse trend is there between CF4 and CCl 4 , justify. PROBLEM 1525

4 Be

7

and 6 C11 are positron emitter while 6 C14 is beta emitter, explain.

PROBLEM 1526 Determine nuclear binding energy of He.

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PROBLEM 1527 Lithium ion being smallest in its group, is expected to have highest ionic mobility and hence solution of its salts would be expected to have a higher conductivity than solution of cesium salts. Explain, why this is not so. PROBLEM 1528 Ionization energy suggest that Cs should be the most reactive metal but the standard reduction potential suggest that Li is most reactive. Reconcile these two observations. PROBLEM 1529 Discuss the colour of alkali metal solution in liquid ammonia. PROBLEM 1530 0.35 g of a metal ( A ) was dissolved in dilute HNO 3 . This solution gave a red colouration to a non-luminous Bunsen-burner flame and on evaporation gave 0.75 g of metal oxide ( B ). A also reacted with hydrogen forming C. On reacting 0.16 g of C with water, a gas D was evolved and a sparingly soluble compound E was formed, which gave a strongly alkaline reaction and required 200 mL 0.1 M HCl for neutralization. Identify A to E. PROBLEM 1531 Which is more heavily hydrated LiCl or MgCl 2 ? PROBLEM 1532 Why do Be salts seldom contain more than four molecules of water of crystallization? PROBLEM 1533 An element ( A ) reacts quitely with water librating a colourless, odourless gas and a solution of ( B ). Passing CO 2 gas through B yields a while precipitate (C ) which dissolves on passing excess CO 2 gas. Precipitate C effervesced with dilute HCl and gave a deep red colouration to a Bunsen burner flame. Heating C gave a caustic white compound ( D ) which when further heated with carbon at 1000°C, gave a solid ( E ) of immense commercial importance. Identify A to E. PROBLEM 1534 A white salt ( A ) evolves a colourless gas (B ) when treated with dilute HCl. Bubbling ( B ) through lime water, a precipitate (C ) formed initially which dissolved forming a clear solution on further passing the gas. A small sample of A was moistened with concentrated HCl, placed on a platinum wire and introduced into a Bunsen burner flame where it caused a green flame colouration. On strong heating, A decomposed giving a white solid ( D ) which turned red litmus blue. Identify A to D. PROBLEM 1535 SnCl 2 is a good reducing agent but PbCl 2 is stable, explain. PROBLEM 1536 AlCl 3 exist as dimer in benzene. While in aqueous solution it remains in the form of hydrated ions, explain. PROBLEM 1537 Discuss the reaction of B2 H 6 with NH 3 under following conditions: (a) Excess ammonia at low temperature. (b) Excess ammonia at high temperature. (c) NH 3 and B2 H 6 in 2 :1 ratio at high temperature. PROBLEM 1538 Complete the following inorganic reactions: (a) B3 N 3 H 6 + HCl → (b) B2 H 6 + H 2O → (c) B2 N 3 H 6 + H 2O → (d) B2 H 6 + EtOH → (e) B2 H 6 + LiH → (f) B2 H 6 + HCl → (g) B2 H 6 + Cl 2 → PROBLEM 1539 An yellowish-white, deliquescent substance ( A ) has its vapour density 133.5. A reacts quickly with water forming a solution B. A sample of B gave a curdy white precipitate (C ) on addition of dilute HNO 3 and AgNO 3 solution, but this readily dissolved on addition of dilute NH 4OH, though a

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white gelatinous precipitate ( D ) was formed in its place. D dissolves in a concentrated solution of NaOH, but passage of CO 2 gas through this solution results in reprecipitation of D. Also compound ( A ) dissolved unchanged in dry diethyl ether and when this solution was reacted with excess of LiH, a new compound E was formed. Identify A to E.

PROBLEM 1540 (a) How SiCl 4 is prepared from SiO 2 ? (b) How SiC is prepared from SiO 2 ? PROBLEM 1541 Complete the following reactions: (a) Al 4C 3 + H 2O → (b) CaC 2 + H 2O → 1100°C

(c) CaC 2 + N 2 →

O

2 (d) SiC + NaOH →

(e) Mg 2C 3 + H 2O →

PROBLEM 1542 Synthesize a linear chain silicones starting with SiCl 4 and ethylmagnesium chloride. PROBLEM 1543 How R 3SiCl acts to control the molecular weight in preparation of polysilicones. PROBLEM 1544 Draw the structures of: 12– (a) tri cyclo-dimethyl siloxane (b) Si 3O 6− 9 (c) Si 6O18

(d) Pyroxene (e) Amphiboles.

PROBLEM 1545 Complete the following reactions: (a) P4 + NaOH → (b) CaNCN + H 2O → (c) N 2 H 4 + I 2 → (d) N 2O + NaNH 2 → PROBLEM 1546 Discuss, how superphosphate and triple superphosphate fertilizers are manufactured? PROBLEM 1547 Complete the following reactions: (a) N 2O 5 + NaCl → (b) N 2O 5 + Na → (c) N 2O 4 + H 2O → (d) NO 2 + Cl 2 → (e) NO 2 + HCl → PROBLEM 1548 Explain why nitrogen exist as diatomic molecule while phosphorus exist as tetra atomic (P4 ) molecule. PROBLEM 1549 Why NF3 is stable while NCl 3 and NI 3 are explosive? PROBLEM 1550 Draw the shapes of P4O 6 and P4O10 . PROBLEM 1551 (a) How nitric acid is prepared by Ostwald’s method. (b) Give one example of each in which HNO 2 acts as an oxidising agent and reducing agent. PROBLEM 1552 Explain the following observations: (a) Liquid oxygen sticks to the pols of magnet but liquid nitrogen does not. (b) The N—O + ion has a shorter bond than does NO. PROBLEM 1553 Describe method how H 2O 2 is prepared by electrolysis of H 2SO 4 . Draw structure of H 2O 2 . Write balanced equations for the reaction of H 2O 2 with: (a) an acidified solution of KMnO 4 , (b) an acidified solution of K 2Cr 2O 7 , (c) aqueous HI, (d) an acidic solution of K 4 [Fe(CN) 6 ].

Problems

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PROBLEM 1554 Draw shapes of the following oxyacid of sulphur and indicate oxidation states of sulphur. (b) H 2S 2O 5 (c) H 2S 2O 4 (d) H 2S 2O 3 (a) H 2SO 3 (g) H 2S 2O 8 . (e) H 2S 2O 6 (f) H 2SO 5 PROBLEM 1555 How sulphuric acid is manufactured by contact process? PROBLEM 1556 Provide two method by which sodium thiosulphate is synthesized. Why thiosulphuric acid can’t be made by adding acid to aqueous solution of thiosulphate salt? PROBLEM 1557 Complete the following chemical reactions: (a) F2 + H 2O → (b) Cl 2 + CO → (c) Cl 2 + NH 3 (excess) → (d) Cl 2 + NaOH → (e) SiO 2 + HF → PROBMEL 1558 Complete the following reactions: (a) Cl 2O + NaOH → (b) HClO 4 + P4O10 → (c) ClO 2 + NaOH → (d) KBrO 3 + F2 + KOH → (e) Cl 2O 6 + NaOH → (f) Cl 2O 6 + HF → PROBLEM 1559 Why it is not possible to produce F2 by electrolysis of aqueous NaF, aqueous HF or anhydrous HF? PROBLEM 1560 (a) Draw structures of OF2 , Cl 2O, O 2 F2 and I 2O 5 . (b) Explain bond angle in OF2 and give a reason why it is different in Cl 2O. (c) Why are the OF bond in O 2 F2 longer than in OF2 while O—O bond in O 2 F2 is short compared with that in H 2O 2 . PROBLEM 1561 Complete the following reactions: (a) XeF2 + HCl → (b) (c) XeF4 + SF4 → (d) (e) XeF4 + H 2O → (f) (g) XeF6 + XeO 3 → (h)

XeF4 + KI → XeF2 + H 2O → XeF6 + H 2O → XeO 3 + XeOF4 →

PROBLEM 1562 Ferric bromide (FeBr 3 ) when dissolved in water, converted into Fe 2+ evolving Br 2 ( g ) while no chlorine gas is evolved when FeCl 3 is dissolved in water. Justify. PROBLEM 1563

When Cl 2 ( g ) is mixed with F2 ( g ), ClF3 ( g ) is produced not the FCl 3 ( g ). Explain.

PROBLEM 1564 Explain why nitrogen forms extra-ordinary stable N 2 at room temperature but phosphorus forms P2 only at very high temperature. PROBLEM 1565 Explain sulphur forms compounds such as SF4 , SF6 , etc while oxygen, of the same group forms only OF2 ? PROBLEM 1566 In laboratory, HF and HCl can be prepared by reacting the metal fluorides and chlorides with concentrated H 2SO 4 but HBr can’t be prepared by the same procedure. Explain. PROBLEM 1567 Discuss the acidic behaviour shown by boric acid in water and explain why it can’t be titrated satisfactorily by NaOH. Why does glycerol or other such syn vicinal diols/triols enhances its acidic properties?

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PROBLEM 1568 Explain why CCl 4 is unaffected by water while SiCl 4 is rapidly hydrolyzed? In CCl 4 unreactive towards superheated steam? PROBLEM 1569

Explain the difference between temporary and permanent hardness in water.

PROBLEM 1570

What is “Boiler-scale” and how it is formed?

PROBLEM 1571

Describe the role of Li 2CO 3 and BaSO 4 in medicinal chemistry.

PROBLEM 1572 Sodium hydroxide has lower melting point (322°C) than sodium chloride (801°C), electrolysis of NaCl is the primary source of Na-metal not NaOH. Explain. PROBLEM 1573 Complete the diagram below indicating the reagents and special reaction conditions needed to produce each substance, using MgCO 3 as the starting material. [E ]

[F ]

[D ]

[A]

[C ]

[B]

Mg(NO 3 ) 2 ← Mg(OH) 2 ← MgO ← MgCO 3 → MgCl 2 → Mg → Mg 3 N 2 [G]

MgHPO 4

PROBLEM 1574 Consider the following sequence of actions and observations (1) A small chunk of dry ice is added to an aqueous Ca(OH) 2 solution and a white precipitate forms initially (2). After a short time, precipitate redissolve. Write chemical equations to explain the observations. PROBLEM 1575 Although, both boric acid and phosphoric acid have three hydroxy groups, phosphoric acid is a weak tribasic acid while boric acid is a weak monobasic acid. PROBLEM 1576

Provide missing reagents below: [A]

[B]

[C ]

[D ]

Na 2 B4O 7 ⋅ 10H 2O → B(OH) 3 → B2O 3 → BCl 3 → B [E ]

→ BF3

PROBLEM 1577 A typical baking powder contain baking soda and alum NaAl(SO 4 ) 2 as its active ingredients. During process, baking powder undergoes a reaction that yields CO 2 . Write a plausible equation for this reaction. PROBLEM 1578 What is the principle that underlies the fact that thallium forms both +1and + 3 ions while aluminium forms only + 3 ion? PROBLEM 1579 Why can’t aluminium cookware be used in cooking strongly acidic food? PROBLEM 1580 In the use of aluminium sulphate in water treatment, the water to be treated is usually kept between pH 5 and pH 8. Why do you suppose this is the case? PROBLEM 1581 Complete the following reactions: (a) N 2 H 4 ( aq. ) + HCl → (b) NO( g ) + O 2 ( g ) → (c) NO 2 ( g ) + H 2O( l) → (d) NH 3 ( g ) + O 2 ( g ) → 200°C

(e) NH 4 NO 3 ( l) →

PROBLEM 1582 Write balanced chemical equation for the following reactions: (a) Action of Al over Fe 2O 3 in presence of electric spark. (b) Action of water over Na 2O 2 . (c) Oxidation of Pb 2+ (aq.) to PbO 2 in acidic medium, by ozone.

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Problems

PROBLEM 1583 Write an equation to represent: (a) The formation of oxygen by the action of water over KO 2 . (b) Oxidation of Cl – to ClO –3 by ozone in acidic solution. PROBLEM 1584

How BF3 is prepared from borax?

PROBLEM 1585 You have available with you elemental sulphur, chlorine gas, sodium metal and water. Using these materials, prepare (a) Na 2SO 3 (b) Na 2SO 4 (c) Na 2S 2O 3 . PROBLEM 1586 Why is it that NF3 has no donor properties, but PF3 forms many complexes with metals? Give one example of such complexes. PROBLEM 1587 Substance (A) is a gas with vapour density 8.5. On oxidation at high pressure and temperature with a platinum catalyst, it gave a colourless gas (B), which rapidly turned brown in air forming a gas (C). B and C were condensed together to give substance (D), which reacted with water forming a compound (E). E turned blue litmus paper red. On treatment of E with acidified solution of KI, gas B was evolved, but when E was treated with a solution of NH 4Cl, a stable, colourless gas (F) was evolved. F did not support combustion, but magnesium continued to burn in it. However, F reacted with calcium-carbide in an electric furnace, forming a solid (G), which was slowly hydrolyzed by water, forming a solution of substance A. Identify A to G. PROBLEM 1588 Explain why P—O bond length in POCl 3 is 1.45 Å whereas the sum of the single bond covalent radii of phosphorus and oxygen is 1.83 Å? PROBLEM 1589

Suggest a reason why PF5 is known but PH 5 is not.

PROBLEM 1590 HCl (g) can be prepared from NaCl and H 2SO 4 . HBr and HI can’t be made in a similar way from NaBr and NaI. Explain why this is so? PROBLEM 1591 Write Lewis structure for each of the followings: (a) HCP (b) PH +4 (c) P2 H 4 (d) PO 3– (e) PF5 4 PROBLEM 1592 Complete and balance each of the followings: (a) P4 + Al → (b) P4 + F2 → (c) P4 + Na → (d) P4 + Se → (e) P4 + Cl 2 → (f) P4 + O 2 → PROBLEM 1593

Oxygen forms double bond in O 2 , but sulphur forms single bond in S 8 . Explain.

PROBLEM 1594 Explain why at room temperature fluorine and chlorine are gases while bromine is liquid and iodine is solid at room temperature. PROBLEM 1595 (a) XeO 2 F2

Suggest shapes of the following species: (d) XeO 4– (c) XeF3+ (b) XeF + 6

(e) XeO 3 F2

PROBLEM 1596 Sodium hydroxide is hygroscopic–that is it absorbs moisture when exposed to the atmosphere. A student placed a pellet of NaOH on watch glass. A few days later, she noticed that the pellet was covered with a white solid. What is the identity of this solid? PROBLEM 1597 PROBLEM 1598

Discuss the reactivity of white P4 . Complete the following reactions: Red hot CuO

(a) NH 3 →

Heat (b) (NH 4 ) 2 Cr 2O 7 →

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PROBLEM 1599 Aqueous Cu(II) sulphate is blue coloured. When aqueous potassium fluoride is added to CuSO 4 solution, a green precipitate is formed. If aqueous KCl is added instead, a bright green solution is formed. Explain, what happens in each case. PROBLEM 1600

Write balanced chemical reaction:

Heat (a) H 3 PO 3 → (c) HI + HNO 2 →

(b) Li 4C + HCl → (d) H 2S + Cl 2 →

PROBLEM 1601 Arrange the following species in increasing order of indicated property: (a) PCl 5 , SiCl 4 , CCl 4 , BCl 3 : Ionic character (b) F – , C 4– , N 3– , O 2– : Ionic radii (c) F, Cl, Br, I: Ionisation energies (d) H 2O, SiO 2 , CO 2 : Acidic character

METALLURGY PROBLEM 1602 Discuss the principle involved in extraction of Mg from sea water. PROBLEM 1603 Discuss the method of extraction of Mg from its dolomite ore. PROBLEM 1604 Why extraction of magnesium is impractical by chemical reduction of its salt? PROBLEM 1605 Mg 2+ ion being smaller than Na + instead, Mg 2+ has very low ionic mobility and electrical conductivity compared to Na + in their aqueous solution. Explain. PROBLEM 1606 (a) How Epsom salt (MgSO 4 ⋅ 7H 2O) is obtained? What is its use? (b) Write a balanced chemical reaction of Mg metal with dilute nitric acid. PROBLEM 1607 How Na 2CO 3 is manufactured by Solvay process? Why this method fails in preparation of K 2CO 3 ? PROBLEM 1608 Discuss the principle of purification of bauxite Al 2O 3 ⋅ 2H 2O by the following processes : (a) Baeyer’s process (b) Hall’s process (c) Serpek’s process. PROBLEM 1609 How Al is extracted from bauxite ore? Describe thermite welding. PROBLEM 1610 Write the reaction of Al metal with dilute HCl, H 2SO 4 and NaOH. Why concentrated nitric acid render Al passive? PROBLEM 1611 Discuss the reactions involved in polymerization of ethene using Zeigler catalyst. PROBLEM 1612 Discuss the principle involved in extraction of iron from its oxide ore, by carbon reduction method. PROBLEM 1613 Discuss the principle of rusting of iron and methods of its protection.

Problems

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PROBLEM 1614 What happens when iron is treated with: (a) Steam (e) Dilute HNO 3 (b) Dilute HCl (f) Concentrated HNO 3 (g) Boiling sulphur. (c) Concentrated H 2SO 4 (d) Anhydrous Cl 2 ( g ) PROBLEM 1615 How ferrous ammonium sulphate hexahydrate (Mohr’s salt) is prepared in laboratory. What are the advantages of using Mohr’s salt as analytical standard of ferrous ion over ferrous sulphate? PROBLEM 1616 How anhydrous and hydrated ferric chloride is prepared in laboratory. What happens when anhydrous ferric chloride is (a) heated above 500°C (b) dissolved in water. PROBLEM 1617 How tin is extracted by carbon reduction method? PROBLEM 1618 What happens when Sn is reacted with: (a) Cl 2 ( g ) (b) dilute HNO 3 (c) conc. HNO 3 (d) conc. H 2SO 4

(e) NaOH

PROBLEM 1619 How copper is obtained from copper pyrite, by self-reduction technique? PROBLEM 1620 When copper is exposed to atmosphere for a longer time, a green mass deposits on its surface. Discuss the chemical reaction involved. PROBLEM 1621 Why copper is passive in dilute HCl or dilute H 2SO 4 . PROBLEM 1622 Discuss the action of pure oxygen gas over copper metal at high temperature. PROBLEM 1623 What happens when copper is reacted with: (Write complete balanced reaction) (a) dil HNO 3 , (b) warm HCl in presence of oxygen, (c) warm H 2SO 4 in presence of oxygen, (d) concentrated HNO 3 , (e) with aqueous Fe 2 (SO 4 ) 3 , PROBLEM 1624 How Pb is extracted from galena (PbS) by self-reduction method? PROBLEM 1625 Discuss the reactions involved in action of (i) air and (ii) pure oxygen on lead. PROBLEM 1626 Lead react with hot concentrated sulphuric acid in the begining but reaction ceases after some time. Explain. PROBLEM 1627 What happens when lead is reacted with: (a) dilute HNO 3 (b) concentrated HNO 3

(c) NaOH

(d) CH 3COOH

PROBLEM 1628 How silver metal is extracted from its sulphide ore? PROBLEM 1629 Discuss the role of AgBr in photography. How does it react with Na 2S 2O 3 solution? PROBLEM 1630 Silver ornament gets tarnished when exposed to atmosphere for a long time, why? PROBLEM 1631 Complete the following chemical reactions: Heat

(a) Ag + Concentrated HNO 3 →

(b) Ag + Concentrated H 2SO 4 →

(c) AgBr + NaCN( aq ) →

(d) AgCl + NH 4OH →

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PROBLEM 1632 How gold is obtained from its ore by cyanide process? PROBLEM 1633 How does Au react with aqua-regia? PROBLEM 1634 Both copper and silver forms monovalent ion like alkalimetal but alkali metals are highly reactive while Cu and Ag are almost inert. What are the reasons for this difference in reactivity? PROBLEM 1635 How KMnO 4 is manufactured from MnO 2 ? PROBLEM 1636 How K 2Cr 2O 7 is manufactured industrially from chrome-iron? PROBLEM 1637 Complete the following reactions: Heat

(a) K 2Cr 2O 7 → (b) K 2Cr 2O 7 + KOH → Heat

(c) K 2Cr 2O 7 + KCl + H 2SO 4 (concentrated) →

PROBLEM 1638 (a) KMnO 4 (b) KMnO 4 (c) KMnO 4 (d) KMnO 4

Write the balanced chemical reaction for the followings: + K 2C 2O 4 + H 2SO 4 → + O 3 + H 2SO 4 → + FeC 2O 4 + H 2SO 4 → + HCl →

PROBLEM 1639 Draw shape of FeCl 3 in (a) gaseous state (b) ether (c) water. PROBLEM 1640 FeCl 3 and FeBr 3 are well known, but FeI 3 has doubtful existence, explain. PROBLEM 1641 How potash alum is prepared? What is its major application?

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Solutions

MOLE CONCEPT 1. Empirical formula : = KAlS 2O 8

        

Al

K

S

O

Elements

10.5

15.1

24.8

49.6

Mass percentage

0.388

0.387

0.775

3.1

Mole ratio

1

1

2

8

Simple ratio

Empirical formula weight = 258 From weight loss information : 54.4 g anhydrous salt ≡ 45.6 g H 2O ⇒

258 g anhydrous salt ≡ 238.89 g = 13.27 mol H 2O

⇒

Empirical formula of hydrated salt = KAlS 2O 8 ⋅ 13H 2O 1.0mole of KClO 3 ≡ 3.0 mole of Zn

2.

3 × 5.104 3 × 5.104 × 65 5.104 mole KClO 3 ≡ mole of Zn = = 8.124 g Zn 122.5 122.5 122.5 3. Apply conservation of moles of silver before and after precipitate exchange reaction as : 2.052 − x 1.8 x = + 143.5 188 143.5 where, x is mass of AgBr in mixed precipitate. ⇒ x =1.064 x 1 1 Also, moles of CuBr 2 = moles of AgBr = × 2 2 188 x 1 (on substituting x) Mass of CuBr 2 = × × 223.5 = 0.6324 ⇒ 2 188 Mass % of CuBr 2 = 34.18 4. Moles of NaCl in sample = 0.01 = moles of AgCl from NaCl in precipitate. Total moles of AgCl precipitate = ⇒ ⇒

2 = 0.01393 143.5

Moles of AgCl from KCl = 0.00393 = moles of KCl Mass of KCl in sample = 0.00393 × 74.5 = 0.2928 g Mass % of KCl in the sample = 29.28

5. Apply conservation of moles of Ag before and after precipitate exchange reaction: 1.94 – x 2.4 x gives x = 1.483 g AgCl in mixed precipitate. + = 188 143.5 188 x Mass of NaCl = ⇒ × 58.5 = 0.6047 g 143.5 1 1.94 – x × 200 = 0.243 g Mass of CaBr 2 = × ⇒ 2 188 Mass % NaCl = 60.47, CaBr 2 = 24.3 ⇒

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6. Apply conservation of mass : If product contain x g SO 2 , mass of SO 3 = (22 − x ) g. x 2 (22 − x ) Computing total mass of S : + = 10 2 5 ⇒ x =12 ⇒ SO 2 = 12 g, SO 3 = 10 g 7. Mass of 1.0 L solution = 1025 g Mass of ethanol in 1.0 L solution = 8 × 46 = 368 g Mass of solvent in 1.0 L solution = 657 g 8 Molality ( m) = × 1000 = 12.17 ⇒ 657 8. 5.0 moles (300 g) of acetic acid is present in 1.0 kg solvent. ⇒ 1300 g solution contain 5.0 moles of solute. ⇒ 1300/1.12 = 1160.7 mL solution contain 5.0 moles of solute. 5 × 1000 = 4.3 ⇒ Molarity ( M ) = 1160.7 9. Let the combustion reaction is : C 8 H18 + nO 2 → xCO + (8 − x ) CO 2 + 9H 2O 1 2650 n = (25 − x ) ; moles of octane = = 23.246 2 114 Computing mass of product = 23.246 × 28 x + (8 – x ) × 44 × 23.246 + 9 × 23.246 × 18 = 11530 ⇒ x =1.125

⇒ n =11.9375

In case of complete combustion : C 8 H18 +

25 O 2 → 8CO 2 + 9H 2O 2

Efficiency = ( n/12.5) × 100 = 95.5% 10. Moles of BaSO 4 = ⇒

1.846 = moles of BaCl 2 ⋅ xH 2O = 7.923 × 10 –3 233 7.923 × 10 –3 × (208 +18x ) = 1.936

⇒

x≈2

11. Let the mixture contain x g CuSO 4 ⋅5H 2O. ⇒

5−x x × 159 + × 120 = 3 249 246

⇒

x = 3.72

⇒ Mass percentage of CuSO 4 ⋅ 5H 2O = 74.4 12. In the two sample mass ratio of Ca to SiO 2 will remain same since both are non-volatile. 2x Before heating, let x% CaCO 3 is present ⇒ % Ca is present. ⇒ (85 − x )% SiO 2 is present. 5 After heating, 51.5% SiO 2 and 41.15% CaO are present. 40 × 41.15 = 29.39% Ca is present. ⇒ 56

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Solutions

Now : mass of Ca/SiO 2 before heating = mass of Ca/SiO 2 after heating. 29.39 2x ⇒ ⇒ x = 49.97% = 5 (85 – x ) 51.5 13. In the reaction of ClO 2 with water, products HClO 3 , HCl as well as reactant ClO 2 contain one chlorine atom each, per formula unit. Hence, total moles of acid produced will be equal to moles of ClO 2 hydrolysed. Therefore, there is no need of balancing the reaction for computation of moles of ClO 2 . Moles of acid = moles of NaOH required = 0.18 = moles of ClO 2 10 Total moles of gas in the flask = = 0.406 0.082 × 300 Mole fraction of ClO 2 =

0.18 × 100 = 44.33 0.406

14. The balanced chemical reaction is : C 6 H 5CH 3 + 2KMnO 4 → C 6 H 5COOK + 2MnO 2 + KOH + H 2O 11.5 × 100 For preparation of 11.5 g C 6 H 5COOK, theoretical yield should be g 71 92 11.5 × 100 Mass of toluene needed = × = 9.31 g 160 71 15. Balanced chemical reaction is : 2MnI 2 +13F2 → 2MnF3 + 4IF5 2 × 309 g MnI 2 ≡ 13 × 38 g F2 13 × 38 12 g MnI 2 ≡ × 12 = 9.59 g 2 × 309

Q

Actual requirement of 100 = 12.78 g 75 0.16 100 Ca = × 40 × = 25.6 100 0.25 0.344 32 × 100 S= × = 41 233 0.115 0.155 14 × 100 N= × = 17.9 17 0.712 F2 = 9.59 ×

16. Mass % of Mass % of Mass % of ⇒ Mass % of C = 15.48 Now : Elements Mass % Mol ratio Simple ratio

Ca 25.6 0.64 1

S 41 1.28 2

N 17.9 1.28 2

C 15.48 1.29 2

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Empirical formula = CaC 2 N 2S 2 , empirical formula weight = 156 Hence, molecular formula = CaC 2 N 2S 2 17. Working in backward direction : In the last step moles of (AgBr + AgI) = moles of AgI 0.4881 – x x 0.5868 ⇒ x = 0.0933 g + = ⇒ 188 235 235 100 0.0933 Mass % of NaI = × 150 × = 29.77 0.2 235 Now subtracting mass of AgI from 1st and 2nd precipitate gives : Mass of (AgCl + AgBr) = 0.3187 g and mass of AgBr = 0.3948 g y 0.3187 – y 0.3948 Again + ⇒ y = 0.245 g = 143.5 188 188 100 0.245 Mass % of NaCl = × 58.5 × = 50 ⇒ 0.2 143.5 Mass % of NaBr = 20.23 18. Let us consider 100 g mixture of NaI and NaCl containing x g NaI.  x 100 − x  142 Mass of Na 2SO 4 =  × =100 ⇒ x = 30.25 + 58.5  2 150 14 3 × 5 = 4.17 g N 2 and × 5 = 0.83 g H 2 17 17 Therefore, initially 9.17 g N 2 and 5.83 g H 2 were present. Since N 2 is limiting reagent, maximum mass of NH 3 = 11.135 g 20. Balance the reaction in terms of ‘ n’ as : 19. 5.0 g NH 3 contains

( CH 2CH 2 — )n 4nXeF2 + —

→ —( CF2CF2 )— + 4nHF + 4nXeF4 n

XeF2 + 2F2 → XeF6 Formula weight of one unit of fluorinated polymer =100 ⇒ for 1.0 kg, n = 10 ⇒ 40 moles of XeF6 would be required and for this 80.00 moles of F2 would be consumed. ⇒ m (F2 ) = 80 × 38 = 3040 g = 3.04 kg 21. Weight loss is due to conversion of NaHCO 3 into Na 2CO 3 : 31 g weight is lost per mole of NaHCO 3 . 0.3 0.3 mol of NaHCO 3 moles of Na 2CO 3 . producing ⇒ 0.3 g wt loss from 31 62 Total moles of carbonate = 15 × 10 –3 3 = 0.01 620 ⇒ 42.4% Na 2CO 3 .

⇒ Moles of carbonate in original sample = 0.015 − Mass of Na 2CO 3 in original sample = 1.06

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Solutions

3 − x 3.76 x 2.45 × 58.5 = 1.0 g + = ⇒ x = 2.45 ⇒ m (NaCl) = 143.5 1435 . 188 188 23. If M is molar mass of (CH 3 ) x AlCl y 22.

m% NaCl = 50%

0.643 x × 16 = 0.222 M 0.643 y m (AgCl) = × 143.5 = 0.996 M 355 . x M = 15x + 27 + 35.5 y = 15x + 27 + = 32.75x + 27 2 m(CH 4 ) =

and dividing :

x = 2, Also y

0.643x × 16 = 0.222 ⇒ 32.75x + 27

⇒

x = 1.98 ≈ 2 ⇒

24. Element

C

H

N

Cl

O

Mass %

25

4.2

9.8

49.6

11.4

Mol %

2.08

4.2

0.7

1.39

0.7

Simple ratio

2.9

6

1

2

1

y=1

Empirical formula C 3 H 6 NCl 2O 25. Mass of AgCl = 0.09 × 143.5 = 12.915 gwhich is 95.77% of total ppt. ⇒ ⇒ ⇒

Total mass of precipitate = 13.485 g and mass of impurity = 0.57 g Mass of NaCl + KCl = 5.9 g 5.9 – x x + = 0.09 ⇒ 58.5 74.5

x = 2.94 g NaCl, 2.96 g KCl

m (Na 2O) = 1.558 g

⇒ m % (Na 2O) = 31.16

m (K 2O) = 1.867 g

⇒ m % (K 2O) = 37.34

26.

Cl 2 + 2KOH → KCl + KClO + H 2O

...(i)

3KClO → 2KCl + KClO 3

...(ii)

4KClO 3 → 3KClO 4 + KCl

...(iii)

Multiply Eq. (i) by 12 and Eq. (ii) by 4 and add Eqs. (i), (ii) and (iii) to obtain 12Cl 2 + 24KOH → 21KCl + 3KClO 4 +12H 2O 852

415.5

1000 g KClO 4 will be produced from 852 × 1000 = 2050.54 g = 2.05 kg 415.5 1

27. 1.598 g TiO 2 loses 0.16 g oxygen therefore 80 g TiO 2 will lose 8 g of oxygen ( 2 mole of oxygen atom). ⇒

Formula of new oxide = TiO

2–

1 2

= Ti 2O 3

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28. Since, the compound contain only one copper atom per formula unit, its formula can be assumed to be as: Cu(NH 3 ) x SO 4 ⋅ yH 2O : M = 159.5 + 17x + 18 y 0.195 Moles of compound present in 0.195 g = M 0.195 Moles of HCl required = …(i) ⇒ x = 30.63 × 0.1036 × 10 –3 M Also M g of compound contain 18y g of water 18 y 0.2 g of compound will contain …(ii) × 0.2 g of water = 0.015 M From Eqs (i) and (ii), 18 y 0.015 M = ⇒ 18 y = 4.6x × 0.2 × M 0.195x 30.63 × 0.1036 × 10 –3 Substituting in Eq. (ii) 18 y × 0.2 = 0.015 ⇒ 61.33x = 159.5 + 21.6x 159.5 + 17x + 18 y

and

x = 4,

y =1

Formula of salt = Cu(NH 3 ) 4⋅SO 4⋅H 2O Molarity of CuSO 4 solution =1.1 M

Percentage yield :

⇒ moles of Cu(NH 3 ) 4⋅SO 4⋅H 2O precipitate (theoretically) = 25 × 1.10 × 10 –3 = 27.50 × 10 –3 ⇒

Theoretical mass of solid = 27.5 × 10 –3 × 245.5 = 6.751g Percentage yield =

⇒

6.127 × 100 = 90.75% 6.751

29. Mass of water produced in the reaction = 0.09912 g 0.09912 g = 11 mg ⇒ mass of hydrogen = 9 0.3283 × 28.8 1 m mol of CO 2 = 2 m mol of NaOH = = 4.73 = m mol of carbon 2 ⇒ mass of carbon in starting compound = 4.73 × 12 = 56.76 mg 65.12 0.225 mass of nitrogen = 28 × × × 1000 = 22 mg 760 0.082 × 298 and Now:

mass of oxygen = 115.2 – (56.76 + 11 + 22) = 25.44 mg Elements C H N O

Mass mg. 56.76 11 22 25.44

m moles 4.73 11 1.57 1.59

Simple Ratio 3 7 1 1

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Empirical formula = C 3 H 7 NO Empirical formula wt. = 73 Molecular formula = C 6 H14 N 2O 2 ⇒ 30. In order to obtain maximum yield from a reaction, the reactants must be supplied in stoichiometric amount so that no reactant should be left unreacted. The balanced chemical reaction is, Pb(NO 3 ) 2 + 2KI → PbI 2 + 2KNO 3 Let x g of KI is taken x x moles of KI = ⇒ moles of Pb( NO 3 ) 2 present = ⇒ 2 × 166 166 5– x x x ⇒ ⇒ x = 2.5 g ⇒ mass of PbI 2 = × 460 = 3.464 g = 2 × 166 330 332 31. Let mixture contain x g XH 4 then (5.628 – x) g X 2 H 6 . x x and, moles of X 2 H 6 = moles of X H 4 = ⇒ M +4 2( M + 4) x …(i) ⇒ 5.628 – x = (2 M + 6) 2( M + 4) Mx 2 Mx …(ii) Also + =5 M + 4 2( M + 4) Solving Eqs. (i) and (ii) we get M = 27.8 32. Let one formula unit of Argyrodite contain ‘a’ mol of silver and b mole of Y . Therefore, moles of a  sulphide per formula unit of mineral must be  + 2b . 2  a Hence, Ag a Y b S  a Ag 2S + bYS + bH 2S  + bH 2 → 2 + 2 b     2

Also, ⇒ ⇒ Solving

moles of H 2 required =

PV = 8.87 × 10 –3 RT

moles of mineral present in its 10 g =

8.87 × 10 –3 b

 a   108a + bM +  2 + 2b 32   1127b = 124a + b( M + 64) 108a 11.88 Also =11.88 ⇒ a = bM bM 108 10 =

8.87 × 10 –3 b

Substituting ‘a’ in Eq. (i), 124 × 11.88bM 1127b = + b( M + 64) ⇒ 108 Empirical formula = Ag 8 YS 6

M = 72.6 g mol –1

…(i)

and

a =8 b

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Mass of uranium in the sample =

33.

1.48 × 238 = 0.894 g 394

Mass % of uranium in the sample = 89.4 Now UO 2 (NO 3 ) 2 + Na 2C 2O 4 + xH 2O → UO 2 (C 2O 4 ) xH 2O ↓ + 2NaNO 3 mmol 3.756 2.985 Here Na 2C 2O 4 is the limiting reagent, therefore, m mol of UO 2 (C 2O 4 ) ⋅xH 2O formed is 2.985. ⇒ ⇒ x=

M(UO 2 (C 2O 4 )) ⋅xH 2O =

1.23 × 1000 = 412 = 238 + 32 + 88 + 18 x 2.985

54 =3 18

35.5 × 0.861 = 0.231g 143.5 mass of iodine in 0.467 g compound = 0.254 g Iodine Chlorine ⇒ weight: 0.254 g 0.213 g mole: 0.002 0.006 simple ratio, 1 3 ⇒ Empirical formula = ICl 3 35. C n H 2n + 2 + O 2 → nCO 2 + H 2O 34. Mass of chlorine in 0.861 g AgCl =

CO 2 + 2NH 2 → H 2 N  CO  NH 2 + H 2O 1000 × 1000 mole = = 16666.67 60 236 × 1000 moles of CO 2 produced = × n = 16666.67 ⇒ n =12.5 14n + 2 Formula of hydrocarbon = C12 H 26 36. Weight of NaHCO 3 dissolved at 60° C = 164 ×

250 = 41 g 1000

Weight of NaHCO 3 dissolved at 0°C = 6.9 g Total weight of NaHCO 3 dissolved = 47.9 g Weight of NaHCO 3 recovered =150 g Total weight of NaHCO 3 present originally = 150 + 47.9 = 197.9 g 197.9 % purity = × 100 = 79.16 250 46 37. Mole of CuSO4 in solution = 159 + 18x 46x mass of H 2Oin solution = 54 + × 18 159 + 18x ⇒

moles of H 2O in solution = 3 +

46x 159 + 18x

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Solutions

⇒

0.05 =

46 159 + 18x 46x 46 +3+ 159 + 18x 159 + 18x

20 × 46 46 (1 + x ) + 3 = 159 + 18x 159 + 18x

⇒ ⇒

3=

46 (19 – x ) ⇒ x = 3.97 ≈ 4 159 + 18x

38. The roasting reaction is, 7 Fe 2 O3 + 2SO2 + O 2 → 2 × 88 = 176 2 2 × 56 + 48 = 160 2FeS

From the above reaction, 16 g weight is lost by 176 g FeS 4 g weight will be lost by

176 × 4 = 44 g FeS 16

⇒ ore contain 44% FeS by weight. 39. Absorbance of a solution is linear function of concentration. ⇒ ⇒

A1 C1 = A2 C 2

or

40 0.001 = ⇒ C 2 = 00007 . M 28 C2

mol. wt. of polymer =

40. Volume of one molecule = Molar mass =

3 = 42857 amu 0.0007 × 0.1

3 2 3 3 3 × 10 –18 cm 3 (10 –6 ) 2 × 3 × 10 –6 cm 3 = a h= 4 4 4 3 3 × 10 –18 × 6.023 × 10 23 × 1.2 = 939 × 10 3 amu 4

41. Volume of smallest cell = πr 2 l = π(60 × 10 –8 cm ) 2 (6000 × 10 –8 cm ) = 6.785 × 10 –17 cm 3 mass of one smallest cell = 7.6 × 10 –17 g ⇒ Molar mass of mother cell = 7.6 × 10 –17 × 24 × 60 × 6.023 × 10 23 = 6.6 × 10 10 amu 42. Mass of CaSO 4 = 0.5 g ⇒

⇒ ⇒

0.5 × 56 = 0.2058 g 136 mass of Mg 2 P2O 7 = 0.5 g 0.5 mass of MgO = × 2 × 40 = 0.18 g 222 0.2058 mass % of CaO on oven dried basis = × 100 = 21.8% 0.9437 mass of CaO =

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0.18 × 100 = 19.07% 0.9437 On air dried basis : CaO = 20.58% MgO = 18%, H 2O = 5.63% 43. Let the sample contain x g Mohr’s salt [FeSO 4 (NH 4 ) 2 SO 4⋅6H 2O] mass % of MgO on oven dried basis =

⇒ Solving Also

0.5 – x 0.75 x ×2+ = 392 132 233 0.23 x = 0.23 g ⇒ Mohr’s salt = × 100 = 46% , (NH 4 ) 2 SO 4 = 54% 0.50 x 0.2 moles of Fe in 0.2 g sample = × = 2.347 × 10 –4 392 0.5

2.347 × 10 –4 × 160 = 18.77 mg 2 44. Smallest volume of AgNO3 would be required when the entire mass is due to highest molecular weight constituent. Hence, for smallest volume, the whole mass should be of BaCl 2 ⋅ 2H 2O 0.3 m mol of BaCl 2⋅2H 2O = × 1000 = 1.229 m mol 244 m mol of AgNO 3 required = 2 × 1.229 = 2.458 2.458 Volume of AgNO 3 required = = 16.38 mL (smallest) 0.15 Largest volume of AgNO 3 would be required when entire mass is due to lowest molecular weight constituent, i.e., NaCl. 0.3 m mol of NaCl = × 1000 = 5.128 = m mol of AgNO 3 required. 58.5 5.128 Volume of AgNO 3 required = = 34.18 mL (largest). ⇒ 0.15 ⇒ mass of Fe 2O 3 obtained on ignition of 0.2 g sample =

ACID-BASE TITRATION Total mmol of AgCl from 20 mL solution =

45.

m moles of AgCl from HCl = 0.8

0.4305 × 1000 =3 143.5

⇒ m moles of AgCl from CaCl 2 = 2.2

⇒ 1.1 m mole of CaCl 2 was consumed for precipitation of oxalate from 20 mL solution. Hence,

total m mol of oxalic acid in 250 mL solution = m% of oxalic acid =

1.1 × 250 = 13.75 20

13.75 × 10 −3 × 90 × 100 = 82.5 1.5

46. The balanced disproportionation reaction is: 2P2O 3 + 6H 2O → PH 3 + 3H 3 PO 4

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Solutions

meq of NaOH added = 6 meq of H 2SO 4 required for back titration = 1.1 ⇒ 4.9 meq of NaOH neutralized 4.9 meq of H 3 PO 4 49 m mol H 3 PO 4 ⇒ Total meq of H 3 PO 4 produced = 49 = 3 2 49 m mol of P2O 3 present originally = × 3 3 ⇒ Mass % of P2O 3 =

98 × 10 –3 × 110 × 100 = 79.85 9 × 1.5

47. m mol of CO 2– 3 in 10 mL = 10 × 0.05 × 2 = 1 In presence of methyl orange 3.235 m mol H + is consumed in which 2 m mol H + would be – + required for CO 2– 3 , hence 1.235 m mol H for HCO 3 ion. ⇒

m % CaCO 3 =

10 × 10 −3 × 100 × 100 = 40 2.5

m % Ca(HCO 3 ) 2 =

6.175 × 10 –3 × 162 × 100 = 40 2.5

48. m mol of H + required to reach the phenolphthalein end point = 1.12 ⇒ 1.12 m mol Na 2CO 3 is present per 10 mL of solution. Now, V = 17 mL and it contain 1.12 m mol of NaHCO 3 produced in titration in association with original NaHCO 3 . Q 5 mL of this solution required 3.53 mL acid to reach the methyl orange end point. 3.53 × 17 = 12 mL acid = 1.92 m mol H + ∴ 17 mL would require = 5 ⇒ m mol of H + consumed for neutralization of original bicarbonate = 0.8 m (Na 2CO 3 )/ litre = 112 × 10 −3 × 106 = 11.872 g m (NaHCO 3 )/ litre = 80 × 10 –3 × 84 = 6.72 g 49. m mol of HCl taken = 7.5 m mol of HCl left unreacted = 5.244 m mol of HCl reacted = 2.256 ≡ 1.128 m mol K 2SO 3 = 0.178 g K 2SO 3 m% K 2SO 3 = 36.6% 50. m mol of NH 3 from glycine = 2.5 – 1.8342 = 0.6658 = m mol of nitrogen. mass of nitrogen = 9.3212 × 10 –3 g = 18.64% Theoretical mass% = 18.66 51. Chemical reactions involved in the entire process are : 7

2Ca 3 (PO 4 ) 2 + Mg → Ca 3 P2 + Mg (PO 3 ) 2 + 3CaO + 2 O 2 Ca 3 P2 + 6H 2O → 3Ca(OH) 2 + 2PH 3 2PH 3 + 4O 2 → P2O 5 + 3H 2O

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⇒ 2.4 g (0.1 mol) Mg will produce 0.2 mol PH 3 which would require 0.4 mol of oxygen. ⇒ Vol. of O 2 needed = 8.96 L, hence vol. of air needed = 42.66 L. at S. T. P. 52. Let the mixture contain x g of Li 2CO 3 , y g of NaHCO 3 . 30x 106 y + + 9.3 − ( x + y) = 7.37 74 168 44x 62 y + =1.93 74 168 From titration information : 2x 2 y 9.3 − ( x + y) + + × 2 = 0.18 74 168 106 32x 62 y − = 0.24 ⇒ 74 168 Solving, Eqs. (i) and (ii) gives x = 2.11g, y = 1.825 g and Na 2CO 3 = 5.365 g ⇒ NaHCO 3 = 19.62%, Na 2CO 3 = 57.7% 4 53. M (acid) = × 1000 = 185.2 8 × 0.27 × 10 ⇒ After heating

...(i)

...(ii)

Formula of acid = C n H 2n O 2 ⇒ M = 14n + 32 = 185.2 ⇒ n =11 55 5 Now 5g acid will produce × 11 = molCO 2 after complete combustion. 186 186 Total moles of NaOH available = 1.0 2 × 55 76 Moles of NaOH left unreacted = 1 − in 500 mL = 186 186 ⇒ Molarity of NaOH after precipitation of Na 2CO 3 = 0.817 Therefore, 0.817 × 10 = 0.5 ×V ⇒ V = 16.34 mL 54. Let 10 mL solution contain x m mol NaHCO 3 and y m mol Na 2CO 3 : ⇒

x + 2 y = 1.24

⇒

and

⇒

x = 1.5 – 1.26

m(NaHCO 3 ) = 0.24 × 50 × 10

–3

y = 0.5

× 84 = 1.008g = 20.16%

m(Na 2CO 3 ) = 0.5 × 50 × 10 –3 × 106 = 2.65 g = 53% m(NaCl) = 1.342g = 26.84% 2NaOH

55. C x H y O 2 + O 2 → xCO 2 → xNa 2CO 3 6.4 M

6.4 x M

6.4 x M

Let 10 mL solution contain a m mol NaOH and b m mol Na 2CO 3 : ⇒

a + b =14.5 a + 2b = 18

Also,

25 M = 32 100

⇒ ⇒

b = 3.5 M = 128

6.4x = 0.35 M

⇒ ⇒

x=

0.35 × 128 =7 6.4

...(i)

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Solutions

Also, 12x + y + 32 = 128 ⇒ y = 12 ⇒ Formula = C 7 H12O 2 M (NaOH) = 1.8 56. The balanced chemical reaction is : Co(NH 3 ) x Cl 3 + xHCl → xNH +4 + Co 3+ + ( x + 3) Cl – 1.58 165.5 + 17x

1.58x 165.5 + 17x

1.58 (x + 3) 165.5 + 17x

1.58 x 23.63 × 1.5 = 165.5 +17x 1000

⇒

⇒

x=6

m(AgCl) = 0.053 × 143.5 = 7.62 g 5x 1 8x 1 77x 1 × + × + × = 098 . ⇒ x = 45 g 100 53 100 50 100 40 meq of Na 2CO 3 = 0.6132 × 40 = 24.528 ⇒ m% Na 2CO 3 = 3.25

57. 58.

meq of NaOH = 23.046 × 40 = 922 ⇒ m% NaOH = 92.2 59. Let the sample contain x mol Na 2CO 3 and y mol NaHCO 3 . 2x + y = 18.1396 × 10 −3

⇒

m mol of NaOH combined with NaHCO 3 = 4.1 ⇒ Total moles of NaHCO 3 = 16.4 × 10 –3 = y ⇒ x = 0.8698 × 10 −3 16.4 × 10 –3 × 84 × 100 = 91.84 1.5 0.8698 × 10 −3 × 106 m % Na 2CO 3 = × 100 = 6.15 1.5 Total NaOH = 0.2 mol,NaOH left unreacted = 0.06 mol

⇒

m % NaHCO 3 =

60.

NaOH combined with CO 2 = 0.14 mol producing 0.07 mol of Na 2CO 3 . Let hydrocarbon is C n H 2n + 2 ⇒ M = 14n + 2 ⇒

n = 0.07 14n + 2

⇒

n = 7, Molecular formula = C 7 H 16

61. Formula = C n H 2n O 2 : M = 14n + 32 Let half-solution contain x m mol Na 2CO 3 and y m mol NaOH. ⇒

x + y = 71.72 2x + y = 123.44

⇒ ⇒

⇒

x = 51.72

Total Na 2CO 3 = 103.44 m mol 2n = 103.44 × 10 –3 ⇒ n = 6 14n + 32

Hence, acid is C 6 H 12O 2 m(NaOH) = 123.44 × 2 × 10 –3 × 40 = 9.87g 62. Moles of NaHCO 3 = 19 × 0.25 × 2 × 10 –3 = 9.5 × 10 −3 = m% NaHCO 3 = 31.92

...(i)

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10 –3 = 8.58 × 10 –3 2 ⇒ m% Na 2CO 3 = 36.38; m% NaCl = 31.7 63. Let N 1 be normality of HCl and N 2 be the normality of NaOH ...(i) ⇒ 20 N 1 = 96 N 2 2 Also, × 1000 + 8 N 2 × 10 = 50 N 1 50 ⇒ 4 +8N 2 =5N1 N 2 = 0.25, N 1 = 1.2 64. Let the mixture contain x m mol K 2C 2O 4 , y m mol KHC 2O 4 and z m mol of H 2C 2O 4 ⋅ 2H 2O. Then, ...(i) 2x + y = 20 ...(ii) y + 2 z = 20 …(iii) 166x + 128 y + 126 z = 2725 166 126 (20 – y) +128 y + ⇒ (20 – y) = 2725 2 2 Solving, y =10.83 ⇒ x = 4.585 Moles of Na 2CO 3 = (26.66 − 9.5) ×

z = 4.585 ⇒

m % KHC 2O 4 = 50.87 m % K 2C 2O 4 = 27.93 m % H 2C 2O 4 ⋅ 2H 2O = 21.2

65. Let sample contain x m mol NH 4 NO 3 and y m mol (NH 4 ) 3 PO 4 0077 . × 2 × 10000 = 2.56 601 ⇒ x = 7.32 and m % NH 4 NO 3 = 58.56 66. Let half solution contain a m mol Na 2CO 3 and b m mol NaOH ⇒

x + 3 y = 15

and

y=

⇒ a + b = 0.55 and 2a + b = 0.8 ⇒ Total m mol of Na 2CO 3 = 0.5 = m mol of CO 2 produced Formula of acid = C x H y O 4 ⇒ moles of CO 2 produced =

a = 0.258b = 0.3 10.38 × 10 –3 ⋅x M

Also, 0.168 g acid = 16.18 × 0.125 × 10 –3 equivalent ⇒ M = 166 ⇒ Also, ⇒

10.38 × 10 –3 x = 0.5 × 10 –3 ⇒ 166 12x + y + 64 = 166 y=6

⇒

Formula = C 8 H 6O 4 M (NaOH) = 0.016 M

67. Moles of Na 2CO 3 = 0.1 × 10 × 11.32 × 10 –3 = 11.32 × 10 –3

x=8

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Solutions

⇒

m % Na 2CO 3 =

m moles of AgCl =

11.32 × 10 –3 × 106 × 100 = 40 3

0.306 × 1000 = 2.13 from 10 mL stock solution. 143.5

⇒ Moles of NaCl in original sample = (21.3 − 11.32) × 10 −3 = 9.98 × 10 −3 ⇒ Mass% of NaCl = 19.46 Total NaOH consumed for 10 mL stock solution = 0.05 × 42.64 = 2.132 ⇒ NaHCO 3 in original sample = (21.32 – 11.32) × 10 –3 = 10 –2 mol. Mass % of NaHCO 3 = 10 –2 × 84 ×

⇒

100 = 28 3

68. m mol of Na 3 PO 4 = 24.4 × 0.1 × 5 = 12.2 ⇒ m % (Na 3 PO 4 ) = 12.2 × 10 –3 × 164 ×

100 = 50 4

Also, if the mixture contain y m mol Na 2 HPO 4 and z m mol NaH 2 PO 4 12.2 × 3 + 2 y + z = 23.572 × 0.5 × 5 = 58.93

Then ⇒ Also,

2 y + z = 22.33

...(i)

142 y + 120 z = 2000

...(ii)

y = 6.9 and z = 8.53

Solving;

m% Na 2 HPO 4 = 6.9 × 142 × 10 −3 ×

⇒

100 = 24.5 4

m% NaH 2 PO 4 = 25.5 69. Weight loss = 5 × 0.265 = 1.325 g Q 106 g weight is lost from 136 g of LiHCO 3 136 × 1.325 = 1.7 g ⇒ m% LiHCO 3 = 34 ⇒ 1.325 g weight will be lost by 106 m moles of H 2SO 4 used for neutralization of Li 2O and Na 2CO 3 = 4 – 0.579 = 3.421 ⇒ Total m mol of Li 2O + Na 2CO 3 = 34.21 Also,

moles of Li 2O = 12.5 × 10 –3

⇒

⇒ moles of Na 2CO 3 = 21.71 × 10 –3

m% (Na 2CO 3 ) = 46 ⇒ m% NaCl = 20

70. Let the mixture contain x g LiHCO 3 , y g NaHCO 3 On heating :

2NaHCO 3 → Na 2CO 3 + H 2O + CO 2 168

106

2LiHCO 3 → Li 2O + H 2O + 2CO 2 136

30

CaCO 3 → CaO + CO 2 100

56

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Mass of residue =

56 30x 106 y + + (5 − x − y) = 2.58 100 136 168

30   56  56 106  − −  x+  y = 0.22  100 136   100 168  ⇒ 0.34x – 0.07 y = 0.22 Also, m mol of NaOH reacted with bicarbonates = 5 – 1.53 = 3.47 Total m mol of LiHCO 3 + NaHCO 3 = 34.7 ⇒ y x + = 34.7 × 10 −3 ⇒ 68 84 Solving, Eqs. (i) and (ii) gives y =1.686g, x =1 ⇒ m% LiHCO 3 = 20, NaHCO 3 = 33.7

...(i)

...(ii)

71. Moles of NaHCO 3 in 1.0 g mixture = 2 × 10 −3 ⇒ m% NaHCO 3 = 16.8 Also moles of HCl consumed by 1 g mixture = 10.53 × 10 –3 Out of this 2 × 10 –3 mol HCl will be used up by NaHCO 3 ⇒ 8.53 × 10 –3 mol HCl will be used up by Na 2CO 3 and CaCO 3 ⇒ Moles of (Na 2CO 3 + CaCO 3 ) = 4.265 ×10 −3 /g of mixture. Also 5 g mixture loses 0.75 g wt. and mixture contain 0.84 g NaHCO 3 ⇒ 0.31 g weight is lost by NaHCO 3 and remaining 0.44 g by CaCO 3 Therefore, 5.0 g mixture contain 0.01 mol CaCO 3 , m% CaCO 3 = 20 Moles of Na 2CO 3 /g of mixture = (4.265 – 2) × 10 –3 = 2.265 × 10 –3 ⇒ m% of Na 2CO 3 = 24 ⇒ m% of NaCl = 39.2 Let the original sample contained ‘a’ m mol of CaCO 3 and b m mol of NaHCO 3 . Therefore, 72. CaCO 3 → CaO + CO 2 2NaHCO 3 → Na 2CO 3 + CO 2 + H 2O b m mol of CO 2 produced = a + 2 CO 2 + NaOH → NaHCO 3 a+

b 2

a+

b 2

NaHCO 3 + NaOH → Na 2CO 3 a+

x

b –x 2

Till phenolphthalein end point m mol of HCl = 5 = x After phenolphthalein end point m mol of NaHCO 3 present in solution = a + ⇒

a+

b = 15 2

b 2 …(i)

Solutions

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Also, from precipitation information: b 0.985 × 1000 = = 5 ⇒ a = 10, b = 10 2 197 ⇒ mass of CaCO 3 = 1g, mass of NaHCO 3 = 0.84 g m% : CaCO 3 = 50, NaHCO 3 = 42, Impurity = 8% 73. In presence of methyl orange, the whole NaOH and Na 2CO 3 are neutralized. ⇒ meq of HCl = 16 × 0.25 = 4 = meq of (NaOH + Na 2CO 3 ) = meq. of NaOH original ⇒ Total meq of NaOH in original 1.0 g sample = 4 × 5 = 20 20 × 40 × 100 mass % of NaOH (original) = ⇒ = 80 1000 Now, let us assume that in 20 mL, x m mol of NaOH has got converted to Na 2CO 3 m mol of NaOH = 4 – x ⇒ In 20 mL, x m mol of Na 2CO 3 = 2 In 2nd titration, HCl used in titration of NaOH + Na 2CO 3 = 5 × 0.1 – 9 × 0.2 = 3.2 x x ⇒ upto phenolphthalein end point, m mol of HCl required = 4 – x + = 4 – = 3.2 2 2 ⇒ ⇒

x =1.6 x 5x Total Na 2CO 3 formed = × 5 = =4 2 2

m mol of NaOH left unreacted = 20 – 4 × 2 = 12 8 × 40 4 × (106 + 18) ⇒ weight of 1.0 g of exposed sample = 1 – + = 1.176 g 1000 1000 4 × 106 × 100 = 36.05% ⇒ weight % of Na 2CO 3 in exposed sample = 1000 × 1.176 74. Molarity (M ) of NH 4SCN solution =

50 × 0.0452 22.98

⇒ m mol of ClCH 2COOH present in beverage = 50 × 0.0452 – 10.43 ×

50 × 0.0452 = 1.234 22.98

⇒ mass of ClCH 2COOH = 1.234 × 94.5 = 116.6 mg 75. Let 20 mL stock solution contain x m mol Na 2C 2O 4 and y m mol H 2C 2O 4 ⇒ and Therefore,

2x + 2 y = 23.34 × 0.04 × 5 = 4.668 2 y = 26.67 × 0.1 = 2.667 ⇒ x =1 and m% of Na 2C 2O 4 = m% of H 2C 2O 4⋅2H 2O =

5 × 10

–3

2

× 134

y =1.3335

× 100 = 33.5

1.3335 × 5 × 10 –3 × 126 × 100 = 42 2

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76. Let the sample contain x m mol (NH 4 ) 2 SO 4 and y m mol NH 4 NO 3 . 1 In 20 mL, m mol of NH +4 ion = (2x + y) 5 1 70 m mol of NaOH reacted with NH +4 = 5 – 9 × × = 3.5 14 30 1 (2x + y) = 3.5 ⇒ 5 466 100 Also x= × = 6.25 ⇒ From Eq. (i) y = 5 233 32 ⇒

mass % of ( NH 4 ) 2 SO 4 =

…(i)

6.25 × 10 –3 × 132 × 100 = 55 1.5

mass % of NH 4 NO 3 =

5 × 10 –3 × 80 × 100 = 26.67 1.5

77. The reaction involved is: [Zn(NH 3 ) 4 ]Cl 2 + 4NaOH → Na 2 [Zn(OH) 4 ] + 4NH 3 + 2NaCl 10 m mol of NaOH consumed = 15 – = 13.33 6 13.33 m mol of complex present = = 3.33 ⇒ 4 ⇒

mass % = 3.33 × 10 –3 × 204 × 100 = 68

Also total m mol of Cl – ions present in final solution = 3.33 × 2 + ⇒ mass of AgCl formed = 8.33 × 10 –3 × 143.5 = 1.195 g 78. From charge balance: x + y = 2z Also if M be empirical formula weight then 1.2 1 × × 1000x = 1.1 M 5 1.2 15 × × 1000 y = 4.95 M 50 Dividing Eq. (iii) by (ii): 3x = y Now, substituting x =1, y = 3, z = 2 Empirical formula: KH 3 (C 2O 4 ) 2 50 × 0.04 = 0.1 20 In 20 mL, meq. of H 2SO 4 left unreacted = 40 × 0.02 = 0.8 ⇒ meq. of Ca 3 (PO 4 ) in 20 mL = 20 × 0.1 – 0.8 = 1.2 ⇒ Total meq. of Ca 3 (PO 4 ) 2 = 1.2 × 5 = 6 6 × 10 –3 310 ⇒ mass % of Ca 3 (PO 4 ) 2 = × × 100 = 15.5% 2 6

79. Normality of original H 2SO 4 solution =

10 = 8.33 6 …(i) …(ii) …(iii)

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80. The chemical reaction involved is: 6NH 4ClO 4 +10Al → 5Al 2O 3 + 3N 2 + 6HCl + 9H 2O 3 1 mole of Al reacted = = 27 9 1 6 1 mole of NH 4ClO 4 present = × = 9 10 15 117.5 100 mass % of NH 4ClO 4 = ⇒ × = 78.33 15 10 1 Also moles of HCl produced = 15 moles of NaOH taken initially = 0.1 ⇒

moles of HCl required to neutralize left over NaOH = 0.1 – Vol. of HCl required =

1 = 0.033 15

33 = 66 mL 0.5

81. Volume of room = 150m 3 = 150 × 10 3 L 150 × 10 3 ⇒ Total mole of CO 2 in the room = × 0.01 = 60.9756 0.082 × 300 4KO 2 + 2H 2O → 3O 2 + 4KOH KOH + CO 2 → KHCO 3 ⇒ moles of KO 2 in original sample = 60.9756 Reaction of KO 2 with H 2SO 4 is: 2KO 2 + H 2SO 4 → K 2SO 4 + H 2O + 32 O 2

Reaction:

moles of KO 2 required to neutralize H 2SO 4 = 2 × 100 × 0.1 × 10 –3 = 0.02 mass of KO 2 required = 82. Equivalent weight of acid =

5000 × 0.02 = 1.64 g 60.9756

3.25 × 1000 = 63.35 ⇒ Molar mass =190 68.4 × 0.75

Formula of acid = C n H 2n –1 (COOH) 3

⇒ 14n – 1 + 135 = 190

⇒ n=4

Formula = C 7 H 10O 6

REDOX TITRATION 83. (a) 2.525 × 10 –3 mol (b) meq of VO 2+ = meq of MnO –4 consumed in last step = 0.86 × 0.02236 × 5 = 9.6148 × 10 –2 Total meq of permanganate taken against Fe 2+ and VO 2+ = 1.4087 ⇒ meq of KMnO 4 used up by Fe 2+ = 1.4087 – 9.6148 × 10 –2 = 1.3125 ⇒ Moles of Fe 2+ titrated with KMnO 4 = 1.3125 × 10 –3

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–2 (c) meq of Fe 2+ consumed with Cr 2O 2– = 2.428 7 = 2.525 – 9.6148 × 10 –3 ⇒ Moles of Fe 2+ consumed with Cr 2O 2– 7 = 2.428 × 10

(d) Mass of V = 9.6148 × 10 –2 × 51 × 10 –3 ⇒

9.6148 × 51 × 10 –5 × 100 = 0.25 2 2.428 × 10 –3 2.428 × 10 −3 moles of Cr = ⇒ m% of Cr = × 52 × 100 = 2.1 3 3×2 m% V =

423.3 × 10 –3 = 5.027 × 10 –4 842 in 20 mL = 1.508 × 10 –3

84. In 20.00 mL solution moles of U 3O 8 = ⇒

Moles of U 4+

meq of U 4+ in 20 mL = 3.016

⇒

meq of MnO –4 require for 20 mL = 0.024 × 27.23 × 5 = 3.2676 meq of Fe 2+ = 3.2676 − 3.016 = 0.2516

⇒ ⇒

Mass of Fe 2O 3 =

0.2516 × 160 × 10 –3 = 20.128 × 10 –3 g 2

⇒ Mass of Fe 2O 3 associated with 100 g U 3O 8 =

20.128 × 10 –3 423.3 × 10 –3

× 100 = 4.755

4.755 × 100 = 4.54 104.755 85. Let sample contain x m mol Pb 3O 4 and y m mol of PbO 2 ⇒

⇒

m% of Fe 2O 3 =

2x + 2 y =

⇒ Also, ⇒

2.7 × 2 × 1000 – 8 × 0.02 × 5 × 25 = 20 134 x + y =10

N (KMnO 4 ) in 2nd titration =

...(i) 5 × 4.48 = 0.4 5.6 × 10

(3x + y) × 2 = 0.4 × 10 × 10 ⇒ 3x + y = 20

… (ii)

Solving, Eqs. (i) and (ii) x = 5 = y ⇒ m(Pb 3O 4 ) = 3.425 g, m(PbO 2 ) = 1.195 g ⇒

m% Pb 3O 4 = 68.5 m% PbO 2 = 23.9 2x = 2 y + z

86. Also,

1.7225 1.7225 y+ z = 10 × 2 × 10 –3 M M 1.7225 1.7225 2y + z = 15 × 2 × 10 –3 M M

Solving Eqs. (i) and (ii) gives y = z ⇒ x =1.5 y Now, substituting y =1, gives Cu 3 (CO 3 ) 2 (OH) 2

...(i) ...(ii)

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Solutions

87. meq. of KMnO 4 = 0.1 × 22 × 5 = 11 = meq. of CaC 2O 4

⇒ m mol of CaCl 2 = 5.5

Mass of CaCl 2 per gram of mixture = 5.5 × 111 × 10 –3 = 0.61g Mass of NaCl per gram of mixture = 0.39 g 1000  1000  + 2 × 6.67 × 5 × 10 –3 × ⇒ – ∆T f = K f  3 × 5.5 × 5 × 10 –3 ×  = 2.92  95  95 ⇒

T f = – 2.92° C

88. Let half solution contain ‘ a’ m mol Na 2CO 3 and ‘b’ m mol NaOH ⇒ a + b =15 and 2a + b = 25 ⇒ Total m mol of CaC 2O 4 = 20 Also,

89.

⇒ a = 10 = m mol of Na 2CO 3

20 × 2 + x = 10 × 06 . × 10

⇒ x = 20 meq of Na 2C 2O 4 134 100 m% Na 2C 2O 4 = 20 × 10 –3 × × = 31.53% 2 4.25 50 × 0.25 M(KMnO 4 ) = = 0.055 45 × 5

⇒ Normality of KMnO 4 in alkaline medium = 0.167 N 25 × 0.1 V= = 15.00 mL ⇒ 0.167 90. I.

meq of KMnO 4 = 3.75 × 0005 . × 5 = 93.75 × 10 −3

⇒

Total meq. of C 2O 42– = 93.75 × 10 –3 × 5 = 0.46875

⇒

m mol of K 3 [Fe(C 2O 4 ) 3 ] ⋅ 3H 2O = 78.125 × 10 –3

II.

meq of MnO –4 = 17.5 × 0.005 × 5 = 0.4375

⇒ Total meq of Fe 2+ ion = 0.875 = m mol of Fe 2+ ⇒ m mol of Fe 2+ from FeCl 3 ⋅ 6H 2O = 0.875 – 78.125 × 10 –3 = 0.7968 ⇒ Mass % of FeCl 3 ⋅ 6H 2O =

0.2155 × 100 = 71.84 0.300

91. Let mixture contain x m mol H 2C 2O 4 ⋅ 2H 2O, y m mol Na 2C 2O 4 and z m mol NaHC 2O 4 ⇒

2 y + z = 40

...(i)

and

2x + z = 60

…(ii)

Also,

126x + 134 y + 112 z = 6100

Solving, Eqs (i), (ii) and (iii), gives z = 20, x = 20 and y =10 ⇒

m% of H 2C 2O 4 ⋅ 2H 2O = m% of Na 2C 2O 4 =

2.52 × 100 = 41.31 6.1

1.34 × 100 = 21.96 ⇒ m% of NaHCO 3 = 36.73 6.1

…(iii)

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92. meq of hypo = 4 = meq of I 2 ⇒ 2 m mol of I 2 is produced from unreacted ICl as ICl + KI → KCl + I 2 ⇒ m mol of ICl combined with oil = 0.5 Also, 0.127 g oil ≡ 0.5 m mol of ICl. ⇒ 100 g oil ≡

0.5 × 100 0.5 m mol of I 2 × 100 m mol of ICl or 0.127 0.127

⇒ Mass of I 2 required for 100 g oil =

100 × 0.5 × 254 × 10 –3 = 100 g 0.127

93. meq of KMnO 4 reacted with nitrite in 10 mL solution = 2 − 1 = 1 ⇒ Total meq of NO –2 = 100 = m mol of NaNO 3 ⇒

⇒

100 × 85 × 10 –3 × 100 = 56.67 15 m% of Mg(NO 3 ) 2 = 43.33 Total m moles of O 2 = 100 + 44 = 144 m mol of NO 2 = 44 × 4 = 176  NO 2  Molar ratio   = 1.22  O2 

m% of NaNO 3 =

94. meq of As 2O 3 taken = 75 × 0.0125 × 4 = 3.75 meq of MnO –4 used up in back titration = 0.1862 ⇒ meq of MnO 2 = 3.5638

⇒

m% of MnO 2 = 3.5638 ×

100 87 × 10 –3 × = 68.9 0.225 2

95. Molarity of original K 2Cr 2O 7 solution = 0.85 × 10 –3 M ⇒ m mol of K 2Cr 2O 7 initially = 2.25 × 10 –3 Tinitial = 41.5 ⇒ (Absorbance) initial = 58.5 T final = 43.5 ⇒ (Absorbance) final = 56.6 Q Absorbance ∝ [concentration] or amount. ∴ 56.6% absorbance correspond to 2.25 × 10 –3 × 56.6 = 2.177 × 10 –3 m mol K 2Cr 2O 7 58.5 ⇒ m mol of K 2Cr 2O 7 reacted with alcohol = 7.3 × 10 –5 meq of alcohol in air = 7.3 × 10 –5 × 6 = 4.38 × 10 –4 meq of alcohol in 56.5 mL of blood = 4.38 × 10 –4 × 2300 = 1.0074 ⇒ meq of alcohol in 100 mL blood = 1.783 46 m(C 2 H 5OH) = 1.783 × = 20.5 mg alcohol content is within the legal limit. 4

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Solutions

96. m moles of BaC 2O 4 lost in washing = 1000 1.5 × 10 –8 = 0.1224 m moles of CaC 2O 4 lost in washing = 1000 2.34 × 10 –9 = 0.0483 meq of MnO –4 used up for oxidising oxalate = 0.05 × 13.94 × 5 = 3.485 3.485 = 1.7425 2 ⇒ Total m moles of oxalate produced during precipitation = 1.7425 + 0.1224 + 0.0483 = 1.9132 Let mixture contain x g Ba(NO 3 ) 2 : 0.3657 – x x + = 1.9132 × 10 −3 ⇒ x = 0.1397 ⇒ m% Ba(NO 3 ) 2 = 38.2 ⇒ 261 164 97. The balanced redox reaction is: 2KO 2 + 6HI → 2KOH + 2H 2O + 3I 2 meq of hypo = 15 × 0.4 = 6 = meq of I 2 ⇒ Total I 2 liberated = 15 m mol ⇒ m mol of oxalate titrated =

⇒ m mol of KO 2 in original sample = 10

m% = 10 × 10 –3 × 71 × 100 = 71%

⇒

1

2Cu + → Cu 2+ + Cu, Cu 2+ + 2I – → CuI + 2 I 2

98.

8.3 × 1000 = 50 166 meq. of I 2 produced from excess KI = 10

m mol of KI taken initially = ⇒ m mol of KI reacted with Cu 2+ = 40

⇒

m mol of Cu 2+ = 20 = m mol of Cu 2O

m% Cu 2O = 20 × 142 × 10 −3 ×

100 = 94.67 3

99. 3Br 2 + 6OH – → 5Br – + BrO –3 + 3H 2O ⇒ ⇒ ⇒

10 3 10 meq of BrO −3 = × 6 = 20 = meq of CaC 2O 4 3

m mol of BrO –3 =

mass of CaC 2O 4 = 20 × 10 –3 × 64 = 1.28 g

⇒ 100. Let sample contain x m mol NaCl.

m% = 85.33

⇒ Initially x m mol of AgCl formed. 2AgCl → 2Ag + Cl 2

Now,

0.6 x

and

3Cl 2 → 5Cl 0.3 x

⇒

0.3 x

–

0.5 x

+ ClO –3

→ 5AgCl 0.5 x

Final AgCl = 0.4 x + 0.5x + 0.9x m mol

⇒

0.9 x × 143.5 × 10 –3 = 1.5

⇒

%NaCl = 11.614 × 10 −3 × 58.5 × 100 = 67.94

⇒ x = 11.614

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101. m mol of KI taken = 2 meq of hypo 50 × 10 –3 ⇒ meq of I – left unreacted = 50 × 10 –3 × 10 = 0.5 ⇒ m mol of I – used up with CuCO 3 = 1.5 Reaction of I – with Cu 2+ is : 2Cu 2+ + 5I – → 2CuI + I –3 ⇒ m mol of Cu 2+ =

2 5

× 1.5 = 0.6

⇒ mass % of CuCO 3 = 0.6 × 10 –3 × 123 ×

100 = 18.45 0.4

102. meq. of Fe 2+ required for excess MnO –4 = 19 × 0.1 × 10

⇒ m mol of excess MnO –4 = 3.8

m mol of MnO −4 reacted with CN − = 16.5 − 3.8 = 12.7 meq of MnO –4 = 12.7 × 3 = 38.1 = meq of CN − ⇒ ⇒

m mol of NaCN = 19.05 m % NaCN = 19.05 × 10 –3 × 49 × 100 = 93.345

103. meq of oxalate reacted with MnO 2 = 200 – 50.73 × 0.6 = 169.562 169.562 = 84.781 ⇒ m mol of MnO 2 produced = 2 ⇒ meq of KMnO 4 used for oxidation of toluene = 84.781 × 3 84.781 × 3 ⇒ m mol of toluene oxidised = = 42.39 6 Mass of toluene = 42.39 × 10 –3 × 92 = 3.8998 g ⇒

density (ρ) = 0.78 g / mL 100 meq of KMnO 4 left unreacted = 38 × 0.2 × = 72.38 10.5 ⇒ m mol of KMnO 4 left unreacted = 14.476 Hence, total m mol of KMnO 4 present in original solution = 99.257 99.257 ≈ 1.0 M ⇒ Molarity of original KMnO 4 solution = 100 104. Let mixture contain x m mol CaOCl 2 and y m mol NaOCl ⇒

2x + 2 y = 015 . × 2 × 10 × 10

⇒

x + y =15

0.287 × 10 × 10 = 20 143.5 Solving, Eqs. (i) and (ii) gives x = 5, y = 10 100 m% CaOCl 2 = 5 × 10 –3 × 127 × ⇒ = 31.75 2 100 m% NaOCl = 10 × 10 –3 × 74.5 × = 37.25 2 Also,

2x + y =

…(i)

3

…(ii)

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Solutions

GASEOUS STATE 8RT RT = 4 × 10 2 ⇒ = 2 × 10 4 ⇒ RT = 2πM × 10 4 πM πM 6 3 9 9 Total K.E. of He = × RT = RT = ⋅ 2π × 4 × 10 −3 × 10 4 = 180 π J 4 2 4 4 12 3 9 9 Total K.E. of Ne = × RT = RT = × 2π × 20 × 10 –3 × 10 4 = 360 π J 20 2 10 10 (360 +180) π Average K.E. per mol = = 807.84 J 1.5 + 0.6

105. Average velocity =

106.

− dP – dP ∝ ( P − 1) ⇒ = K ( P – 1) dt dt −dP = Kdt ⇒ P −1

⇒

Now, Now,

∫

P0

t P −1 − dp = Kt = K ∫ dt ⇒ ln 0 P −1 P −1 0

d  − mRt    = K ( P − 1) dt  VM 

Also, ⇒

P

K=

RT (– dm/ dt ) 0.0821 × 300 × 1.0 = = 7 × 10 –7 3 VM ( P –1) 7.28 × 10 × 28 (171.8 – 1) ln

171.8 − 1 = 7 × 10 −7 × 10 × 24 × 3600 P −1 ln

170.8 = 1.83 P –1

⇒ P = 94.33 atm.

nRT n2a 0.5 × 0.082 × 300 0.25 × 3.63 − = 22.06 atmosphere − 2 = V − nb V 0.5 − 0.5 × 0.04267 0.25 99b 108. V − b = RT = 100(0.011075V − b) = 1.1075V − 100b = 921b ⇒ V= 0.1075

107. P =

⇒ ⇒ 109. n (butane) =

920 b = RT ⇒ b = 24.33 cm 3 mol –1 = 4 ×

4 π r 3 × 6.023 × 10 23 3

r = 13.4 × 10 −9 cm = 134 pm 1 × 40 = 1.626 0.082 × 300

⇒

Mass of butane = 94.308 g

Mass of Ar = 94.308 × 19 = 1791.852 g n(Ar) = 44.8 Final P = 110.

46.426 = 28.55 bar 1.626

Cl 2O 7 → Cl 2 + 3.5O 2 1−α

α

3.5 α

…(i)

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Now, let x be the mole fraction of Cl 2 in the gaseous mixture being analyzed. rO2 71 0.6 ⇒ = 3.5 = ⇒ x = 0.115 rCl 2 32 x ⇒ Mole fraction of Cl 2O 7 = 0.285 rCl 2O7 1 − α Also, = rCl 2 α N  M  r  111.  1  =  1  2  r2  n = 1  N 2  0  M 1  112. (a)

r rO2

= 1.33 =

Vm =

32 M

n/2

71 0.285 = ⇒ α = 0.2 183 0.115

N  =  1  N2 n

⇒

25 10  22  =   75 90  20 

n/2

Solving n = 23

M = 18 amu.

18 = 50 L 0.36 1 × 50 PV = = 1.22, repulsive force is dominating. Z= RT 0.082 × 500

3 3 KT = × 1.38 × 10 −23 × 1000 = 2.07 × 10 −20 J 2 2 2 × 0.082 × 300 nRT 113. ( P + K ) (V − nb) = nRT ⇒ K = −P = − 11 = 1.615 bar 4 − 2 × 0.05 V − nb Pb Pb 114. ⇒ 0.5 = Z =1 + RT RT 0.5 RT ⇒ b= = 0.5 × 0.082 × 273 = 11.193 L mol –1 P a Also, ⇒ a = TB ⋅ Rb = 107 × 0.082 × 11.193 TB = Rb (b) KET =

a = 98.2 bar L2 mol –2

⇒ 115.

Also, Also,

He

N2

2 mol

6 mol

nHe 1 400 2 = = × nN 2 3 200 3

nHe + nN 2 = 8 ⇒ n He = P 1 = 300 × 2 200 × 16

24 16 , nN2 = 5 5

⇒ P = 1.066 atm

5

116. Applying concept of uniform pressure : nN 2 nO2 150 = = ⇒ nN 2 = 36.16, nO2 = 5.16 (1 − 0.6 3 ) (0.6 3 − 0.3 3 ) (0.3) 3

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Solutions

⇒ W = 300 + 36.16 × 28 + 5.16 × 32 = 1477.6 g After adding 50 g H 2 : 175 36.16 5.16 = 3 = 3 3 3 (1 − rB ) ( rB − rC ) rC Solving, rB = 57.6 cm and rC = 28.8 cm 117.

rmix 32 = = 0.45 ⇒ M = 158 rO2 M rPCl 3

Also, ⇒

rCl 2

PCl 5 1–α

rPCl 5 rCl 2 n2a V2

nPCl 3 71 = 0.72 = nCl 2 137.5

71x + 0.72x × 137.5 + (1 − 1.72x ) × 208.5 = 158

Now,

118.

=

= 002 .

=

1–α α

⇒

x = 0.268

PCl 3 + Cl 2 α

71 0.539 = 208.5 0.268

α

⇒ α = 0.225

⇒ a = 0.02 × (22.5) 2 = 10.125 atm L2 mol –2

V − nb = nRT ⇒ b = V − RT = 22.5 − 0.082 × 273 = 114 cm 3 mol –1 TB =

a 10.125 = = 1083.12 K Rb 0.082 × 0.114

119. 300 × 1 = n × 1.072 × R × 473 600 V = n × 1.375 × R × 273 ⇒ V = 0.37 L a 8a 120. Tc = = 282 and Pc = = 50 27Rb 27b 2 Dividing

282 8a 27b 2 282R = × ⇒ b= = 0.0578 L mol −1 a 50 27Rb 50 × 8

Substituting b in Tc : a =

282 × 27 × 0.082 × 0.0578 = 4.5 atm L2 mol –2 8

121. The second Virial coefficient = b 2 = 2 × 10 –2 ⇒ b = 0.1414 L mol −1 nRT 0.082 × 300 = = 4.92, V real = Vid + nb = 5.0614 L 5 P 4 122. b = 4.42 × 10 −2 L mol −1 = 44.2 cm 3 mol −1 = 4 × πr 3 × 6.023 × 10 23 3 Videal =

⇒ r = 16.36 × 10 −9 cm 123. Vc =

⇒ Distance of closest approach = 2r = 32.72 × 10 −9 cm = 327.2 pm

V 44 = 97.77 cm 3 mol –1 ⇒ b = c = 32.6 cm 3 mol –1 0.45 3

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TC =

8a = 300 ⇒ a = 2.7 atm L2 mol –2 27Rb

RT + B = 2.058 L. Also, for the given equation : P 10( − 0.1814) PB =1 + = 0.918 Z =1 + RT 0.082 × 273 a From van der Waal’s equation, for Z < 1, Z = 1 − VRT a = 0.082 ⇒ a = 3.77 bar L2 mol –2 ⇒ VRT rX 62 40 125. = =2 ⇒ M = 60 amu. rAr 38 M

124. V =

2

 100  nRT n2a 100 × 0.082 × 313 P= − 2 = −  ⋅ 3.6 = 11.16 bar 100  44 × 5  V − nb V 44 (5 − × 0.044) 44

126.

nRT 100 0.082 × 313 = × = 11.666 bar 44 5 V

Pid =

Percentage deviation =

Pid − Preal × 100 = 4.33% Pid

∂( PV ) =0 ∂P −B 2.879 × 10 −2 B + 2CP = 0 ⇒ P = = = 96 atm 2C 2 × 14.98 × 10 −5

127. At minimum in PV − P curve, ⇒

RT PV V α α  β 128. P = − ⇒ = − = 1 −  2 2 V − β TV RT V − β RT V  V  ⇒

−1

−

α RT 2V

β β2 α α  β2 PV 1 β = 1 + + 2 + ...− = + − 1 + ...  + RT V V V  RT 2V RT 2  V 2

Comparing with Virial equation : B =β −

α RT

2

and TB =

α Rβ

129. V. P. = 0.6 × 23.76 = 14.256 mm Moles of gas in 1.0 L dry air =

1 = 0.0409 0.082 × 298

Moles of N 2 = 0.0409 × 0.79 = 0.0323 ⇒ mass of N 2 = 0.9044 g Moles of O 2 = 0.0409 × 0.21 = 0.0086 ⇒ mass of O 2 = 0.2752 g

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Solutions

Moles of H 2O( v ) in 1.0 L moist air =

14.256 = 0.76 × 10 –3 760 × 0.082 × 298

⇒ Mass of H 2O( v ) = 0.0138 g Adding masses of all the gases gives = 1.1934 g / L  n2a  1  130. T=   P + 2  (V − nb)  = 623.57 K nR   V   P t − dP KP − dP K 131. = ⇒ ∫ = ∫ dt dt P M M 0 P 0

Kt 4K P  ln  0  = ⇒ ln 4 = = 2K P M 4 10 K ln = ⇒ PHe = 7.07 atm P 2 Kt Kt P  P  and ln  0  ln  0  = =  P  He  P  CH 2 4

⇒

Also,

4

 P0    =  P  He

⇒

2

 P0     P  CH

⇒ 4

10 100 = 2 7.07 PCH 4

⇒ PCH 4 = 8.4 atm 132. Pideal (V − b) = RT for 1.0 mol ⇒

Pideal =

0.082 × 273 = 438.9 0.075 − 0.024

⇒

Z=

600 = 1.367 (Repulsive force) 438.9

NH 3 ( g ) + HCl( g ) → NH 4Cl( s)

133. Moles

0.08 0

0.26 0.18

⇒

0 0.08

Q = 0.08 × 4300 = 3440 J 3440 = 955.55 Q = nC v ∆T ⇒ ∆T = 0.18 × 20

Also,

Pfinal

T final = 1255.55 K 0.18 × 0.082 × 1255.55 = = 10.3 atmosphere 1.8

134. Let P1 be the pressure in methane chamber and P2 be the pressure in He chamber at any instant t. Then : − dP1 K1 P1 K 2 P2 K1 P1 K 2 P2 = − = − dt 4 2 16 4 Also, at equilibrium, and

− dP =0 dt P1 = P2 = 3.7

⇒

K 1 = 2K 2

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− dP1 K1 K = ( P1 − P2 ) = 1 (2P1 − 7) dt 4 4

⇒ 4

∫

⇒

5

135. r.m.s. ( s) = 136. 2NH 3

− dP1 K = 1 (2P1 − 7) 4

ds 3RT ⇒ = M dT N 2 + 3H 2

1.6 – 2x

∫ dt

Solving,

t=

0

3R 3R 1 3R = 1558.87 ms –1 ⋅ = =2 ⇒ S = M 2 T 2 MS 4M

1.6 × 0.082 × 500 = = 32.8 2

Also,

32.8 1.6 = ⇒ x = 0.383 48.5 1.6 + 2x

n( NH 3 ) = 0.834, n( N 2 ) = 0.383, n( H 2 ) = 1.149 P(dry oxygen) = 729.9 mm ⇒ V (dry oxygen) =

137.

2 ln 3 = 87.88 second K1

3x = 1.6 + 2x

x

Pinitial

t

Q P1 + P2 = 7

729.9 × 30 = 29.15 mL. 751

PV = constant, V = 25.82 mL T 138. Let x mol gas is present in cold flask and y mol in hot flask. Now,

⇒

applying

2x = 3 y and

x + y =1

Solving:

3 2 x = , y= 5 5

3 0.082 × 200 × = 9.84 atmosphere. 5 1 139. Boyle’s law : PV = C (constant) P=

C P

⇒

V=

⇒

K =−

−C V  ∂V  = 2 =−    ∂P  n , T P P

⇒

1  ∂V  1 = = 1.0 atm –1   V  ∂P  n , T P

140. ∆m = mcold − m hot = [ ncold − n hot ] M =

PVM R

1 1 T − T  2  1

Solving : ∆T = T2 − T1 = 3 ∂( PV ) =0 ∂P ∂( PV ) P = +V ∂P ( ∂P / ∂V )

141. The minimum in curve will occur at Also,

…(i)

for 1.0 mole, van der Waals’ equation can be written as: P= ⇒

RT a − 2 V −b V

− RT 2a ∂P = + ∂V (V − b) 2 V 3

…(ii)

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Solutions

 ∂P  Substituting   in Eq. (i) and equating to zero gives  ∂V  ( bRT – a ) V 2 + 2abV – ab 2 = 0 Substituting a, b, R and T gives − 0.647V 2 + 0.087V − 1.38 × 10 −3 = 0

…(iii)

Solving, Eq. (iii) gives V = 0.0185 L mol –1

and 0.01158 L mol –1 .

The first solution yields a negative pressure on substitution into Eq. (ii), but usingV = 0.1158 gives P=

0.082 × 273 1.362 − = 165.56 atmosphere 0.1158 − 0.032 (0.1158) 2

142. Steam : 24.06 Benzene : 5.464 Toluene : 18.00 Neon : 0.2107 (Higher the intermolecular force of attraction, higher will be ‘a’) 143. Toluene : 0.1463 Steam : 0.0305 Benzene : 0.1154 Neon : 0.0170 (Larger the van der Waal’s radius, higher will be ‘b’)

THERMOCHEMISTRY 144. Heat consumed by water = 450 × 4.18 × 75 = 141.075 kJ Heat consumed by copper = 500 × 0.38 × 75 = 14.25 kJ Total heat required = 155.325 kJ 141.075 % heat used to raise temperature of H 2O = × 100 = 90.82 155.325 Fe 2O 3 + 2Al → Al 2O 3 + 2Fe;

145.

50 g

∆H = – 851.5 kJ / mol

25 g

Here, Fe 2O 3 is limiting reagent. Hence, 146.

– 851.5 × 50 = – 266.1 kJ 160 ∆H = Σ ∆H comb (reactants) − Σ ∆H comb (products) Q=

= − 1300 − 2 × 286 + 1560 = − 312 kJ 147. Writing the reactions given as follows : 2NH 4Cl( s) → 2NH 3 ( g ) + 2HCl( g ) 2NH 3 → N 2 + 3H 2 N 2 + 4H 2 + Cl 2 → 2NH 4Cl( s)

∆H = 352 kJ ∆H = 92.22 kJ ∆H = − 628.86 kJ

Now add the above equations : H 2 ( g ) + Cl 2 ( g ) → 2HCl( g ) ⇒

∆H °f HCl( g ) = – 92.32 kJ / mol.

∆H = − 184.64 kJ

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148. Writing the given reactions as follows : NO 2 + O 2 → NO + O 3 O 3 → 1 2

3 2

O2

O 2 → O( g )

∆H = + 200 kJ ∆H = −142.7 kJ ∆H = + 249 kJ

Adding above reactions gives : NO 2 → NO + O( g )

∆H = 306.3 kJ

149. Performing a + b – 2c gives : 4H 2 + 2O 2 → 4H 2O ⇒

H2 +

1 O 2 2

→ H 2O

∆H = −1090 ∆H = − 2725 . kJ …(i)

Also a + b + c gives : 3CH 4 + 2O 2 → H 2O + 3CO + 5H 2 5 2

5H 2 + O 2 → 5H 2O

and

∆H = − 349 kJ ∆H = − 1362.5 kJ

Adding the above two equations gives : 9

∆H = −1711.5 kJ

3

∆H = − 570.5 kJ

3CH 4 + 2 O 2 → 6H 2O + 3CO ⇒

CH 4 + 2 O 2 → 2H 2O + CO

150. N 2 ( g ) + 3H 2 ( g ) 151.

2NH 3

− 92 = 941.3 + 3 × 436 − 6 B. E. (N – H) ⇒ B. E. (N—H) = 390.25 kJ 306 = 314 + 3 × 121 − 3 B. E. (P – Cl) ⇒ B. E. (P—Cl) = 123.67 kJ

8 = 314 + 3 × 216.5 − 3 B. E. (P – H) ⇒ B. E. (P – H) = 318.5 kJ 152. Multiply the 2nd reaction by 2 and subtract from 1st : Fe( s) + 2H 2O( l) + 2SO 2 ( g ) → FeS 2 + 2H 2 + 3O 2 987 = ∆H °f ( FeS 2 ) + 2 × 285.83 + 2 × 296.81 ⇒ ∆H °f FeS 2 = − 178.28 kJ / mol Also, −137 = − 178.28 − 2H °f ( H 2S) ⇒ ∆H °f (H 2S) = − 20.64 kJ / mol 153.

+ 3H2

;

∆H = 987

∆H = − 119 × 3 = − 357

– 357 = – 156 – ∆H °f (benzene) ⇒ ∆H f ° (benzene) = 201 R. E. = ∆H °f (exp) – ∆H °f (theoretical) = – 152 kJ/ mol 154.

trans-2-butene → cis-2-butene

∆H = + 095 . kcal

cis-2-butene → 1-butene 1-butene + 6 O 2 → 4CO 2 + 4H 2O

∆H = +1771 . kcal ∆H = − 649.8 kcal

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Solutions

Adding above three equation gives: trans-2-butene + 6O 2 → 4CO 2 + 4H 2O CH 3 HC==CH + 6O 2 → 4CO 2 + 4H 2O( l) H 3C H 2O( l) → H 2O( v )

CH 3

CH==CH

CH 3

∆H = − 647.079 kcal

∆H °V = +11 kcal

+ 6O 2 → 4CO 2 + 4H 2O( v )

∆H = − 636.079 kcal

– 636.079 = 8 × 98 + 2 × 80 + B. E. (C==C) + 6 × 118 – 8 × 196 – 8 × 110 B. E. (C==C) = 159.921 kcal. B. E. (C== C) = x, B. E. (C  C) = y ⇒ x − 2 y = − 72

155. Let

6C(gr) + 3H 2 ( g ) →

…(i)

(g )

∆H = 79 = 6 × 715 + 6 × 218 − 3 ( x + y) − 6 × 415 − 152 ⇒ x + y = 959 From, Eqs. (i) and (ii) x = 615.33, y = 343.67 kJ. (g)

156.

…(ii)

(g) + 3H2(g)

∆H = 6 × 415 + 3 × 348 − 3 × 600 − 3 × 436 = 426 kJ C 6 H 6 ( g ) → C 6 H 6 ( l) C 6 H12 ( l) → C 6 H12 ( g ) C 6 H 6 ( l)Theo. → C 6 H 6 ( l) Exp. Adding the above equations yields : Cyclohexane (l) → C 6 H 6 ( l) + 3H 2 157. Let B. E. of C  C= x, C== C = y and C ≡≡C = z y − 2x = − 86 z − x − y = −148 y + 436 − x − 2 × 415 = − 132 ⇒ y − x = 262 From Eqs. (i) and (iii) x = 348 kJ, y = 610 kJ, Substituting in Eq. (ii) gives, z = 810 kJ 158. Heat evolved on combustion of one unit of CH 2 = – 158

∆H = − 30.8 kJ ∆H = 33.0 kJ ∆H = −152 kJ ∆H = 276.2 kJ

⇒

…(i) …(ii) …(iii)

⇒ ∆H comb (C10 H 22 ) = ∆H comb (CH 4 ) + 9 × ∆H comb (CH 2 ) = – 210 – 9 × 158 = − 1632 kcal

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3C 2 H 2 ( g ) → C 6 H 6 ( g )

159.

∆H = 3 × 930 − 3(615 + 348) = − 99 kJ

C 6 H 6 ( g ) → C 6 H 6 ( l) ⇒

∆H = 45 kJ

3C 2 H 2 ( g ) → C 6 H 6 ( l)

⇒

∆H = − 54 kJ = ∆H °f C 6 H 6 ( l) – 3 × 75

∆H °f C 6 H 6 ( l) = 171 kJ

⇒

R.E. = 49 − 171 = − 122 kJ / mol

160. Ca

2+

(g )

+

C 2– 2

(g )

∆H = 95 ∆H = 1733

C 2 (g )

U0

∆H = − 614

Ca( g )

2C( g )

∆H = 179

Ca( s)

+

∆H = 1436 ∆H = – 60 kJ

2 C(gr) → CaC 2 ( s)

179 + 1733 + 1436 − 614 + 95 + U 0 = − 60 ⇒ U 0 = – 2889 kJ 161. Required combustion reactions are : C 3 H 8 + 2O 2 → C 2 H 2 + CO + 3H 2O ∆H = − 2220 + 1300 + 285 = − 635 kJ 3 CH 4 + O 2 → CO + 2H 2O ∆H = − 890 + 285 = − 605 kJ 2 10 90 For 100 g L.P.G., Q = (– 635) + (– 605) = – 1676.98 kJ 16 44 162. Total O 2 consumed in 1.0 hour = 400 × 60 = 24000 mL = 24 L. n(O 2 ) =

1 × 24 = 0.944; C 6 H12O 6 + 6O 2 → 6CO 2 + 6H 2O 0.082 × 310

Energy produced = 45312 . kJ, available energy = 113.28 kJ, distance travelled = 1.1328 km. 164. From combustion data : ∆H °f (1, 3-butadiene) = − 4 × 394 − 4 × 285 + 2841 = + 125 ⇒ 165.

From bond enthalpies : 2 × 348 − 615 = 81 kJ = ∆H °f (1,3-butadiene)–130 ∆H °f (1,3-butadiene) = 211 R.E. = 125 – 211= − 86 kJ / mol 2NH 3 ( g ) → N 2 H 4 ( l) + H 2 ( g ) ∆H =142 kJ

N 2 H 4 ( l) → N 2 H 4 ( g ) Adding the above two equations yields : 2NH 3 ( g ) → N 2 H 4 ( g ) + H 2 ( g ) ⇒ ⇒

160 = 2 × 393 − B. E. (N–N) –436 B. E. (N – N) = 190 kJ/ mol

∆H =18 kJ ∆H =160

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Solutions

166. – 56 = – 285 – ∆H °f (OH – ) ⇒ ∆H °f (OH − ) = − 229 Also for

HF + OH – → F – + H 2O

∆H = − 320 − 285 + 229 + 329 = − 47 kJ 167. Writing the given reactions as follows : 1

2FeO( s) + 2 O 2 → Fe 2O 3 ( s)

∆H = − 284 kJ

2Fe( s) + 4H + → 2Fe 2+ + 2H 2

∆H = – 175.6

3

Fe 2O 3 ( s) → 2Fe( s) + 2 O 2

∆H = 8214 .

2H 2 + O 2 → 2H 2O

∆H = – 570

Adding the above equations yields : 2FeO( s) + 4H + → 2Fe 2+ + 2H 2O ⇒

∆H = − 2082 .

FeO( s) + 2H + → Fe 2+ + 2H 2O

∆H = − 104.1 kJ

168. Q = 1316.7 × 2.36 = 3107.412 J 3107.412 × 1000 J = – 51.79 kJ / mol 60 (C 2 H 5 ) 2 S( g ) + S( g ) → (C 2 H 5 ) 2 S 2 ( g ) ∆H = − 202 + 147 − 223 = − 278 kJ ∆H Neut = −

169. For

B. E. (S – S) = + 278 kJ / mol 170. Let B.E. C – C = x, C ≡ N = y and C − H = z ⇒

1

2C(gr) + 2H 2 + 2 N 2 → CH 3CN( g ) 1

∆H = 88 = 2 × 717 + 2 × 436 + 2 × 946 − ( x + y + 3 z ) …(i) x + y + 3 z = 2691 kJ 2C(gr) + 3H 2 → C 2 H 6 ∆H = − 84 = 2 × 717 + 3 × 436 − ( x + 6 z ) or x + 6 z = 2826 kJ ⇒ x = 2826 − 6 × 410 = 366 kJ / mol y = 2691 − 366 − 3 × 410 = 1095 kJ/ mol 171. CH 3COOH( g ) + 2O 2 → 2CO 2 + 2H 2O( g ) or

∆H = 3 × 410 + 348 + 728 + 352 + 463 + 2 × 498 − ( 4 × 728 + 4 × 463) = − 647 kJ ⇒ CH 3COOH( l) + 2O 2 → 2CO 2 + 2H 2O( l)

∆H = − 647 + 52 − 82 = − 677 kJ

⇒ ∆H reaction = − 677 + 860 = 183 kJ 172. For CH 3OCH 3 ( g ) : – 348 = – 2 × 94 – 3 × 68 – ∆H °f ⇒ ∆H °f = − 44 kcal/ mol ⇒ For 173.

C 2 H 5OH( l) → CH 3OCH 3 ( g )

∆H = − 44 + 66 = + 22 kcal / mol

BH 3 ( g ) → B( g ) + 3H( g )

∆H1°

B2 H 6 ( g ) → 2B( g ) + 6H( g )

∆H 2°

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∆H 1° = ∆H f° B( g ) + 3∆H f° H( g ) − ∆H f° BH 3 ( g ) = 563 + 3 × 218 − 100 =1117 kJ ∆H 2° = 2∆H f° B( g ) + 6∆H f° H( g ) – ∆H f° B2 H 6 ( g ) = 2 × 563 + 6 × 218 – 36 = 2398 kJ ⇒ mean B H bond energy in BH 3 =

∆H1° = 372.33 kJ/mol 3

mean B H bond energy in B2 H 6 ∆H 2° = 399.67 kJ/mol 6 Also, B2 H 6 has four terminal B H sigma covalent bond and two 3-centered 2-electron bond. =

⇒ 2398 = 4 × 372.33 + 2x ⇒ x = 454.34 kJ/mol. Therefore, average B H bond energy of the bridged B H  B bond = Also, BE ∝

454.34 = 227.17 kJ/mol 2

1 ; Hence, terminal B H bonds are shorter than bridged B H bond. Bond length

174. Given: ∆H ° = −120 kJ

+ H2

⇒

∆H ° = − 120 × 3 = − 360 kJ

+ 3H2

Using combustion data:

° [Benzene + 3H 2 – Cyclohexane] = − 215 kJ ∆H ° = ∆H combustion

Resonance energy = ∆H ° (from bond energy) – ∆H ° (combustion) = −145 kJ 7893 = 171.587 46 7025 moles of octane in 1.0 L = = 61.62 114 C 2 H 5OH + 72 O 2 → 2CO 2 + 3H 2O

moles of ethanol in 1.0 L =

175.

Also:

∆H = − (2 × 394 + 3 × 286) + 278 = − 1368 kJ mol –1 C 8 H18 + 25 O 2 → 8CO 2 + 9H 2O 2 ∆H = − (8 × 394 + 9 × 286) + 208.4 = − 5067.6 kJ mol –1 ⇒ Heat produced from 171.587 mol ethanol = − 171.587 × 1368 = − 234.731 × 10 3 kJ Heat produced from 61.62 mol octane = − 61.62 × 5067.6 = − 312.26 × 10 3 kJ

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Solutions

⇒

312.26 × 10 3

= 1.33 234.731 × 10 3 i.e., automobile running on gasoline will gas 1.33 times farther than the automobile running in ethanol, on combustion of 1.0 litre of fuel. factor =

176. The combustion reaction is: C 3 H 8 + 5O 2 → 3CO 2 + 4H 2O 10 50 mol of propane, mole of O 2 would be required. i.e., for 44 44 50 Therefore, vol of O 2 required = × 0.082 × 303 = 28.234 L. 44 100 Vol. of air required = × 28.234 = 134.45 L. ⇒ 21 10 Heat produced in the combustion reaction = × ( −394 × 3 − 4 × 286 + 104) = − 2222 kJ 44 Q = ms∆T

Since, ⇒

2222 × 10 3 = 8000 × 4.18 ∆T

⇒

∆T = 66.45 ⇒ T2 = 96.45° C.

177. Multiplying both Eqs (i) and (ii) by 2 and finally adding them yields

Adding:

N 2 ( g ) + 2O 2 ( g ) → 2NO 2 ( g )

∆G° =103.56 kJ

2NO 2 ( g ) → N 2O 4 ( g )

∆G° = − 5.77 kJ

N 2 ( g ) + 2O 2 ( g ) → N 2O 4 ( g )

…(iii)

∆G° = 97.79 kJ

178. The combustion reaction is: 1 CO( g ) + O 2 ( g ) → CO 2 ( g ) 2 1 ∆H ° (from bond enthalpies) = 743 + × 498 − 2 × 743 = − 494 kJ 2 ∆H ° (from formation enthalpies) = − 394 + 111 = − 283 kJ The above values of enthalpy indicates that in carbonmonoxide, the bond between carbon and oxygen is stronger than the same in CO 2 . 179. The formation reaction is: N 2 + 3H 2 2NH 3 For reaction of 0.1 mol N 2 , 0.3 mol H 2 is also reacted and 0.2 mol NH 3 formed. ⇒

BE (entered) = 0.1 × 944 + 0.3 × 436 = 130.8 kJ

Energy released on formation of NH 3 = 0.2 × (3 × 388) = 232.8 ⇒ Energy transferred to surrounding = 232.8 − 130.8 = 102 kJ.

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Heat produced on neutralization = 1316.7 × 2.36 = 3107.412 J

180.

moles of HCl neutralized = 0.15 × 0.4 = 0.06. 3107.412 ∆H (neutralization) = = 51.8 kJ/mol of HCl. 0.06

⇒

CuSO 4 ( s) + ( aq ) → CuSO 4 ( aq )

181. Given:

CuSO 4 ⋅ 5H 2O( s) + ( aq ) → CuSO 4 ( aq ) Subtracting the second equation from first equation yields: CuSO 4 ( s) + 5H 2O → CuSO 4 ⋅ 5H 2O( s) 182. (a) For the desired equation: (i)–(ii) CaCl 2 + 2H 2O → CaCl 2 ⋅ 2H 2O 183. Carrying out the following operations and rewriting the equations: (i) × 2 and reversing (ii) reversing (iii) as such (iv) × 2 2NaCl( s) → 2Na( s) + Cl 2 ( g ) H 2SO 4 ( l) → H 2 ( g ) + S( s) + 2O 2 ( g ) 2Na( s) + S( s) + 2O 2 ( g ) → Na 2SO 4 ( s) H 2 ( g ) + Cl 2 ( g ) → 2HCl( g ) Adding the above equations gives the desired equation: 2NaCl( s) + H 2SO 4 ( l) → Na 2SO 4 ( s) + 2HCl( g ) 184. The required reaction is: Ag( s) + 12 Cl 2 ( g ) → AgCl( s)

∆H ° = − 66.04 kJ ∆H ° = −11.495 kJ ∆H ° = − 54.545 J ∆H ° = 37.628

∆H ° = 821.2 kJ ∆H ° = 810.54 kJ ∆H ° = −1381.5 kJ ∆H ° = −184.42 kJ ∆H ° = 65.82 kJ ∆H ° = ?

(i) + (ii) + 2 × (iii) – (iv) gives 2Ag + Cl 2 → 2AgCl ∆H ° = − 324.4 − 30.56 − 2 × 92.21 + 394 = − 145.38 kJ 1 Ag( s) + 2 Cl 2 ( g ) → AgCl( s) ∆H ° = – 72.69 kJ.

⇒

185. Adding the theory of covalent bonding, the structures of N 6 ( g ) can be predicted as: N N N

N

N

N

N

N

N

N

I

II

N N

Here I has delocalized π-electrons as well as aromaticity. N N

N 3N2 N

N N

∆H° = 1072 kJ mol–1

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Solutions

∆H ° (from bond enthalpies) = 3 × 944 − 3 (163 + 409) = 1116 kJ mol –1 . The observed ∆H °f is 44 kJ mol –1 less than that calculated from bond enthalpies indicating presence of resonance. Hence, I structure is most probable. 186. If all energy available for muscular activity is consumed in jogging, energy available from 0.5 kg fat for jogging = 500 × 38 × 0.7 = 13300 kJ 13300 minimum jogging hour = = 6.65 hr. 2000 3.24 187. Total energy absorbed = 42.6 × = 3 kJ 46 3 1 Energy absorbed in one second = = kJ 60 × 10 200 1 m 2 = 50 cm 2 . 200 188. Adding the two given thermochemical equations (i) and (ii) yields Surface area of beaker =

+ 3H2

∆H° = –208 kJ

…(a)

Had there been no resonance energy in cyclohexadiene, the enthalpy of (ii) reaction would have been − 170 − 70 = − 240 kJ. This indicates that independent hydrogenation of one double bond gives off on an average 120 kJ of heat. Therefore, theoretically. + H2

∆H°Theoretical = –120 kJ

⇒

° − ∆HTheo ° = 82 = R. E. (benzene) − R. E. (cyclohexadiene) ∆H exp

⇒

R. E. (benzene) = 152 kJ mol –1 .

THERMODYNAMICS 189. − W = nRT ln

V2 − nb 0.5 – 0.5 × 0.02 = 0.5 × 8.314 × 300 ln = – 1747.8 J V1 − nb 2 – 0.5 × 0.02 3

190. ∆H = ∆E + P∆V + V∆P : ∆E = nCV ∆T = 2 × 8.314 × 100 = 1247.1 J P1 = 4.92 atm and P2 = 4.1 atm ⇒ P1 ∆V = 4.92 × 3 = 14.76 L atm and V2 ∆P = 8(4.1 − 4.92) = − 6.56 L atm ⇒ P1 ∆V + V2 ∆P = 8.2 L atm = 831.48 J ⇒

∆H =1247.1 + 831.48 = 2078.58 J

⇒

W =1747.8 J

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191. ∆H = ∆E + P1 ∆V + V2 ∆P = 2 × 8.314 × 500 + [4.92 × 3 + 8 (8.2 − 4.92)] × 101.4 = 6.236 × 10 3 J 192.

T V  100  20.8 ln 2 + 8.314 ln 2   T1 28  V1  For irreversible adiabatic expansion :  C + ( P2 / P1 )  T2 =  V  T1 = 217.76 K CP   ∆S =

…(i)

V2 P1T2 30 × 217.76 = = = 2.1776 V1 P2T1 10 × 300

Also,

Substituting in Eq. (i) gives ∆S = − 0.692 JK –1 193. ∆S = nCV ln

V T2 373   + 8.314 ln 2 = 0.17 JK –1 + nR ln 2 = 002 . 12.6 ln   300 V1 T1

194. Applying conservation of heat to : ∆S

∆S

1

2

∆S

1 2 3 H 2O( s) → H 2O( l) → H 2O( l) ← H 2O( l) q q q

0° C

0° C

3

TK

363 K

20 10 [6000 + 75.42 (T − 273)] = × 75.42 (363 − T ) 18 18 ⇒

T = 306.48 K

Also,

∆S 1 =

10 6000 × = 12.21 JK −1 18 273

∆S 2 =

306.48 10 × 75.42 ln = 4.85 JK −1 273 18

∆S 3 =

20 306 × 75.42 ln = − 14.31 JK −1 18 363

∆S system = ∆S 1 + ∆S 2 + ∆S 3 = 2.75 JK −1

⇒

195. Process involved are : H 2O( l) → H 2O( l) → H 2O( s) → H 2O( s) ∆S – 10°C

∆ S1

∆S 1 = C P ln ∆S 2 =

0° C

2

0° C

∆S 3

– 10°C

T2 273 = 2.814 JK –1 = 75.42 ln 263 T1

− 6000 263 = − 21.98 JK –1 and ∆S 3 = 37.2 ln = – 1.388 JK –1 273 273 ∆S = ∆S 1 + ∆S 2 + ∆S 3 = – 20.554 JK –1

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Solutions

Zn + H 2SO 4 → ZnSO 4 + H 2

196.

100 × 8.314 × 300 = 3.84 kJ 65 ∆V = 0 ⇒ − W = 0

− Wirr = ∆ng RT = In sealed vessel,

197. P ∝ d ⇒ d (final) = 2.5 m. Also P = kd ⇒ k = 2 atm m –1 − dWrev = P dV = k . d. ⇒ −W =

kπ 2

2.5

∫

d 3 dd =

0.5

π 2 kπ 3 d dd = d dd 2 2

2 atm m –1 × π kπ × 39 m 4 [(2.5) 4 − (0.5) 4 ] = 8 8 = 30.63 atm m 3 = 3.1 × 10 6 J

198. A : P = 10 bar, T = 300 K, B: P = ? T = 300K 3 C : P = 2 bar, T = 250 K, CV = R 2 Connecting reversible adiabatic points B and C : γ 1−γ

T PB = PC  C   TB  −W

⇒

AB

− WBC − WAC

 250  =2   300 

A

B

T –5 2

= 3.15 bar

C V

10 = 2.88 kJ 3.15 = CV (T1 − T2 ) = 1.5 × 8.314 × 50 = 623.55 J = 3.50355 kJ

= 8.314 × 300 ln

199. Process is irreversible : (a) (b)

4  2 × R 2R  − W = Pext ∆V = 2 atm  −  × 273 = 2 × × 8.314 × 273 = 3631.55 J  2 5 10  ∆H = ∆E = 0

∴ ∆T = 0

200. In reversible adiabatic process : TV γ – 1 = constant ⇒

V  T1 = T2  2   V1 

γ −1

= T2 (2) 2/ 5 ⇒ T2 = 225.84 K

5

∆E m = 2 × 8.314 × 72.16 = 1499.84 J 7

∆H m = 2 × 8.314 × 72.16 = 2099.78 J V −b RT dV ⇒ − W = RT ln 2 V −b V1 − b 0.2 − 0.03 = − 10.155 kJ − W = 8.314 × 300 ln 10 − 0.03 W = + 10.155 kJ

− dW =

201.

⇒

338

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202.

a   RT − dW =  − 2  dV  V V  ⇒

− W = RT ln

 1 V2 0.384 × 0.15 1 = – 1095 J + a  −  = − 1152.56 + V1  V2 V1  10 –3

⇒

W = + 1095 J

203. Work will be done only between A – B ⇒

−WAB = P∆V = 0.8 × 20 L atm = 1622.4 J 0.8 × 20 Initial temperature TA = = 390.25 K 0.5 × 0.082

1

TB = 2T A = 780.5 K TC = 975.625 K ⇒ ∆E AB = 0.5 × 1.5 × 8.314 × 390.25 = 2433.4 J ∆E BC = 0.5 × 1.5 × 8.314 × 195.125 = 1216.7 J ⇒ ∆E AC = ∆E AB + ∆E BC = 3650.1 J q = ∆E − WAB = 4461.3 J and ∆H = ∆E + P ∆V = q 204. At constant n and V , P ∝ T ⇒ T2 = 3030 K ⇒ 205. − WAB

20 L

q = ∆E = CV ∆T = (75.42 − 8.314) × 2727 = 183 kJ ∆H = C P ∆T = 205.67 kJ A(1.0 atm, 20 L) = P∆V = 20 L atm = 2028 J

WBC = 0 − WCA = 2 × 8.314 ln

0.5 = − 11.52 1

Since process in cyclic, ∆U and ∆H = 0 and q = − W = 2016.48 J 206. In adiabatic irreversible expansion : 0 = CV (T2 − T1 ) + Pext (V2 − V1 )  C + R ( Pext / P1 )  Solving for T2 : T2 =  V  T1 = 263.74 K  CV + R ( Pext / P2 ) 

CV (300) = 21.52 + 8.2 × 10 −3 × 300 = 23.98 JK −1 C P (300 K) = CV + R = 32.294 JK −1

V

40 L

B(1.0 atm, 40 L)

C(0.5 atm, 40 L)

V

− W = nCV (T1 − T2 ) = 3.22 × 20.785 = 2293 J 10 × 1 n= = 0.4 0.082 × 300 ∆U = ∆H = 0 − W = 0.4 × 8.314 ln 10 = 7.66 J

⇒

B

P

− WA ↔ A = 2016.48 J

207. Process is isothermal

C

p 0.8 A

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Solutions

208. dG = VdP − SdT at constant P , dG = − SdT ⇒

∫ dG = − 36.36 ∫ dT − 20.79 ∫ ln TdT T

∆G = – 36.36 (T2 – T1 ) – 20.79 [T ln T – T ]T 2 1

= − 909 − 20.79 (1543.18 – 1399.73) = − 3891.33 J CV = 20.266 +1.76 × 10 –2 T

209.

dE = CV dT = (20.266 + 1.76 × 10 −2 T ) dT ⇒

∆E = 20.266(T2 − T1 ) +

1.76 × 10 −2 (T22 − T12 ) = P2 (V1 − V2 ) = R 2

  P2  T1 − T2  P   1

1.76 × 10 −2 T22 + 28.58 T2 – 7869.48 = 0 2

⇒ Solving, T2 = 255.3 K Also, ⇒

dT dP dT dP −R = (28.58 +1.76 × 10 –2 T ) −R T P T P T P ∆S = 28.58 ln 2 +1.76 × 10 –2 (T2 – T1) + R ln 1 = 2.22 JK −1 T1 P2

dS = C p

∆U = ∫ CV dT = ∫ (20.266 +1.76 × 10 –2 T ) dT = 20.266 (255.3 – 300) +

1.76 × 10 –2 [(255.3) 2 – (300) 2 ] = − 1124.32 J 2

∆H = ∫ C p dT = ∫ (28.58 +1.76 × 10 −2 T ) dT = 28.58(255.3 – 300) +

1.76 × 10 −2 [(255.3) 2 − (300) 2 ] = − 1496J 2

210. − W = 2 × 30 L atm = 6084 J ⇒ ∆E = (20000 − 6084) J = 13.916 kJ V dV 211. −W = ∫ PdV = ∫ 10 ⋅ = 10 ln 2 = 10 × ln 10 L atm = 23 L atm = 2332.2 J V V1 q = 420 + 2332.2 = 2752.2 J 212.

∆E = n∫ CV dT = 3 ∫ (30 + 14 × 10 −3 T ) dT = 3 × 30 (T2 − T1 ) + 42 × 10 −3 (T22 − T12 ) = 9000 + 2940 = 11.94 kJ − W = (T2 − T1 ) nR = 3 × 8.314 × 100 = 2494.2 J q = 11.94 × 10 3 + 2494.2 = 14.4342 kJ

213. Applying adiabatic conditions between C and A.  500  P =   300 

–2.5

× 5 = 1.39 atm.

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WAB = 0

Q ∆V = 0

A(500 K, 5 atm)

3 WBC = − 8.314 × 300 ln = − 1918.8 J 1.39 3 WCA = × 8.314 × 200 = 2494.2 2 W = 575.4 J  500  PC =    300 

214.

P B (300 K, 3 atm)

– 2.5

V

× 5 = 1.39 atm

5 × 1.39 = 2.32 atm 3 5 = − 8.314 × 500 ln = − 3192 J 2.32 =0 = 1.5 × 8.314 × 200 = 2494.2 J

PB = ⇒

WAB WBC WCA

A(500 K, 5 atm) P

B(500 K, PB ) C(300 K, PC )

W = − 697.8 J 215. Applying adiabatic condition between A and D  300  P =   200 

⇒ 216.

C(300 K, P )

V

– 2.5

× 5 = 1.82 atm

A(300 K, 5 atm)

W1 = − PV1 = − nRT1 = − 300 R 5 W2 = − nR × 600 ln = − 606.4 R 1.82 W3 = 1.82 (VC − VD ) = 400 R W4 = CV × 100 = 150 R W = − 356.4 R = − 2.963 kJ

P

 RT RT  . RT − Wirr = P2 (V2 − V1 ) = P2  −  = 08 P1   P2 − Wrev = RT ln

2 P

C(600 K, P)

D(200 K, P)

V

P = (2) – 2.5 × 5 = 0.88 atm 5 WAB = − R × 500 ln = − 868.63 R 0.88 1 WBC = 0.88 (500R − 250R ) × = 250R 0.88 WCA = 1.5R × 250 = 375 R ⇒ W = − 243.63 R = – 2.026 kJ 217.

B(600 K, 5 atm)

A(500 K, 5 atm) P B(500 K, P) C(250 K, P)

V A(500 K, 10 bar)

P 2.0

V

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Solutions

Adding : ⇒ 218. (a)

RT ln

2 + 0.8RT = 4200 P P = 1.62 atm

PB = (2) – 2.5 × 10 = 177 . atm

A

WAB = − 1.5 R × 250 = − 375 R WBC = − 250 R ln

1.77 = − 142.75 R 1

T

(b) ⇒

 250  P =   500 

− 2.5

W = − 500R ln

= 5.65 atm

V

10 − 1.5R × 250 5.6

T

A

= − 664.9 R = − 5.528 kJ V1 = 0.082 × 300 = 24.6 L,

219.

0.082 × 500 = 1.138 atm 36 ∆H = ∆E + P1 ∆V + V2 ∆P = 1.5 R × 200 + [11.4 + 36 × 0.138] × 101.4 = 4.154 kJ P2 =

dG = C P dT − TdS − SdT = C P dT − T (3 × 10 −3 dT ) − (1.5 + 3 × 10 −3 T ) dT

220.

= (C P − 1.5) dT − 6 × 10 −3 T dT ⇒

∆G = (C P − 1.5) (T2 − T1 ) −

221.

∆S = C P ln

222.

∆V = 0 ⇒ dS = CV ⇒

223.

6 × 10 −3 (T22 − T12 ) = 3377 J 2

T2 P 5 1 + R ln 1 = 3.5 R ln + R ln = 9.1 J 3 2 T1 P2 dT dT dT = (12 + 28 × 10 −3 T ) = 12 + 28 × 10 −3 dT T T T

∆S = 12 ln

T2 40 × 1 + 28 × 10 −3 (T2 − T1 ), T2 = = 487.8 K T1 0.082

∆S = 12 ln

487.8 + 28 × 10 −3 × 187.8 = 11.09 JK –1 300

A → B ∆G ° = − 3 × 1000 − T × 20 = − 9000 J A → C ∆G ° = − 3600 − 10T = − 6600 J

224.

C

B

W = − 517.75 R = − 4.304 kJ

∆G ° indicates that B is more stable than C. dG = V dP − S dT = V dP − (10 + 12 × 10 −3 T ) dT

B(500, P)

C(250, 1) V

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∆G = V ∆P − 10 ∆T − 6 × 10 −3 (T22 − T12 )

⇒

⇒

∆G = 24.6 ×

225.

V=

0.082 × 300 = 24.6 L 1

P2 =

400 4 = atm. 300 3

1 × 101.4 − 10 × 100 − 6 × 10 −3 [( 400) 2 − (300 2 )] = − 588.52 J 3 dT dT = 2(19.686 + 31 × 10 −3 T ) dS = nCV T T T ∆S = 2 × 19.686 ln 2 + 31 × 10 −3 (T2 − T1 ) = 14.43 JK –1 T1

∆S system = nR ln

226.

V2 = 2 × 8.314 ln 2.5 = 15.236 J V1

4.17 × 10 3 = − 13.9 J 300 ∆S univ = 1.336 J 3 ∆S = R ln 3 + 2R ln = 15.87 JK −1 2 T A =122 K and TB = 244 K ∆S surr = −

227. 228. ⇒ WBC = 0,

nR = − 244 R 1 WCA = 2R × 122 ln 2 = 169.12 R W = − 74.88 R = − 622.5 J

A

B

P

WAB = − 1(T2 − T1 )

C

V

b 2 (T2 − T12 ) = 7.4 kJ / mol 2 b ∆H = ∫ C P dT = ∫ (33.314 + bT ) dT = 33.314(T2 − T1 ) + (T22 − T12 ) = 9.063 kJ/ mol 2 3 × 0.082 × 300 V1 = = 36.9 L 2 3 × 0082 . × 500 V2 = = 41 L 3 We can construct a path as:

229. ∆E = ∫ CV dT = ∫ ( a + bT ) dT = a (T2 − T1 ) +

W

1 A (2.0 bar, 36.9 L, 300 K) → B (2.0 bar, 41 L, T )

W2

→ C (3 bar, 41 L, 500 K) 2 × 41 × 300 T= = 333.33 K 2 × 36.9

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Solutions

Then, ⇒

W1 = P ∆V = 2 × 4.1 × 101.4 = 831.48 J 3

and W2 = 0

q = 7.4 × 10 + 831.48 = 8.23 × 10 J 3

dV  dT dV  dT   +R dS = n CV  = n  ( a + bT ) T + R V  V T      V  T ∆S = n  a ln 2 + b(T2 − T1 ) + R ln 2  = 59 J V1  T1  R 230. WAB = − 5(500 − 400) = − 100 R 5 WBC = 0 Connecting adiabatic points A and D. P Applying P 1− γ T γ = K gives TD = 210 K = TC ⇒

A

B C

PC = 2.1 atm

D

2.1 WCD = − 210 R ln = − 155.8 R ⇒ 1 WDA = 1.5 R × 190 = 285 R W = WAB + WCD + WDA = 29.2 R = 242.76 J 231. TD = 388 K, PB = 2 bar, WAB = − 500 R ln

5 = − 458.14 R 2

V

A

D P B

WCD = 1.5R × (388 − 250) = 270 R W = − 251.14 R

C

= − 2.088 kJ 232. According to second law of thermodynamics: ∆S v = ⇒ (b)

⇒

V

∆H v Tb

∆H v = ∆S v × Tb = 85 × 353 = 30.005 kJ mol –1 . Since vaporization process is endothermic, heat lost by surrounding in vaporization of 100 g benzene is: 30005 Q= × 100 = 38468 J 78 Q 38468 ∆S surrounding = − = − = − 108.97 JK –1 . 353 T

∆H v = ∆S v ⋅Tb = 85 × 329 = 2796 5 J mol –1 100 × 27965 (b) Heat gained by surrounding = = 48215.5 J 58

233. (a)

Increasing in entropy of surrounding =

48215.5 = 146.55 J 329

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234. For a reaction: ∆GR° = Σ ∆Gf° ( Products) − Σ ∆G °f ( Reactants) ⇒

∆G° (i) = − 2 × 200 + 762 = + 362 kJ ∆G° ( ii) = − 396 + 762 = + 366 kJ Since, both ∆G° are positive, a reaction with less positive ∆G° will be more likely to occur. Therefore, reduction of TiO 2 will be more favourable through reaction (i). 235. ∆H r° = − 111 + 242 = 131 kJ ∆S r° = 190 + 131 − (5.7 + 70) = 245.3 JK –1 ⇒

∆G ° = ∆H ° − T∆S ° = 131 × 10 3 − 300 × 245.3 = 57.41 kJ

Positive value of ∆G° indicates that reaction is non spontaneous at 300 K. Increasing temperature will decrease ∆G° and after certain value of T , ∆G ° will become negative, i.e., reaction will turn from non-spontaneous to spontaneous. That specific temperature at which reaction turns from non-spontaneous to spontaneous can be determined by equating ∆G° = 0 as: 131000 = 534 K 0 = 131 × 10 3 − T × 245.3 ⇒ T = 245.3 i.e., at temperature above 534 K, reaction will become spontaneous. 236. At 300 K, ∆G ° ( cis-2-butene → trans-2-butene ) = − 3 kJ mol –1 ∆H ° ( cis-2-butene → trans-2-butene) = − 4.2 kJ mol –1 ⇒

∆S ° ( cis-2-butene → trans-2-butene ) = =

⇒ Also,

∆H ° − ∆G ° T

−4.2 + 3 × 1000 = − 4 JK –1 300

∆G° (1) at 400 K = − 4.2 × 10 3 + 400 × 4 = − 2600 J ln K (1) = −

∆G ° (1) 3000 = 8.314 × 300 RT

⇒

K (1) = 3.33 at 300 K

and

K (1) = 2.20 at 400 K ∆G° (2) = − 8 kJ mol –1 at 300 K

⇒ ⇒

ln K (2) = −

∆G ° (2) 8000 = 8.314 × 300 RT

K (2) = 24.7 at 300 K.

Now, let at 400 K, mixture contain x% trans-2-butene and y% 2-methyl propene. x 18 ⇒ At 400 K, K (1) = 2.2 = = 100 − x − y 82 − y ⇒

y = 73.8%

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Solutions

⇒

K (2) at 400 K =

⇒ ⇒

y 73.8 = =9 100 − x − y 8.2

∆G° (2) at 400 K = − 8.314 × 400 ln q = − 7.3 kJ ∆G° (2) at 300 K = − 8000 = ∆H ° (2) − 300 ∆S ° (2) ∆G° (2) at 400 K = − 7300 = ∆H ° (2) − 400 ∆S ° (2)

⇒

∆S ° (2) = – 7 JK –1 ∆H ° (2) = − 8000 + 300 ( −7) = – 10.1 × 10 3 J K (3) =

Also,

24.7 = 7.41 3.33 ∆G° (3) = − 8.314 × 300 ln 7.41 = − 4995 J 9 K (3) = =4 2.2 ∆G° (3) = − 8.314 × 400 ln 4 = − 4610 J −4995 = ∆H ° (3) − 300 ∆S ° (3) −4610 = ∆H ° (3) − 400 ∆S ° (3)

⇒ At 300 K:

K (3) =

⇒ At 400 K: ⇒ ⇒ ⇒

K (2) K (1)

∆S ° (3) = – 3.85 JK –1

and

∆H ° (3) = – 6.15 kJ.

237. For the decarboxylation reaction: CH 3COOH CH 4 + CO 2 Also, above 65 K, reaction is spontaneous ⇒ ∆G° = 0 at 65 K ⇒

0 = 15.7 × 10 3 − 65 ∆S °

⇒

∆S ° = 241.54 JK –1 .

∆H ° = −394 − 74.8 + 484.5 = 15.7 kJ

For a spontaneous reaction ∆G° < 0. Since, ∆H ° > 0, spontaneity of reaction is due to ∆S ° > 0. 238. Since,

∆G = ∆G ° + RT ln Q = RT ln Q − RT ln K Q ∆G = RT ln   K

⇒ In the present condition: Q= ⇒

0.66 × 1.2 = 4.06 0.25 × 0.78

2540  Q  ∆G = = 15.275 × 10 –2 ln   =  K  RT 8.314 × 2000

Q =1.165 ⇒ K = 3.48 K Q Q > K , reaction will proceed in backward direction as: ⇒

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H2

0.25 + x

+

H 2O( g ) +

CO 2

0.78 + x

K = 3.48 =

0.66 − x

CO

1.2 − x

(1.2 − x ) (0.66 − x ) (0.25 + x ) (0.78 + x )

Solving : x = 0.02. ⇒ PH 2 = 0.27 atm, PCO2 = 0.80 atm, PH 2O = 0.64 atm. PCO = 1.18 atm. 239. For the above reaction: K p = pCO 2  K p (2)  ∆H °  T2 − T1  = ln    R  T1T2   K p (1) 

Also, ⇒

ln

 1830 ∆H °  250 =  = 4.394  22.6 R  973 × 1223 

⇒

∆G ° = − RT ln K p° = − 8.314 × 973 ln

Also,

∆H ° =173.88 kJ mol –1 . 22.6 = 28.43 × 10 3 J 760

∆H ° − ∆G ° 173.88 − 28.43 = × 1000 = 149.5 JK –1 mol –1 . 973 T 240. The minimum value of ∆S will occur at ∆G = 0 for spontaneous reaction. ∆S ° =

⇒

19000 = 55.07 JK –1 . 345 T − Tc 2200 − 760 Efficiency of engine = h = = 0.6545 Th 2200

⇒

0 = 19000 − 345 ∆S °

241.

∆S ° =

⇒

2 × 3.1 × 1000 × 5510 = 299666.67 J 114 Heat converted into work = 299666.67 × 0.6545 = 196132 J Total heat produced =

⇒

Q = mgh h=

196132 = 16.67 m 1200 × 9.8

242. Since, expansion occurred at constant temperature, ∆S = nR ln

V2 1 3.0 = × 8.314 ln = 0.36 JK –1 V1 32 0.75

Since, this is case of free expansion, Pext = 0.

⇒

− W = Pext ∆V = 0,

q =0

Also, since, ∆T = 0, ∆H = ∆E = 0. ∆S = nC v ln

243. Q n=

V T2 + nR ln 2 V1 T1

3 × 550 × 10 −3 3.5 × 0.73 × 300 = 464.5 K = 0.067 and T final = 0.082 × 300 3 × 0.55

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Solutions

⇒

464.5 5 0.73 R ln + 0.067 R ln = 0.767 JK –1 300 2 0.55 5 ∆E = nC v ∆T = 0.067 × R × ( 464.5 − 300) = 229 J 2 ∆H = ∆E + P∆V + V∆P = 229 + [3(0.73 − 0.55) + 0.73 × 0.5] × 101.4 = 320.767 J. ∆S = 0.067 ×

244. Under isothermal condition, V2 10 = 0.133 × 8.314 ln = 1.33 JK –1 3 V1

∆S = nR ln ∆G = nRT ln

P2 3 = 5 × 3 × 101.4 ln = − 1831.25 J 10 P1

245. (a) In case of adiabatic reversible expansion, dq rev = 0 ⇒ ∆S = 0. (b) In case of irreversible adiabatic expansion: 0 = Pext (V2 − V1 ) + nC v (T2 − T1 ) T2 = ∆S = C v ln

Pext (V1 − V2 ) −3 × 8 + T1 = + 1000 = 805 K nC v 0.0821 × 1.5

T2 V 805 + R ln 2 = 1.5 × 8.314 ln + 8.314 ln 2 = 3.06 JK –1 . 1000 T1 V1

(c) In case of free adiabatic expansion Pext = 0 ⇒

0 = Pext ∆V + nC v (T2 − T1 )

⇒

T1 = T2 ∆S = R ln

⇒

V2 = 5.76 JK –1 V1

246. Since, the two gases have same heat capacities, equal amounts, final temperature will be (T1 + T2 ) / 2 = 323 K. Also, final P =1 atm, i.e., partial pressure of each gas in the final mixture will be 0.5 atm. Now,

⇒ 247.

∆S 1 (for cold gas ) = C P ln

1 323 + R ln 0.5 273

∆S 2 (for hot gas) = C P ln

323 1 + R ln 373 0.5

 323 × 323  –1 ∆S = ∆S 1 + ∆S 2 = 2R ln 2 + C P ln   = 12.03 JK . 273 373 ×   dV dV dT dT + nR = n (C p − R ) + nR V V T T  dV  dT  = n  16.6 + 14.8 × 10 −2 dT − 9 × 10 −5 TdT  + R   V  T 

dS = nC v

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⇒

 T V  9 × 10 −5 ∆S = n 166 . ln 2 + 148 . × 10 −2 (T2 − T1 ) − (T22 − T12 ) + R ln 2  T1 V1  2  = n [5.585 + 29.6 − 10.8 + ( −2.965)] = 42.84 JK –1 . dG = − SdT

248.

at

∆P = 0

= − (25.1 + 29.3 ln T ) dT , on integration ∆G = − (25.1 − 29.3) ∆T − 29.3 (T2 ln T2 − T1 ln T1 ) Substituting, T1 = 298, T2 = 348. . ∆G = − 9750 J 249. Since, the process is irreversible adiabatic expansion: 0 = ∆E + Pext ∆V = ∆E + Pext (V2 − V1 ) Also, dE = nC v dT = (18.8 + 0.021 T ) dT 0.021 2 ∆E = 18.8 (T2 − T1 ) + (T2 − T12 ) 2 T   Substituting in Eq. (i), 18.8 (T2 − T1 ) + 0.0105 (T22 − T12 ) + R  T2 − 1  = 0  2

…(i) …(ii)

Substituting, T1 = 400 K, solving yields T2 = 352.5 K. dT dT dV dV Now, dS = C v = (18.8 + 0.021T ) +R +R T T V V T V Integrating: ∆S = 18.8 ln 2 + 0.021 (T2 − T1 ) + R ln 2 T1 V1 = 18.8 ln

352.5 352.5 × 10 + 0.021 (352.5 − 400) + R ln 400 400 × 5

= 1.338 JK –1 mol –1

250.

. 0021 [ (352.5) 2 − ( 400) 2 ] = – 1268.3 J 2 dH = ( R + C v ) dT = (27.114 + 0.021 T ) dT

From Eq. (ii),

. (352.5 − 400) + ∆E = 188

⇒

∆H = 27.114 (T2 − T1 ) +

0.021 2 (T2 − T12 ) = – 1663.22 J 2

∆G = nV ∆P = 5 × 0.089 × 99 L-atm = 4467.18 J

251. For isoentropic process: 0 = nC P ln

T2 P + nR ln 1 T1 P2

⇒

P  5 T R ln  2  = R ln 2 T1  P1  2

Solving, P2 = 1.474 atm. 252. Under isothermal condition: ∆S = nR ln P1 / P2 where P1 = initial pressure, P2 = final pressure Let, flask A has N 2 at 4 bar and flask B at 2 bar. After mixing final pressure in the combined system = (8/ 3) bar.

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Solutions

⇒

∆S A = R ln

12 8

and

∆S B = R ln

6 8

∆S A + ∆S B = ∆S = 0.98 JK –1 . ∆S system =

253. At boiling point,

∆H vap Tb

=

349 × 10 3 = 111.4 JK –1 3133

Heat change in surrounding = − 349 kJ ∆S surr = − ⇒

349 × 1000 = − 99.71 JK –1 3500

∆S univ = ∆S sys + ∆S surr = 111.4 − 99.71 = + 11.69 JK –1 .

ATOMIC STRUCTURE V nh = = 6.5 × 10 15 rps. 2π r 4π 2 r 2 m  4  1 1  4 1 1  255. (a) ν = KZ 2  2 − 2  ⇒ K  − 2  = K  2 − 2  ⇒ in He, required transition in 4 → 2.   1 2  n1 n2   n1 n2 

254. Number of revolutions per second =

 9 9  1 1  (b) K  −  = K  2 − 2   1 16   n1 n2 

⇒ n1 = 3, n2 = 12

 1 256. E ext = 2.18 × 10 −19 1 −  × 6.023 × 10 23 = 116.71 kJ/ mol H  9 D. E. = 116.71 × 2.67 = 311.62 kJ/ mol H 2 1 PV = = 0.04 n= . RT 0082 × 300 ⇒ 257.

T. E. = 0.04 × 311.62 + 0.08 × 116.71 = 21.8 kJ

n( n − 1) = 10 ⇒ nB = 5 Hence, n A = 3 2  1 1  ∆E = KZ 2  2 − 2  = 3.868 eV when  n1 n2  KZ 2 = 54.4 eV = I. E.

⇒ 258. ν =

n1 = 3, n2 = 5

6.626 × 10 −34 h = = 7.28 × 10 6 ms –1 mλ 9.1 × 10 −31 × 10 −10 λ=

259. ∆V =

h 2meV

h = 8.78 × 10 –11 ms –1 4πm∆x

⇒ V=

h2 2λ 2 me

= 150.7 V

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350

1 8 = RH ⋅ λ 9

260. EThreshold =

⇒ λ =105.14 nm

6.626 × 10 −34 × 3 × 10 8 hc 6.626 × 10 −34 × 3 × 10 8 = E = , incident λ 105.14 × 10 −9 230 × 10 −9

K. E. =

6.626 × 10 −34 × 3 × 10 8  1 1  –18 −  = 1.026 × 10 J  −9   105.14 230 10  1 1  25.7 = KZ 2  2 − 2  = KZ 2  n1 n2 

261.

1 1   − 2 4 n 

…(i)

1 1  8.7 = KZ 2  − 2  9 n 

…(ii)

Solving, Eqs. (i) and (ii) gives : n = 5 KZ 2 = 122.38 = I. E.

Also,

1   13.6 1 − 2  = 10.93 + x  n 

262.

1   54.4 1 − 2  = 49.18 + x,  n 

where x is work-function.

Solving : n = 4, x = 1.82 eV ∆H E hc 263. ∆H = 248 × 10 3 J ⇒ ⇒ λ = 482.76 nm = = 4.12 × 10 −19 J = λ molecule N A 264. ∆H reaction (O 3 → O + O 2 ) = 249 − 142 = 107 kJ/ mol λ=

hc ⋅ N A = 1119 nm ∆H

4 × 10 −17 hc = 121 = 3.313 × 10 −19 J ⇒ Number of photons = λ 3.313 × 10 −19 λ 2λ 3 hc hc hc 266. λ1 = = + ⇒ λ2 + λ3 λ1 λ 2 λ 3 h 3RT 267. r. m.s. ( c) = = 7.38 × 10 −11 m. = 1351.67 ms −1 ⇒ λ = M mc hc 268. Energy of photon = = 3.313 × 10 −19 J λ E ( required for melting ) = n ∆H = 166.67 kJ 265. E =

Number of photons = 5.03 × 10 23 . Also, average energy required/molecule =

6000 = 9.96 × 10 −21 J NA

Average number of H 2O molecules melted/photon = 33.

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Solutions

269. 3rd I.E. =

KZ 2

⋅ N A = 11817 kJ/ mol ⇒ IInd I.E. = 19612 − (520 + 11817) = 7275 kJ/ mol. n2 h 270. ∆V = = 19.85 × 10 −6 ms −1 . 4π m ∆ x KZ 2 3 271. E = 2 = K b ⋅ T ⇒ T = 20.636 × 10 6 ° C. 2 n 272. Maximum number of emission lines = n( n − 1)/ 2, where n is orbit number ⇒ nB = 5. Possible values of n A are 2, 3 and 4. Since only one photon of energy greater than 3.066 eV is emitted, n A must be 2 and only n = 5 to n =1 transition would emit photon of energy greater than 3.066 V. Ionization energy ( KZ 2 ) : 1 1  3.066 = KZ 2  −   4 25 

⇒

KZ 2 (IE) = 14.6 eV

273. We know that average energy of an oscillator is hν E = hν / kt e −1 hν 6.627 × 10 −34 × 1014 = = 4.8 kT 1.38 × 10 −23 × 1000

Q

E =

⇒

⇒

e 4. 8 = 121.5

6.627 × 10 –34 × 1014 = 5.5 × 10 −22 J 120.5

.

1 1  K.E. = h (ν − ν 0 ) = hc  −  λ λ0 

274. ⇒

1 E 1 where E = 1.5 eV = 1.5 × 1.6 × 10 −19 J = 2.4 × 10 −19 J = + λ hc λ 0

⇒

2.4 × 10 −19 1 1 = + 34 8 – λ 6.626 × 10 × 3 × 10 23 × 10 –8

Solving, λ = 1815 . × 10 −7 m = 1815 Å λ′ − λ =

275. ⇒

λ (1 − cos θ ) where λ′ = wavelength of scattered beam = 0.22 Å. mc

λ = λ′ −

6.626 × 10 −34 h (1 − cos θ ) = 2.2 × 10 –11 − (1 − cos 45° ) = 0.2129 Å. mc 9.1 × 10 −31 × 3 × 10 8

276. de-Broglie equation is: λ= or

V=

h2 2meλ 2

=

h h = mv 2meV

⇒

2meV =

h2 λ2

(6.627 × 10 −34 ) 2 2 × 9.1 × 10 −31 × 1.6 × 10 −19 × (3.8 × 10 –11 ) 2

=1044 volt.

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mvr =

277. Since,

nh 2π

⇒

v=

nh × Z 6.626 × 10 −34 × 2 hZ nh = 1.457 ×10 6 ms –1 = = = 2πmr 2πma 0 n 2 2πma 0 n 2π × 9.1 × 10 −31 × 53 × 10 −12 × 3

⇒

h=

h 6.626 × 10 −34 = = 5 × 10 −10 m = 5 Å. mv 9.1 × 10 −31 × 1.457 × 10 6

278. The Heisenberg’s uncertainty equation is h ∆x⋅∆p = ⇒ 4π ⇒

∆v ( min ) =

∆x⋅∆v =

h 4πm

h 6.626 × 10 −34 = 0.58 ms –1 = 4πm ∆x ( max ) 4π × 9.1 × 10 −31 × 10 −4

279. The uncertainty in velocity is ∆v = 0.001 km hr –1 = 2.778 × 10 –4 ms –1 . ⇒ The uncertainty in position: ∆x =

6.626 × 10 −34 h = = 3.796 ×10 –34 m. 4πm∆v 4π × 500 × 2.778 × 10 –4

The uncertainty in measurement of position ( ∆x = 3.796 × 10 −34 m ) is wholly negligible as compared to mass and velocity of moving automobile. hc 6.626 × 10 −34 × 3 × 10 8 280. Energy of incident photon = = = 9.94 × 10 –16 J –12 λ 200 × 10 Energy of emitted electron =

1 1 mv 2 = × 9.1 × 10 −31 (2 × 10 7 ) 2 = 1.82 × 10 −16 J 2 2

⇒ Binding energy of electron = Energy of incident photon – Energy of emitted electron = 8.12 × 10 –16 J = 5075 eV. 281. Stopping potential is the maximum kinetic energy of photoelectron: ⇒

K.E. (max) = eV = 0.5 × 1.6 × 10 −19 = 8 × 10 −20 J Energy of photon =

hc 6.626 × 10 −34 × 3 × 10 8 = = 7.835 × 10 −19 J −9 λ 253.7 × 10

Work-function = 7.835 × 10 −19 − 8 × 10 −20 = 7.035 × 10 −19 J = 4.4 eV. 282. The maximum uncertainty in position of electron inside the nucleus = 10 −14 m. ⇒ minimum uncertainty in momentum ( ∆ p) = 283. Average

T.E. =

h = 0.527 × 10 −20 kg ms –1 . 4π ∆ x

3 3 kT = × 1.38 × 10 −23 × 300 = 6.21 × 10 −21 J 2 2

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Solutions

⇒

p (momentum) = 2mE = 2 × λ=

h 6.626 × 10 −34 = 1.46 × 10 −10 m = 1.46 Å. = p 4.540 × 10 −24

Potential energy =

284.

10 –3 × 6.21 × 10 –21 = 4.54 × 10 −24 kgms –1 N0

(1.6 × 10 –19 ) 2 e2 = 7.69 × 10 −19 J = 4πε 0 r 4πε 0 × 3 × 10 −10

285. Probability (P) between 0 and 2a 0 = ∫

2a 0

ψ12s dτ where dτ = 4πr 2 dr.

0

∴

P =∫

2a 0

0

=−

( πa 03 ) −1

e

−2 r/a 0

4

2

4πr dr =

a 03

2a 0

∫0

−

e

2r a0

r 2 dr

 –4 e − 4 1  a 03 −4 a 03  4  3 −4 3 −4  3e + −  = 0.7624 a e a e e 2 = − 4 + + −   0 0 4 4 4 4   a 03 

286. The radial distribution function for 1s orbital is R 2 r 2 = 4πr 2 ψ12s ⇒

⇒

For maximum,

d −2 r/a 0 [ 4πr 2 ( πa 03 ) −1 e ]=0 dr

⇒

r

r

0

0

2r −

Probability = ∫ ψ12s dτ = ∫ ( πa 03 ) −1 e

287.

P=

4 a 03

r

∫0 r

2 −2 r/a 0

e

d[ R 2 r 2 ] =0 dr

2r 2 = 0 or a0 −2 r/a 0

r = a0 .

4πr 2 dr

dr

Integrating by parts and substituting r = 10 −12 m and a 0 = 53 × 10 −12 m. P = 9 × 10 −6 .

⇒ v rms =

288.

3RT 3 × 8.314 × 298 = 1360 ms –1 = M 4 × 10 −3

de-Broglie equation: λ=

h 6.626 × 10 −34 = × 6.023 × 10 23 = 7.34 × 10 –11 m. mv 4 × 10 −3 × 1360

3  1 Transition energy = 2.18 × 10 −18 1 −  = 2.18 × × 10 −18 J = 1.63 × 10 −18 J.  4 4

289.

Translational energy = ⇒

T=

3 kT = 1.63 × 10 −18 2 2 × 1.63 × 10 –18 3 × 1.38 × 10

−23

= 78.744 × 10 3 K .

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290. The reduced mass of moving electron me ⋅ m positron m = e (µ ) = me + m positron 2 rn = n 2

⇒

4π ε 0 h 2 ( me / 2) e 2 Z

= (1.0589 × 10 −10 m ) n 2

r1 = 1.0589 × 10 −10 m = 1.0589 Å.

⇒

CHEMICAL BONDING 291. CO 2 : O==C==O Linear, non-polar molecule. COS : S==C==O Linear, polar molecule. Ο  Trigonal planar ion with sp 2 -hybridized carbon. It exists C CO 2− 3 : O −

in three resonating forms.

O − Cl

SiCl4 :

Si Cl

Cl

Tetrahedral molecule with sp 3 -hybridized silicon.

Cl

Pentagonal bipyramidal with sp 3 d-hybridized phosphorus. Equatorial bond angle = 120°, Axial-equatorial bond angle = 90° .

Cl Cl PCl5 : Cl

P Cl H

NH4+

:

Cl +

N H

H

H

PH +4 :

Similar to NH +4 .

PCl +4 :

Similar to NH +4 . Cl Cl Cl

PCl6– :

P Cl

SF6 :

Tetrahedral ion with sp 3 -hybridized nitrogen. All bond angles are equal and 109° 28′ .

Cl Same as PCl −6 .

Cl

–

Square pyramidal ion with sp 3 d 2 -hybridized phosphorus.

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Solutions

F

F IF7 :

F

72°

F

Pentagonal bipyramidal with sp 3 d 3 -hybridized iodine.

I F F

••

292. SO 2 :

Angular (V-shaped) with sp 2 -hybridized sulphur.

S

O

F

O

NH3 :

N H

Trigonal pyramidal with sp 3 -hybridized nitrogen. H

H ••

H 2O :

Angular (V-shaped) with sp 3 -oxygen.

O

••

H

H

PCl3 :

P Cl

Trigonal pyramidal with sp 3 -phosphorus. Cl

Cl

NCl 3 : Same as PCl 3 . 293. (a) If every things are same, and central atoms are forms same group, bond angle decreases on descending down the group. NH 3 > PH 3 > AsH 3 > SbH 3 > BiH 3 (b) H 2O > H 2S > H 2Se > H 2Te 294. AsH 3 < H 2Se < PH 3 < H 2O < NH 3 .

F 295. ClF3 : F

T-shaped with sp 3 d-hybridized chlorine. (In trigonal bipyramidal geometry, lone pairs are stable at equatorial positions).

Cl F O  C

COCl 2 : Cl Cl 2O :

Triangular planar with sp 2 -carbon. Cl

Two shape can be predicted as : ••

••

O

Cl ••

••

Cl

Cl

V-shaped with oxygen at centre

Cl

••

• O •••

V-shaped with chlorine at centre

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356 ••

OF2 :

V-shaped with sp 3 -oxygen.

O

••

F

F Cl

Cl

ICl4– :

–

I

Square planar with sp 3 d 2 -iodine.

Cl

Cl

F F

F

IF5 :

I F

Square pyramidal sp 3 d 2 -iodine.

F

O Cl IOCl4– :

Cl

Square pyramidal with sp 3 d 2 -iodine.

I Cl

Cl Cl Cl

IOCl4+ :

O

••

+

I

Trigonal bipyramidal with sp 3 d-iodine. Cl

Cl

296. N −3 :

–

+

•• −

N ≡≡ N → N •• Linear with sp-hybridized nitrogen. I

I3– :

••

–

Linear with sp 3 d-hybridized central iodine.

I I F F

BrF5 :

F

Square pyramidal with sp 3 d 2 -bromine.

Br F

F

F IF4– :

F

–

Square planar with sp 3 d 2 -iodine.

I F

F F

BF4– :

B F

F F

–

Tetrahedral with sp 3 -boron.

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Solutions

ICl +2 :

   

••

I

••

Cl

Cl

+

   

V-shaped with sp 3 -iodine.

Cl –

ICl2–

I

:

Linear with sp 3 d-iodine.

Cl F F PF5 :

F

P

Trigonal bipyramidal with sp 3 d-phosphorus.

F F

297. Relative electron withdrawing power of nitrogen is more in NCl 3 than in NOCl therefore homolytic cleavage of N—Cl bond would be difficult in NCl 3 . F – + F As AsF2+ : 298. AsF5 : F As SnCl3– : Sn F F Cl F Cl Cl F Trigonal pyramidal with sp3-Sn

V-shaped with sp2-As

Trigonal bipyramidal

F F

F

NOF : O==N

SO32– :

S O

2

Angular with sp -N

F

TeF5– :

O O

Te F

F

Square pyramidal with sp3d2-Te

Tetrahedral

Cl

–

GeF3–

:

Ge F

SCl2 :

F F 3

299. ClF4+ :

F

F

SbCl6– :

F FClO3 :

F

Cl

V-shaped with sp -S

+

Cl

Sea Saw

S Cl

Cl O

O O

Cl

Cl

3

Trigonal pyramidal with sp -Ge

F

–

Tetrahedral with sp3-Cl

–

Sb Cl Cl

Cl

Octahedral with sp3d 2-Sb

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358

O F2ClO+ :

+

SeO32– :

Cl F

Se O

O

F

O Tetrahedral

Trigonal pyramidal

300. (a) BF3 (µ = 0) = PCl 5 (µ = 0) < POCl 3 < SCl 2 < ICl 3 (b) XeF2 (µ = 0) = SnCl 4 (µ = 0) < PCl 4 F < XeO 3 < SF4 301. Li 3 N < Li 2O < LiF (Fajan’s rule) 302. (a) Al 2 (CO 3 ) 3 < BeCO 3 < MgCO 3 < CaCO 3 < BaCO 3 < Na 2CO 3 (b) CaSO 4 < CaSO 3 < CaCO 3 (c) Al 2 (SO 4 ) 3 < MgSO 4 < SrSO 4 < BaSO 4 (d) FeC 2O 4 < CaC 2O 4 < CaCO 3 < K 2C 2O 4 303. Symmetrical repulsion by the two lone pairs from axial positions cancels the effect of one-another and bond angles remains intact. 304. AlCl 3 < GaCl 3 < MgCl 2 < CaCl 2 < BaCl 2 < NaCl (Fajan’s rule). 305. (a) NH 3

(b) PF3

(c) AsF3

306. (a) H 2S < PH 3 < H 2O < NH 3

(d) AsCl 3 (b)

GeH –3

(e) H 2O

< SiH –3 < CH –3 F

F

F

Cl 307. (a) F

Cl (b) Cl

P

(c) Br

P

Cl

Cl F

Cl

–

I [Cl

P

Cl F

F

(d)

Cl

Cl]

+

(e)

I

3

V-shaped with sp -iodine

Cl

Linear

308. PCl 3 F2 (µ = 0) < PF3Cl 2 < PBrF2Cl 2 309. CCl 4 < CHCl 3 < CH 2Cl 2 < CH 3Cl F O 310. (a) O Cl F F Trigonal bipyramidal with sp3d 2-Cl

O F

F (b)

Xe F

F

Square pyramidal with sp3d 2-Xe

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Solutions

O Cl

Cl (c)

+

–

I (d)

I Cl

Cl

Cl

Cl

V-shaped with sp3-I

Square pyramidal with sp3d 2-I

311. In solid state, PCl 5 remains as dipolar ion as [PCl 6 ]– [PCl +4 ] Cl Cl

Cl

Cl

–

P

P Cl Cl

Cl Cl

Cl

Cl

+

Cl Cl

In gas phase PCl 5 remains as isolated gas molecule :

Cl

P Cl

Cl

312. Due to back bonding of a lone pair from p-orbital of oxygen to vacant d-orbital of Si ( pπ − d π bonding). 313. (a) Linear (b) Square planar (c) Octahedral (d) T-shaped (e) Square pyramidal (f) Sea-saw (g) Trigonal bipyramidal CH3 314. (a) F3C

Cl

I

Cl

CH3

CF3 (b) H3C

P CH3

(h)

CF3

Cl Cl

I

Cl Cl

CF3

P CH3

CF3

Greater electronegativity when bonding through axial positions. Pπ − Pπ back bonding in BF3 gives some double bond character, which is absent in BF4− . Pπ − d π back bonding occurs between Si and Cl, which is absent in CCl 4 . BF3 < BCl 3 < BBr 3 < BI 3 Lone pair of nitrogen in N(SiH 3 ) 3 is involved in pπ − d π back bonding, not available for donation to a Lewis acid. H H H Br 320. C==C C== C H Br Br Br

315. 316. 317. 318. 319.

µ≠0

µ=0

321. d (µ = 0) = b(µ = 0) < c < a 322. Due to larger size of 3 p-orbital with Si, there is fewer chance of sidewise overlap giving π-bonds.

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360

F 323.

F

F

N==N

N==N

µ≠0

µ=0

F

324. Due to free rotation about C—C sigma bond, 1,2-dichloro ethane acquire the anti (most stable) conformation which is non-polar. In dichloroethane, rotation is restricted. Cl H H Cl Cl C==C H H Cl H H due to restricted rotation, µ ≠ 0

anti conformation of 1,2-dichloro ethane (µ = 0)

325. In ClHC==C==CHCl, the two C—Cl bonds are in perpendicular plane and hence the molecule is polar. In HClC==C==C==CHCl, the two C—Cl bonds are in the same plane and individual C—Cl dipoles are at 180°, cancelling one another giving zero dipole moment. Cl Cl Cl H H Cl 326. C==C C==C C==C H H H Cl H Cl I

II

III

Dipole moment: II < III < I. N 327. (a)

N

N N

(b)

N

N

N N

N

N N

N

328. NaF has higher lattice energy than NaCl. Lattice energy is inversely proportional to square of the interionic distance. Since, fluoride ion is smaller in size than chloride ion, NaF will have higher lattice energy. 329. The NO 2 exist in two equivalent resonance form as: ●

O

N

●

O

O

N

●

O

O

N

O

both N—O bonds are identical, therefore equal bond energies. • • • N ==O

It has a one sigma and one pi-bond and bond order is two whereas in NO 2 , the N  O bond order is less than two. Therefore, N  O bond energy in NO is greater than in NO 2 . 330. Both NO and NO 2 have one odd (unpaired) electron. When the two molecule approach each other, unpaired electron may be shared to form a covalent bond as: ON + NO 2 → ON NO 2 both N have complete octet.

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Solutions

O

O N—N

Shape of N2O3 :

O

F O 331. (a)

+

O==S

(b)

S

Cl F

Triangular planar

F S

F

One S-is sp3d and other is sp3 hybridized.

332. N has higher electronegativity than C, therefore, energy of orbitals of N is less than that of carbon and energy level diagram will be as follows:

C(2p)

N(2p)

C(2s)

333. (a) (b) (c)

N(2s)

SiF4 is tetrahedral, non-polar molecule while PF3 is trigonal pyramidal, polar molecule. Therefore, PF3 will be more soluble in polar solvent. SF6 is square bipyramidal, non-polar molecule while SF4 is a seasaw shaped, polar molecule. Therefore, SF4 will be more soluble in polar solvent. IF5 is a square pyramidal, polar molecule while AsF5 is a trigonal bipyramidal, non-polar molecule. Therefore, IF5 will be more soluble in a polar solvent.

334. π-bond

σ-bond

CHEMICAL EQUILIBRIUM 335. K =

0.5 × 0.2 2 = 9 0.5(0.2 + x )

⇒ x = 0.7 mol

δ-bond

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336. H + B 0.1 0.08

HB

0.1 0.08

337. NH 4Cl( s)

0 0.02

Initial At equilibrium ⇒ K c = 3.125.

NH 3 ( g ) +

HCl( g )

3.66 × 10–2 3.66 × 10−2 (0.154 – x ) (3.66 × 10–2 – x )

moles at equilibrium K c = 134 . × 10−3 after adding 2.0 g NH3.

⇒ K c = 1.34 × 10 −3 = (0.154 − x )(3.66 × 10 −2 − x ). Solving, x = 0.163 and

0.026

x cannot be greater than 3.66 × 10 −2 , hence it is 0.026. Therefore, on adding 0.2 g NH 3 , 0.026 moles of both HCl and NH 3 will combine forming equal amount of NH 4Cl( s) before equilibrium was re-established. Thus, amount of NH 4Cl( s) at new equilibrium =1.431g; % dissociation = 28.45. 2

 0.63 + 2x  338. K = 6.05 =  ⇒ x = 0.134 ⇒ [H 2 ] = [I 2 ] = 0.366, [HI] = 0.898.   0.5 − x  339. (CH 3 ) 3 CCl (CH 3 ) 2 C==CH 2 + HCl 0.2 – x

x

Molarity

x

K x2 KC = P = 0.084 = RT 0.2 − x 340. CO 2 ( g ) + C( s)

⇒

x = 0.094

2CO

79.2 − x

56 x 44

56 x − x = 2.3 ⇒ x = 8.433 g 44 PCO( g ) = 6.286 atm, PCO2 ( g ) = 26.38 atm ⇒ K P (1000 K) = 1.49 atm 56 At 1100 K : x − x = 5.3 ⇒ x =19.43 g. 44 PCO( g ) = 15.93 atm, PCO2 ( g ) = 24.5 atm; ⇒ K P (1100 K) = 10.35 atm K (2) ∆H  T2 − T1  Also = ln P  ⇒ ∆H = 177.25 kJ  K P (1) R  T1T2 

At 1000 K

341. N 2 + 3H 2 1− x

1 − 3x

2NH 3 Total 2x

2(1 – x )

⇒

KP =

0.85

⇒

(1 − 3x )3 p 2

≈

16x 2 P2

= 64x 2

x = 9.68 × 10 −3 , x NH 3 = 9.78 × 10 −3 .

⇒ 342. H 2O( g )

16x 2 (1 − x )

1

H 2 ( g ) + 2 O 2 ( g ) Total moles 0.15

0.075

1.075

K P (T ) = 4.66 × 10 −2

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Solutions

 4.66 × 10 −2  57.8 × 10 3  T − 1000  = ln    −11   1000 T  2  8.7 × 10 

⇒

H2 + I 2

343.

x 2 3– x

1−

2HI

x 2 3– x

3−

x 2x

x2 4x 2 = x  (3 − x ) 2  x  1 −   3 −   2  2

⇒

⇒ T = 3284 K

⇒ x = 1.5, K = 4.

344. Iodine in the gaseous state = 0.0386 mole; Iodine in solid state = 0.0114. Minimum H 2 ( g ) would be required when solid I 2 ( s) has just been exhausted and the following equilibrium is maintained : H 2 (g ) + I 2 (g ) 2HI ( g ) moles at equilibrium

0.0386

x

20 =

⇒

0.0228 2

(0.0228) x (0.0386)

⇒ x = 0.67 × 10 −3

⇒

moles of H 2 required = 0.012 0.0492 345. I 2 + I – I −3 K = = 745.57 L mol −1 −3 1.299 × 10 × 0.0508 Adding water will drive the reaction in backward direction and solubility will decrease. 346. A(g ) B (g ) + C (g ) 0.46 − x

⇒

3=

x

(0.46 + x )(0.082) × 300 5

K P (300 K) =

⇒

x

 0.15     0.61 

⇒ x = 0.15

2

 0.31    0.61

× 3 = 0.357

K P (320 K) = 1.145 K P (350) ∆H    20 30 1.145 ∆H  = = ln    and ln  R  300 × 320  1.145 0.357 R  350 × 320 

Similarly, Now,

K P (350) = 5.12, P0 (350) = 2.64

Solving, ⇒ Also,

5.12 =

2

P 2.64 − P

⇒ P =1.92

⇒

α(350 K) =

0.15 = 0.326 0.46 0.4012 × 100 = 123.06% % increase = 0.326 α(300 K) =

1.92 = 0.7272 2.64

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364

347. P(NH 3 ) =175 bar

175 = 43.75 bar 4 = 3 × 43.75 = 131.25 bar

⇒ PN 2 =

and

PH 2

⇒

KP =

(175) 2 (43.75)(131.25)

3

= 3.096 × 10 −4 bar −2

348. Let r be the ratio of P (Cl 2 ) to P (Br 2 ). Thus, P (Cl 2 ) = rP ( Br 2 ) ⇒ Pt = P (Cl 2 ) + P ( Br 2 ) + P ( BrCl) = rP ( Br 2 ) + P (Br 2 ) + P (BrCl) = (1 + r ) P ( Br 2 ) + P ( BrCl) P − P (BrCl)  P − P (BrCl)  PBr2 = t , P (Cl 2 ) = r  t ⇒    r +1 r +1 ⇒

KP =

P (BrCl) 2   Pt − PBrCl    Pt − P (BrCl)     r   r +1   r +1  

=

P (BrCl) 2 2

[ P − P (BrCl)] r t  ( r + 1)   Taking logarithm of each side and differentiating w.r.t. r : ln K p = 2 ln [ P (BrCl)] − ln r − 2 ln [ Pt − P (BrCl)] + 2 ln ( r + 1)    − ∂P (BrCl)  1 1 1  ∂P (BrCl)  1 0=2   − −2  +2   r  r +1 ∂r ∂r P (BrCl)   Pt − P (BrCl)   ∂P ( BrCl) = 0 and solving for ‘ r’ gives r =1 ∂r Hence, stoichiometric amounts of Cl 2 and Br 2 generate maximum yield of BrCl. 349. F2 2F Assigning

0.06 − x

2x

2

 2x   0.06 + x  4x 2 × 2.07 = 9.5 × 10 −3 ⋅ KP =  P =    2 −4  0.06 + x   0.06 − x  36 × 10 − x ⇒ x = 2.03 × 10 −3 ⇒ mole fraction of F = 0.065, mole fraction of F2 = 0.935 350.

N 2 (g ) + O 2 (g ) 0.8 − x

2NO( g )

0.2 − x

4x 2 K= = 4 × 10 −4 (0.8 − x )(0.2 − x )

(2x )

⇒ x = 3.95 × 10 −3

⇒ mole fraction of NO( g ) = 0.08, N 2 ( g ) = 0.796, O 2 ( g ) = 0.196 351.

Br 2 + 3F2 0.25 1.497

K=

0.75 1.497

(0.497 × 1.497) 2 0.25(0.75) 3

2BrF3 0.497 1.497

= 5.248

After increasing equilibrium pressure = 2 bar, reaction will shift in forward direction and let x mole of Br 2 reacted further.

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Solutions

Then new partial pressure : Br 2 =

0.25 − x 0.75 − 3x × 2; F2 = ×2 1.497 − 2x 1.497 − 2x BrF3 =

0.497 + 2x ×2 1.497 – 2x

Substituting these partial pressure in equilibrium constant expression and solving gives x = 0.061. Therefore, new compositions would be: Br 2 = 0.189, F2 = 0.567, BrF3 = 0.619 moles. 352. Ag 2CO 3 ( g ) Ag 2O( s) + CO 2 ( g ) K P = 0.0095 bar at 120°C. P (CO 2 ) at 120° C = K P = 0.0095 which is less than 0.01 bar, the partial pressure of CO 2 in air. Therefore, Ag 2CO 3 ( s) will not effervess at 120°C. 353. At equilibrium : 2 A + B 3C + 2D 0.4 1.7 0.9 16 . K c = 6.86 P 354. PCl 5 PCl 3 + Cl 2 K P = 3 Let x mole of Cl 2 is added out of which y mole reacted with PCl 3 , giving y mol PCl 5 . Also, at new equilibrium n =12. ⇒ 6 + x − y = 12 ⇒ x − y=6 2 20 P 2P (2 − y) and and = ⇒ y= x= 3 3 3 3(2 + y) 355. In the combined system partial pressure of the gases before any reaction occurred are : 250 100 NO = 0.46 × = 0.3285; O 2 = 0.86 × = 0.2457 350 350 Here, NO is limiting reagent : Partial pressure after 1st step : NO = 0, O 2 = 0.0814, NO 2 = 0.3285 Now, let p-be the decrease in partial pressure of NO 2 due to equilibrium. ⇒ 0.0814 + 0.3285 − p +

p = 0.37 ⇒ 2

p = 0.0798 ⇒ K P = 0.645 atm –1

356. Let us consider P0 be the initial partial pressure of each Cl 2 ( g ) and F2 ( g ). Then at equilibrium partial pressure of each gas would be : PCl 2 = P0 − 0.12, PF 2 = P0 − 0.16 3.2 =

⇒

P (ClF) 2 (0.2) 2 = P (Cl 2 ) P (F2 ) ( P0 − 0.12)( P0 − 0.16)

Solving, gives P0 = 0.253 and 0.026, where the second value is not acceptable. ⇒ K p for second equilibrium = 357. Cl 2 + F2 0.0665

0.0665

2ClF 0.267

(0.04) 2 (0.133) (0.093) 3

K =16.12

= 14.95

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366

When Br 2 is added :

Cl 2

0.075 − x

+ F2

2ClF

0.075

16.12 =

0.25 2

(0.25) 0.075(0.075 − x )

⇒ x = 0.023

Cl 2

+ Br 2

Kc =

(0.046) 2 = 0.528 0.052 × 0077 .

0.07 − x

2BrCl

0.1 − x

2x

2NOBr. Let initial partial pressure of NO = p0 and decrease in partial pressure 358. 2NO + Br 2 before the 1st equilibrium be p1 . Then : p1 + p1 + 1 − p0 = 2.12 2

p0 − p1 + 1.25 − ⇒

p1 = 0.26 and

Kp =

(0.26) 2 ( P0 − 0.26) 2 (1.12)

Similarly, if p2 be the total decrease in partial pressure of NO before the second equilibrium was established then p0 − p2 + 21.75 − ⇒

p2 = 0.5 and

Equating the two K P ; P0 = 0.69 and 359. Before addition of Br 2 ( v ) :

Cl 2 ( g ) 0.492 bar

⇒

−

p2 + p2 + 1 − p0 = 22.5 2

KP =

(0.5) 2 ( p0 − 0.5) 2 (21.5)

K P = 0.326. + I 2 (g )

K1

2ICl 2x

0.03

x

0.522 + x = 0.767 ⇒ x = 0.245 and

K 1 = 32.4

After adding Br 2 ( v ) Cl 2 ( g )

0.247 + q − p

I 2 (g ) + 0.03

Also, ⇒

K1

+ I 2 (g )

Br 2 ( g )

K2

0.492 – 0.1476 − p

2IBr ( g ) 0.2952

P (Br 2 ) = 0.2 ⇒ K2 =

2ICl( g ) 0.49 − 2q

0.03

p = 0.1444 2

(0.2952) = 14.52 0.03 × 02

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Solutions

Cl 2 ( g )

+

0.247 + q − p

⇒

2ClBr ( g )

0.492 − 0.1476 − p

K1 = 32.4 =

Also,

K2

Br 2 ( g )

2p 2

(0.49 − 2q ) (0.03) (0.1026 + q )

q = 0.05 ⇒ K 3 =

(0.2888) 2 = 2.73 (0.1526)(0.2)

4x 2 ⇒ 2x ( moles of AB ) = 1.87 = 50 (1 − x )(2 − x ) 361. m mol of I 2 reacted with hypo = 0.785 360. K =

⇒ At equilibrium : m mol of H 2 = I 2 = 0.785, m mol of HI = 5.93 2

 0.785  –3 K =  = 17.52 × 10  5.93  362. Let the equilibrium partial pressure of NO 2 = H 2O( v ) = 2 p1 Equilibrium pressure of N 2 = p2 and equilibrium pressure of NH 3 = 2 p3 ⇒ Equilibrium pressure of H 2 = 2 p2 − p3 . Then, N 2 H4

1 − p1 − p2 − p3

+ 3O 2

2NO 2 + 2H 2O

1 − 3 p1

2 p1

2 p1

N 2 + 2H 2

N 2 H4

1 − p1 − p2 − p3

2 p2 − p3

p2

N 2 H4

N2

1 − p1 − p2 − p3

2 p2 − p3

+ 2NH 3 2 p3

K1 = 3 K p (2) = ? K P (3) = ?

9 1 bar and 2 p3 = bar 19 38 3 11 and = − p2 PO2 = bar 4 38

2 p1 =

Given, ⇒

PN 2H 4

⇒

K1 = 3 =

9    19 

4

3

 11    (0.75 − p2 )  38 

Solving p2 = 0.233

Using values of p1 , p2 and p3 gives p(N 2 H 4 ) = 0.517, p(N 2 ) = 0.233 and p(H 2 ) = 0.4528 p(NH 3 ) = 0.0263 bar. ⇒ K p (2) = 9.24 × 10 −2 bar 2 and K p (3) = 2.95 × 10 −3 . 2NOCl

363. P: mole :

0.64

2NO 0.24

15.2 × 10–3

5.7 × 10−3

+

Cl 2 0.12 2.85 × 10−3

K P = 16.875 × 10 −3 bar .

Total = 23.75 × 10 −3

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368

Let at this stage x mole of Cl 2 is added, out of which y mole react back. Also volume is increased to 1.5 litre but at same temperature and pressure, therefore new total mole = 23.75 × 1.5 × 10 –3 = 35.625 × 10 –3 . A new equilibrium : mole Cl 2 = 2.85 × 10 –3 + x − y NO = 5.7 × 10 −3 − 2 y and

NOCl = 15.2 × 10 −3 + 2 y

Adding moles and equating to total moles gives x − y = 11.875 × 10 −3 Therefore, at new equilibrium, mole-fractions are: Cl 2 = (2.85 +11.875) × 10 −3 /35.625 × 10 −3 = 0.4133 NO =

5.7 × 10 −3 − 2 y 35.625 × 10 −3

and

NOCl =

15.2 × 10 −3 + 2 y 35.625 × 10 −3

Substituting in expression of K p gives y = 1.108 × 10 −3 ⇒ x = 12.983 × 10 –3 mol N 2O 4

364.

At P =1, ρ = 2.33 : Applying,

2NO 2

Kp =

4α 2 p 1−α2

ρRT M where M = molar mass of N 2O 4 . = 1+α P

⇒ α = 0.6 and K p = 2.25. Let β-be the degree of dissociation at new equilibrium. Then, M 2.33RT = 1.6 1

and

M 5.08RT = 1+β P

⇒ 1 + β = 0.734 P

4β 2 P 4β 2 = (1 − β)(1 + β) 0.734(1 − β)

⇒ β = 0.468 and

At P =1.5, using K p ; α = 5.22 and ρ =

92 × 1.5 = 3.685 g/ L 1.522RT

⇒ 2.25 =

P = 2 bar.

P2 2(1 − P ) where, P is partial pressure of H 2 ( g ) at new equilibrium. 0.356 = 0.21 ⇒ P = 0.356 and α = 1.644 PV 366. (a) n = = 0.05, i.e., out of 0.03 mole of PCl 5 , 0.02 mol has decomposed into PCl 3 and Cl 2 RT giving 0.05 mole of gaseous mixture. 2 ⇒ α = = 0.67 3

365. H 2S( g )

H 2 ( g ) + S( g )

K p = 0.099 =

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Solutions

(b) From the information in part ‘ a’; K p = PCl 5

⇒ 367. I:

0.8 =

PCl 3 + Cl 2 p

p2 2.73 − p

⇒

0.17

0.07

0.03

K1 = 7.563 × 10 . H 2 ( g ) + CO 2 ( g )

H 2O( g ) + CO( g )

0.17 + y

− y

+ y

n( H 2 ) = 0.01 K1 = 7.563 × 10 −2 =

x = 0.06 ⇒

K2 =

0.09 = 9; 0.01

0.09(0.03 – y) 0.01(0.17 + y)

y = 2.83 × 10 −2 K3 =

0.17 + 2.83 × 10 –2 0.03 – 2.83 × 10 –2

2NOBr n = 0.01 +

x

x/ 2

x PV = 2 RT

x = 0.0074

If 5 g of NOBr is taken, initial mole =

= 116.64

2NO + Br 2

0.01 – x

Solving,

Co( s) + CO 2 ( g )

0.03 + x = 0.09 ⇒

Solving,

368.

Co( s) + H 2O( g ) +x

K3

CoO( s) + CO( g )

⇒

0.03 − y

0.03 + x

K2

−x

Also,

0.03

–2

CoO( s) + H 2 ( g )

Hence,

p =1.13 atm.

1.13 = 0.42 2.73 H 2 ( g ) + CO 2 ( g ) H 2O( g ) + CO( g )

0.07 − x

Also

p

α=

⇒ II :

= 0.8

1−α2

2.73 − p

P:

⇒

α 2P

⇒ α=

x = 0.74 0.01

5 = 0.045 110

 α Moles at equilibrium = 0.045 1 +  = 0.045 × 1.37 = 0.06165  2

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370

nRT = 0.753 atmosphere. V 4HCl + O 2 2Cl 2 + 2H 2O P=

369.

0.405 0.905

0.12 0.905

Equilibrium

Kp =

0.19 0.905

4

(0.19) (0.905) 1 ⋅ (0.12) 4 (0.405) P

Also

0.19 0.905

mole fraction

P = P0 (0.905) = 0.905 atm.

4

 0.19   1  −1 =  = 15.5 atm .    0.12   0.405  N 2O 5

370.

N 2O 3 + O 2 x− y

4−x

N 2O 3

Kc

x− y

K = 4.5 =

x+y

N 2O + O 2

( x + y)( x − y) 4−x

x+ y

y

and

x + y = 4.5

5 17 and x = 3 6 4.5 × 5 × 6 Kc = = 6.428. 3×7 y=

⇒

371. Let initial volume = 1 L. N 2O 4 0.487

2NO 2

0.0475

K c = 4.633 × 10 −3 After compression moles : N 2O 4 = 0.487 + x; NO 2 = 0.0475 − 2x ⇒ Solving, x = 0.041

Kc = and

(0.0475 − 2x ) 2 × 2 = 4.633 × 10 −3 (0.487 + x )

0.0068

The 1st value of x is rejectable, therefore [NO 2 ] = 0.0678, [N 2O 4 ] = 0.9876 372. Let the initial partial pressure of NOCl be p0 . 2NOCl p0 − p1

2NO + Cl 2 p1

p1 = 1.2 ⇒ 2 At equilibrium, additional pressure of Cl 2 = 7.8 Total P = 1 +

p1 2

p1 = 0.4

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Solutions

Therefore, at new equilibrium : 2NO + Cl 2

2NOCl p0 − 0.4 + p2

0.4 − p2

8−

p2 2

p2 = 8.9 ⇒ p2 = 0.2 2 Now, equating equilibrium constant for the above two stages : Total P = 9 −

0.2(0.4) 2

=

( p0 − 0.4) 2

2

 p − 0.2  7.9 × 0.04 ⇒  0  = − 0.4 0.2 × 0.16 p   0

7.9(0.2) 2 ( p0 − 0.2) 2

⇒

p0 = 0.493

K P = 3.7 bar. 373. The equilibrium reaction is : PCl 3 ( g ) + Cl 2

PCl 5 ( g ) Kp =

α

2

1−α2

P = 1.78 ⇒ P = 4.8 (Pressure due to PCl 5 , PCl 3 and Cl 2 )

⇒ Now,

PSolvent = 0.7 P (PCl 5 + PCl 3 + Cl 2 ) 0.1(1 + α ) = P (solvent) nSolvent

⇒ nSolvent = 0.022,

mass = 3.388 g.

374. At equilibrium P (Cl 2O 7 ) =10, P (O 2 ) = 70, P (Cl 2 ) = 20 ⇒ K P =

100 (20) 2 ( 70) 7

= 3.035 × 10 −14

A when gases are taken in equimolar ratio, mole-fractions are : Cl 2O 7 = 0.06, Cl 2 = 0.545, O 2 = 0.395 ⇒ KP =

(0.06) 2 2

(0.545) (0.395)

7

⋅

1 P7

= 3.035 × 10 −14

⇒ P = 115 bar.

375. Molar ratio of the gases coming out of flask initially will be equal to their rate of effusion. PCl 5 1−α

PCl 3 + Cl 2 α

α

Here, α = degree of dissociation. r (Cl 2 ) 137.5 0.53 = = ⇒ r ( PCl 3 ) 71 x where x = mole fraction of PCl 3 in the gaseous mixture coming out initially Hence, mole fraction of PCl 5 outside of flask = 0.09 r ( PCl 5 ) 1 − α 0.09 71 Now, = = ⇒ α = 0.77 r (Cl 2 ) 208.5 0.53 α Kp =

(0.77) 2 = 2.577 bar. 0.23

⇒ x = 0.38

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372

A ( s)

376.

2p 2 p − 0.2 p

Case I Case II

KP = 4 p3

+ C (g )

2B ( g )

p p + 1 − 0.1p

K P = (1.8 p) 2 (0.9 p + 1) = 4 p 3 ⇒

P = 2.988 bar P (case II ) = 2.7 p + 1 = 9.06 bar 377. The exchange equilibrium is : [Ca - EDTA]2– + Pb 2+ [Pb - EDTA]2− + Ca 2+ K=

{[Pb - EDTA]2– }[Ca 2+ ] 2–

{[Ca – EDTA] }[Pb

2+

]

=

1018 10

5 × 10

= 2 × 10 7

The very high K value for the exchange reaction indicate that all Pb 2+ will be converted into [Pb – EDTA]2− , but since concentration of both Ca 2+ and [Cd – EDTA]2− are very high, will remain unchanged. Therefore : Ratio

[Pb – EDTA]2– [ Pb

2+

]

=

K[Ca – EDTA]2– [Ca

2+

]

= 2 × 10 7

1 = 8 × 10 6 2.5

378. From the thermodynamic informations : ∆H ° = 90.5 kJ and ∆S ° = 285.5 JK –1 ⇒ K = 0.143 = p 2 ( p = partial pressure of NH 3 ) ⇒ p = 0.378 bar and P (eqm) = 2P = 0.756 bar. 379. ∆G ° (reaction) = 2 × 52 − 98 = 6 kJ = − RT ln K ⇒ K = 0.09 ⇒

∆G ° = 4.85 kJ = – RT ln K

Also, for

N 2O 4 K=

4α

2

1−α2

2NO 2

P = 0.09 ⇒ α = 0.1483

Also, since reaction is occurring at constant pressure and temperature : v ∝ n ⇒ % volume increase would be 14.83. 5 9 Let at new equilibrium partial pressure of AB ( g ) = x and of AB 2 ( g ) = y bar.

380. AB 2 ( g ) + A ( s)

⇒

2 AB ( g )

K=

5 x2 = y 9

Also x + y = 0.4 Solving, Eqs. (i) and (ii) gives x = 0.27; y = 0.13  0.13  Volume % of AB 2 ( g ) =   × 100 = 32.5  0.4  Volume % of AB ( g ) = 67.5.

…(i) …(ii)

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Solutions

381. Since, volume of container is 1.0 litre, concentrations of each species will be equal to their moles. Now, setting the equilibrium table N 2 ( g ) + 3H 2 ( g ) 2NH 3 ( g ) 1 − x 3 − 3x 2x − y NH 3 ( g ) + H 2S( g )

K p = 8 × 10 –3 = K c ( RT ) –1

NH 4 HS( g ),

y ⇒ K c = 0.4592 y also 2x − y = 0.9 K c = 0.4592 = (1 − y) (2x − y)

2x − y

⇒

1− y

1− y = 2.42 y

⇒

⇒

y = 0.3

and

(2x − y) 2

=

x = 0.6

(0.9) 2

⇒

K c (for the 1st reaction) =

⇒

K p (for 1st reaction) = K c ( RT ) –2 = 3.55 × 10 –4

27 (1 − x ) 4

27 (0.4) 4

= 1.17

[ H 2S] = 1 − y = 0.7 M 382. Let P0 be the initial pressure of N 2O 4 ( g ), then: N 2O 4 ( g )

2NO 2 ( g )

P0 − 0.3 P0

0.6 P0

Kp =

18 P0 35

…(i)

On doubling volume, the initial pressure will be halved. N 2O 4 ( g ) 2NO 2 ( g ) P0 –P 2

2P

2

⇒

⇒ ⇒

Kp =

 2P  4P 2 1 =   P02 P0  P0   2P  P0 −P  1 − 2 P0  2 

=

α2 18 ⋅ 2P0 = P0 1−α 35

{from Eq. (i)}

35 α 2 + 9 α − 9 = 0 α=

− 9 + 81 + 36 × 35 70

= 0.3945

⇒ 39.45% N 2O 4 dissociated. 383. ∆H ° (Reaction) = ∆H °f ( N 2O 5 ) − ∆H °f (O 3 ) − 2∆H °f ( NO 2 ) = 11 − 143 − 2 × 33 = − 198 kJ Q Reaction turned spontaneous to non-spontaneous at 1175 K implies that at this temperature ∆G° = 0. i. e.,

− 198 × 10 3 − 1175 ∆S ° = 0

⇒

∆S ° = −168.5 JK –1

Now at 500 K, ∆G° = − 198 × 10 3 − 500 ( − 168.5) = − 113.75 × 10 3 Also,

∆G ° = − RT ln K = − 113.75 × 10 3

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374

⇒

ln K = 27.36 and

K p = 7.65 × 1011

384. (a) ∆H ° (Reaction) = ∆H °f ( Products) − ∆H °f ( Reactants) − 55 = ∆H °f ( N 2O 5 ) − 2 × 33.2

⇒ Also,

∆H °f (N 2O 5 ) = + 11.4 kJ

⇒

∆S ° (Reaction) = S ° (Products) − S ° (Reactants) − 227 = S ° (N 2O 5 ) – 2 × 240 – 12 × 205

⇒

3

S ° (N 2O 5 ) = 355.5

3

∆G° = − 55 × 10 + 298 × 227 = 12.646 × 10 J (b) Q ∆G° > 0, reaction is non-spontaneous at 298 K. (c) Q ∆G° > 0, increasing temperature will drive the reaction in backward direction. kP CH 3 CH 3 H CH 3 1 385. C==C C==C ∆G °1 H H H CH 3 ∆G °2

∆G °3

k P2

k P3

H

H 3C C==C H 3C

H

∆G1° = 62.97 − 65.85 = − 2.88 kJ = − RT ln K P1 ⇒

ln K P1 =

2880 = 0.693 ⇒ K P1 = 2 8.314 × 500

∆G2° = 58.07 − 65.85 = − 7.78 kJ = − RT ln K P2 ⇒

ln K P2 =

7780 8.314 × 500

⇒ K P2 = 6.5

Now let us consider initially there was 1.0 mol of cis-2-butene only and at equilibrium, there are x-mol trans-2-butene and y mol 2-methyl propene. y x ⇒ K P1 = K P2 = = 2, = 6.5 1− x − y 1− x − y Solving, x = 0.21 and y = 0.6842 ⇒ At equilibrium, mixture contain 21% trans-2-butene, 68.42% 2-methyl propene and rest 10.58% cis-2-butene. 386. Let x mol PCl 5 ( g ) decomposes into PCl 3 ( g ) and Cl 2 ( g ) as: PCl 5 ( g ) 0.02 − x

At equilibrium.: ⇒

PCl 3 ( g ) + Cl 2 ( g ) x

x

Total moles of gaseous species = 0.02 + x + 0.015 = 0.035 + x Partial pressure : PCl 5 ( g ) =

0.02 – x × 1.843 0.035 + x

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Solutions

Cl 2 ( g ) = PCl 3 ( g ) = ⇒

K P = 0.4 =

x × 1.843 0.035 + x

x 2 × 1.843 (0.035 + x ) (0.02 − x )

Solving, x = 0.01 ⇒ Total moles at equlibrium = 0.035 + 0.01 = 0.045 0.045 × 0.082 × 500 V= = 1.0 L. ⇒ 1.843 387. Let initially, there are 100 moles of gaseous mixture. Then SO 2 + 12 O 2 SO 3 15

20

65

After changing temperature to T, let x mole of SO 2 is consumed. Now at new equilibrium : SO 2 = 15 − x O 2 = 20 − x/ 2 SO 3 = 65 + x x  From the given condition : 65 + x = 0.8 100 −  ⇒ x =10.7  2 Therefore, partial pressures at new equilibrium are : 75.7 4.3 14.65 atm, SO 2 = , O2 = SO 3 = 94.65 94.65 94.65 75.7 94.65 94.65 × × = 44.75 94.65 4.3 14.65 22600 21 Also, ∆G ° = − RT ln K P = − 22600 + 21 T ln K P = − = 3.8 ⇒ RT R Solving, T = 429.7 K. 388. Initially : CO( g ) + H 2O( g ) CO 2 ( g ) + H 2 ( g ) 1− x 1− x x x ⇒

K P (T ) =

 x  K =9=   1 − x 

2

⇒ x=

3 4

1 3 and moles of CO 2 = H 2 = . Now, let us assume that 4 4 CO 2 ( g ) flask contained x mole of CO 2 ( g ). Then CO( g ) + H 2O( g ) CO 2 ( g ) + H 2 (g ) 1 1 1 3        3 Also y =  + y  + y  + x − y  − y     4 4 4 4 4 0.5 (0.5 − x ) K =9= ⇒ x=4 ⇒ (0.5) (0.5) At this equilibrium, moles of CO = H 2O =

Hence, volume of CO 2 ( g ) flask = 4.00 litre.

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H2

I2

2HI

0.5 − x

0.5 − x

2x

0.8 − y

0.8 − y

2y

389.

+

2

 2y   2x  K =   =  0.8 − y   0.5 − x  ⇒ 390. For

2x = 0.15 V 2y ⇒ =? V ⇒

2

⇒

y =1.6 x

2 y 3.2x = × 0.15 = 0.24 M V 2x C( s) + CO 2 ( g ) 2CO( g ) KP =

⇒ ⇒

2 PCO = 100 PCO2 = 63 PCO2

PCO2 = 0.63 and PCO = 6.3 atmosphere P (total) = 6.3 + 0.63 = 6.93 atmosphere.

IONIC EQUILIBRIUM 391. For weak acid HA, the ionization equilibrium is : HA

C (1 − α ) +

[H ] = 10

Also, ⇒

Ka =

−2

H+ + A − Cα

Cα

= Cα = 0.1 α

⇒ α = 0.1

Cα 2 = 1.1 × 10 –3 1−α

Now, let us assume that 100 mL 0.1 M HCl is added to V mL of the above weak acid. m mol of HCl added = 100 × 0.1 = 10 m mol of A − in the solution = x m mol of HA = 0.1 V If, in presence of HCl, x m mol of HA is ionized, m mol of H + in the solution = x + 10 m mol of A − in the solution = x m mol of unionized HA = 0.1V − x ⇒ and ⇒

Ka =

+

[H ][ A − ] (x + 10)(x ) = = 1.1 × 10 −3 [ HA ] (100 + V )(0.1V − x ) [H + ] = 2 × 10 −2 = 2 × 10 −2 x = 1.1 × 10 −3 (0.1 V − x )

x + 10 100 + V ⇒ V =192 x

…(i)

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Solutions

Substituting in [H + ] expression : 2 × 10 −2 = ⇒ 392. Let dibasic acid be H 2 A :

V + 1920 192(100 + V )

V = 540 mL H2 A HA −

H + + A 2−

[ A 2− ] = Ka 2 = 3 × 10 −6

⇒

[H + ] = [HA − ] + [ A 2− ] =

and ⇒

H + + HA −

[H + ] =

1.4 × 10 –6 +

[H ]

Ka1 [ H 2 A ] [H + ]

+ 3 × 10 −6

+ [ A 2− ]

⇒ [H + ] = 1.18 × 10 −3 pH = 2.926

393. Enthalpy of ionization of formic acid is 15 kJ/mol ⇒

K ( 47° C) 15 × 10 3 = K (27° C) 8.314

ln

  20  = 0.376   300 × 320 

K a ( 47° C) = 2.47 × 10 −4

⇒

[H + ] = K a C = 4.97 × 10 −3 394. [D + ] = 10 −7 +

1.35 × 10 −15 +

⇒ pH = 2.3

⇒ [D + ] = 1.12 × 10 −7

[D ] 395. The hydrolysis equilibrium is :

OCl – + H 2O Kh = =

⇒ pD = 6.95

HOCl + OH −

Kw 10 −14 = = 3.33 × 10 −7 K a 3 × 10 −8 [HO – ][HOCl] [OCl – ]

=

[HOCl] [OCl – ]

=

3.33 × 10 −7 [ HO – ]

= 0.528

⇒ % (by mole) HOCl = 34.55, OCl − = 65.45. 396.

PV = 4.64 × 10 −4 RT moles of NaOH used up = moles of CH 3COOH produced from decomposition = 7.8 × 10 −4 n(dimer) =

⇒ Moles of dimer dissociated = 397.

7.8 × 10 −4 = 3.9 × 10 −4 2

⇒ α=

3.9 × 10 −4 4.64 × 10 −4

[CO 2 ] = 2.28 × 10 −3 × 3.2 = 7.296 × 10 −3 M [H + ] = K1 [CO 2 ] = 5.53 × 10 −5

⇒ pH = 4.25

= 0.84

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398. [OCl – ] = 0.67 :

OCl – + H 2O

HOCl + OH –

[OH − ] = K h C = 4.72 × 10 −4 ⇒

pOH = 3.325 and pH = 10.675 [S ] pOH = pK b + log ⇒ [ S ] = 0.5 [B ]

399.

m mol of NH 4Cl = 100 m mol of NH 3 = 50 After adding 7.5 mmol KOH; m mol of NH 4Cl = 92.5 m mol of NH 3 = 57.5 92.5 ⇒ pOH = pK b + log = 4.9 57.5 ⇒ pH = 9.1 and ∆ pH = 0.1. 400. mol of CH 3COOH = 0.25, mol of CH 3COONa = 0.30 0.3 = 4.78 0.25 306.25 pH = pK a + log = 4.799 243.75 290 pH = pK a + log = 4.745 260 pH = pK a + log

(a) (b) (c) 401. K a =

Cα 2 ; Solving gives α = 0.27 1−α

⇒ van’t Hoff factor ( i) = 1 + α = 1.27 − ∆T f = iK f m = 1.27 × 1.86 × ln

402.

⇒ ⇒

1.99 × 10 −15 3.16 × 10 ln

−16

=

0.5 × 1000 = 1.26 935.5

 ∆H  10   8.314  293 × 303 

K w (50° C) 3.16 × 10 −16

=

⇒ ∆H =136.2 kJ

 136.2 × 10 3  30   8.314  293 × 323 

K w (50° C) = 5.7 × 10 −14 pD = 6.62

403. MOH has lower solubility ( S = 10 −6 ) and will precipitate first of pH = 8 404. K sp = 1.6 × 10 −14 = [Fe 2+ ][OH – ]2 (i) At pH = 8, [OH − ] = 10 −6 , [Fe 2+ ] = Solubility = 1.6 × 10 −2 (ii) At pH = 6, [OH − ] = 10 −8 , [Fe 2+ ] = 160 indicates very high solubility.

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Solutions

Ca 2+ + 2F −

CaF2

405.

F – + H+

HF

Let ‘α’ be the fraction of total fluoride ion produced in F − form Ka [F – ] [F – ] α= = − = + 2S [F ] + [H ] K a + [H + ] (a) At pH = 7, [H + ] = 10 −7 1.8 but < 7.2, only CoS will precipitate. At pH > 7.2, MnS will start precipitating. 457. The cell reactions are: H 2 → 2H + + 2e 2Ag + + 2e → 2Ag

E ° = 0.8 V

H 2 + 2Ag + → 2H + + 2Ag E = 0.589 = 0.8 –

+ 2

[H ] 0.059 log 2 [Ag + ]2

S = 1.325 × 10 –4

⇒ [Ag + ] = 2.65 × 10 –4

and 2 S = 2.65 × 10 −4

K sp = [Ag + ]2 [C 2O 42– ] = 4 S 3 ⇒

E ° = 0.8 V

K sp = 4 S 3 = 9.3 × 10 –12 .

⇒

458. For the dissolution reaction: Zn(OH) 2 + 2OH – 0.1 − 2x

x

K = 0.588 =

Zn(OH) 2– 4

x (0.1 − 2x )

2

K = K sp ⋅K f = 0.588

x

=

x 10

–2

− 4x 2 − 0.4 x

x = 0.0048 M, i.e., 0.0048 × 250 = 1.2 m mol Zn(OH) 2 dissolved. 150 mmol of Zn(OH) 2 taken initially = = 1.515 99 ⇒ mmol Zn(OH) 2 left undissolved = 1.515 − 1.2 = 0.315 ⇒ mass of Zn(OH) 2 left undissolved = 31.185 mg 459. Reaction for dissolution of AgCl in ammonia is : Solving:

AgCl + 2NH 3 x

0.1 − 2x

K = 4 × 10 –3 =

⇒

Ag(NH 3 ) +2 + Cl – 0.1 + x

x

x (0.1 + x ) (0.1 − 2x ) 2

K = K sp ⋅K f = 4 × 10 –3

=

0.1 x + x 2 10 −2 + 4x 2 − 0.4 x

⇒

0.1 x + x 2 = 4 × 10 –5 + 1.6 × 10 –2 x 2 − 1.6 x × 10 –3

or

0.984 x 2 + 0.1016 x − 4 × 10 −5 = 0

⇒

x=

−0.1016 + (0.1016) 2 + 16 × 10 −5 (0.984) 2 × 0.984

460. The order in which halide ions should be added is :

= 3.92 × 10 –4 M

Cl – than Br – than I – .

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Solutions

When Cl – is added to 0.1 M, [Ag + ] = When Br – is added to 0.1 M, [Ag + ] =

K sp ( AgCl) 0.10 K sp (AgBr)

= 1.8 × 10 −9 M

0.10 K sp ( AgI)

= 5 × 10 −12 M

= 8.4 × 10 −16 0.10 The [Ag + ] in equilibrium with AgCl in 0.1 M Cl − is1.8 ×10 –9 M, which is more than 5 × 10 −12 M ([Ag + ] required to initiate precipitation of AgBr when [Br – ] = 0.1 M). Hence, the solution containing [Ag + ] = 1.8 × 10 –9 in equilibrium with AgCl in 0.1 M Cl – , will give precipitate of AgBr when Br – is added to 0.1 M. Had we added Br – to 0.1 M before Cl – , [Ag + ] would have been 5 × 10 −12 M and on making this solution 0.1 M in Cl – in the subsequent step would not have yielded AgCl precipitate. Hence the order. 461. As we know, a buffer solution shows its best buffering action near to its ionization constant value. Therefore, the above buffer can best be prepared by taking H 2 PO 4– and HPO 2– 4 . When I – is added to 0.1 M, [Ag + ] =

(a) pH = pK 2 + log

[ HPO 24– ] [H 2 PO 4– ]

= 7 ⇒ log

[ HPO 24– ] [ H 2 PO 4– ]

= − 0.22 and

[HPO 2– 4 ] [ H 2 PO 4– ]

= 0.6

(b) In 50 mL buffer, [H 2 PO 4– ] = 0.10 M = 5 m mol ⇒ [HPO 2– 4 ] = 0.06 M = 3 m mol m mol of NaOH added = 20 × 0.1 = 2 This added NaOH will neutralize 2.0 m mol of H 2 PO –4 ⇒

m mol of H 2 PO 4– = 5 – 2 = 3 m mol of HPO 24– = 3 + 2 = 5 pH = pK 2 + log

5 3

= 7.44

462. From the titration curve it is concluded that the 1st end point correspond to completion of the following reaction : Na 2CO 3 + HCl → NaHCO 3 + NaCl ⇒ m mol of HCl consumed = 11 × 0.115 = 1.265 = m mol of Na 2CO 3 ⇒ mass of Na 2CO 3 = 1.265 × 106 × 10 –3 = 0.134 g Second end point corresponds to the following reaction : NaHCO 3 + HCl → NaCl + H 2O + CO 2 Volume of HCl consumed = 35 − 11 = 24 mL Out of this 24 mL, 11 mL HCl would be consumed by NaHCO 3 produced from Na 2CO 3 . ⇒ m mol of original NaHCO 3 = (24 – 11)0.115 = 1.495 ⇒ mass of NaHCO 3 = 1.495 × 84 × 10 –3 = 0.1256 g 0.134 × 100 = 26.8 0.5 0.1256 × 100 = 25.12 ⇒ m% of NaHCO 3 = 0.5 m% of KCl = 48.08%. ⇒ m% of Na 2CO 3 =

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10 −3 = 8.165 × 10 −2 M 0.15 i.e., to prevent precipitation, [H 2 PO 4– ] should be less than 8.165 × 10 −2 M.

463. K sp = 10 −3 = [Ca 2+ ][H 2 PO –4 ]2

Also

⇒ [H 2 PO –4 ] =

H + + H 2 PO 4–

H 3 PO 4 K a1 =

K a [ H 3 PO 4 ] 7 × 10 –3 × 0.25 [H + ] [H 2 PO 4– ] ⇒ [H + ] = 1 = [H 3 PO 4 ] [H 2 PO 4– ] 8.165 × 10 −2 = 2.14 × 10 −2 M

i.e., to prevent precipitation, [H + ]> 2.14 × 10 –2 , pH = 1.66. i.e., at pH below 1.66, no precipitation of Ca(H 2 PO 4 ) 2 will occur. 464. (a) Addition of HNO 3 will neutralize OH – in solution driving solubility equilibrium in forward direction and this will increase concentration of Fe 3+ in solution. Also, as Fe 3+ comes in solution due to neutralization of OH – with HNO 3 , it will form complex with SCN – present in solution due to high K f value, but still, the overall effect will be increase in concentration of Fe 3+ in solution. (b) Addition of NaOH will drive both equilibrium (i) and (iii) in backward direction and it will decrease concentration of Fe 3+ ion in solution. (c) Addition of NaSCN will drive equilibrium (ii) in forward direction and concentration of Fe 3+ ( aq ) will decrease. (d) Addition of Fe(OH) 3 ( s) will have no effect on Fe 3+ ( aq ) concentration since active mass of former is unity. 465. The dissociation equilibrium of the complex is: [Cu(CN) 4 ]2−

Cu 2+ ( aq ) + 4CN – ( aq )

As strong acid is added, following equilibrium will be established parallely : 1 = 1.67 × 10 9 H + + CN – HCN K = Ka The above equilibrium will lie predominantly to right, removing most of the cyanide ion from 1st equilibrium. This will increase concentration of Cu 2+ in solution. 4 × 1000 466. m mol of NaCl added = = 68.38 58.5 m mol of Ag + in solution = 25 Now:

AgCl( s) + 2NH 3 ( aq ) x − 0.2

Ag(NH 3 ) +2 ( aq ) + Cl – ( aq ) 0.1

K = K sp ⋅K f = 2 × 10 −3 = ⇒

2.735 × 10 x − 0.2

x =13.875 M

468. The solubility equilibrium is: K = K sp ⋅ K f

0.2735 −2

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Solutions

Co(OH) 3 + 4SCN – 1 − 0.4

At the given conditions : Q=

(0.1) (10 –7 ) 3

K = 1.6 × 10 −41

Co(SCN) –4 + 3OH – 10–7

0.1

= 7.716 × 10 −22 >> K (1.6 × 10 –41 )

(0.6) 4

Therefore, precipitation must occur. 469. [OH − ] of buffer = 2 × 10 −5 ⇒ [H + ] = 5 × 10 −10 5 × 10 −10 x 0.1 − x −3 −10 Solving : x = 8.6 × 10 M 0.1 − x 5 × 10 x Now, on adding equal volume of 0.1 M Ba 2+ ion, concentration of Ba 2+ ion in final solution will become 0.05 M and that of carbonate ion will become 4.3 × 10 −3 M. HCO –3

Now:

K a 2 = 4.7 × 10 −11 =

H + + CO 2– 3

–4 K IP = [Ba 2+ ][CO 2– >> K sp , precipitation will occur. 3 ] = 2.15 × 10

⇒

470. For the equilibrium :

HCO –3

K a 2 = 4.7 × 10 −11 = Also

H + + CO 2– 3

[H + ][CO 2– 3 ] [HCO 3– ]

= 10 −8

[CO 23– ] [ HCO 3– ]

⇒ [HCO –3 ] = 213 [CO 2– 3 ]

2– –4 [HCO –3 ] + [CO 2– 3 ] = 0.1 ⇒ [CO 3 ] = 4.67 × 10

[Co 2+ ] =

⇒

[Co 3+ ] =

1.4 × 10 −13 4.67 × 10 −4 1.6 × 10 –44 4.67 × 10

−4

= 3 × 10 −10 M = 3.4 × 10 −41 M

471. The hydrolysis reaction is : C 6 H 5 NH +3 + H 2O

C 6 H 5 NH 3OH + H +

C (1 − α )

Cα

Cα

K 10 −14 = 2 × 10 −5 Kh = w = K b 5 × 10 −10

Also K h =

Cα 2 1−α

⇒ 2 × 10 −5 = α=

⇒

10 −3 α 2 1−α

⇒ α 2 + 2 × 10 −2 α − 2 × 10 −2 = 0

−2 × 10 −2 + (2 × 10 −2 ) 2 + 8 × 10 −2 2

[H + ] = Cα = 10 −3 × 0.13 = 1.3 × 10 −4 pH = 3.88

472. As, acetic acid is added, it will react quantitatively with OI – as: CH 3COOH + OI – CH 3COO – + HOI m moles: 1 2 0 0 0 1 1 1

= 0.13

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K=

K a [ acetic acid ] = 10 6 > very large value. K a ( HOI)

Now the conjugate base of weaker acid (HOI) will hydrolyze predominantly as : K OH – + H 2O HOI + OH – K h = w = 5 × 10 −4 Ka Concentration of OI – left unreacted after reaction with CH 3COOH = 5 × 10 –3 Concentration of HOI produced by reaction of CH 3COOH with IO – = 5 × 10 –3 Now, if x MIO – is hydrolyzed then : IO –

+

H 2O

HOI

5 × 10−3 + x

5 × 10–3 − x

K h = 5 × 10 −4 =

(5 × 10

⇒

x=

5 × 10 x + x 2 5 × 10 −3 − x

2 pH = 10.63

ClC 6 H 4 NH 3OH + H +

C (1 − α )

Cα

K h = 2.5 × 10 −3 =

2

Ka = when 75% ionized

+

Kh =

Kw = 2.5 × 10 −3 Kb

⇒ α 2 + 0.25 α − 0.25 = 0

−0.25 + 0.0625 + 1 = 0.39 2

[H + ] = Cα = 3.9 × 10 –3

⇒

= 4.22 × 10 −4

Cα

−2

2

Cα 10 α = 1−α 1−α

α=

and

–

+

and

(not negligible)

pH = 2.4

+

[H ][dnp ] [H ][25] [H ] ⇒ pH1 = 3.48 = = 75 3 [dnp]

[H + ][ 75] = 1.1 × 10 −4 25 [H + ] =

Mg(OH) 2

1.1 × 10 −4 3

⇒ pH = 4.53

Mg 2+ ( aq ) + 2OH – ( aq )

K sp = [Mg 2+ ][OH – ]2 ⇒

=

x −3

−5.5 × 10 –3 + 3.025 × 10 −5 + 10 −5

ClC 6 H 4 NH +3 + H 2O

475. For

+ x )x

5 × 10 −3 − x

⇒ pOH = 3.37 and 473. The hydrolysis reaction is :

⇒

−3

x 2 + 5.5 × 10 −3 x − 2.5 × 10 −6 = 0

⇒

474.

OH –

+

⇒ 15 . × 10 −11 = 0.1 [OH – ]2

[OH – ] = 1.224 × 10 –5

This is the minimum concentration of OH – required to begin the precipitation of Mg(OH) 2 . Therefore, [OH – ] ≤ 1.224 × 10 –5 M.

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Solutions

Now for

NH +4 + OH –

NH 3 + H 2O Kb =

[NH +4 ][OH – ] 2 × 10 –5 × 0.1 ⇒ [NH +4 ] = = 0.16 M [NH 3 ] 1.224 × 10 –5

i.e., to maintain [OH – ] equal to 1.224 × 10 −5 M or less, the concentration of ammonium ion in solution must be greater than or equal to 0.16 M. 476. The equilibria existing are: Ba 2+ ( aq ) + SO 2– 4 ( aq )

BaSO 4 ( s)

x+ y

y

SO 2– 4 + H 2O

HSO –4 + OH –

Kb =

Kw = 8.33 × 10 −13 Ka

The above K b value indicates that tendency of hydrolysis of SO 2– 4 is very less. Also, Here x M

HSO –4

HSO –4

H + + SO 2– 4

0.25 − x

x

x+ y

is ionized and y M BaSO 4 dissolved. K sp = 10 –10 = y( x + y)

⇒

K a = 12 . × 10 −2 = ⇒

83 . × 10 –9 =

y(0.25 − x ) x

…(1)

x ( x + y) 0.25 − x

…(2)

⇒ x >> y and 1.2 × 10 −2 =

x2 0.25 − x

10 −10 10 −10 ≈ = 2.24 × 10 −8 M x+ y x 10 −5 Solubility in water = K sp = 10 −5 ⇒ factor = = 446 2.24 × 10 −8 477. Initial mmol of Na 2CO 3 = 500 × 0.1 = 50 Now, let us assume that x mmol of HCl is added to adjust the desired pH. Then Na 2CO 3 + HCl → NaHCO 3 + NaCl Solving x = 4.46 × 10 –3

⇒

y=

50 − x

0

x

x

The leftover Na 2CO 3 will hydrolyze to give the desired pH as : CO 2– 3 + H 2O Kh =

HCO –3 + OH –

Kw [OH – ][HCO –3 ] 10 −4 x = 2 × 10 −4 = = K a2 50 − x [CO 23– ]

100 1 = ×V 3 3 ⇒ V =100 mL i.e., to prepare the desired solution, 100 mL of the given HCl will have to be added and finally solution will be diluted to 1000 mL. ⇒

x=

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478. Initial (M) : Final (M) : ⇒

Ag + + 2CN – Ag(CN) –2 0.03 0.1 ≈0 0.1 − 0.06 0.03 K f = 25 . × 1018 =

[ Ag(CN) 2 ]− +

– 2

[Ag ][CN ]

=

0.03 +

[Ag ] [0.04]2

[Ag + ] = 7.5 × 10 –18 M

⇒

479. (a) For CaF2 : 3.4 × 10 –11 = [Ca 2+ ][F – ]2 = [Ca 2+ ] ( 0.05) 2 ⇒ [Ca 2+ ] required to begin precipitation of CaF2 = 13.6 × 10 –9 M 2+ For CaCO 3 : 9.5 × 10 –9 = [Ca 2+ ][CO 2– ][0.02] 3 ] = [Ca 2+ ⇒ [Ca ] required to begin precipitation of CaCO 3 = 4.75 × 10 –7 N i.e., higher concentration of Ca 2+ is required to start precipitation of CaCO 3 , CaF2 will precipitate first. (b) When CaCO 3 start precipitating, the required [Ca 2+ ] is 4.75 ×10 –7 . To commence simultaneous precipitation of CaF2 and CaCO 3 at this stage, the minimum [F – ] can be evaluated as:

3.4 × 10 −11 = 4.75 × 10 −7 [F – ]2 [F – ] = 8.46 ×10 –3 M.

⇒ 480. (a) For

PbI 2 ( s)

Pb 2+ ( aq ) + 2I – ( aq )

∆G ° (at 17° C) = − RT ln K sp = − 8.314 × 290 ln ( 7.47 × 10 –9 ) = 45.1166 kJ and

∆G ° (at 27° C) = −RT ln K sp = − 8.314 × 300 ln (1.4 × 10 −8 ) = 45.1056 kJ

Now

∆G ° (at 17° C) = 45116.6 = ∆H ° – 290 ∆S ° ∆G ° (at 27° C) = 45105.6 = ∆H ° – 300 ∆S °

From Eqs. (i) and (ii), 11 = 10 ∆S ° ⇒ ∆S ° = 1.1 JK –1 Substituting, ∆S ° = 1.1 in Eq. (i): ∆H ° = 45116.6 + 290 × 1.1 = 45.4356 kJ mol –1 (b) ∆G ° (at 77° C) = ∆H ° – 350 × 1.1 = 45.8206 kJ = − RT ln K sp ⇒

K sp = 1.45 × 10 −7 = 4 S 3

481. Normality of solution = Q

2.54 × 2 = 0.0318 N 159.5 K ×1000 N

Λ eq = =

Also

⇒ S = 3.3 × 10 –3 M

⇒ k (specific conductance)

91 × 0.0318 = 2.89 × 10 –3 Ω –1cm –1 1000

1  A =k    l R

⇒ R=

1 1 l  1  =   = 346 Ω. k  A  2.89 × 10 –3  1

…(i) …(ii)

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Solutions

482. For a weak electrolyte: Λc 10 =α = = 0.042 238 Λ∞ For NH 3 :

NH +4 + OH –

NH 3 + H 2O C (1 − α )

Kb =

Cα −2

2

Cα

2

10 (0.042) Cα = = 1.8 × 10 –5 1−α 1 − 0.042

483. The ionization reaction of acetic acid is: CH 3COO – + H +

CH 3COOH C (1 − α )

Cα

K a = 2 × 10 –5

α 2 + 2 × 10 –3 α − 2 × 10 –3 = 0

⇒ α=

⇒ Now ⇒ Also ⇒ 484. G =

Cα

Cα 2 10 –2 α 2 = = 1−α 1−α

−2 × 10 −3 + 4 × 10 −6 + 8 × 10 −3 2

= 0.0437

Λc = 0.0437 Λ∞ Λ c =17.043 cm 2 Ω –1eq –1 k ×1000 Λc = C Λ C 17.043 × 0.01 k= c = = 1.704 × 10 –4 Ω –1cm –1 . 1000 1000

1 1  A = k   = k ⋅ , where K is cell constant ( l/A ).  l R K K = Rk = 25 × 0.0116 = 0.29 cm –1

⇒

For acetic, acid solution, K = 0.29 cm –1 = kR ⇒ ⇒

k (CH 3COOH) = Λm =

0.29 cm –1 = 1.463 × 10 –4 Ω –1cm –1 1982 Ω

k × 1000 1.463 × 10 −4 = × 1000 = 1.463 cm 2 mol –1 Ω –1 . 0.1 C

485. Initially, only NaOH was present at 0.1 M. ⇒

Λ m (NaOH) =

k × 1000 0.022 × 1000 = = 220 Ω –1cm 2 mol –1 . C 0.1

After adding equal volume of HCl, NaOH will be neutralized completely giving NaCl of concentration 0.05 (since volume is doubled, concentration of salt will be halved).

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0.0056 × 1000 =112 Ω –1cm 2 mol –1 . 0.05 On further adding HCl, no neutralization will occur and conductivity will be due to NaCl and HCl 2 2 together. Now the concentrations of NaCl = × 0.05 and concentration of HCl = × 0.05 M. 3 3 C Now : k = k ( NaCl) + k ( HCl) = ( Λ NaCl + Λ HCl ) × 1000 1000 × 0.017 ⇒ × 3 = 112 + Λ HCl 2 × 0.05 ⇒

Λ NaCl =

Λ HCl = 510 − 112 = 398 Ω –1cm 2 mol –1

⇒ 486. For NaCl solution :

k × 1000 …(i) = 126.5 C Here the 1st objective is to determine k (specific conductance) of NaCl and then C (NaCl). 1  A k Also; conductance …(ii) G= =k   =  l K R Λm =

Λm =

For KCl solution : ⇒

k (KCl) =

k ×1000 C

138.3 × 0.02 = 2.766 ×10 –3 Ω –1cm –1 1000

Now from Eq. (ii), K (cell constant) = kR = 2.766 × 10 −3 × 85 = 0.235 cm –1 Now; conductance of water ⇒

GH 2O =

Conductance of

NaCl( aq ) = GNaCl( aq ) =

⇒ Conductance due to NaCl alone = ⇒

1 9200 1 7600

1 1 − = 22.88 × 10 –6 Ω –1 7600 9200

k (NaCl) = GK = 22.88 × 10 –6 Ω –1 × 0.235 cm –1 = 5.38 × 10 −6 Ω –1cm –1

Substituting, in Eq. (i) : 126.5 = ⇒ ⇒

5.38 × 10 −6 × 1000 C

C (NaCl) = 4.25 ×10 –5 M Volume of swimming pool =

Number of moles of NaCl 500 = Molarity 58.5 × 4.25 × 10 −5

= 2.011 × 10 5 L

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ΛC 195 × 0.05 = = 9.7 × 10 −3 Ω –1cm –1 1000 1000 1 1.5  A = 0.0291 Ω –1 (G ) = = k   = 9.7 × 10 −3 ×  l 0.5 R

487. k (specific conductance) = Conductance

R = 34.36 Ω =

⇒ ⇒ 488. Since, Λ eq ⇒

V I

5 V = = 0.1455 A. R 34.36 k × 1000 K × 1000 where K is cell constant. = = N RN K × 1000 Λ eq ( KCl) = = 138.3 85 × 0.02 K × 1000 Λ eq ( PbCl 2 ) = = 106 70 N

⇒

I=

70 N 138.3 = 106 85 × 0.02

N = 3.168 × 10 −2

⇒

⇒ Molarity of PbCl 2 in its saturated solution =

N = 1.58 × 10 −2 2

K sp = 4 M 3 = 1.6 × 10 −6 .

⇒

ELECTROCHEMISTRY 0.15 × 8 × 3600 faraday 96500 Volume of F2 ( g ) produced =11.2 F = 0.5 L (b) Volume of O 2 ( g ) produced = 5.6 F = 0.25 L 490. Partial pressure of O 2 = 722 − 19.83 = 702.17 torr. Q = It =

489. (a)

⇒ Moles of O 2 ( g ) produced =

25 × 10 −3 702.17 × = 9.548 × 10 −4 760 0.082 × 295

Gram equivalents of O 2 = 3.82 × 10 −3 = 491. Total surface area =11.2 cm 2 Mass of Ag = 11.76 × 10 –3 g

⇒ I = 0.2 ampere

⇒ V = 11.2 × 10 −4 cm 3 ⇒ g eq. of Ag = 1.088 × 10 –4 =

⇒ 492. The cell reaction is :

I × 30 × 60 96500

t =105 seconds. 2Sn

2+

→ Sn 4+ + Sn

Initial moles of SnCl 2 =

19 = 0.1 190

It 96500

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0.119 = 0.001 = moles of SnCl 4 produced. 119 Moles of SnCl 2 left unelectrolysed = 0.1 − 0.002 = 0.098 m(SnCl 2 ) 0.098 × 190 = = 71.34. m(SnCl 4 ) 0.001 × 261

Moles of Sn deposited =

⇒

Moles of Br 2 required =

493.

5 × 10 5 94

10 6 It 20 × 10 3 = = t 94 F 96500 t = 14.258 hr. pH (from Henderson equation) = 0.28, = 4.188 Gram equivalent of Br 2 required =

494. E ° cell

E = E ° − 0.059 log [H + ][Cl − ] = E ° + 0.059 pH = 0.527 V 495. The spontaneous cell reaction is Cu + Br 2

Cu 2+ + 2Br –

E ° = 0.73 V

0.059 log (0.05)(0.1) 2 = 0.827 V 2 In order to reverse the cell reaction, minimum voltage requirement = 0.827 V After deposition of 2.827 g Cu : m mol of CuBr 2 left = 50 − 44.87 = 5.13 ⇒ [CuBr 2 ] = 0.01026 M 0.059 E cell = 0.73 − log (0.01026)(0.02052) 2 = 0.888 V ⇒ 2 Thus, at this stage, minimum voltage required to reverse the reaction is 0.888 V. nE ° F 496. For I2 + I− I 3− E ° = − 0.001 V ⇒ ln K = ⇒ K c = 0.925 RT At the beginning : E cell = 0.73 −

⇒

0.925 =

[I –3 ] −

0.25[I ]

⇒ 0.925 =

x 0.25(0.25 − x )

Solving, x = 0.158 M. Tl → Tl + + e

497. Cd Cell reaction :

2+

0.34

+ 2e → Cd − 0.40

2Tl + Cd 2+ → 2Tl + + Cd E = − 0.06 −

E ° = − 0.06

+ 2

[Tl ] 0.059 log = − 0.036 V 2 [Cd 2+ ]

498. The cell reaction before adding KCl is : M + H + → M + +

1 H2 2

E = − 0.4 = E ° − 0.059 log

[M + ] [H + ]

Solutions

⇒

Join @unacademyplusdiscounts 401 E ° (M + / M) = + 0.4177 V [Cl − ] = 0.1 M

After adding KCl, Cell reaction :

M + Cl − → MCl + e H + + e →

1 2

H2

M + Cl − + H + → MCl + 2 H 2 1

⇒ ⇒ ⇒ E ° for MCl ⇒

− 0.05 = E ° − 0.059 log

1 [Cl ][H + ]

E ° = 0.05 ⇒ E ° ( MCl / M ,Cl – ) = − 0.05 M + + Cl – = − 0.05 − 0.4177 = − 0.4677 V = 0.059 log K sp K sp = 1.18 × 10 −8 . 0.987 = 0.8 − 0.059 log

499. ⇒

–

[H + ] = 5.4 × 10 −4 M Ka =

[H + ] 0.8

[H + ]2 = 2.09 × 10 −4 [HOCN]

500. [Ag + ] in a saturated solution of AgCl = 1.4 × 10 −9 [Ag + ] in a saturated solution of AgBr = 3.3 × 10 −10 The spontaneous direction of reaction will increase concentration of Ag + in AgBr/KBr solution and decrease Ag + in AgCl/KCl solution. Therefore, spontaneous cell reaction is Ag → Ag + ( R ) Ag + ( L) → Ag Ag + ( L) → Ag + ( R ) E = 0 − 0.059 log

[Ag + ]R [ Ag + ]L

0.76 = − E ° Zn 2+ / Zn −

501. ⇒

[Zn 2+ ] 0.059 log 2 [H + ]2

E ° Zn 2+ / Zn = − 0.76 V 0.24 = − E ° Ni 2+ / Ni −

⇒

= 0.037 V

E ° Ni 2+ / Ni = − 0.24 V

[Ni 2+ ] 0.059 log + 2 2 [H ]

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Zn + Ni 2+

⇒ For

Zn 2+ + Ni

1.5 − x

E ° = 0.52 V =

x

[Zn 2+ ] x 2.303RT = 1.99 ×1017 log ⇒ 2+ 1.5 − x 2F [Ni ]

Such a large value of equilibrium constant indicate that all Ni 2+ has been quantitatively reduced to Ni. Amount of Zn consumed = 1.5 × 0.75 = 1.125 moles E = E° −

502. (a)

[Mg 2+ ] 0.059 0.059 log log = 3.14 V = 3.17 − 2 + 2 2 [ Ag ]

(b) As the reaction proceed, [Ag + ] will decrease and [Mg 2+ ] will increase. m mol of Ag + initially present = 34.6 m mol of Ag + present finally = 34.6 − 11.11 = 23.48 [Ag + ] = 0.0678 M

⇒

m mol of Mg + produced = 5.55 [Mg + ] = 0.119 M 0.059 0.119 log = 3.128 V E = 3.17 − 2 (0.0678) 2

⇒

E = E° −

503. (a)

[Zn 2+ ] 0.059 0.059 0.2 = 1.1 V log log = 1.1 − 2+ 0.2 2 2 [Cu ]

(i) On adding NH 3 ( aq ) to Cu 2+ solution, complex Cu(NH 3 ) 24+ will be formed reducing [Cu 2+ ], E cell will decrease (ii) On adding NH 3 ( aq ) to Zn 2+ solution, complex Zn(NH 3 ) 24+ will be formed reducing [Zn 2+ ], E cell will increase. (b) After adding NH 3 , following equilibrium will establish: Cu 2+ + 4NH 3 0.68 = 1.1 −

Now,

Cu(NH 3 ) 2+ 4

[Zn 2+ ] 0.059 log 2 [Cu 2+ ]

[Cu 2+ ] = 1.16 × 10 −15 M

Solving,

Such low concentration of Cu 2+ indicate that almost all Cu 2+ has been consumed in complex formation. [Cu(NH 3 ) 2+ 4 ] = 0.1 M, [NH 3 ] = 1.1 M

⇒

Kf = 504. E ° =

∆S ° ∆H ° T− nF nF

0.1 4

(1.1) × 1.16 × 10 –15

= 5.88 × 10 13 .

Solving for E ° at 25°C and 20° gives

∆S ° = − 30.88 JK −1

and

∆H ° = − 77.23 kJ and

∆G° = − 68.03 kJ

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Solutions

505. For the given cell, Nernst equation can be written as : E = C + 0.059 pH, where C is a constant ⇒ 0.129 = 0.059(6.86 − pH1 ) ⇒ pH1 = 4.68 0.059 506. 0.348 = 0.13 − log [Pb 2+ ] ⇒ [Pb 2+ ] = 4.075 × 10 −8 2 K sp = [Pb 2+ ][ F − ]2 = 4.075 × 10 −8 507. The spontaneous cell reaction will be : Zn 2+ (T f = − 2) E =0−

Zn 2+ (T f = − 0.9)

0.059 0.9 = 0.01 V log 2 2

508. The discharging reaction is : Pb + PbO 2 + 2H 2SO 4 → 2PbSO 4 + 2H 2O 23.92 ⇒ m(PbSO 4 ) produced = × 2 × 302 = 60.7 g 238 23.92 g eq. of PbO 2 reduced = × 2 = 0.201 238 ⇒ 0.201 × 96500 = 3 × t ⇒ t = 6465.5 second = 1.8 hour 2ClO –3

509. E ° for

ClO –4 + ClO –2 = – 0.03 V

0.1 − 2x

x

x

K = 9.66 × 10 −2

⇒

2

Therefore,

. × 10 966

−2

 x  =  ; Solving x = 0.019 M  0.1 − 2x 

0.035 = 0.77 − 0.059 log

510.

[Fe 2+ ]

⇒

[Fe 3+ ]

NiO 2 + 4H + + 2e

2e + NiO 2 + 2H 2O

–

Adding

Ni 2+ + 4H 2O

Ni(OH) 2 + 2OH

512. Q

E ° = −1.678 V E ° = − 0.49 V

Ni(OH) 2 + 4H + + 2OH –

2H 2O Subtracting :

[Fe 3+ ]

= 2.87 × 10 12

Ni 2+ + 2H 2O

511.

[Fe 2+ ]

Ni 2+ + 2H 2O dE ° ∆S ° = dT nF

2H + + 2OH – Ni(OH) 2 + 2H +

E ° = − 2.168 V K1 = 4.5 × 10 −74

K 2 = 10 −28 K = K1 / K 2 = 4.5 × 10 −46

⇒ ∆S ° = − 0125 . × 2 × 96500 = − 24.125 kJ K −1

∆G ° = − nE ° F = − 2 × 0.0372 × 96500 = − 7179.6 J ⇒ ∆H ° = − 7196.43 kJ

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513. Let x be the concentration of Ca 2+ ion in diluted sea water. 1 0.059 log − 0.053 = E ° − 2 x 0.059 50 and log − 0.0422 = E ° − 2 50x + 0.05 0.0108 =

⇒

50x + 0.05 0.059 log 2 50x

⇒ x = 7.55 × 10 −4

[Ca 2+ ] in sea water = 10x = 7.55 × 10 −3 M

⇒

514. Moles of Cl 2 ( g ) produced =

9.6 × 10 3 = 82.05 58.5 × 2 P=

515. E = E ° − 0.059 log 516. E = 0 − 0.059 log

[H + ] +

[Ag ] [Ag + ]L

[ Ag + ]R

nRT n2a − 2 = 21.66 atm. V − nb V

= 0.8 − 0.059 log

0.22

= 0.26 V 1.6 × 10 −10 [ Ag + ]L ⇒ 0.4312 = − 0.059 log 0.1

[Ag + ]L = 4.9 × 10 −9

⇒ K sp = [Ag + ][Cl − ] = 4.9 × 10 −9 × 0.05 = 2.4 × 10 −10

E = 0.28 −

517.

[H + ]2 RT = 0.6885 ⇒ [H + ] = 2.57 × 10 −7 ln pH 2 2F

or

pH = 6.6

518. For the given cell : E = E ° + 0.059 pH ⇒ 0.112 = E ° + 0.059 × 4

…(i)

0.3865 = E ° + 0.059 pH

…(ii)

Solving, Eqs. (i) and (ii) gives pH = 8.65 519. E = E ° − 0.059 log

1 [Ag + ]

= 0.8 + 0.059 log K sp = 0.51 V

520. Voltage drop due to resistance by the circuit = 0.55 V I=

⇒

∆V = 0.055 A R

Q = 055 . × 96500 = 53.075 kJ Ag(CN) −2

521.

0.1 − x

3 × 10 −19

Ag + + 2CN − x

x (0.05 + 2x ) 0.05x = ≈ 0.1 − x 0.1

0.05 + 2x

⇒ x = 6 × 10 −19

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Solutions

E = 0.8 − 0.059 log 522. E ° Au + / Au − 0.059 log

1 +

[Au ]

= E ° Ag + / Ag − 0.059 log

1 [Ag + ]

= − 0.27 V

1 [Ag + ]

Substituting values of E ° and [Ag + ] = 0.1 M gives [Au + ] = 1.21 × 10 −16 2x

0.1 − x

K = 5 × 10 −39 =

1.21 × 10 −16 [CN − ]2 0.1

0.48 =

⇒ [CN − ] = 2.02 × 10 −12 M.

Zn + 2Ag +

Zn 2+ + 2Ag.

E1 = 1.52 = E ° −

[Zn 2+ ] 0.059 log 2 [Ag + ]12

E 2 = 1.04 = E ° −

[Zn 2+ ] 0.059 log 2 [Ag + ]22

523. The cell reaction is

⇒

Au + + 2CN −

Au(CN) –2

Now,

[Ag + ]12 0.059 log 2 [Ag + ]22

⇒ [Ag + ]2 = 7.31 × 10 −10

[Cl − ] = 0.2 M ⇒ K sp = 1.46 × 10 −10 524. (a) For

2Ag + + S 2–

Ag 2S E° =

0.059 log K sp = − 1.425 V 2 2Ag + + 2e

2Ag

E ° = 0.8 V

…(ii)

Adding, Eqs. (i) and (ii) gives : Ag 2S + 2e

2Ag + S 2−

0.766 = 0.896 −

⇒

[S 2− ] = 3.92 × 10 −5 M

⇒

E ° = − 0.652 V

0.059 1 log 2− 2 [S ]

(b)

525. Cell reaction is : H 2 ( g ) + I 2 ( s)

2H + + 2I −

E ° = 0.535

0.755 = 0.535 − 0.059 log [H + ][I – ] ⇒ pH = 2.72

526. For the cell reaction : 1 2

H 2 + AgCl

…(i)

Ag + H + + Cl –

0.47 = 0.23 − 0.059 log [H + ][Cl – ] ;

pH = 3.07

0.5 = 0.23 − 0.059 log [H + ][Cl – ];

pH = 3.58

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If there is x mol of acid and y mol of NaOH in its 20 mL, then y 2.5 y and 3.58 = pK a + log 3.07 = pK a + log x− y x − 2.5 y 0.51 = log

⇒

pK a = 3.07 − log

and

527. Faraday of electricity passed = ⇒ 528. 2I –

2.5( x − y) x − 2.5 y

y = 3.89 ⇒ K a = 1.3 × 10 −4 6.6 y

1.25 × 1.1 × 3600 = 0.0513 96500

meq of KMnO 4 required = 51.3 ⇒ I 2 + 2e F =

⇒ x = 7.6 y

M (KMnO 4 ) =

I × 3 × 60 It = = 9.3 × 10 −3 96500 96500

⇒ I = 4.98 A

Total meq of I − = 9.3 + 36.3 × 0.1 = 19.93 ⇒ 529. [Ag + ]R = 8 × 10 −13

⇒ E = 0 − 0.059 log

530. For the cell reaction

2Hg + 2Cl − + 2H +

− 0.52 V = − 0.28 − 0.059 log For salt R 2 NH 2Cl : K h =

51.3 1 × = 0.41 M 25 5

M (KI) = 0.16

[ Ag + ]L

= − 0.71 V [ Ag + ]R Hg 2Cl 2 + H 2 1 +

[H ][Cl – ]

⇒ [H + ] = 8.5 × 10 −5 M

Kw [H + ]2 = ⇒ K b = 1.36 × 10 −7 K b [ R 2 NH 2Cl]

531. Since mass of Sn increasing, Sn-electrode is working as cathode and X-metal electrode as anode and electrons are flowing from X-electrode to Sn-electrode in the external circuit. The half cell reactions are : Sn 2+ + 2e X Cell reaction :

n+

+ ne

nS 2n+ + 2 X

Sn

E ° = − 0.14 V

X

E ° = − 0.78 V

nS n + 2 X

n+

Now applying the Nernst equation: E= ⇒

0.65 =

⇒

0.01 =

⇒

E ° – 0.0591 [ X n + ]2 log 2n [Sn 2+ ]n 1 0.64 − 0.0591  −2  log 10 − n log   2n 2 0.0591 (2 − 0.3 n) 2n n=3

Solving :

° = 0.64 V E cell

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Solutions

532. (a)

The spontaneous cell reaction is : Zn + 2Ag + ( aq )

(b) (c)

° = E cathode ° ° = 0.8 − ( −0.76) = 1.56 V E cell − E anode Applying Nernst’s equation: 1.6 = 1.56 −

⇒

log

⇒ (d)

Zn 2+ ( aq ) + 2Ag( s)

[ Zn 2+ ] + 2

[Ag ] [Zn

2+

[Zn 2+ ] 0.059 log 2 [Ag + ]2

= − 1.356 ⇒

] = 4 × 10

−4

[Zn 2+ ] 10

−2

…(i)

= 4 × 10 −2

M

As we add KI to cathode chamber, some Ag + will precipitate out as: Ag + + I – → AgI

The above reaction reducing [Ag + ] from cathode chamber. This will reduce E cell according to Nernst’s equation (i). 533. In the above conditions, two Galvanic cell can be constructed as: (i) Al/Cu and (ii) Fe/Cu. ° /Cu = + 0.34 + 1.76 = 2.1 V and E Al

° /Cu = + 0.34 + 0.44 = + 0.78 V E Fe

° , cell (i) is more likely to produce higher voltage. Q E cell depends predominantly on E cell E cell (i) = 2.1 −

[Al 3+ ]2 0.059 0.059 10 –6 = 2.1295 V log = 2.1 − log 6 6 10 –3 [Cu 2+ ]3

E cell (ii) = 0.78 − ⇒ Cell :

[Fe 2+ ]2 0.059 1 0.059 log = 0.78 − log = 0.75 V 2+ 2 2 0.1 [Cu ]

Al Al 3+ ( aq ) Cu 2+ ( aq )Cu will produce greater voltage.

534. Q = It = 430 × 60 × 60 = 1.548 × 10 6 C Q = 16.04 = number of equivalents of Mg deposited. 96500 24 mass of Mg deposited = 16.04 × = 192.5 g 2

⇒ number of faraday passed = ⇒

(d) Number of gram equivalents of MgCl 2 required to be electrolyzed = ⇒ ⇒

Q=

10 6 . 47.5

10 6 × 96500 = It = 500 t 47.5

t = 4.063 × 10 6 sec = 1128.6 hours.

535. (a) 2Al +6 2 +12e → 4Al ⇒ 12 moles of electron will produce 4 moles of Al. 1250 × 1.25 × 3600 (b) Number of faradays passed = = 58.29 96500

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⇒ Mass of aluminium metal deposited = 58.29 ×

27 = 524.6 g = 0.5246 kg. 3

° 536. (a) E cell = E ° (IrCl 63– /Ir, Cl – ) – E ° (CO 2 , H 3O + /HCOOH) = + 0.97 V (b) For thermodynamic spontaneity of a cell reaction, ∆G° < 0. Also ∆G ° = − nE ° F < 0 since, E ° > 0 ⇒ Reaction is thermodynamically spontaneous. 537. (a) The cell reaction is :

Al + 3Ce 4+ → Al 3+ + 3Ce 3+

…(i)

° = E ° (cathode) − E ° ( anode) = + 1.443 + 1.76 = 3.203 V E cell ° ° E cell = E cell −

[Ce 3+ ]3 [Al 3+ ] 0.059 8 0.059 log = 3.23 V log = 3.203 − 4 3 + 3 3 (6) 3 [Ce ]

(b) Since, Ce +4 is in limited quantity. ∆G ° = − nE ° F = − (1 mol e per mol Ce 4+ ) (3.203) × 96500 = − 309 kJ/mol Ce 4+ ⇒

∆G° ( total) = − 309 × 12 = − 3708 kJ

(c) Initially [Al Also

3+

] = 8 M, [Ce 4+ ] = 6 M, [Ce 3+ ] = 1 M

total mole of Ce 4+ present initially = 6 × 2 = 12 total mole of Al 3+ present initially = 8 × 2 = 16 total mol of Ce 3+ present initially = 2

As the concentration of Ce 4+ drop to half, 6 mol of Ce 4+ will be converted to Ce 3+ . 8 ⇒ [Ce 4+ ] = 3 M, [Ce 3+ ] = 2 M = 4 M Also, for 3 mol Ce 4+ reduction one mol Al 3+ comes in solution. ⇒ for 6 mol Ce 4+ reduction, two mol Al 3+ comes in solution. [Al 3+ ] final = 9 M

⇒ ⇒

E cell = 3.203 −

( 4) 3 (9) 0.059 log = 3.177 V 3 (3) 3

538. (a) First of all, we need to determine E ° for Cu 2+ /Cu half-cell. E° Given:

Add: ⇒ Note :

+

e + Cu ( aq ) → Cu( s)

∆G°

+ 0.522 V

− 0.522 F

Cu 2+ ( aq ) + e → Cu + ( aq ) + 0.158 V

− 0.158 V

Cu 2+ ( aq ) + 2e → Cu( s)

° = − 0.68 F ∆G ° = − 2 E cell

° = + 0.34 V E cell Here E° is determined using ∆G ° = − nE ° F relationship since, E ° is an intensive

property and can’t be manipulated algebraically.

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Solutions

Now, for the Daniell cell: Zn + Cu 2+ ( aq )

Zn 2+ ( aq ) + Cu( s)

° = E cathode ° ° = 0.34 + 0.76 = 1.1 V E cell − E anode Cu 2+ = 2 × 4 = 8

(b) Initial moles of ⇒

moles of Cu 2+ finally = 2 × 0.4 = 0.8

⇒

moles of Cu 2+ deposited = 8 − 0.8 = 7.2

⇒ ⇒

moles of Zn 2+ produced during deposition of Cu 2+ = 7.2 final moles of Zn 2+ in anode chamber = 2 × 4 + 7.2 = 15.2 finally [Zn 2+ ] =

⇒ ⇒

° − E cell = E cell

15.2 = 7.6 M 2

[Zn 2+ ] 0.059 7.6 0.059 = 1.062 V. log = 1.1 − log + 2 0.4 2 2 [Cu ]

539. (a) Ni(OH) 3 ( s) is cathode: Ni(OH) 3 ( s) + e → Ni(OH) 2 + OH – Cd(OH) 2 + 2e → Cd + 2OH –

…(i) …(ii)

2 (i) – (ii) 2Ni(OH) 3 ( s) + Cd → 2Ni(OH) 2 + Cd(OH) 2 (b) E ° ( Nicad ) = 0.49 + 0.81 = 1.3 V E ° for MgCl 2 → Mg + Cl 2 is − 3.74 V, –ve because the reaction is being driven in a thermodynamically non-spontaneous direction. To make the reaction proceed as written, the nicad would have to supply atleast 3.74 V. The voltage calculated for nicad battery is 1.3 V, so that nicad battery can’t drive this electrolysis. It would take 3 nicad batteries in series to have enough voltage to do this: 3 × 1.3 = 3.9 V > 3.74 V. (c) Q = 0.1 × 12 × 3600 C = 4320 C number of faradays = Q/96500 = 0.0447 F ⇒ mass of Mg deposited = 0.0447 × 12 = 0.54 gram. ⇒ 0.059 (d) Nernst equation is: E = E° − log Q. 2 All the reactant in nicad battery is solid and the products are also solid and not part of Q. Therefore, as the cell approach equilibrium, there is no change in Q and no change in voltage. This is a fundamental idea on the design of most modern batteries. 540. (a) The balanced cell reaction is: PbO 2 + H 2SO 4 + CO → PbSO 4 + H 2O + CO 2 E ° = E ° (PbO 2 /PbSO 4 ) – E ° (CO 2 /CO) = 1.685 + 0.12 = 1.905 V (b) Initial moles of H 2SO 4 = 1.3 × 3000 ×

40 1 × = 15.918 100 98

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Initial moles of H 2SO 4 = 1.2 × 3000 ×

20 1 × = 7.347 100 98

moles of H 2SO 4 consumed = 8.571 ⇒ Q It is evident from the given cell reaction that per mole of H 2SO 4 consumed involves transfer of two moles of electrons, total electron transferred =17.142. ⇒ Q(withdrawn) = 17.142 × 96500 C Ampere-hour = Q/3600 = 459.5 (c) The Nernst equation tells us that voltage depends on Q(reaction quotient). PCO2 In the given cell Q = . [H 2SO 4 ] PCO 541. (a) Glu → Gln + 2e + 2H + (

CH(OH) →

C==O + 2H + + 2e) E ° = − 0.29 V…(i)

4– Fe(CN) 3– 6 + e → Fe(CN) 6

Cell-reaction:

(i) + 2 (ii) Glu + 2[Fe(CN) 6 ]

3–

E ° = + 0.69 V …(ii) +

→ Gln + 2H + 2[Fe(CN) 6 ]4– E ° = + 0.40 V

(b) Q E ° = 0.40 > 0, ∆G° < 0, cell is spontaneous. (c) Nernst equation is: E = E° −

[Gln][H + ]2 {[Fe(CN) 6 ]4– }2 . 0059 log 2 [Glu] {[Fe(CN) 6 ]3– }2

Increasing concentration of glucose will increase cell-potential. ° > 0 ⇒ M → Zn and X → Ni and net cell reaction is: 542. (a) For spontaneous cell reaction, E cell Zn + Ni 2+ ° − E cell = E cell (b)

Zn 2+ + Ni

° = + 0.51 V E cell

[Zn 2+ ] 0.059 0.059 log = 0.51 − log 0.01 = + 0.569 V 2+ 2 2 [ Ni ]

During use, [Zn 2+ ] increases and [Ni 2+ ] decreases by the same factor. Therefore, when [Zn 2+ ] = 1.0 M, [Ni 2+ ] = 0.01 M

0.059 1 log = + 0.451 V 2 0.01 Note : Equal volume of electrolytes have been considered in the two half-cells. (c) If salt bridge is removed, electrolysis will stop immediately due to accumulation of Zn 2+ ion around Zn-electrode making solution positively charged. This will prevent further oxidation of Zn into Zn 2+ ion. Hence, cell voltage will become zero. nE ° F 2 × 0.55 × 96500 543. (a) ∆G ° = − nE ° F = − RT ln K ⇒ ln K = = = 42.84 RT 8.314 × 298 E cell = 0.51 −

⇒

K = 4.046 × 1018 .

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Solutions

E° 1 O (g ) 2 2

(b) Given:

+

+ 2H + 2e → H 2O

+1.23 V

+ H 2O → H 2O 2

− 0.55 V

1 O (g ) 2 2

O 2 ( g ) + 2H + + 2e → H 2O 2

Adding: ⇒

∆G° − 2 × 1.23 F + 0.55 F

∆G ° = − 2E ° F = − 1.91 F

E ° = 1.91/ 2 = 0.955 V. Q = It = 100 × 1.5 × 3600 = 5.4 × 10 5 C

544. ⇒

Cu 2+ Also,

Q = 5.596 = number of gram equivalent of Cu 2+ deposited mass of 96500 63.5 deposited as Cu = 5.596 × =177.673 g 2 vol. of O 2 ( g ) produced at anode at STP = 5.596 × 5.6 = 31.34 L.

number of faradays =

initial moles of Cu 2+ = 4.0

Also;

final moles of Cu 2+ = 4.0 − ⇒

molarity of CuSO 4 in final solution =

5.596 = 1.2 2

1.2 = 0.6 M 2

Cu( s) → Cu + ( aq ) + e

545. (a) Cu adding

2+

E ° = − 0.52 V

+

( aq ) + e → Cu ( aq )

Cu( s) + Cu

2+

E ° = 0.16 V

+

( aq )

E ° = − 0.36 V

2Cu ( aq )

Negative E ° value indicates that the above reaction is non-spontaneous. Also ∆G ° = − nE ° F = − RT ln K nE ° F − 0.36 × 96500 = − 14.02 ⇒ K = 8.14 × 10 –7 = ln K = ⇒ 8.314 × 298 RT (b) The reaction is combination of: Cu( s) + Cu 2+ ( aq ) 2Cu + ( aq ) + 2Br – ( aq ) Adding : Cu( s) + Cu 2+ ( aq ) + 2Br – ( aq )

2Cu + ( aq )

K = 8.14 × 10 −7

2CuBr( s)

K=

2CuBr( s)

K=

1 K

sp 2

8.14 × 10 –7 k sp 2

Now, E ° for the given cell is 0.5147 V. ⇒ ⇒

E ° F = RT ln K

⇒ ln K =

K = 5.086 × 10 8 =

0.5147 × 96500 = 20.04 8.314 × 298

8.14 × 10 −7 K sp 2

⇒ K sp =

8.14 × 10 –7 5.086 × 10

8

= 4 × 10 –8 .

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412

° = 1.36 + 0.76 = 2.12 V E cell

546. (a) ⇒

. − E cell = 212

(0.04) [Zn 2+ ] . 0059 0.059 log log = 2.12 − = 2.025 V – 2 2 2 [Cl ] (5 × 10 –3 ) 2

(b) On adding ammonia, following reaction will go to right quantitatively: Zn 2+ ( aq ) + 4NH 3 Initial Eqn. ⇒ Now :

K f = 7.8 × 10 8 = E = E° −

Zn(NH 3 ) 2+ 4

0.04

1M

0

x

1 − 0.16

0.04

[Zn(NH 3 ) 2+ 4 ] 2+ [Zn ][NH 3 ]4

=

(0.04) x (0.84)

4

K f = 7.8 × 10 8

⇒ x = [ Zn 2+ ] = 10 −10 M

[Zn ] 0.059 10 −10 0.059 2.12 – log = = 2.28 V log 2 2 (5 × 10 −3 ) 2 [Cl – ]2 2+

547. The galvanic cell reaction is : 2Ag + + Cu and

E = E° −

[Cu 2+ ] 0.059 0.059 log log [Cu 2+ ] + 0.059 log [Ag + ] = E° − + 2 2 2 [Ag ] E = K + 0.059 log [Ag + ]

⇒ where ⇒ ⇒

2Ag( s) + Cu 2+

K = E° −

0059 . log [Cu + ] = constant in the given problem. 2

0.382 = K + 0.059 log [Ag + ]1 0.372 = 0.059 log

[Ag + ]1 +

[Ag ]2

and 0.01 = K + 0.059 log [Ag + ]2 ⇒

0.01 [Ag + ]2

= 2.0187 ×10 6

[Ag + ]2 (final) = 4.95 × 10 –9 M

⇒

Also, initial mmol of Cl – (added) = 2 × mmol of Ca 2+ = 2 × 0.02 × 250 = 10 Initial mmol of Ag + = 250 × 0.01 = 2.5 ⇒ mmol of Cl – left after precipitation = 7.5 ⇒ [Cl – ] final = 0.03 M ⇒

K sp ( AgCl) = [Ag + ]2 (0.03) = 4.95 × 10 –9 × 0.03 = 1.485 × 10 –10 . Q = It = 2.5 × 35 × 60 = 5250 C

548.

Q = 0.0544 = number of equivalent of metal deposited. 96500 3.06 Equivalent mass of metal = = 56.25 ⇒ Molar mass =112.5 amu. ⇒ 0.0544 549. (a) The spontaneous cell reaction is: ⇒ Number of faradays =

2Sn 2+ ( aq ) + O 2 + 4H +

2Sn 4+ ( aq ) + 2H 2O

° = 1.08 V E cell

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Solutions

Applying Nernst equation : 2

. − E = 108

4

. 0059  0.01   1  log     = 0.8735 V  0.10   10 –4  4

At any instant of uses, if x M Sn 2+ concentration is decreased, [Sn 4+ ] will increase by the same amount.   0.01 + x  2  0.059 16 log   0.8 = 1.08 − ⇒  × 10  4   0.10 − x   4+ Solving x = 0.0966 M ⇒ [Sn ] = 0.1066 M; [Sn 2+ ] = 0.0034 M 550. E1 = E ° + 0.059 pH1 and E 2 = E ° + 0.059 pH 2 ⇒ E 2 − E1 = 0.16 = 0.059 (pH 2 – pH1 ) ⇒ pH 2 = 12.11 551. The electrolysis reaction is : 2Cu 2+ + 2H 2O → 2Cu + 4H + + O 2 4 × 30 × 60 Number of faradays of electricity passed = 96500 = 0.0746 = number of gram equivalent of H + produced. 0.0746 molarity of H + = ⇒ = 0.373 ⇒ pH = 0.43 0.20 552. Iron will be deposited first. The cell reaction would be : (b)

Fe 2+ + H 2O → Fe + 2H + + 12 O 2 E ° = − 0.44 – 1.23 = − 1.67 V ⇒

E = − 1.67 –

[H + ]2 0.059 log = − 1.4 V 2 [Fe 2+ ]

⇒ A minimum of 1.4 V would be required for the onset of electrolysis. 553. Reduction half-reaction is : M (+aq ) + e → M( s) E1 = E ° − 0.059 log ⇒

1 [M + ]1

= E ° + 0059 . log [M + ]1

E 2 = E ° + 0.059 log [M + ]2 = E ° + 0.059 log 10 −2 [M]1+ = E1 + 0.059 log 10 –2 = E1 − 0.118 V

i.e., lowering applied potential by 0.118 V. ° = 1.66 – 0.44 = 1.22 V E cell

554.

E cell = 1.2 = 1.22 − ⇒

 [Al 3+ ]2  log  2+ 3  = 2.033 [Fe ] 

[Al 3+ ]2 0.059 log 6 [Fe 2+ ]3

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414

[Al 3+ ]2

⇒

1.25 × 10

⇒ moles of Al

3+

= 108 ⇒ [Al 3+ ] = 1.16 × 10 –4 M

–10

in sauce = 1.16 × 10 –4 × 0.1 = 1.16 × 10 –5

⇒ mass of Al dissolved = 1.16 × 10 –5 × 27 = 3.132 × 10 –4 g ⇒ vol. of of Al-foil dissolved =

3.132 × 10 –4 = 1.16 × 10 −4 cm 3 2.7

V 1.16 × 10 –4 = = 1.16 × 10 −3 cm 2 . thickness 0.1 1500 555. mol H 2 required 1500 = n(28.8 − 2) = 26.8 n ⇒ n = 26.8 ⇒ 2n faraday of electric current would be required 1500 ⇒ Q =2× × 96500 = 8.5 t. ⇒ t = 353 hours. 26.8 556. The electrodes reaction are : ⇒ area of holes =

Zn → Zn 2+ + 2e

0.76 V

+

2H + 2e → H 2 Cell reaction:

0.00 V

Zn + 2H + → Zn 2+ + H 2

E ° = + 0.76 V

Now, applying Nernst equation : E = 0.72 = 0.76 – For HIO 3 :

0.059 0.1 log + 2 2 [H ]

⇒ [H + ] = 0.0664 M

HIO 3

H + + IO –3

0.1 − x

x

Ka =

x

(0.0664) 2 x = = 0.13 01 . − x 0.1 – 0.0664 2

557. mmol of Cu 2+ electrolyzed in 30 min = (0.1 – 0.0528)100 = 4.72 meq of Cu 2+ deposited = 9.44 = number of mF. Q Q = It ⇒

9.44 × 10 –3 × 96500 = I × 30 × 60 ⇒ I = 0.5 A

i.e., meter is showing only 80% of the actual current. 558. The spontaneous cell reaction is : 2Cr 2+ + Ni 2+ E = 0.11 V = 0.16 − ⇒

2Cr 3+ + Ni +3 2

E ° = + 0.16 V

[Cr ] 0.059 2.56 × 10 –4 0.059 log = 0.16 − log 2 2 [Ni +2 ] [Cr +2 ]2 [Ni +2 ] [Ni 2+ ] = 5.16 × 10 –6 M.

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Solutions

559. (a) The electrode reactions occurring are: 2H 2O + 2e → H 2 + 2OH –

: at cathode

2H 2O → 4H + + O 2 + 4e : at anode

(b)

(c)

Since, phenolphthalein is colourless below a pH value of 8 and pink above a pH values of 10. At the start of electrolysis, both the solutions will be colourless. As electrolysis proceeds, cathode compartment will become basic and will acquire pink colouration while anode compartment will become more and more acidic and remains colourless. If electrolysis is stopped at any stage, the two compartments, cathode and anode compartment, will have same number of gram equivalent of hydroxide and hydronium ions. Therefore, if the content of two compartments are mixed, the net result will be formation of a neutral solution and it will be colourless. As electrolysis proceed, HO – will be produced and will be neutralized with HCl added from out side. At end point when the whole HCl is neutralized, solution will acquire alkaline nature if electrolysis is carried on further and will become pink coloured.

⇒

Q=

25 × 10 −3 × 8 × 60 F = 1.24 × 10 –4 = equivalents of HCl in 10 mL. 96500

⇒

Molarity of HCl =

1.24 × 10 –4 = 0.0124 M 0.01

HgI 2– 4 + 2e

560. Given :

Hg + 4I – Hg 2+ + 2e

Hg Adding :

E ° = − 0.04 V

HgI 2– 4

Hg

2+

+ 4I

E ° = − 0.85 V –

E ° = − 0.89 V

Writing Nernst equation : E = − 0.89 −

E = 0 = − 0.89 −

At equilibrium ⇒

[Hg 2+ ][I – ]4 0.059 log 2 [HgI 2– 4 ]

K d = 6.768 × 10 –3 =

1 Kf

0.059 log K d 2 ⇒ K f = 1.47 × 10 30 .

CHEMICAL KINETICS WU =5 WPb

561.

U 238 →

N0 − N

Also,

⇒ Pb 206

NU 5 × 206 : = N Pb 238 ⇒

N

λt = ln

N0 N0 − N

N 0 − N 1030 = 238 N …(i)

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Q

N 235 = N 0 − N 1030

⇒

N0 238 + 1030 1268 = = N0 − N 1030 1030 t=

⇒

1 1.54 × 10

−10

ln

1268 = 1.35 × 10 9 year. 1030

7 80 p = 240 mm ⇒ p = mm 2 7 200 200 and 140 K = ln 15 K = ln ⇒ 80 200 − p′ 200 − 7 Solving p′ = 84.5 mm 7 ⇒ PTotal ( after 140 min ) = 200 + p′ = 495.75 mm. 2 563. Let after 30 min, partial pressure of 1,3-butadiene be p mm. 55 30 × 60 × 2 × 10 −4 = ln ⇒ p =16.63 ⇒ 55 − p 562. P0 = 200 nm

After 15 min : P0 +

⇒

p(cyclobutene) = 38.37 mm [cyclobutene] = 1.45 × 10 −3 M

564. (a)

ln

K 2 E a  T2 − T1  =   K1 R  T1T2 

Also, ln K1 = ln A − ⇒ (b)

Ea RT1

8.5 × 10 −11 2.1 × 10

−11

=

Ea   10   8.314  300 × 310 

⇒ E a = 108.1 kJ

108.1 × 10 3 8.314 × 300

A = 13.97 × 10 7

⇒ ln 2.1 × 10 −11 = ln A −

ln K ( 47° C) = ln A − Kt = ln

Ea R × 320

0.15 1.65 × 10 −7 λ=

565. (a) (b)

⇒ ln

⇒

⇒ K ( 47° C) = 3.155 × 10 −10 S −1 ⇒ t = 6.53 × 10 11 second

ln 2 = 6.32 × 10 −3 min −1 t1/ 2

5.6 × 60 × 6.32 × 10 −3 = ln

100 X

⇒

X = 12%

6.32 × 10 −3 t = ln 100 ⇒ t = 12.14 hours

(c) 566. 1 H 3 in fresh 10 g H 2O :

x = 5.35 × 10 6

x = 8 × 10 −18 10 × 2 × 6.023 × 10 23 − x 18 ⇒

5.35 × 10 6 ln 2 × 40 = ln 12.3 N

⇒

N = 5.6 × 10 5 1 H 3

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Solutions

2N 2O 5 → 4NO 2 + O 2

567.

p0 − p

2p

PT = p0 +

p 2

3 p 2

5 p0 = 584.5 mm ⇒ p0 = 233.8 mm 2 3 After 30 min : PT = 284.5 = 233.8 + p ⇒ p = 33.8 2 233.8 1 ln K= = 5.2 × 10 −3 min −1 ⇒ 30 200 233.8 After 1.0 hour; 60 × 5.2 × 10 −3 = ln ⇒ p′ = 62.71 233.8 − p′

After infinite time PT =

3 p′ = 327.87 mm. 2 568. Mass of X decayed in 20 days = 0.75 g = mass of He ( g ) produced. PT = p0 +

⇒

Volume of He (S.T.P.) = ln

569.

Kc E − Ec = uc K uc RT

ln K c = ln A − ⇒

⇒ ln 500 =

90 × 10 3 8.314 × 400

and

0.75 × 22.4 = 4.2 L 4

(106 − E c ) × 10 3 8.314 × 310 ln K uc = ln A −

 K (127° C)  106 × 10 3 90 × 10 3 = − ln  c   K uc (27° C)  8.314 × 300 8.314 × 400

⇒

⇒ E c = 90 kJ 106 × 10 3 8.314 × 300

K c (127° C) = 5.05 × 10 6 K uc (27° C)

570. Let p0 be the initial partial pressure of A2 ( g ). A2 ( g )

2 A(g )

p0 − p1 p0 − p1

I. at 30 min II. at 60 min

2 p1 2 p2

I.

PTotal = p0 − p1 + 2 p1 + 700 − p0 = p1 + 700 = 760 ⇒

II.

PTotal = p2 + 700 = 800 ⇒ p2 = 100 p0 p0 and 60 K = ln 30 K = ln p0 − 60 p0 − 100

Now, Solving :

p0 = 180 mm

After 75 min : 75 K = ln

⇒ K = 13.5 × 10 −3 min −1

180 180 − p3

⇒

⇒ t1/ 2 = 51.28 min.

p3 =114.68 ⇒ PT = 700 + p3 = 814.68 mm

E af = 11420 × 2.303R = 218.66 kJ

571. ⇒

p1 = 60

E ab = E af − ∆H = 115.66 kJ mol −1

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1 N 2O 5 → 2NO 2 + O 2 2

572.

1−α

2α

α 2

Let x be the mole fraction of O 2 r (NO 2 ) 4 32 0.5 = = 46 r (O 2 ) x r (N 2O 5 ) 1 − α 46 0.5 − x = = 2α 108 r (NO 2 ) x Kt = ln

Now,  1 573. m = m0    2 ⇒

1 1−α

⇒

ln 2 1 1 = ln t1/ 2 15 0.69

⇒ x = 0.15 ⇒ α = 0.31 ⇒ t1/ 2 = 27.17 min.

n

where, n = number of half-lives.  1  1 2  1 3  1 4  1 5  m = 5 1 + +   +   +   +    = 6.66 mg  4    4  4  4  4 

574. Meq I 2 taken = 15, meq of I 2 reacted with hypo = 7 ⇒ meq of I 2 consumed by SO 2 = 8 = Meq of SO 2 ⇒ m mol of SO 2 produced = 4 ⇒ After 4.0 hours, total m mol of gases =14 14 × 10 −3 × 0.082 × 400 = 2.296 atm 0.2 ln 2 1 10 = ln ⇒ t1/ 2 = 5.43 hour t1/ 2 4 6

P=

ΣE i K i 180 × 10 3 × 1.2 × 10 −2 + 200 × 10 3 × 3 × 10 −2 = 194.28 kJ = ΣK i 1.2 × 10 −2 + 3 × 10 −2 576. Let at equilibrium degree of conversion be x. x = 0.16 ⇒ x = 0.138 ⇒ 1− x

575. E =

Also,

K1 = 0.16 ⇒ K 2 = 6.25 K1 K2

Now,

dx = K1 (1 − x ) − K 2 x = K1 (1 − 7.25x ) dt

⇒ ⇒

dx = K1 dt 1 − 7.25x 1 1 ln = K1 t 7.25 1 − 7.25x

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Solutions

For half equilibrium reaction : t =

1 ln 7.25K1

1 1 − 7.25 ×

x 2

= 290 second

K (Ca) t1/ 2 ( Ar ) n(Ca) = =8= = n( K ) = 52.64 K (Ar) t1/ 2 (Ca ) n(Ar)

577.

K ( K ) = K (Ca ) + K (Ar ) = 5.198 × 10 −10 y −1 ⇒

t=

1 100 ln = 1.23 × 10 9 years K 52.64

α 0 = α ( t = ∞ ) − α ( t = 0) = 35° ⇒ 10 K = ln

578.

20 K = ln

and

35 35 = 2 ln α 25

35 25

⇒ α =17.85

0 ⋅ R after 20 min = 17.85 + 5 = 22.85° . m mol of reactant consumed after 20 min = 10 ×

17.15 = 4.9 m mol. 25

m mol of alkene = 2.94 = m mol of Br 2 = 0.15 V ⇒ V = 19.6 mL ∆H R = − 3.7 kJ and

579.

E af − E ab = − 3.7 ⇒ E ab = 53.5 kJ

 (T )  E   5 ln  1/ 2 1  = a    (T1/ 2 ) 2  R  333 × 338 

580.

E=

581.

E=

⇒

E a = 351.13 kJ

ΣE i K i E1 K1 + E 2 K 2 + E 3 K 3 = ΣK i K1 + K 2 + K 3

K1 10 = K 2 25

Also,

and

and

K1 10 = K 3 15

E1 K1 + 2.5K1 E 2 + 1.5K1 E 3 = 79 kJ K1 + 2.5K1 + 1.5 K1

1 = 0.223 hr. −1 0.8 0.8 K (37° C) = ln = 0.693 hr. −1 0.4

K (27° C) = ln

582.

⇒

ln

 10 0.693 E a  =   0.223 R  300 × 310  K (T ) = ln

⇒

ln =

⇒ E a = 87.63 kJ

0.4 = 1.386 hr. −1 0.1

1.386 87.63 × 10 3  T − 310  =    310T  0.693 8.314

⇒ T = 316.45 K

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20K = ln

583. ⇒

ln

p0 p0 − p1

and

p0 p0 − p2

40K = ln

p0 p0 = 2 ln p0 − (2.875 − p0 ) p0 − (2.5 − p0 ) p0 p02 = 2 p0 − 2.875 (2 p0 − 2.5) 2

⇒ Solving, p0 =1.5625, 2 If

p0 = 2 then 60K = ln pB 1 = pC 15

584. ⇒

60( K1 + K 2 ) = ln

2 3 = 3 ln 2− p 1.5

⇒

p =1.156,

⇒

⇒

p0 = 15 × 40 + 40 = 640

pC =15 pB

640 640 − P

PT = 3.156 atm.

⇒ P = 546.17 = pB + pC = 16 pB

⇒ pB = 34.135 pC = 512.025 mm of Hg. 585. Initial m mol of cyclobutene =10 Let after 20 min, x m mol cyclobutene isomerized. m mol of cyclobutene left = 10 − x and m mol of diene formed = x m mol of Br 2 required after 20 min = 10 − x + 2x = 10 + x = 16 10 20K = ln ⇒ x =6 ⇒ 4 If y m mol of cyclobutene isomerized after 30 min. 10 30K = ln 10 − y From Eqs. (i) and (ii) y = 7.47 ⇒ m mol of Br 2 required = 10 + y = 17.47 ⇒ Vol. of bromine solution required = 17.47 mL 586. A → A ′ α α 0 − α1 − α1 ⇒ α 0 − 2α1 = 40 ⇒ α1 = 0 − 20 α0 −α2

2 α0 − 12 ⇒ α 0 − 2α 2 = 24 ⇒ α 2 = 2

−α2 10K = ln

⇒

α0 0.5α 0 + 20

and 20K = ln

α0 0.5α 0 + 12

Solving, α 0 = 59° ⇒

10K = ln

59 49.5

and

40K = ln

59 59 − α 3

⇒ α 3 = 29.77

⇒ Optical-rotation after 40 min = α 0 − 2α 3 = 59 − 29.77 = − 0.54° .

…(i)

…(ii)

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Solutions 2

10 e

587.

−

14000 RT

3

= 10 e

⇒ Rate constant K1 = K 2 = 0.464 hr. ⇒

−

20000 RT

Solving, T = 313.42 K

−1

K1 t = ln

n0 ( A2 ) n0 ( A2 ) − n1

⇒ n1 = 0.37

K 2 t = ln

n0 ( B 3 ) n0 ( B 3 ) − n2

⇒ n2 = 0.37

⇒ Total moles of gases after 1.0 hr. =1.37 +1.74 = 2.11 P = 5.42 atm. 588. Let α be the degree of dissociation after 12 hr. ⇒ P = P0 x1 From Raoult’s law. 20.69 10 10  10  = = ⇒ α = 0.3 ⇒ 12K = ln   ⇒  7 24 10 + 1 + 2α 11 + 2α 30K = ln

1 1−β

where β is degree of dissociation after 30 hr. ⇒ β = 0.6 and moles of solute after 30 hr. = 1 + 2β = 2.2  10  P = 24  ⇒  = 19.67 mm of Hg.  12.2  589.

R2 = 4 = (2) z R1 y

R3  3  3 = 0.668 =      2  2 R1

⇒ Z(order w.r.t. C ) = 2 2

R4  4 = 10.65 = (2) x ×   × (2) 2  3 R3

⇒

y (order w.r.t. B) =1

⇒ x(order w.r.t. A) =1

Rate Law : R = K [ A][ B][C ] 2 590.

0.42 × 10 −3 dN = KN = 9.88 × 10 4 = K × × 6.023 × 10 23 dt 347 ln 2 ⇒ K= = 1.35 × 10 −13 sec −1 t1/ 2 ⇒ t1/ 2 = 5.13 × 1012 sec = 1.628 × 10 5 year

591. If x mol of U 238 and y mol of U 235 has decayed, then presently U = 100 − ( x + y) 100 − ( x + y) 50 − x Also, = 50 − y ⇒ = 99 100 50 − y

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Also,

t ( K 2 − K1 ) = ln

⇒

t=

ln 99 = 5.5 × 10 9 year. K 2 − K1 K1

A

592.

50 − x = ln 99 50 − y

1−x

K2

K1 7 = K2 3

B, x

dx  10  = K1 (1 − x ) − K 2 x = K1 1 − x   7  dt 7 1 = K1 t ln 10  10  1 − x   7 

⇒

…(i)

Solving at t =1 gives K1 = 0.309 hr. −1 After 4.0 hr, ln

594. log

Kf Kb

10K1 t 1 = ⇒ x = 0.58 ⇒ mole % of A after four hour = 42% 10 7 1− x 7

= 5.457 −

5800 ∆S ° ∆H ° = − T 2.303R 2.303RT

⇒ ∆S ° = 104.48 JK −1 , ∆H ° = 111 kJ

595. Normality of chromic acid =1.2 N Normality of KI = 1 N meq of KI left = 10.4 × 1.2 = 12.48

Ist flask :

meq of KI ≡ H 2O 2 = 20 − 12.48 = 7.52 Initial meq of H 2O 2 = 25 ×

2.5 11.16 = 11.16 ⇒ K = ln 5.6 7.52

IInd flask : Let x meq of H 2O 2 left undecomposed. 3K = ln

11.16 x

⇒ x = 3.41 = meq of KI consumed

meq of KI reduced with chromic acid =16.59 Vol. of chromic acid required = 13.825 mL 596. R = K [H + ][ester] = K ′ [ester ] where K ′ = K [ H + ] = 3.478 × 10 −4 s −1 ⇒

t=

1 100 ln = 17 minutes 70 K′

597. Let the initial mole percentage of A is x and that of B is (100 − x ). Let after ‘ t’ time, p mole of A and q mole of B are converted into C. ⇒

2 × 10 −2 × 30 = ln

x x− p

⇒

x = 2.22 p

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Solutions

5 × 10 −3 × 30 = ln

and

100 − x 100 − x Q p + q = 25 = ln 100 − x − q 75 − x + p

= ln

100 − x 75 – 0.55x

2 × 10 −2 × 50 = ln

After 50 min:

36 36 − p

5 × 10 −3 × 50 = ln ⇒

⇒

64 64 − q

p = 22.74%

⇒ q =14.2%

mole percentage of C after 50 min = p + q = 36.94 = 71 sec.

t

1/ 2 Required nuclei ratio O14 →

598.

solving x = 36

1− x

1− x = 0.25 1− y

t1/ 2 = 124 sec.

O15 →

1− y

ln 2 1 t = ln 71 1− x

1− y 2 × 71 × 124 ln 2 1  1 1 = 332.226 sec. t = ln ⇒ ln 2  − = ln 4 ⇒ t =  t = ln  71 124  1− x 124 − 71 124 1− y ln

599.

Kc E − Ec = uc K uc RT

Now,

ln

⇒ ln 10 =

80 × 10 3 − E c 8.314 × 300

⇒ E c = 74.25 kJ

K c (127° C) 74.25 × 10 3  100  =   8.314 K c (27° C )  300 × 400  K c (127° C) t (27° C) = 1706.6 = 1/ 2 K c (27° C) t1/ 2 (127° C)

⇒

100 ln 2 = t ⇒ t = 84.4 seconds 25 42.2 Xe (CF3 ) 2 ( g ) → Xe ( g ) + C 2 F4 ( g ) + F2 ( g )

⇒ t1/ 2 (127° C) = 42.2 sec. ⇒ K t = ln 600. Total mole :

0.005 − x

After time t : n =

x

2.63 × 0.1 PV = = 0.0107 = 0.005 + 2x RT 0.082 × 300

ln 2 0.005 t = ln 30 0.005 − 2.85 × 10 −3

⇒ K1

→ β Ac

x

x = 2.85 × 10 −3

⇒

601.

x

K2

→ α

K1 [β] =9= K2 [α ]

⇒ t = 36.53 sec.

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nα =

Also,

PV = 6.5 × 10 −3 RT

nβ = 9nα = 58.5 × 10 −3 ⇒ Moles of Ac decayed in 5.0 hours = nα + nβ = 0.065 ln 2 0.44 ⇒ t1/ 2 = 21.68 years ⇒ × 5 = ln t1/ 2 0.44 − 0.065 t

= 1.5

1/ 2 602. NO → After 5.0 hour

x − y + 1.5 − x − z = 0.225

x− y

t

NO 2

1.5 − x − z

Also,

= 2.5

1/ 2 →

⇒

ln 2 x × 5 = ln ⇒ 1.5 x− y ln 2 1.5 − x × 5 = ln 2.5 1.5 − x − z

⇒

z = 1.275 − y

…(i)

x =10 x− y

…(ii)

1.5 − x =4 1.5 − x − z

…(iii)

Solving Eqs. (i), (ii) and (iii) x = 1.0 g and

y = 0.5 g

2/ 3

K K  2 K = 1 2 ⇒ E = ( E1 + E 2 − E 3 ) = 153.34 kJ 3  K3  604. Since, absorbance of solution is directly proportional to the concentration of coloured species A1 C1 A (absorbance) ∝ C ⇒ = A2 C 2

603. Q

⇒

C1 0.600 = 0.200 4.00 × 10 −5

⇒ C1 = 1.2 × 10 −4 M

Also, decomposition of X follow first order kinetics: C A 0.600 Kt = ln 0 = ln 0 ⇒ K × 35 = ln = ln 3 0.200 C A 0.600 Kt = ln = ln 80 0.0075 ln 80 t = 35 = 139.6 minutes. ⇒ ln 3 605. (a)

The decomposition reaction is following 1st order kinetics, since, for a 1st order reaction: [ A ]0 Kt = ln [ A]

where [ A ]0 is initial concentration and [ A ] is concentration of reactant left unreacted at any time “t ”. ⇒ Kt = ln [ A ]0 − ln [ A ] ⇒ ln [ A ] = ln [ A ]0 − Kt i.e., A plot of ln [ A ] vs time is a straight line with negative slope.

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Solutions

Rate = K [H 2O 2 ]

Hence, rate law: (b) From the graph, at t = 0

ln [ A ]0 = 0 ⇒ [ A ]0 =1.0 At t = 2800 min, ln [ A ] = − 3 3 = 1.07 × 10 −3 min −1 ⇒ −3 = 0 − 2800 K ⇒ K = 2800 3 At 2000 min, ln [ A ] = − × 2000 = − 2.142 ⇒ [ A] = 0.1173 M 2800 606. Let P0 be the partial pressure of each Cl 2O and N 2O 5 in the beginning. Therefore, the total pressure after infinite time would be 2P0 + P0 / 2. 5 ⇒ P0 = 750 and P0 = 300 mm of Hg. 3 Now, let us consider “ p” be the decrease in partial pressure of both of the reactants after time “t ”, then: 2Cl 2O + 2N 2O 5 → 2NO 3Cl + 2NO 2Cl + O 2 300 − p

10 K = ln

At 10 min.

300 − p

300 300 − p

p

and 600 +

p

p = 650 ⇒ 2

p/ 2

p =100

 3 10 K = ln    2

⇒

30 K = ln

At 30 min.

…(i)

300 300 − p

…(ii)

Solving Eqs. (i) and (ii) yields p = 211. Total P = 600 + p/ 2 = 705 mm of Hg. 607. The rate expression for a general n th order reaction is: 1  1 1  Kt =  n −1 −  n − 1  [ A ] [ A ]0n − 1   1  1 1 1 1 ⇒ (2 n − 1 − 1) = − 400 K = 1   n −1 n −1 − n − 1  (0.5) n − 1 n 1 [ A ]0  [ A ]0 964 K = ⇒

 1  1 1 1 1 ( 4 n − 1 − 1) = − 1   n −1 n −1 n −1 − n − 1  (025 n 1 . ) [ A ]0  [ A ]0

964 K 4n − 1 − 1 = 2 n − 1 + 1 ⇒ 2 n − 1 = 141 . = (2)1/ 2 = 2.41 = n − 1 400 K −1 2

Substituting, n =1.5 in Eq. (i) gives 400 K = ⇒

…(i)

2( 2 − 1)

0.6 K = 2.67 × 10 −3 sec –1 ( M ) − 1/ 2

⇒ n=

…(ii) 3 2

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608. From the given graph, half-life is determined to be 20 sec. ln 2 ⇒ K= 20 0.4 ⇒ At 10 sec. K × 10 = ln [ A] [ A ] = Concentration of reactant left unreacted after 10 sec. ln 2 0.4 and [ A ] = 0.28 M × 10 = ln ⇒ 20 [ A] Rate = K [ A ] =

Hence, at 10 min:

ln 2 × 0.28 = 9.7 × 10 −3 mol L–1 sec –1 . 20

609. The Arrhenius equation is: log K (s –1 ) = log A −

Ea 2.303 RT

⇒ From the given graph,

Ea 10 4 = 3 2.303 R

and slope

log A = 3

⇒ E a = 63.82 kJ

Therefore, at 500 K, 63.82 × 10 3 = − 3.66 2.303 × 8.314 × 500 1 100 ⇒ t = ln = 7485.75 sec = 2.08 hrs. K 20

log K (S –1 ) = 3 − ⇒

K = 2.15 × 10 −4 S –1

610. For a 1st order reaction: R = K [ A] ⇒

[ A ]0 [ A ]0 and 30 K = ln [ A ]10 [ A ]30 [ A ]10 = 0.239 ⇒ K = 1.2 × 10 −2 min –1 . 20 K = ln [ A ]30

10 K = ln

Also ⇒

[ A ]30 min

similarly, ⇒ 611. For reaction:

R10 [ A ]10 min = = 1.27 R 30 [ A ]30 min

[ A ]2 hrs

= 2.94 and [ A ]30 min =

[ A ]2 hr = 0.237 M and A

a − xe

R 30 = 0.6975 M K

R 2 hr = 2.84 × 10 –3 mol L–1 min –1 .

P

xe

K =2=

xe a − xe

⇒ xe =

2 a 3

Also, at any instant: d[ P ] dx = = K f ( a − x ) − K b x = K b (2a − 3x ) dt dt

…(i)

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Solutions x

∫0

⇒ Now,

dx = Kb 2a − 3x

t

∫0

⇒ Kb t =

dt

1 2a ln 3 2a − 3x

x e 2a ln 2 = ; Kb t = 2 6 3 ln 2 2 ln 2 2 ln 2 t= = = × = 18.48 min. 3 Kb 3 K f 3 2.5 × 10 −2 x=

at

⇒

100 5 = ln 80 4 80 In presence of 1st catalyst: = ln 2 30 KC = ln 1 40 40 In presence of 2nd catalyst: 10 KC = ln = ln 4 2 10 EUC − EC KC 1 KC 1 2 ln 2 1 = = 6.21 ⇒ ln ⇒ = ln 6.21 = KUC RT KUC ln(5 / 4)

612. In absence of catalyst:

60 KUC = ln

Solving, EC = 70.44 kJ; Similarly, EC = 70.975 kJ. 1 2  −dN  –1 613. (a) Activity  = λN = 12.5 count min   dt  ⇒

N =

12.5 12.5 t1/ 2 12.5 × 5730 × 365 × 24 × 60 = = = 5.4 × 1010 λ ln 2 ln 2

Total number of C-atom in 1 g = ⇒

%

14

1 . × 10 22 × 6.023 × 10 23 = 502 12

C per gram of carbon =

5.4 × 1010 5.02 × 10 22

× 100 = 1.082 × 10 −10

 −dN  (b) Initial activity   = λN 0 = 12.5  dt  0  −dN  Activity after 50,000 years =   = λN  dt  Also

λt = ln

N0 N

⇒

⇒

 −dN     dt  N = 12.5 N0

N ln 2 × 50,000 = ln 0 5730 N

N0 = 423.4 Substituting in equation (i): N 1 −dN . × = 125 = 2.95 × 10 −2 count per minute per gram. 423.4 dt (c) 12.5 = λN 0 N0 N 1 5730 ln 1.785 = 4794 years =1.785 ⇒ t = ln 0 = 7 = λN ⇒ λ ln 2 N N ⇒

…(i)

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COLLIGATIVE PROPERTIES 614. (a) − ∆T f = 0.6 = K f ⋅ m ∆Tb = K b ⋅ m ⇒ (b) ⇒

0.6 K f 1.86 = = ∆Tb K b 0.52

− ∆T f = 0.6 = K f m = 1.86 × x1 =

n1 = n1 + n2

⇒ ∆Tb = 0.167

n2 × 1000 ⇒ 18n1

n2 = 5.8 × 10 −3 n1

1 = 0.9942 n 1+ 2 n1

⇒

P = P0 x1 = 23.75 × 0.9942 = 23.61 torr. 0.6 (c) = 0.32 ≈ Molarity m= 1.86 π = MRT = 0.32 × 0.082 × 293 = 7.688 bar. 60 615. Mole fraction of benzene = 78 = 0.638 ⇒ Mole fraction of toluene = 0.362 60 40 + 78 92 V. P. = 0.638 × 93.4 + 0.362 × 26.9 = 69.327 mm n 616. + 0.744 = 1.86 ⋅ 2 × 1000 ⇒ n2 = 0.1 250 752.7 = 760 ⋅

Also,

n1 n1 + 0.1

⇒ n1 =10.31

⇒ Mass of ice formed = 250 − 10.31 × 18 = 64.42 g. −∆ T f = 1.86 × 617. V.P. of solution = ⇒

0.1 × 1000 = 1 ⇒ T f = − 1° C. 185.58

0.25 × 0.082 × 300 × 760 = 21.638 mm. 18 × 12 21.638 = 23.75 ⋅

n1 n1 + n2

⇒

n2 = 9.76 × 10 −2 n1

 n  1000 = 2.82 ∆Tb = K b ⋅ m = 0.52 ×  2  × 18  n1 

618.

619.

x + 6 10 − x + 0.77 180 60 = ⇒ x = 5.843 g x 10 − x 0.58 + 180 60 ∆Tb = iK b m ⇒ 0.15 = 3 × 0.5 × m ⇒ m = 0.1

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Solutions

Now,

Pb(NO 3 ) 2 + 2NaCl 0.1 0

0.2 0

→ PbCl 2 + 2NaNO 3 0 0.1

0 0.2

Now, the solution contains two salts NaNO 3 and PbCl 2 . − ∆T f = 0.83 = K f (2 × 0.2 + 3S ) where S is molar solubility of PbCl 2 . S = 1.54 × 10 −2

⇒ K sp = 4S 3 = 1.46 × 10 −5 .

620. In mixture of A and B if mole fraction of A = x, then : 400x + 500(1 − x ) = 470 ⇒ x = 0.3 In final mixture, if combined mole fraction of A and B = y, then 470 y + (1 − y)600 = 496 ⇒ y = 0.8 Therefore, in the final mixture, mole fractions are : A = 0.24, B = 0.56, C = 0.2 ⇒ Mole fraction in vapour phase are : A=

x A ⋅ PA0 0.24 × 400 = = 0.1935, P 496

B=

x B PB0 0.56 × 500 = = 0.5645 P 496

C = 1 − (0.1935 + 0.5645) = 0.242 Moles of solute =

621.

Mass of H 2O = 0.582 = 3 × 0.52 ×

⇒

⇒

38.2 202 + 18x

38.2 × 18x + 250 202 + 18x 38.2 × 1000

 38.2 × 18x  (202 + 18x )  250 +  202 + 18x  

x =10, formula = Na 2 B 4 O 7 ⋅ 10H 2O 10.2 − x  x  RT ×4+ π = π (NaCl) + π (sucrose) ⇒ 7.32 = 2 × 342  58.5 

622. ⇒

342 × 8x + 58.2(10.2 − x ) =

7.32 × 58.5 × 342 0.082 × 296

⇒ x = 2.03 g

m% NaCl = 19.9 623. In benzene solvent : 1.15 = 5.12 × In diethyl ether :

1.5 × 1000 ⇒ M × 36

1.55 = i × 2.02 ×

0.48 × 1000 185.5 × 10

M =185.5 amu.

⇒ i =3

Hence, per molecule of complex, three particles are produced on dissociation in diethyl ether. Also, in one mole complex, mass : C = 185.5 × 0.6488 ≈ 120 H = 185.5 × 0.0545 ≈ 10 ⇒ Molecular formula = M (CH) 10

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n2 × 1000 100 n 1 = K b ⋅ 2 × 1000 W1

0.26 = K b ⋅

624.

…(i) …(ii)

Solving Eqs. (i) and (ii), W1 = 26 g 10 625. Mole fraction of urea = 60 = 0.0322 10 +5 60 20 180 Mole fraction of glucose = = 0.01 20 + 10 180 Q mole fraction of glucose is less, vapour pressure above the glucose solution will be higher than the vapour pressure above urea solution. To establish equilibrium, some H 2O( v ) molecule will migrate from the glucose side to urea side in order to make the solutions of equal mole fraction. 10 20 60 180 = ⇒ 10 20 +5+ x + 10 − x 60 180 Solving, x=4 So, now in urea solution mass of water = 90 + 72 = 162 g 100 m% = × 100 = 6.17 162 nH 2O 626. At equilibrium : 23.6 = 24 × ⇒ nH 2O = 9.833 ⇒ m(H 2O) =177 g. nH 2O + n urea ⇒ Mass of H 2O( v ) migrated from the glucose solution = 177 − 90 = 87 g. n1 n2 ⇒ 23.6 = 24 × = 16.95 × 10 −3 ⇒ n1 + n2 n1 Also, 18n1 + 180n2 = 113 ⇒ n1 + 10n2 = 6.277 Solving, Eqs. (i) and (ii), n1 = 5.367 and n2 = 0.09 ⇒ mass of glucose =16.2 g. m% glucose in original solution = 8.1 64.6 = 70 ⋅

627. ⇒ 628.

n1 n1 + n2

where

…(i) …(ii)

n1 = moles of (methanol + ethanol)

n2 x 80 − x = 0.0836 ⇒ n1 = 1.9938 = + n1 32 46

⇒ x = 26.78 g CH 3OH

1 0 1 0 PB + PT = 80 ⇒ PB0 + PT0 = 160 2 2

…(i)

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Solutions

Mole fractions of benzene and toluene in the vapour phase : PB0 P0 and xT = T 160 160 Now, these mole fractions will be the mole fractions of benzene and toluene in the liquid phase of condensate. xB =

PB0 P0 0 0 ⋅ PB0 + T ⋅ PT0 = 100 ⇒ PB 2 + PT 2 = 160 × 100 160 160 Solving, Eqs. (i) and (ii) ⇒

PB0 = 120 torr and

…(ii)

PT0 = 40 torr.

 25   25      629. V.P. of mixed solvents =  32  × 94 +  46  44 = 73.48 mm  25 + 25   25 + 25   32 46   32 46  n1 where, n1 = moles of solvent =1.324 Now, 58 = 73.48 n1 + n2 ⇒

n2 = moles of solute = 0.35 2 × x 11.85 − x n2 = 2 × moles of NaCl + moles of urea = + = 0.35 ⇒ x = 8.7 g 58.5 60 630. ln

P2 ∆H  T2 − T1  =   P1 R  T1T2 

⇒ 631. π = CRT

ln

and

V.P. of H 2O at 100°C is 760 mm

 760 40.6 × 10 3  50 =   P1 8.314  373 × 323 

⇒ P1 = 100.15 mm

⇒ C = 6.286 × 10 −2 M

⇒

P = 55.324 ×

Also,

2 = 1.86 ×

55.55 55.55 + 6.286 × 10 –2

6.286 × 10 −3 × 1000 w1

= 55.26 mm

⇒ w1 = 5.84

Initial amount of water = 100 − 2.15 = 97.85 Mass of ice formed = 97.85 − 5.84 = 92.01 g 2.3 = 5.1 ×

632.

1.3 = 5.1 ×

MA

1 1000 × + 2M B 20

…(i)

MA

1 1000 × + 4M B 20

…(ii)

Solving, Eqs. (i) and (ii) M A = 25.58 and

M B = 42.64

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633. Mol. wt. of acid : ⇒

0.2 = 15.1 × 0.1 × 10 −3 M 0.168 = (1 + α ) × 1.86 ×

0.558 = i × 1.86 × 0.1

634.

635. In benzene i = In water :

1 2

⇒ − ∆T f = 2.16 =

⇒ ⇒

31.78 = 31.82 ×

⇒ Formula = [Co(H 2O) 5 Cl]Cl 2

10 1000 × 10 118.5

⇒

M =118.5

⇒ α = 0.115

M 2+ + 2Cl −

55.55 55.55 + 3S

2S

S

⇒ S = 2.33 × 10 −2

K sp = 4S 3 = 5.06 × 10 −5 5.93 × 10 −3 = ( x + 1) × 0.52 ×

637.

⇒ α = 0.196

1 10 1000 × 5.12 × × M 100 2

− ∆T f = 1.75 = (1 + α ) × 1.86 × Solubility = S

M =132.45

1 1000 × 132.45 100

⇒ i =3

MCl 2

636.

⇒

0.25 1000 × 10 M

M = 23x 100 From Eqs. (i) and (ii) x = 20.3 ≈ 20 Formula of protein = H 20 P Mol. wt. = 2300 × 20 − 23 × 20 + 23 = 45563 amu 100 638. ⇒ M =123.46 754.5 = 760 ×  100 5  18  +   18 M  Also,

⇒

754.5 = 760 ×

100  100  α  2  78  + 1 −  ×  2  123.46   78 

α = 0.846 ⇒ % dimerization = 84.6 n 639. 3.31 = 5.12 × 2 × 1000 ⇒ n2 = 12.93 × 10 −3 20 Let the mixture contain x g cyclohexane. x 1.32 − x + = 12.93 × 10 −3 ⇒ x = 0.64 g 84 128 1 640. 0.25 = K f ⋅ × 1000 ⇒ K f = 3 60 × 200 1.5 Also, 0.2 = 3 × × 1000 ⇒ M y =180 M y × 125

…(i) …(ii)

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Solutions

and 99 = 100 ×

n1 n1 + n2

⇒

⇒ K f =3=

M 1 RT f 2 ∆H f

=

Also,

K b = 3.6 =

∆H v

=

119.5 × 10 −3 × 8.314 × (335) 2 ∆H v

∆H v = 30.97 kJ mol –1

Now,

ln

⇒

ln

P2 ∆H v  T2 − T1  =   P1 R  T1T2 

 760 30.97 × 10 3  12 =   P1 8.314  335 × 323 

P1 = 502.8 mm of Hg. 2.5 1000  α − ∆T f = 2 = 1 −  × 14 × × ⇒ α = 0.925  2 94 100 Mass of solution =102.5 g ⇒ V =

⇒

643.

⇒ ∆H f = 34.1 kJ mol –1

0.4 × 1000 ⇒ K b = 3.6 128 × 25

M 1 ( RT f0 ) 2

⇒

642.

1.5 = 0.825 180

125 = 151.51 amu 0.825

151.51 × 10 −3 × 8.314 × (285) 2 ∆H f

0.45 = K b ⋅

641.

⇒

⇒ n1 = 99n2 = 99 ×

M (Solvent) = 0

∴

n2 1 = n1 99

m =116.5 mL f

2.5 1000 = 0.228 M × 94 116.5 α Kc = = 360.6 2C (1 − α ) 2

Initial molarity =

n2 × 1000 ⇒ n2 = 0.041 300 Let mixture contain x g anthracene, then 0.7 = 5.12

6−x x + = 0.041 ⇒ x = 2.677 g. 178 128 After cooling to 4.5°C, − ∆T f = 1 = 5.12 × ⇒ 644.

w1 = 209.92 g

0.041 × 1000 w1

⇒ mass of solid benzene produced = 90.08 g

1.9 = (1 + α ) × 1.86 ⇒ α = 0.0215 Molarity =1.098 ⇒ K a =

Cα 2 = 5.18 × 10 −4 . 1−α

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SOLID STATE 645. Number of A per unit cell =1 1 7 = 8 8 Empirical formula = AB 7/ 8 = A8 B7 646. Number of A per unit cell =1 Number of B per unit cell = 7 ×

1 ⇒ empirical formula = AB 5/ 2 = A2 B5 2 31 7 647. Number of atoms per unit cell = + 3 = 8 8 Also, in face centred cubic lattice, a = 2 2r Number of B per unit cell = 5 ×

Packing fraction (φ ) =

31 × 4πr 3 8 × 3(2 2r ) 3

648. Volume occupied by atoms = 2 ×

= 0.717

4 8π 63 3 π (R 3 − r 3 ) = R × 3 3 64

Packing fraction (φ ) =

⇒

8π × 63R 3 × 3 3 3 × 64 × 64 × R 3

= 0.669

 1 1  16 649. Number of atoms per unit cell = 6 −  +  =  6 2 3 Volume of unit cell = 24 2r 3 650. Density ( s) =

nM Na

⇒ 2.7 =

3

⇒ Packing fraction =

16πr 3 3 × 24 2r

n × 27 6.023 × 10 8.27 =

651.

23

× ( 4.05 × 10 −8 ) 3

3

×

4 = 0.658 3

⇒ n = 4, FCC

4 × 128 6.023 × 10 23 a 3

⇒

a = 46.8 × 10 −9 cm = 4.68 Å = 2[ r (O 2− ) + r (Cd 2+ )]

⇒

r(Cd 2+ ) = 1.1 Å

652. Density =

2M 6.023 × 10

23

× ( 4.47) 2 × 10.89 × 10 −24

Formula of salt is Hg 2Cl 2 . 4M 653. 4.269 = 23 6.023 × 10 × (6.93) 2 × 10 −24

⇒

= 7.15 ⇒

M = 212

Solving x = 3 ⇒ Mass of sulphur required =

96 × 1500 = 672.89 g 214

M = 468.52

Formula Co 2S x

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Solutions

654. Given

h 2 r = 3.35 Å = 3.35 × 10 −8 cm. =2 2 3 3

 3 1  Volume of unit cell = 24 2r = 24 2  × × 3.35 × 10 −8  = 2.93 × 10 −22 cm 3  2 2  3

density =

6 × 12 6.023 × 10

23

× 2.93 × 10 −22

= 0.4 g / cc

655. In HCP, number of atoms per unit cell = 6 Volume of unit cell = 24 2 r 3 Packing fraction = 656. h = 2r− and A = 6 3r−3

3 × 24 2r

3

=

π 3 2

= 0.74

⇒ V = 12 3r−3

Packing fraction = 657. Given a − 2r = 60 and

6 × 4π r 3

3 × 4π( r+3 + r−3 ) 12 3r−3

= 0.64 ( r+ / r− = 0.414)

for BCC, 4r = 3a ⇒ a −

Density (ρ) =

3 a = 60 ⇒ a = 448 pm 2

2 × 48 6023 × 10 .

23

= 1.77 g/cc

× ( 4.48) 3 × 10 −24

658. For Fe 95O100 , if there are x Fe 2+ , (a)

2x + (95 − x )3 = 200 ⇒ x = 85 Fe 2+

⇒

Fe

3+

85 17 = 15 3

=

(b) 5% of cation are vacant. 659. (a) 5% of anion sites are vacant. (b)

Density (ρ) =

(c)

Density (ρ) =

4 × 128 6.023 × 10

23

× ( 4.7) 3 × 10 −24

4 × 127.2 6.023 × 10 23 ( 4.7) 3 × 10 −24

= 8.81g/cc.

= 8.136 g/cc.

660. Number of cylindrical atoms per unit cell = 3 Packing fraction = 661. Edge not covered by atom = a − 2r also in BCC, 4r = 3a

3πr 2 l 6 3r 2 l

= 0.906

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Therefore, edge not covered = a −

2 − 3 3  a =a 2  2 

2− 3 = 0.134 2 In FCC 2( r+ + r− ) = a = 4.8 Å ⇒ r+ + r− = 2.4 Å 3  rB   4 ( 4rA3 + rB3 ) π   4 +    = 0.753 Packing fraction = π = 3 (2 2rA ) 3 12 2   rA    2 2 πR + πr π Packing fraction = = [1 + (0.414) 2 ] = 0.92 2 4 (2R ) 4 Edge length of unit cell = r 3 16 16 ⇒ Area of a surface of unit cell = r 2 = × (96) 2 × 10 −20 cm 2 = 4.9152 × 10 −16 cm 2 3 3 6 ⇒ Number of unit cell on surface = = 1.22 × 1016 4.9152 × 10 −16 ⇒ Fraction of edge not covered by atoms =

662. 663.

664. 665.

Volume of unit cell = a 3 =

64 3 3

(96) 3 × 10 −30 = 1.089 × 10 −23 cm 3

Volume of solid = 6 cm 3 ⇒

Total number of Ca present =

4×6 1.089 × 10 −23

= 2.2 × 10 24

Number of Ca present on surface = 1.22 × 1016 × 2 = 2.44 × 1016 ⇒ Fraction of Ca lying on the surface = 666. 2.32 = 667.

2.44 × 1016 2.2 × 10

24

= 11 × 10 −9

2× M

⇒ M =1268 ⇒ x = 47 6.023 × 10 (1.22) 3 × 10 −21 4 × 78 ⇒ a = 54.6 × 10 −9 cm = 546 pm 3.18 = 6.023 × 10 23 a 3 23

668. Let us consider 100 cc of solid Ar : mass =159 g Volume of 159 g liquid Ar =

159 = 112 mL 1.42

Q Packing fraction of solid Ar = 0.74, actual volume occupied by Ar = 74 mL. ⇒ % empty space in liquid phase =

112 − 74 × 100 = 33.92% 112

a a  669. If the side length of unit cell is ‘ a’. Coordinate of centre of guest atom B will be  , , 0 . 2 4 

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Solutions

1/ 2

 a  2  a  2  a 5 ⇒ R + r =   +    =     2 4 4   3a Also, in BCC R= 4 r 5 ⇒ r = 0.29R 1+ = ⇒ R 3 3 4  3 r3  3 3 3π  1  r   Packing fraction = π  2R +  ⋅ = 2 +    = 0.684 3  2  64R 3 16  2  3   670. a − 2r = 34 pm 4r = 3a

Also, a−

⇒

54 × 2 3 a = 54 ⇒ a = = 395.6 pm 2 2− 3

Density (ρ) = 671. a = 2R

and

2 × 23 6.023 × 10

23

× (3.956) 3 × 10 −24

= 1.23 g/cc.

3a = 2R + 240 ⇒ ( 3 − 1) a = 240 ⇒ a = ρ=

209 6.023 × 10

× (3.28) 3 × 10 −24

23

240 3 −1

= 9.83 g / cc

672. Volume of unit cell = 4.6 × 8 × 5.7 × 10 −24 cm 3 = 2.0976 × 10 −22 cm 3 ⇒

2.93 =

n × 100 6.023 × 10

23

× 2.0976 × 10 −22

⇒ n = 3.7 ≈ 4

673. (a) Number of Na + = 3, Number of Cl − = 3 Packing fraction =

1 4π (3r+3 + 3r−3 ) × = 0.59 3 16 2r−3

1 7 = 2 2 1 7 Number of Cl − = 4 − = 2 2 4π 7 3 1 ⇒ Packing fraction = × ( r+ + r−3 ) × = 0.64 3 2 16 2r−3

(b) Number of Na + = 4 −

 1 5 (c) Number of Na + = 4 − 1 +  = , Number of Cl − = 4  2 2 Packing fraction =

4π 3

1 5 3 3 = 0.77  r+ + 4r−  × 2  16 2r−3

= 327.84 pm.

…(i) …(ii)

…(i)

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SURFACE CHEMISTRY 1.36 × N A = 3.65 × 1019 22400 M Molar volume (H 2 ) = ρ M ⇒ Volume of one H 2 molecule = = 4.743 × 10 −23 cm 3 ρN A

674. Number of molecules of H 2 in 1.36 cc =

 3V  radius =    4π 

1/ 3

= 2.246 × 10 −8 cm

A (H 2 ) = πr 2 = 1.583 × 10 −15 cm 2 Surface area covered on Cu per gram = 3.65 × 1019 × 1.583 × 10 −15 = 5.78 × 10 4 cm 2 675. Let V (mL) of solution is required 0.016 V N (acid) = × 6.023 × 10 23 = 9.637 × 1018 V 1000 Area covered = 9.637 × 1018 V × 0.2 × 10 −14 cm 2 = 19274 V = 500 cm 2 V = 25.94 × 10 –3 mL 676. Total A = 25000 m 2 = 25 × 10 7 cm 2 , r = 1.5 × 10 −8 cm A ( NH 3 ) = πr 2 = 7.068 × 10 −16 cm 2 ⇒

N =

25 × 10 7 −16

= 3.53 × 10 23

7.068 × 10 moles of NH 3 = 0.587 Vol. of NH 3 at STP required = 13153.5 mL 100 N = × 6.023 × 10 23 = 2.69 × 10 21 22400

⇒ 677.

A/molecule = 0.16 × 10 −14 cm 2 Total area covered by N 2 ( g )/gram of catalyst = 1.6 × 10 −15 × 2.69 × 10 21 = 43.04 × 10 5 cm 2

REACTION MECHANISM 682. (a) O 2 NCH 2CH •• < CH 2 •• < CH 3CH •• < C 6 H 5CH •• < (C 6 H 5 ) 2 C •• (b) O2N—

Cl—

+

+

+

+

+

+

(c) (aromatic)

H3C—

+

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Solutions

+ +

(d) O

O

+

O

O

+

O +

O +

O O O O     − − − (e) CH 2 —C—NH 2 < CH 2 —C—OCH 3 < CH 2 —C—CH 3 < CH 2 —C—H −

O  < CH 2 —C—CH 2OCH 3 −

683. (a) (b) (c) (d) (e)

H 2O C(4). A, B, D are all more acidic than “C” because they are all phenols and the conjugate base that is formed upon deprotonation can be stabilized by resonance. C is incapable to delocalize negative charge via resonance. B is most acidic, because in addition to all the resonance forms available in A and D, the charge of anion can also be delocalized through the  NO 2 group as: O–

O

O

O

– – N O

+

O –

N O

+

O –

N O

+

O –

O

–

N O –

+

O –

(most stable)

O

–

N O –

+

O

N O –

+

O

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D is less acidic than A because the two OCH 3 groups on ring make the system more electron rich, which inturn destabilize the conjugate base relative to A. O–

O

O –

OH3C

H3CO

OCH3

CH3O

CH3O

O

O–

O–

OCH3

–

–

Others +

CH3O

OCH3

–

OCH3

CH3O

OCH3

CH3O

(destablizing the conjugate base) l

l

N

703. l

Basic

l

N

Non-basic

H

Lone pair of the basic nitrogen is fully available with nitrogen, i.e., not involved in delocalization while lone pair of non-basic nitrogen is part of aromatic delocalization as: N l

N +

l

N H

–

–

N +

–

N +

N

N

N

H

H

H N – +

N H

704. III < II < IV < I O–

Negative charge is on electronegative atom and stabilized by resonance. This is the most stable conjugate base of the four acids.

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Solutions O–

The negative charge on electronegative atom but not stabilized by resonance. This anion is then less stable than phenoxide ion but more stable than the anions produced from II and III. CH2–

Negative charge is on electropositive carbon but stabilized by resonance.

CH2CH2–

Negative charge is on electropositive carbon, and not stabilized by resonance, hence least stable. 705. (i) (ii)

Equilibrium favour left-side since amide nitrogen is not basic because the lone pair on nitrogen is delocalized through carbonyl groups via resonance. Therefore, the reaction will not proceed in the forward direction. Equilibrium favours left side. The positive charge will be better stabilized on pyridine nitrogen as: +

+

N

N

H

H

+

+

N H

N H

+

N H

(more stabilized) +

+

CH 3 —C ≡≡ N—H ←→ CH 3 —C== NH (less stabilized) (iii) Equilibrium favours left side because the positive charge will be more stabilized on nitrogen bonded to three ethyl groups. The inductively electron withdrawing nature of three trifluoro methyl groups will destabilize the ammonium salt on the right side of reaction. 706. 2,6-ditertiarybutyl pyridine is more basic than pyridine since t-butyl group stabilized the protonated ammonium salt by inductive effect. However, nucleophilic strength depends on its rate at which it attack to an electrophile. In case of di-tertiarybutyl pyridine, the alkyl group adjacent to the nitrogen, hinder the lone pair of nitrogen to attack at an electrophilic center. Therefore, it is weaker nucleophile than pyridine.

R N

l

l

R

O ⇒ Posing steric hindrance to the attack of lone pair.

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py py

H xz-plane

707

H

pz

xy-plane

C–———C———C

Br

Cl

pz

Br  C  H and Cl  C  H are two perpendicular planes of the molecule. 708. I > III > IV > II 709. Equilibrium favours left side since on right side we have strong conjugate base HO – and strong acid, they react to produce more stable species H 2O and a weak conjugate base O O–   CH 3O  C CH==C  OCH 3 . K eq = 10 −3 also predict equilibrium to favour left side. 710. Reaction (a) will proceed to right. For a spontaneous acid-base reaction, a strong acid must react with a strong base to produce weak acid weak base. +

s

In (a) PhCH 2 NH 3 ( pK a = 9.33) is stronger acid than CH 3CH 2 NO 2 ( pK a =10) and CH 3C HNO 2 is stronger base than C 6 H 5CH 2 NH 2 . The opposite is case in reaction (b). 711.

(a)

(b) HO—

—NO2

N+ H (c)

O

O

(d) MeO

OMe

Conjugate base is resonance stabilized. N+ H

COOH

(e)

(h) PhCOOH

(f)

(g)

(i) p-nitrophenol. OH NH3Cl

712. (a) O2N

COOH

(c)

(b) COOH

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Solutions

713. (a) p-methoxyaniline, (b) pyridine, (c) CH 3CH 2 NH 2 , (d) 2,4,6-trinitro-N-N-dimethylaniline. (d) N,N-dimethylaniline, (b) H 2 N  C— NH 2 , (c) PhNH 2 , 714. (a) CH 3 NH 2 ,  NH (e) m-nitroaniline. 715. (a) IV < II < I < III (b)

+

CH 3CH 2 OH 2 + CH 3CH CH 3  S– and 2HSO 4 –

CH 3CH 2OH + CH 3 CH CH 3  SH

H 2SO 4 + SO 2– 4

Major components at equilibrium are: CH 3CH 2OH, CH 3CH(SH)CH 3 , HSO 4 – . 716. (a) is better site of protonation, since it gives a resonance stabilized conjugate acid: CH3 N

+

CH3

+

N—H

N

CH3

CH3 N

N—H

CH3

N—H

+

CH3 CH3 +

N

N—H

CH3 (Complete octet of all atoms)

717. Oxygen, since it gives resonance stabilized oxonium ion as: H + O O O

OH

OH

+

H+

N +

H+

N

N

+

N

N

H

no resonance stabilization

(Resonance stabilization)

OH

+

N Complete octet of all the atoms.

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718. (a) (i) A > D > B > C (ii) C > A > B (b) D > C > E > B > F > A (c) B > E > C > A > D (d) C > B > A > D (e) B > C > D > A (f) A > C > D > B (g) B > A > E > D > C (h) (i) C >D > B > A (ii) E > C > D > B > A (i) B > E > D > C > A (j) B > C > D > A > E CH3

l l

N

CH3

CH3

CH3

CH3

l l

N

CH3

+

N

CH3

CH3

δ+

N

+

719. (i)

(ii)

δ–

O

(Resonance hybrid)

–

+

B

–

O

–

O

O

δ–

δ+

+

B

B

B

δ+

(Resonance hybrid)

O

O –

l l

(iii)

l l

–

N

N

N

Oδ–

O

–

CH2

(major)

+

H2N

H2N +

(i)

H2N

O–

O–

O H2N

H2N

+

H2N

+

(ii) –

δ–

(Resonance hybrid)

720. The resonance forms of (i), (ii) and (iii) are: H2N

δ–

N

+ –

–

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Solutions +

H2 N

H2 N –

(iii)

721. 722. 723. 724.

725. 726.

The resonance forms of three compounds reveal that compound (i) has maximum double-bond character in C—N bond and compound (iii) has minimum double bond character in C—N bond. Therefore, the order of C—N bond-lengths in three compounds is: (i) < (ii) < (iii). Order of acid strength is: III > II > I. I > II > III. For carbocations, more stable the carbocation, weaker will be electrophile. Hence, order of electrophilic strength is: II > I > III (aromatic). (i) To left since ammonia is weaker acid than water. (ii) To right since in aqueous medium, secondary amine is stronger base than tertiary. (iii) To left since hydroxide ion is stronger base than methylamine. (a) B > C > A, (b) A > C > B, (c) A > B > C. Strength of a base depends on stability of its conjugate acid. For all the above bases, their conjugate acids formed after protonation acquire very high stability from resonance as: H I

N

H+

N

H

N+

N

N N

II

N +

+

N

H+

N

+

N

N

N

H

H2N

NH2

NH2

NH III

H +

NH2 + H+

H2N

NH2

+

H2N

NH2 NH2 H2N

Here all three resonance structures are equivalent.

+

NH2

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–

+

l l

N

727.

N

+

N—

–

+

N

+

N—

+

N

+

N—

+

N—

+

+

N

N

+

N—

+

N— –

–

(ii) H a : Conjugate base is aromatic. 728. (i) H c > H a > H b . (iii) H b : Electron withdrawing group increases acidity. (iv) b: Methoxy group decreases electrophilic character of carbonyl carbon by resonance effect. 729. (iii) > (ii) > (iv) > (i). +

NH2 N

730. H2N

+

+

+

NH3

NH2

NH3 COOH

H2N

H

N

COO–

H

at pH < 2.18

pH between 2.18 – 9.09

+

NH2 N

H2N

COO–

H

NH2

NH

NH2 H2N

N

COO–

H

pH between 9.09 – 13.2

pH > 13.2

H N 731. (a)

(b) p-nitroaniline > aniline. O

N H

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Solutions

COOH

+

732. (a) H3N

NH3

NH3

+

+

COO–

+

(c) H3N

COO–

+

(b) H3N

COO–

(d) H2N NH2

NH2

733. 3 < 4 < 1 < 2. 734. Conjugate base is resonance stabilized as:

CH 3  CH==CH  CHO

O–  CH 2  CH==CH  CHO ←→ CH 2 ==CH  CH==CH

735. The relative strengths of ammonium ions as shown above lies in stability of conjugate base. Stronger acid produce more stable conjugate base and hence lower pK a value. Also, since aromatic amines are weaker base than aliphatic amines, conjugate acid of former will be stronger than the conjugate acid of later. ⇒ A > B: acid strength.

STEREOCHEMISTRY

736.

Cl

Cl

Cl

H Cl

H H

H H

H cis-1,3-dichlorocyclopentane (1)

Cl

Cl

(2)

(3) trans-1,3-dichlorocyclopentane

Here (2) and (3) are non-superimpossible mirror images of one another, hence they are enantiomers. Structure (1) is meso isomer and it is diastereomers of both (2) and (3). Br Br Br Br 737.

H Cl

H Cl

H H

H H

H

H

Cl

Cl

(1)

(2)

(3)

(4)

1 and 2, 3 and 4 are enantiomers while 1 and 3, 1 and 4, 2 and 3, 2 and 4 are diastereomers.

738. (a) Enantiomers (b) Identical (c) Diastereomers (d) Identical 739. 13.9x − 13.9(100 − x ) = − 300 ⇒ x = 39.2, Hence, 39.2% of the mixture is (+) 2-butanol and 60.8% is (–) 2-butanol.

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H

740. (a)

Br

H

F

Cl

Br

H

Cl

Br

C==C

Cl

C==C II

I

H

C==C

F

IV

C==C V

F

H

Br

F

F

H

Cl

Br

Cl

C== C

Br

III

Cl

C== C

F

VI

I and II, III and IV, V and VI are pair of geometrical isomers. Cl Cl (b)

Cl H

C==C

Cl Cl

H C==C HH

H

H

C==C

H C==C HH

Cl H

H

C==C

Cl Cl

(ii)

(i)

H C==C HH

Cl H

(iii)

II and III are enantiomers while I is optically inactive, meso form. CH 3 CH 3 Cl CH 3     (c) Six isomer : CH 3 —CH—CH 2CH 2Cl, CH 3 CH—CH—CH 3 CH 3 —C—CH 2CH 3 ,  Cl (two stereo isomers), CH 2Cl  CH 3 —CHCH 2 CH 3 (two stereo isomers). CH 3  741. HC ≡≡C—CH—CH== CH 2

CH 3  CH 3CH 2 —CH—CH 2CH 3 B

A

Br 742.

A 743. CH 3 CHCH 2CH 3  OH ( A)

( B)

CH3CH2 CH3

B C CH 3 —C —CH 2CH 3  O

C==N

CH3

+

C2H5

CH3CH2 CH3

(C)

C

C==N CH3

C2H5

H

CH3 NHCH3 (D)

H

C CH3 NHCH3

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Solutions

744. Addition of bromine at carbon-carbon double bond proceed via a cyclic bromonium ion intermediate followed by nucleophilic attack of bromide ion as: Br CH3 C H 2 5 Br + CH3 C2H5 Br + Br – Br C 2H 5 C————C CH3 CH3 C2H5 CH3 C2H5 CH3 C2H5 Br Mixture of enantiomers obtained in equal amount giving racemic mixture. Catalytic hydrogenation occurs at metal surface giving syn hydrogenation. H C2H5 CH3

C==C

C 2H 5

H2/Pt

CH3

H

CH3 C2H5 C2H5 CH3

COOH Cl

745. A

B D

COOH C

746. CH 3CH 2 —CH—CH==CH 2  CH 3

CH 3CH 2 CHCH 2CH 3 CH 3CH 2 CHCH 2CH 2Cl   CH 3 CH 3 B

A

C

Cl  CH 3CH 2CH CH CH 3  CH 3

Cl  CH 3CH 2 CCH 2 CH 3  CH 3

CH 3CH 2 CHCH 2 CH 3  CH 2Cl

CH 3CH 2 C== CHCH 3  CH 3

D

E

F

G

CH 3CH 2 CCH 2 CH 3  CH 2 H

747. A: Achiral, B: Achiral, C: Meso, D: Meso, E: Meso, F: Meso, G: Chiral, H: Chiral, I: Achiral. 748. (a) Distereomers, (b) Identical and meso. 749. Compound I is optically inactive since it contain a plane of symmetry. Compound II is enantiomeric since it does not contain plane of symmetry, hence chiral. Also compound I is polar while II is non polar.

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πz πy 750. In allene, the two pi-bonds are in perpendicular Cl planes, hence the two terminal H  C  H planes are also perpendicular. Therefore, if allene is substituted xz property it becomes chiral as: C–———C—–——C The two terminal Cl  C  H planes are perpendicular to one-another and Cl and H at one H terminal are equidistant from either Cl or H from other carbon. This is non superimposable on its mirror image, hence optically active. 751. (a) II and IV are chiral, hence optically active. (b) I and III are achiral, possess plane of symmetry, hence optically inactive. (c) There is no enantiomer pair, both II and IV are identical structure. (d) I and II, II and III, I and III are pair of distereomers. 752. (a) Both I and II are optically, but they are not mirror image of one another, hence, they distereomers. (b) Distereomers. (c) Enantiomers. (d) I and V are enantiomers. (e) IV and VI are distereomers. CH3 CH3 CH3 753.

+ HCl

xy H

are

H

Cl +

H

Cl

H

H Chiral (two stereoisomers)

Cl

Chiral (two stereoisomers)

Therefore, in the above addition reaction four stereoisomers (two pair of enantiomers) will be produced. CH3 CH3 CH3 CH 3 CH 3 H CH3 CH3   H 754. H 3C  C 2  C 3  CH 3 60° 60°   CH3 H H H CH3 H H CH3 CH3 H CH3 CH3 2,3-dimethylbutane CH 3

I

II

III 60°

CH3 H

H3C 60 °

60 °

60 °

I ← II ← III ←

H CH3 IV

CH3

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Solutions I

I III

III Energy : I > III > II > IV

II

II

E

IV

0

60

120

180

240

300

360

Angle

are 755. I: CH 3  C ≡≡C  CH 2  COOH and CH 2 ==CH  CH==CH  COOH isomers. CH 3  C ≡≡C  CH 2COOH and HC ≡≡C  CH 2CH 2COOH are positional isomers. H

functional

H

HC ≡≡C

C ≡≡CH stereoisomers.

COOH + HOOC CH 3

CH 3

O  II: CH 3CH 2CH 2CH 2  C  NH 2

O  and CH 3  CH  CH 2CH 2  C  H  NH 2

functional isomers. O  CH 3CH 2CH 2CH 2  C  NH 2

O  and CH 3CH 2CH 2  C  NHCH 3 are positional isomers. CH 3

CH 3 H

and NH 2 CH 2CH 2CHO

H2 N

H CH 2CH 2CHO

are stereoisomers.

756. Only two stereoisomers, since H 2O will attack to the intermediate carbocation from a side anti to the methyl group on adjacent carbon. H3 C

CH3

CH3

H3 C

H3 C

CH3

l l

H+

+

OH2

(bottom side attack)

OH

Enantiomeric (two stereomers)

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456

757. The stereocenters are labeled by stars: H N *

*

O

O O

I

The distereomer pairs are: H N

H N O

O

O

O

and O

O

I

I

758. In acid catalyzed hydration of alkene, a planar carbocation is generated in the 1st step. The nucleophile (H 2O) attack in the next subsequent step from a side where the steric hindrance is minimum.

H+

I

H2O –H+

+

CH3

CH3

OH CH3

(Pair of enantiomers)

H+

+

H2O

II

–H+ CH3

CH3

OH CH3

(Pair of enantiomers)

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Solutions

H+

H2O

III

OH

–H+

+

CH3

CH3

CH3

(Pair of enantiomers) ⊕

H+

IV

CH3

CH3 (No chiral carbon in generated on protonation)

759.

+

••

CH 2 ==CH  N HCH 3 ←→ CH 2  CH== NHCH 3 N-methylethenamine

H

+

CH3 +

C== N –

C2H5 CH3 60°

760. H H

CH3 CH3 I

C== N CH 3

CH 2

(These are geometrical isomers) C2H5 C2H5 H CH 3 H 60°

H CH3 II

CH3

H

+

–

H

H 2C

H

H

60°

CH3 CH3

CH3

C2H5 H

H3C H

CH3 III

CH3

IV 60°

CH3

H

C2H5

H CH3

CH3 V

Energy: I > V > III > II > IV Stability: I < V < III < II < IV 761. (a) This compound contain a symmetry plane, hence optically inactive. (b) This compound contain ‘one’ chiral carbon, hence enantiomeric.

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458

Chiral carbon

Enantiomers

(c) This compound contain a symmetry plane, hence, optically inactive. (d) This compound contain ‘one’ chiral carbon, hence, enantiomeric. Chiral carbon

O OH

O HO

Enantiomers

CH 2OH

H HOH 2C 762.

OH

HO

H

HO

H

HO

H

≡≡

CHO

HO

H

HO

H

HO

H

HO

H CHO

CHO 180°

→

CHO

H

OH

HO

H

H

OH

HO

H

H

OH

HO

H

H

OH CH 2OH

HO

H CH 2OH

II

V

Enantiomers Also II → D-sugar. V → L-sugar. Sugar I, III and IV are all L-sugar. Any pairs of two structures other than II and V are distereomer pair. 763. This compound has two chiral carbon, and a double bond capable of showing geometrical isomerism giving rise to total eight different configurations possible for the molecule as shown below: O O

H

HO H

H Enantiomeric

COOH

H H

OH H

Enantiomeric

COOH

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Solutions

O

O

H

H

COOH H

OH

COOH H

H

Enantiomeric

O

O 764. (a)

OH

–

+

NH2

NH2

no stereo-isomerism

O

–

O

(b)

N

CH3

+

N

CH3

CH3 CH3

no stereoisomerism –

O

O (c)

+

N

N

H

H

(exhibit geometrical isomerism)

O

O (d)

H Enantiomeric

N

H

H

–

+

N H

H

(exhibit geometrical isomerism)

765. (a) Both are similar structures. (c) They are enantiomers.

(b) They are positional isomers. (d) They are constitutional isomers. O

OH

OH

: All have chiral carbon

766. I I

keto

enol

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460

OH

O

OH

II II

keto

enol

All have chiral carbon. O

OH

OH

III III

keto

enol

All have chiral carbon.

767. (a) I and II are identical. (b) I and II are identical. (c) I and II are distereomers. (d) I and II, I and III are pairs of distereomers. II and III are enantiomer pair. 768. Both has nitrogen as their chiral center. Ion II recemizes in aqueous solution due to the existence of following equilibrium: H CH

+

N

H2O + H3C

CH2

l l

H3O+ +

C2H5

N H3C

CH C2H5

CH2

(achiral)

769. It has one chiral center and can exhibit enantiomerism as: C2H5 H

C2H5

and O

O

O

O

are enantiomers.

H

770. The cis and trans forms of 3-methoxy cyclohexanol are: OH OH OCH3

H

H

H cis

H

trans

OCH3

O–

OCH3 OCH3

OCH3 cis-3-methoxycyclohexanol

CH3I SN2

base

H H

H H meso form (optically inactive)

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Solutions

O– H trans-3-methoxycyclohexanol

base

H H

O–

  OCH3   CH3I  OCH3    

OCH3

H H

OCH3 +

H

OCH3

OCH3

H Enantiomers (Racemic mixture)

H

771. (a) Tautomers (Isomer), (b) Conformers, (c) Resonance structures, (d) Positional isomers.

HYDROCARBONS O  772. Ph—CH 2 —C ≡≡C—CH 3 PhCH 2CH 2 CCH 3 A

B

PhCH 2COOH PhCH 3 D

E

CH3

O  PhCH 2 CCH 2CH 3 C

PhCH 2Cl F

CH3

CH3 Cl

Cl2 AlCl3

+ Cl

E

OH

773.

C

CH

A B OH COOH

OH

OH

C—CH3

CH—CH3

O

OH C

(i) KCN (ii) H3O+

O

D O

O COOH

774.

O

CHO A

B

COOH C

D

O

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462

CH3 775.

• CHCH3

A

Labelled carbons are chiral

CH3

Et

•

H

Et

H

CH3

Enantiomeric (two structures)

Enantiomeric (two structures)

CH3

CH3 CH

C

CH2—C—CH3

CH3

776.

Br A

B

CH3 COOH

CH—CH—CH3 Br

777.

E

D

C

—CH2—C

—CH2—CH2—CH2—

C— O

A

CH3 —CH2COOH

CH2Br

—COOH O

C

D

E

F

CH3 778.

—CH2OCH2CH2CH—CH==CH2 A

O

B

CH3 —CH2OCH2CH2CHCH2CH2OH B

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Solutions

CH3 —CH2OH OH

HO

O

D

C

E RO•

SO 2Cl 2 → ROSO 2Cl + Cl •

784.

RH + Cl • → R • + ClH R • + SO 2Cl 2 → R  Cl + SO 2Cl • Br

Br

Br

785. B

A

D

C

F

E

Cl Cl

786. A

B

C

787. A

B

Cl

CH3 CH3   788. (a) CH 3  C  C H  CH 2I  Cl I  (c) C 6 H 5  C H  C HCH 3  Br

C

Cl

D

Cl

D

Cl

Cl

E Cl

E

Br  (b) C 6 H 5  C H  C H  CH 3  Cl CH3 Br   (d) C 6 H 5  C  C (C 6 H 5 ) 2  I

G

F

F

G

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464

Cl

789.

(a)

(b)

+

ring expansion

+ H+

Cl

+ +

ring expansion

+ H+

Cl Cl

+

CH3

790.

CH3

C H2SO4

CH3

B2H6

OH

H2O2/NaOH

OH

791. (a) (c) 792. (a) (b) (c)

(b) CH 3CH 2CH 2OH (CH 3CH 2CH 2 ) 3 B 2-methyl-3-pentanol (d) 2-methylcyclohexanol. 1-butene + BH 3 / H 2O 2 / NaOH → 1-butanol 1-butene + dil. H 2SO 4 /heat → 2-butanol 2-bromo-2,4-dimethylpentane + alcoholic KOH → 2,4-dimethyl-2-pentene 2,4-dimethyl-2-pentene + BH 3 /H 2O 2 /NaOh → product. CHO

O3

H2-Pt (A)

Zn-H2O

2 CH3CHO + (D)

(E)

CHO

H2-Pt (C)

(B) H+/KMnO4

CH3CH2CH2COOH

(F) + CH COOH 3

793. Ozonolysis products shows that A is a symmetrical diene: 794. Cold, dilute, alkaline KMnO4 solution adds on C==C producing cis-diols. One of the stereomer A produces meso cis-diol therefore it is confirmed that the starting compound is a symmetrical alkene. Also oxidative ozonolysis of the starting compound gives no acid as product, it indicates that there is no hydrogen attached to doubly bonded carbon. From these finding it is now concluded that the starting compound has following structural formula-3,4-dimethyl-3-hexene Also: A + Cold, dil-KMnO4 → C(Meso) therefore, A is cis isomer and B is thus trans one. Thus the various compounds are: A: cis-3,4-dimethyl-3-hexene. B: trans-3,4-dimethyl-3-hexene. C: Meso-3,4-dimethyl butan-3,4-diol. D: 2-Pentanone. Cl

795.

(A)

(B)

(C)

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Solutions

796. (a)

OH

H2SO4

B2H6/H2O2

heat

NaOH

B2H6/H2O2

(b)

OH

NaOH

OH

KMnO4 +

H Br2

(c)

heat

CH3

Br2

KOH Br Alcohol

KOH

Br

Alcohol

Br Ph

Ph Br

Br2

(d)

Ph C

+

Na

hν

Ph Pd/H2/BaSO4

H

(e)

H

Na/liq NH3

(1) Na/heat

NBS

H Br

H

(2) CH3Br

H

H Mg/ether D2O

D

H

H

Na/liq NH3

(f)

KCN

(g) Br

Br

BrH

+

(CH2=CH)2Cu

Br2

H OH2

Br+ HOOC

Br

–H+

O

HOOC O

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466 Br2

(h)

Product

OH2

797. (a) +

+

H

+

–H

•• OH ••

OH

O +

–H Me-shift

•• OH •• OH

+

•• OH ••

H 2O +

–H

(b) IV < II < III < V < I B : Et 2 NCHCl 2 798. A : Et 3 N + CCl –2 Ph

Ph

Ph

Ph

+

+

H

799. (a)

C : Et 2 N  CHO

+

Cl

Cl

+

H

(b)

•• O ••

O

Cl

+

resonance stabilized

(c) HOCl

C6H5CH

HO– + Cl

CHCH3

•• O ••

•• O +

Cl

+

Cl

+

OH +

C6H5CH

CHCH3 Cl

stabler carbocation

HO–

C6H5CH

CHCH3 Cl

•• OH ••

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Solutions

CH2Cl

Cl

Cl

800. A

C

B

E CH2Cl

D

Cl

CHO

801.

H 3C

H O

G

F

J

CH3 H

H

O

H 3C CH3

H 3C

I OH

CH3

CH3COOH

OH OH B

A

H 3C

C

CH3 D

CH3 H H 3C

CH3 Cl

Cl

F

E

802. A: 5-methyl-2,4-heptadiene; B: CH3CHO E: 2-hydroxyethanoic acid. 803. A: 3-methyl-1-butene; B: 3-methyl-2-butanol;

C: 2-butanone;

D: Glyoxal

C:3-methyl-2-butanone CH3

H2

H 804. CH3

C

C

C

C

C

CH3 H

H

CH3

Pd/BaSO4

(A)

CH3

Pt

CH3

H 3C H

Na/liq NH3

H

H

H

( Optically Inactive C )

H

H 3C

O3

H CH3

H H 3C

H ( O A) D

H

H HOOC

C CH3

CHO E(OA)

H

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468

CH2Cl 805. A

Cl

B Cl

H

D

C

H H

H

Cl

OH OH E

F

G

C2H5

806.

C2H5 HOOC

COOH

B

A

C

D

O 807. A

B C Cu

Cu(I)

Br A

+

B

808. HO

Br

E=A C

809. (A)

(D)

OH

(B)

F

ONa (C)

OMe

MgBr (E) OH Br

(F)

(G) Br

(H)

(I) Br

+

H

D=A

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Solutions

OH (J)

C Na

+

(K)

Br (L) Br Br

CH2Br

810. Br

Br B

A

811. A

Br C Br

Br E H

D

D

C

B CH2Cl

Cl

E

Cl OH

Cl

812. (a)

(b)

(c)

Cl

H

813. B

A

CHO O C O

814.

A

Br Br

B

OH Br

C

Br

D

O

815.

816. A

B C OH

817. (a)

+

H

–H2O

+

Me-shift

+

–H+

E

OH

OH

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470

Cl

Br

Br2

(b)

Cl

Br

Br

+

Br Br Br

OH2

OH

CH3

CH3 HOOC

818.

H

CHO H3C

CHO

H 3C

C

B

A

H

CH3

CH3 H D

H 3C

COOH

I E

COOH

O

Br

OH

819. B

A

C

D

E

820. (a)

+

+ H+

(b) (i)

+

–H

+

+ Br (ii)

H Bu–

C

COOH

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Solutions Cl (iii)

Cl

Cl

N

– OC(CH3)3

N

H

H 821. (a)

B

A H

(b) HOCH2

Br

Br

Br

Br

CH2OH Y

X

Z

Br

822. A

C

B COOH COOH D

O

O

COOH OH

COOH

E

F H

H

Br + Br HO OH

(b) H H

823. (a)

CH3

CH3

CH3

O

O

O

(e)

(d)

(c)

H

CH3

CH3

H

H H

OH (f)

OH H

Br (g)

H Br

OCH3 (h)

H Cl

OH

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472

H CH3

—OCH3

824. H

NaOH syn hydration

CH3

trans

—OCH3

OH

BH3/H2O2

C

C

CH3

H CH3

H CH3

CH3

CH3

NaOH syn hydration H3C

—OCH3

H

—OCH3

OH

BH3/H2O2

C

C

H O —C

825. CH3—

COOH

CH3—C—CH2CH2—CH—CH2COOH

C—CH3

B A

COONa

COOH

CH—CH2COONa

CHCH2COOH

CH2CH2COONa

CH2CH2COOH D

C

H 826. (a) Br

CH 3

H

Br H Br

Br H H

CH 3

(enantiomeric)

H Br CH 3

Br Br Br

H Br Br

(enantiomeric)

CH 3

(meso)

Br H H

(enantiomeric)

CH 3 Br Br Br Br

H H (d) H H

CH 3

CH 3

CH 3

CH 3 H H Br H

Br Br

CH 3

(meso)

CH 3 H Br H H

Br Br

H (b) H H

(enantiomeric)

CH 3

(c)

H Br Br

CH 3

CH 3

(meso)

Br H

CH 3

Br H H Br

H Br Br H CH 3

(meso)

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Solutions 827.

(CH2)10C

OMe

O

B

A

C

CH3 HO H

D

Ph

H CH3

E

Br

+

Br 828. Starting compound

+

Br2

H—O HO

–H+

Product

l l

829. (a) (i) HgSO 4 /H 2SO 4 (ii) NaBH 4 (b) (i) H 2SO 4 /heat (ii) Br 2 /MeOH (c) (i) C 2 H 5ONa/C 2 H 5OH (ii) HOOC  CH==CH  COOH 830. (a)

CH3 CH ———CH (b) 3 3 CH3

—CH3

831. If the number of double-bonds are same, heat of hydrogenation depends on degree of substitution at carbon-carbon double bonds. More the number of alkyl substituents at carbon-carbon double bond, more stable will be the alkene towards hydrogenation, hence smaller will be the heat of hydrogenation. On that basis : C—least substituted, highest heat of hydrogenation. A—most substituted, least heat of hydrogenation. H+ OH –H2O

832.

+

H2C

H –H+

I

+

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474

+

–H+

H+

H+

H

+

+

+

H–-Shift

–H+

II

OMe

833. (a)

(b)

(c) Br

Ph

(racemic)

OCH3

HO (racemic) OH

Ph

(d)

(e)

+

H

H

Br 834.

C2H5ONa

RCO3H

O

C2H5OH

HC

CNa

O– CH3I

C Cl 835. (a)

(c)

(b) D

CH

HgSO4 H2SO4

Product

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Solutions

+

+ H+

836.

+

I

I

Cl

Cl–

Cl +

II

Cl–

II

+

Cl

Cl

837. A

B H

H

COOH

C

H

H

D CN

E

F

O

G

OH

C2H5 C2H5

A

839. CH 3  C ≡≡C  CH 2  COOH (A)

O  H  C ≡≡C  CH 2  C  OCH 3 C

CH3 840. (a)

Br2 heat

COOH

HOOC

838. O

B

O

C

CH 3  H  C ≡≡C CH  COOH B CH 3C ≡≡CCH 3 D

CH3 Br

C2H5ONa

CH3

CH3 MCPBA

O

C2H5OH

OH CH3 (i) CH3COONa (ii) H3O+

OAc

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476 Cl Cl2 hν

(b)

C2H5ONa

(i) CHCl3

C2H5OH

(ii) NaOH

Product

O CH3I

Na/∆



Na ∆

(c) H  C ≡≡C  H → → CH 3C ≡≡CH → CH 3C ≡≡C – Na + → H 3O +

→ Product CH 2   Na → CH 3  C  C  CH 3 Heat  CH 2

HBr

(d) CH 3  C ≡≡CH → CH 3  C==CH 2  Br

H

H O

l l

O+

CH3

O

H l l

+

CH3

+

O

CH3

H+

841. (a)

O –H+

(b)

O

O

+

+ NH3

H3N H +

HO

O

N

–

HO H

–

OH

N

HO

H

O

NH2

O

H +

HO

N

O

–

(c)

OH OH

H OH

N

HO

OH

H

O

O+

Br2

Br +

–H+

Br

O Br

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Solutions

H+

842.

+

H +

HSO4–

H

+ H2SO4

H (3° carbocation)

H

+

OCH3

l l

Cl2

843. (a)

+

Cl

O

H

CH3

–H+

H

Product

Cl H

(b)

Br l l O—H

+ Br2 H

HO–

H +

O

+

H+

(c)

O H

O

OH

H

O—H H

H O

Br

Br

+

O

–H+

l l

O— OH

OH

O OH

OH

H

–H+

+

H+ l l

O

O

CH3

OH CH

OH

Br

CH3

CH3

OH

CH2—C—CH3

CH—CH—CH3

C—CH3

O

O

H

844.

Br B

A

OH

CH3 CH2—C

OH

C

CH3 CH2—C—CH2OH

CH2

O

H D

E

F

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478

+

CH3 + H+

845.

+ I H+

I– A

I

+

I– +

B +

Br–

+H+

846.

+

Br

H– shift

Br Br–

+

+

Br–

Br

847. This compound has a chiral carbon. Therefore can be resolved into enantiomers. Treatment of this compound with concentrated sulphuric acid bring about isomerization into a more stable compound but without any chiral carbon, hence optical activity is lost as: +

H+

H– shift

2° carbocation

CH3

CH3

+

3° carbocation

–H+ No chiral carbon

CH3 OH

848. A

C

B +

849.

+

+ H+ +

Protonation at any other side will produce a less resonance stabilized carbocation.

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Solutions

ALKYL HALIDES 850. (a) n-iodoheptane since iodide ion is a better leaving group. (b) Sodium ethoxide since it is a stronger nucleophile than ethanol (c) 1-butyltosylate, due to less steric hinderence. (d) triarylbutoxide with ethyl bromide. (e) with-2-bromopentane. 851. Both the reactions are examples of S N 2 reaction. Since aniline is a weaker nucleophile than cyclohexylamine, former undergoes slower substitution reaction. CH3

Cl 852.

CH3 CH3

H3C

CH3

H3C

CH3

H3C CH3

B

A

C CH3

CH3 CH3

H2C

CH3CH2CH—COOH E

D Cl 853.

H

H

H H3C

CH3 CH3 H3C B

A H

H H

H D

C

H

CH3 CH3

CH3

CH3

H3C

CH3 E

854. (a) Methylbromide undergo halogen exchange with NaI forming methyl iodide. Iodide being a better leaving group then bromide, rate of hydrolysis is increased. (b) The predominant reaction in this case is elimination (E-2) which depends on concentration of both alkyl halide and base. Increasing concentration of base increases the rate of E-2 reaction decreasing concentration of alkyl halide. (c) NaCN is ionized completely giving ambident (CN − ) nucleophile. Since carbon is better electron donor than nitrogen, bonding occurs from carbon side. AgCN being covalent, CN − is not totally free for attack and attack primarily occurs from nitrogen side giving isocyanide.

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855. (a)

OH

H+ –H2O

CH2+

Cl– Cl

CH2+

CH2Cl Cl–

(b)

856. (a) (b) (c) 857. (a) (b) 858. (a)

H+ OH

–H2O

+

Cl– Cl

(3° carbocation)

C—F bond is the strongest carbon-halogen bond, undergo very difficult heterolysis. Latter forms a more stable carbocation. t-butylchloride form a more stable carbocation. AgNO 3 forms insoluble AgX driving reaction in forward direction. More acidic solvent increases the rate of heterolysis facilitating formation of carbocation. CH2CH2CH2OCH3 (b) CH 3 CHCH 3 ( A ) CH 3 CHCH 3 ( B )   OTs N3 C H CH3S CH3

859. III < II < I 860. (a) Ethanol being a very weak nucleophile can’t activate S N 2 reaction, and CH 3 Br being very less reactive in S N 1reaction, neither S N 1nor S N 2 take place. Adding excess of C 2 H 5ONa involve S N 2 reaction and ether is formed. (c) Et 3 P is a better nucleophile than Et 3 N. N3 CH2 OCH2CH 861. (a) (b) I Br

(c) (d) MeO—

N—H

(e)

Br

862. Tertiary butyl being a bulky base abstract H-from less hindered β-carbon giving this alternative product.

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Solutions

OTs

OMe

863.

;

CH3ONa SN2

OTs

OTs

OTs

OTs CH3ONa

E-2

OTs

1,5-hexadiene,

864.

1,4-pentadiene

N

N

H

CH3

2-methyl-1,4-pentadiene,

3-methyl-1,4-pentadiene

N

N

H

H

H

H

(b) CH3——OH

865. (a) HO——CH3

(c)

C4H9

C4H9

CH3 CH3 H H

(racemic)

CH3 CH3——H (d) CH3——H C2H5

866. (a) First

867. (a)

CH3 CH3O——H (e) H——CH3 C2H5

(b) Second (c) First

Br

+

Heat

Br

+

OH

H2O –H+

CH3 (b)

H2O

Heat +

2H

+

H 3C H + OH

CH3 H3C H OH

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482

868. (a) Though neopentylbromide is primary, bulky tertiary butyl group poses very large steric hindrance to the attack of bulky nucleophile N −3 . (b)

Br H——CH3 + N3–

N3 CH3——H

S N2

D D (c) Rate will double (d) Rate will double (e) not related (f) Racemization occur through carbocation intermediate. 869. (a) n-iodoheptane (b) C 2 H 5ONa (c) n-butyltosylate CH3I Product Excess

NaOH

—OH

870. (a) HO—

(d) isopropoxide with C 2 H 5 Br.

NH

TsCl 3 (b) C 6 H 5CH 2OH → → Product. CH CH CH SNa

3 2 2 TsCl (c) C 6 H 5OH → → Product.

(d)

B2H6

Cl

KOH

Product.

H2O2/OH–

871. Allyl bromide produces a resonance stabilized carbocation justifying its greater reactivity in S N 1 reaction. Also carbon β to Br has an empty p-orbital which share some of the shared electron pairs of pentavalent transition state formed in S N 2 reaction, increasing its stability and hence reactivity. 872. (a) CN − being weaker nucleophile than H − . (b) 3° carbocation is more stable than 1°. (c) Geminal diols are unstable.

+

873. (a)

–H+ +

•• OH ••

Product

O H

(b)

—Br

+ Br –

O –

O

874. (a) CF3– < CH 3O – < CH 3S – < F –

(b) CH 3COO – < CH 3SO 3– < CF3SO 3–

Br Br Br

875. A

B

Br

Br C

D

E

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Solutions

CH2Br H——CH3 H3C——H CH2Br

CH2Br H——CH3 H——CH3 CH2Br F

G COOH

876. Br A

B

(CH3)2CHOH

877. O A

C

H

CH3—CH—CH2CH3

B

CH3 C

D

HOOC C

CH3 H

OH C

CH3

C

C

H CH3

D

H

E

Cl C—CH3

CH3CH

878.

CH3 A

B

Br

C O

O

879.

COOH A

E

B

C

D

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484

880.

O I A

I

NC

I

I

OH D

C

B

Br

Br

881. OH A

C

B C H SNa

TsCl 2 5 882. (a) → →

TsCl NaI (b) → →

PhCH CH I

CH C ≡≡ CNa

NaH 2 2 (c) → →

3 (d) n-butylbromide →

NH

3 TsCl (e) → →

O 883.

S O

O CH3

D

H2N—CH— CH2OH  CH==CH2

→

(f) NaI

I

Rate = 1

mesylate

O S O

O CF3

NaI

I

Rate = 5000

triflate

In both mesylate and triflate, the leaving group anion is stabilized by resonance. In the case of triflate, the three fluorides provide additional stabilization to the anion via electron withdrawing inductive effect. O– O O O     – Mesylate: O  S  CH 3 ←→ O ==S  CH 3 ←→ O ==S  CH 3 ≡≡ O  S  CH 3     O O O O – O– O O F O      – Triflate: O  S  CF3 ←→ O ==S  CF3 ←→ O ==S  CF3 ≡≡ O  S  C  F      O F O O O – (more stable)

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Solutions

O

O 884.

1.

O

HOCH3

Ag+

+

OCH3 +

–H+

O

O

O

Ph

Ph

OCH3

Ph

Planar carbocation

O 2.

Ag+

O

O OCH3

HOCH3

+

O

O

O

+

O

–H+

O O

Ph

Ph

Ph a resonance stabilized acyl oxonium ion

(c) SN–1, CH 2 ==CH  CH 2 Br.

885. (a) SN–2, CH 3CH 2 Br, (b) S N 1, (CH 3 ) 3 CCH 2 Br, H 886. (a)

n-Bu

(c)

CH3—CH—CH2—

Et

Ph

Cl

(b)

C==C

CH3

CH3 Et

iPr (d)

CH3

(e)

CH(CH3)2 (f)

H

OH Et

H

COOCH3

Et CH3

(g)

(Retention of configuration)

(Racemic mixture)

CH3

CH 3CH 2

887. (a)

(b)

CH3

OH (c)

C CH  CH 2CH 3  CH 3

H 3C

CH3

(Enantiomers)

SCH(CH3)2 (d)

Ph (e)

CH3

ClCH 2

Ph Et

H (Retention of configuration)

(f) CH 2 == C C(CH 3 ) 3

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486 (g)

CH 3 CH 3 Cl    + heat H– -shift 888. CH 2 == C  CH 2  CH  CH 3 → CH 2 ==C  CH 2  CH  CH 3 ←→ CH 3 CH 3  +  + CH 2 == C  CH  CH 2CH 3 ←→ CH 2  C== CH  CH 2CH 3 X

Y

CH 3 CH 3   + CH3OH X + CH 3OH → CH 2 == C  CH  CH 2CH 3 → CH 2 ==C  CH CH 2CH 3 + CH 3OH 2   H 3C  O  H OCH 3 +

H CH 3   CH3OH Y + CH 3OH → CH 3  O  CH 2 C== CHCH 2CH 3 → +

CH 3  + CH 3OCH 2  C== CHCH 2CH 3 + CH 3OH 2 KOH

I

889.

NBS

—Br

EtOH NBS

KOH EtOH

—Br

KCN(aq)

Product

890. Since, the tosylate is tertiary, E–2 and S N 1 are equally probable: F

F PhNH2 E-2

H3C

OTs

CH3 F

SN1

H3C

F

PhHN +

: Net result is racemic mixture.

NHPh Enantiomeric

CH3 Enantiomeric

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Solutions

C3H7 H 891. H

Br

Br C4 H 9

H

H

C3H7 H

C==C

C4H9

H

C4H9

H

H

C3 H 7

C2H5OK

Major alkene

More stable conformer

H

C3 H 7 H

Br

Br

H

C3H7

H

Repel each other

C4 H 9

H

C3H7

C4H9

H

H

H

C2H5OK

C==C

C4 H 9

Minor alkene

Less stable conformer

CH2CH3 892. (a) H CH3CH2CH2O

C

CH3

OCH3 (c)

(b)

CH3 +

CH3

OCH3

CH3

OCH3

(Racemic mixture)

Br

893.

O B

A

D

C

OMe 894. (a)

(b) no reaction

H

Cl

(c)

OEt

Ph (d) CH3CH2CH

C—CH3

CN

(e)

(f)

ONa + CH4

(cis + trans)

Br Br

895. +

l l

H—N

N

–

H N

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488 Br

H N

H

+

N

l l

– H+

N

896. Esters are usually water insoluble, if acetic acid is used, it gives homogeneous medium driving reaction in forward direction. 897. (a) 2-chlorobutane, (b) 2-chloro-2,3-dimethylbutane, (c) 1-methylchlorocyclopentane, (d) 1-ethyl-2-methyl chlorocyclohexane. D 898. (a)

D (b)

C NC

CH3

D

C

NC

H

H

899. (a)

OH

(b)

OH

(c)

OH

(d)

OH

(d)

OH

C

+

CN

H3 C TsCl

KI

TsCl

CH3CH2NH2

CH3 H

CH3CH2COOH H+/heat TsCl

NaHS

TsCl

CH3ONa

900. Only II can be used for successful synthesis of Grignard reagent, rest all contain acidic proton and will react with R – (from Grignard reagent) forming alkane. Cl O– O O – O O + OCH3 901. (a) + Cl– Cl (b) Br

–

+ OCH3

Br

OCH3

O

O– O –Br–

OCH3

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Solutions

O –

+ OH

(c)

O

O–

HO

O

HO

O– O

H2O

Product

O

HO

O–

O —Br

902. (a)

Mg

—Br

s (MgBr)+

SN2

H (b)

+

H–-shift

H+

—OH

H

+

–H2O

H– – shift

+

Br–

Product

+

ring expansion

903. (a) (CH 3 ) 3 CI is more reactive since I – is a better leaving group than Cl – . (b) I is more reactive since β H is less sterically hindered compared to II. CH(CH3)2 CH(CH3)2 CH(CH3)2 Cl Cl CH3O– CH3O– H H E-2 904. (a) E-2 H H (achiral) Enantiomers

(b) As shown in (a), the predominant elimination product is achiral. (CH3)2HC (c)

H

Cl H

CH(CH3)2 OCH3 CH3

O–

SN2

CH(CH3)2 OCH3

H

H H

H

Enantiomers

CH3O–

H Cl H

CH(CH3)2

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490

CH3 905.

I

(CH3)3COK

II

(CH3)3COH

(CH3)3COK (CH3)3COH

II is more reactive towards elimination reaction since the bulky base will abstract a H from side of least steric crowding giving more stable alkene. OCH3

+

OCH3

Cl

OCH3

+

Resonance stabilization

906. (a) (b) (c) (d) (e)

907.

First is more reactive since it produces a more stable carbocation: Second is more reactive since it produces a resonance stabilized allylic carbocation. First is more reactive since it produces a more stable benzylic carbocation. Second is more reactive due to the same reason as with (b). Second is more reactive due to the same reason as with (a).

I

–I+

+

Cl– CH3OH –H+

I III

Me– -shift Cl– +

CH3OH –H+

II IV

ALCOHOLS AND ETHERS 908. Ethylene oxide being a three membered ring, suffers from very large angle strain and therefore highly unstable whereas THF is a five membered ring, it is stable. O CH3ONa H2SO4 909. Ph—CH —CH 2OCH 3 ← Ph—CH—CH 2OH Ph  → CH OH CH 3 3OH   OH OCH 3

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Solutions

O CH2OH 910. (a)

CH2OH

N2H4

H2SO4

O3

Heat

Zn-H2O

NaOH/Heat

LiAlH4

OH

Br (b)

aqueous

Heat

+ Br2

KOH +

H+

–H+

Heat

+

O

O

O3

+

Zn/H2O

Y

X

X

Conc. H2SO4

LiAlH4

Y

Product.

Heat

H2CO

NaOH

CH2

O

Heat

O3

O

O

Zn-H2O

CH3

A

B CH3 —CH2 E

—CH2D

—CH2MgBr

—CH2Br

911.

C CH3

C

—CH2—C—OH CH3 D

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492

912. At higher temperature, intramolecular dehydration is entropy favoured.

913. 2-butanol

2-butanone

2-bromobutane

B

C

A

OMgBr D

CH 3 CH 3   Ph—C—CH 2CH 3 ( B ) Ph—C== CHCH 3 (C ) CH 3COOH( D ).  Br

914. Ph—CH —CH —CH 3 ( A )   OH OH

O NBS

915. (a) Ph—CH2—

Ph—CH—

CCl4

KOH

O3

Ethanol

Zn–H2O

+ PhCHO

Br

H 3O

(b)

C2H5ONa

Ag2O Br2/Heat

O

C2H5OH

O

CN

Br NaOH

KMnO4

CaO Heat

OO

O

H

H2SO4

O COOH

O

Br

Ag2O Br2 Heat

O O

C2H5ONa

KMnO4

C2H5OH

H2SO4

COOH

O

O3

N2H4/OH–

Zn-H2O

Heat

CH2 O

(CH3)3COK

Ag2O

(CH3)3COH

Br2 Heat

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Solutions

H3O+

(c) O

CH3COCl

OH

H2O

OH

O

O O3

H2SO4

Zn-H2O

Heat

NaBH4

OCOMe OCOCH3

O

OCOCH3 H

SOCl2

H3O+

CH3CH2CHO + O

Zn Heat

OCOCH3 PBr3

H

NaBH4

Br

O

OH OH

Br

916. O A

C

B

OH

E

D

F

H

G

917. (a)

OH

H+

+

–H+

–H2O

(b)

OH

•• OH

OH

+ H+

+

O –H+

+

(c)

+ OH + H

+ CH2

–H2O

H-shift –H+

+

CH3

–H+

+

–H

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494

OCH3

OH H+

(d)

+ CH3OH

H2O

NaBD4

H2O

+ (CH3COO)2Hg

918. (a)

O

OH D

OH

(b)

SOCl2

Mg

Cl

A

Ether

H

PCC

CrO3

H3O+

A

H2SO4

O

O

OH  H3 PhMgBr (c) CH 3COCl → → C6 H 5 CH —C— 3 excess  C6 H 5 O+

O

O

O

MeMgBr

(d)

O

H3O+

OH

C2H5ONa

SOCl2

C2H5OH

OH

OH OH

B2H6

CH3MgBr

H2O2/OH–

OH

H3O+

Cl NaCl

TsCl

919. (a)

SN2

(b) Ethylbromide being a primary halide undergo predominantly S N 2 reaction while cyclohexyl bromide being a secondary halide, undergo E-2 reaction. O

OH

920. (a) A

OH

O (b)

H

H B

A

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Solutions

O (c)

+ CH2O

OH

(d) (i)

O (ii)

OH

COCH3

OH

921.

A

922.

H

H

CH3

Br

B

H3O+

H

CH3

C

OH CH3 H OH

H CH3

CH3

O

OH

CH3 H

H3O+

H

(Racemic mixture)

CH3 H OH

CH3

(meso diol)

923. Chlorination of alcohols with HCl proceeds by S N 1 mechanism, rearranged products are also obtained. Chlorination of alcohol with SOCl 2 proceed by S N 2 mechanism, no rearranged product is obtained.

924. (a)

ONa +

(b)

ONa + PhCH2Br

(c)

—ONa + CH3CH2Br

Br

Product.

Product.

Product.

(d) CH 3CH 2ONa + CH 2 == CH—CH 2 Br → Product. 925. The major product is 2-chloro-2-methylbutane. 926. (a) (CH 3 ) 2 C==CHCH 2Cl gives a resonance stabilized tertiary carbocation as: + +

Allyl chloride gives primary carbocation and therefore less stable than the above mentioned carbocation.

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496

(b) In acidic medium alcohol undergo isomerization via carbocation intermediate as: H 2O –H+

H+ –H2O

OH

+

+

HO

Ph 927. (a)

Ph

H+

Ph

+ +

–H2O

OH OH

Ph

Ph •• O—H •• Ph

+

OH Ph –H+

Ph O

OH +

H+

(b)

+

–H+

–H2O

H+

(c)

+

+

–H2O

OH OH

—— OH ••

OH

H+

(d)

+O

O

–H+

+

O

–H+

HO

HO

H

928. PhCH(OH)CH 2 Ph

PhCOCH 2 Ph Y

X

R 929. (a)

R

HBr Peroxide

(CH3)3COH

Br (b)

OH O

R

(CH3)3COK

H+ –H2O

–H+ +

O

CH2

+

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Solutions OH

CHO

(c) (i)

CHO

OH

A

(ii)

OH

CH3—CH—CH2SPh

(iii)

OH C

B

OH D

OH (iv) E

930. C 6 H 5 —CH 2 —CH —CH 3  OH

C 6 H 5 —CH—CH 2OH  CH 3

( A)

CH 3  931. CH 3 —CH— CH —CH 3  OH

( B)

CH 3 CH 3   CH 3 —CH CH 2CH 2OH CH 3 —C—CH 2CH 3 ( B)  OH

( A)

(C )

CH 3 CH 3   CH 3 —CH CH 2COOH CH 3 —C—CH 2CH 3 (D )  Br

CH 3  CH 3 —CH—COOH (F )

(E )

OH 932. A

CH 3  933. CH 3 —CH —C—CH 2CH 3  CH 2 ( A)

O C

B

CH 3  CH 3 —CH —CH—CH 2CH 3  CH 2OH

CH 3  CH 3 —CH —CH—CH 2CH 3  COOH

( B)

(C )

CH 3  CH 3 —CH— C —C 2 H 5 ;  O

CH 3  CH 3 —CH— CH —CH 2CH 3  OH

(D )

(E )

CH 3 CH 3

C== CHCH 2CH 3 (F )

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498

936. 1,3-butandiol (A) 3-ketobutanoic acid (B). +

Methyl shift

H+

934.

H2O –H+

+

OH (Major)

Cl 935.

*

Cl

–

OC2H5

O I

937. (a)

* CH2—CH—CH2—OC2H5

* CH2—CH—CH2—OC2H5

O

O –

—OH + HCl

D2O

Mg

—Cl

—D

ether

O H2SO4

(b) CH 2 == CH 2 + Cold, dilute, alkaline KMnO 4 → CH 2 —CH 2 → Heat   OH OH

O

CH 3  Heat 938. (a) CH 3 —CH—CH 3 + Br 2 → (CH 3 ) 3 CBr CH 3  (b) (CH 3 ) 3 CBr (from a) + C 2 H 5ONa → CH 3 —C== CH 2 C2H5OH E-2

Peroxide (c) (CH 3 ) 2 C==CH 2 + HBr → (CH 3 ) 2 CHCH 2 Br CH ONa SN 2

3 (d) (CH 3 ) 2 CHCH 2 Br → (CH 3 ) 2 CHCH 2OCH 3

(e) (CH 3 ) 2 CHCH 2 Br + CH 3COONa → (CH 3 ) 2 CHCH 2OCOCH 3 (f) (CH 3 ) 2 CHCH 2 Br + KCN( aq ) → (CH 3 ) 2 CHCH 2CN O 939.

OH

Cl COOH

—O—C— A

B

D

C

CH2OH Cl F

COOH

G

H

I

E

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Solutions

OH

OH

940. A

B

C

I

O 941. A

O B

I

C H O

2 2 942. (a) 2-methyl propene + B2 H 6 → Product

NaOH

CH COOT

3 (b) 2-methyl propene + B2 D 6 → Product.

CH COOT

3 (c) 2-methyl propene + B2 H 6 → Product.

C2H5Br NaH NaOH (d) 2-methylpropene + B2 H 6 + H 2O 2 → (CH 3 ) 2 CHCH 2OH → → Product.

OH

943.

H+ –H2O

+

CH2

–H+

+

CH3O

CH3O 944. HO

CH

CH—CH2OH

CH—CH—CH2OH,

HO

Br

X

Br

A O HO—CH2—CHO

CH3O

O

C6H5—C—O

CH

B C

CH—CH2O—C—C6H5,

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500 CH3O

I

HO

CH—CH2Br,

CH

HO

CHCH2OH

CH

D

E

CH3O CH—CH2OH

CH

CH3O F

945. O

O A

C

Cl

Cl B

CH3 CH3 946.

C—CH2CH3

C

C

CH3

CH3 and

H

OH A

CH3

B

O

OH

—C—CH3

CHCH3

—CH—CH3

C

E

D

CH 3 CH 3   947. CH 3CH 2 —CH—CH== C—CH—CH 2CH 3  CH 3

CH 3  CH 3CH 2 — CH— C—CH 3  O

A

CH 3 CH 3   CH 3CH 2 —CH— C—H CH 3 —CH 2 —CH—COOH  O C

H C

C

D

B

CH 3  CH 3CH 2 —CH—CH —CH 3  OH E

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Solutions

CH 3  CH 3 —CH 2 —C== CHCH 3

CH 3CHO

CH 3 —C —CH 2CH 3  O

F

G

H

CH3

CH3 —CH—CH2OH

948.

CH3

—CH—COOH

A

C

B

OH

O

C—CH3

D

H

H O

O

E

F OH CHO

OH

949. (a)

I

(b)

(enantiomers)

(c)

(d)

OCH3 HO

OH (g) H

(f)

(e)

CH2CH3

SCH3

(h) HO

OH OH OH

(j) (enantiomers)

(i)

OCH2CH3

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502

950. (a) PCC, (b) NaBH 4 , H 3O + , (c) Pd/C, (d) Mg, Ethyleneoxide, H 3O + , PBr 3 , (e) TsCl, CH 3ONa, (f) SOCl 2 , Mg/ether, CH 2O, H 3O + , (g) Ethylene glycol/ H + , LiAlH 4 , H 3O + , (h) CH 3 MgBr, H 3O + . Br Mg

951.

C6H5MgBr A

O

O

H

H

OH PCC

A H3O+

CH3MgBr

OH

Product

CH3CH2ONa

OCH2 CH3

952. (a)

H3O+

NaH

Br

(b)

+

Br + CH3 Br

OH l

+

(c)

S Br–

CH3

l

S O

(d)

(e)

O

enantiomers OH OH

CH3 (f)

OCH3

(g)

O

OH

SCH2 CH3

O (i) O

—S— O

(h)

(j)

OH CH3 CH3

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Solutions

Si

O 953.

O Si

+ CF3CO3H

F–

O

H3O+

O + (CH3)3SiF

O

O–K+ +

954. A : CH 3 MgBr/Ether, H 3O ; D:

B:

OH

(A) +

NaOCH 3 , H 3O , NaH/CH 3CH 2 Br;

Hg(OAc) 2 , CH 3OH, NaBH 4 ; E : CH 3CH 2SNa/H 3O +

F:

C:

mCPBA; SO2

MeO

955. In case of reaction 1, the mechanism is a series of protonation and S N 2 reactions as: H O

+

CH3

O

I

CH3

H+

+ CH3OH

I– +

H+

CH3—OH2

I–

CH3I + H2O

In the case of reaction 2, the mechanism is also a series of protonation and S N 2 reaction. However, phenol is not converted to iodobenzene because of protonated phenol can’t be substituted by iodide ion (I – ) due to strong C  O bond in phenol: H O+

O CH3

OH

CH3

H+

I–

+

others

H

O

H

+ CH3I

O+

–

– Carbon-oxygen bond acquire partial double bond character, can’t be substituted under SN2 conditions.

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504

O

OH CH2N2

OH K2Cr2O7

956. (a)

O

OCH3

H3O+ (1) CH3CH2MgBr (2eq)

Product

(2) H3O+

OH

(b)

N3

NaN3

TsCl

NH2

H2/Pd

O Cl

OH

(c)

CH2

PCC

PPh3

OH

TsCl

C–Na+

CH

PCC CH2Cl2

(d)

Product

O

OH

Product

C

H 3O + C

CH

CH

NaBH4

Pd BaSO4

mCPBA

Product

O 957. (a)

—D

(b) OH

H

H (c)

(d) SCH3

(e)

CH2CH3 +

(enantiomers)

OH CH3

OH CH2CH3

CH3

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Solutions

OH

OH

I

(g)

(f) HO

OH

HO OCH2CH3

(h)

O 958. (a) (1) PCC, (2) CH 3 Li, (3) H 3O + ; (b) HI/Red P; (c) (1) SOCl 2 , (2) Mg/ether; (3)  , (4) H 3O + ; (d) (1) PCC, (2) NaBH 4 , (3) H 3O + . H 959.

H

CH3

CrO4

OH

H+

2–

B–

O

O

O—Cr—OH

Slow

O

C—H bond cleavage occur in slow rate determing step. Therefore, replacement of H by D will slow down the rate of reaction. O H

D

C:

OH

PCC CH2Cl2

D

C  H bond will be broken in preference to C  D bond. O

O 960.

O H

l l

+

O —S—

+ Cl—S—

H

O

— + Cl–

O H

O Pyridine

O —S— O

+ N Cl–

—+

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506

Cl– H O N O—S—

+

+

Cl

60°C

O H O

+

N

–

+ O—S—

+

O

Higher temperature promotes SN–2 reaction of tosylate by chloride ion generated in the first step of reaction. Ot-Bu O

961. (a)

(b)

SCH2CH3

(c)

S

OCH3

O OCH3 O

O (d)

Br (f)

(e)

+ O

OH N3

O OCH3

(g)

O

(h)

(i)

OH

O

OPh O (j)

(k) O

O

OH

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Solutions

HO

Br

O

N3

NaN3

NaOH

962. (a)

HO

CH3I

NaH

Product

H3O+ O

HO CH2

(b)

CHLi

mCPBA

NaH CH3I

LiAlH4

NaH

CH3I

Product

from a

OH

O H

963.

H 3O +

Br

PCC

Mg/ether O

H 3O +

O

CH3CH2ONa

HO

OCH2CH3

CH2N2

HO

OCH2CH3

mCPBA

O

964. I > II > III > IV. OCH2CH3

965. (a)

D (b)

(c) OH

O

Br Ph (d)

(e) H

OH CH3 CH2CH3

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508 OCH2CH3

OH

HO

OH 966.

(a)

(b)

(c)

(enantiomers) (Major)

(Major)

OH

O

Br (d)

(e)

OH

OH

(b)

PCC CH2Cl2

CH2N2

NaH

EtONa

(c)

H3O+

CH3Li

PCC CH2Cl2

967. (a)

NaH

CH3I

Product

Product CH3I

O OH PBr3

(d)

PhSNa

mCPBA

Product

968. There are two problems with the route outlined above. First, because the above alcohol is tertiary, it will be difficult to make the tosylate, which is a large, sterically demanding group. If you can make it, the second step is also difficult as it require SN–2 reaction of a tertiary tosylate by H. This system is very hindered and will not readily undergo susbtitution reaction. Elimination will be a serious competition reaction. A better method would be as follows: S OH

O—C—SCH3 NaH CS2

CH3I

Bu3SnH

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Solutions OH

Cl K2Cr2O7

969. (a)

SOCl2

NH3

H2SO4

Br

(b)

NaOH

O

POCl3

COOH

K2Cr2O7

Product

Na/NH3

H3O+

EtOH

H2SO4 O

OH

OH (c)

TsCl

NaCN

CH2N2

NH2

LiAlH4

CH3COCl

OCH3 970. (a)

—Cl

(b)

Product

O

H

Product

CH3 H

(c) H

(d)

971. (a)

CH3 (e) Br

NaOH EtOH

mCPBA

O NaCN H 2O

Product

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510

(b)

—OH

C. H2SO4

NBS/hν CCl4

—Br

Mg ether

—OH

PCC CH2Cl2

—MgBr

O

OH NaBH4

TsCl OH

CH3

Cl

973. O

(c)

OH

CH3

O

PhS–

OH

OH

Ph

(b) CH3—C—CH2CH3

972. (a)

(d)

I

O–

O

O

O

H2O

Product

OH

O

O

O

O

–

SPh

Product

OH

C H ONa

Hg(OAc)2

C2H5OH

H 2O

NaBH

2 5 4 975. (a) PhCH 2CH 2CH 2 Br → PhCH 2  CH==CH 2 → →

O OH    PCC Ph  CH 2  CH  CH 3 → Ph—CH 2 —C—CH 3 CH2Cl 2

I

OH

 H 3O + I Na Pd CH 3C ≡≡CH → CH 3C ≡≡C – Na + → → CH 3C ≡≡C  C  CH 3 → Product heat CaCO3 /PbO  CH 2 Ph

O (b)

B2H6

—CH2OH

H2O2/NaOH

Cl2

PCC CH2Cl2

—C—H I

Cl Mg ether

AlCl3

H 3O +

I

OH —CH—

OH Cl2

AlCl3

—CH—

—Cl

PCC CH2Cl2

Product

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Solutions

O SOCl 2

C2H5ONa

mCPBA

(c) CH 3CH 2CH 2CH 2OH → → → CH 2CH 2  CH  CH 2 C2H5OH

OH  PCC → → CH 2CH 2 CH  CH 2CN → Product NaCN

H 2O

CH2Cl 2

NBS/hν

Mg

CCl 4

THF

976. CH 2 ==CH  CH 3 → CH 2 ==CH  CH 2 Br → CH 2 ==CH  CH 2 MgBr OH  → → CH 2 ==CH  CH 2  CHCH 2CH 3 H 3O +

Propanal

CH3I

NaH

Product ← ← Conc. H2SO4

977. (a)

mCPBA

PhO–

H 2O

Product

heat

CH3 O (b)

CH3

OH HO–

H3C

O

Br NaNH2

OH PBr3

H 2O I

OH

C

NH3

Br

CH

OH Na heat

I

C2H5—C

+ H+

978. O

C—CH2—CH—C2H5

PCC CH2Cl2

Product

HOCH2CH2—CH—CH2—CH3 O+

CH—CH3 +

H

H– Shift

Br HOCH2CH2—C—CH2CH3 CH2CH3

Br–

+

HOCH2CH2—C—CH2CH3 CH2CH3

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512

C H ONa

Hg(OAc)

C2H5OH

H 2O

NaBH

2 5 2 4 979. (a) CH 3CH 2CH 2CH 2 Br → CH 3CH 2CH==CH 2 → 

OH  CH 3CH 2  C  CH 3  CH 2CH 3

(b) PhCH 2CH 3

O OH   PCC ← CH 3CH 2  C  CH 3 ← CH 3CH 2  CH  CH 3 (i) C2H5MgBr

CH2Cl 2

(ii) H2O

Br  C2H5ONa mCPBA → Ph  CH  CH 3 → Ph  CH==CH 2 → NBS/hν

C2H5OH

CCl 4

O HBr

Ph   → Product B2H6

H2CrO4

SOCl 2

(c) (CH 3 ) 3 C  CH==CH 2 → (CH 3 ) 3 C  CH 2CH 2OH → → H2O2 /NaOH

O  → (CH 3 ) 3 C  CH 2  C  AlCl 3 PhI

—I

H2SO4

→ Product SO3

PCC CH2Cl 2

(d) CH 3CH 2OH → CH 3CHO(I) Br

OH O

NaOH

OH

PBr3

H 2O

CH3—CH—CH2Br

NaNH2 I

CH3—C H 3O +

C–Na+ Product

CH 3 CH 3 O   C2H5ONa PBr3 Mg  (e) CH 3 —CH—CH 2 Br → → CH 3 —CH—CH 2CH 2CH 2OH → → Et 2O

C2H5OH

CH 3  (i) KCN mCPBA → CH 3 —CH—CH 2 —CH== CH 2 → Product + (ii) H3O

CH 3CH 3   980. (a) CH 3  C  CH  CH 2CH 3  * Br (Racemic)

+ CH3C

(b) O

–

OMgBr

CH

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Solutions CH3 SEt

OH H (c)

CH3 SEt

OH H

+

CH 3  (d) CH 3  CH  C  CH 3   NHPh CH 3

(Racemic)

O S (e)

(f) Br

OEt

Br 981. (a)

OCH3

O Mg Et2O

OH

H2O

OH

NBS/hν CCl4

Br ONa

Product Br (b) Cl

(CH3)3COK (CH3)3COH

HO

KOH

NBS/hν CCl4

∆

Br

NaOH

NBS/hν CCl4

H2O

OH

OTs TsCl

982. (a)

CH3

CH3 OTs

CH3

Cl NaCl Et2O, SN2

SN2

(from a)

CH3 I

NaI

(b)

Cl NaCl SN2

CH3

CH3

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514

Cl

Cl

NBS/hν CCl4

983. (a)

Cl

CH3ONa

OCH3

Br

CH3

O

Mg/Et2O

Br

Product Br

C2H5OK

NBS/hν CCl4

(b)

H 2O

NBS/hν CCl4

C2H5OH

PhS–

Product

Cl (c)

C2H5ONa

CHO

Br Mg

NBS/hν CCl4

Et2O

CH3 TsCl

(d)

CHO

C2H5OH

I H 3O +

CHO I KCN

mCPBA

O

EtOH

OH

984. (a)

OH

PCC CH2Cl2

O

Product

OH

Br

CH3

EtOK

Br

NBS/hν CCl4

Product

Ph

H3O+

Ph

PhMgBr

H+ EtOH

OH

OMgBr OH

(b)

Ph

OH (as in a)

OH

H2SO4 heat

Ph

O (c)

—Ph

O3 H2O2

O

HO–, H2O2

O

Ph–C—CH2CH2CH2—COOH

from (b)

B2H6

OH Ph

EG/H+

O

Ph—C—CH2CH2CH2COOH

O H2SO4

Ph

Ph—C—(CH2)3—CH2OH

LiAlH4 H3O+

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Solutions

Ph

Ph

Ph

–H2O

H+

985.

OH2

OH

—CH3

(a planar carbocation)

+

(+)-2-phenyl-2-butanol

CH3OH

Ph

+

H3C—C

attack from both side giving racemic product

(±)

OCH3

Br  986. CH 2 ==CH  CH 2CH 2OH ( A ) CH 2  CH  CH 2  CH 2OH ( B )  Br Br

O D

O C

O E

OH 987.

H

Br

988.

PPh3

OTs TsCl

H

NaOH SN2

Product H3O+

mCPBA

Acetone

Product

H 989. (a)

O

Cl

Cl2

Et2O

hν

OH

Mg H 3O

CN TsCl

O (b)

mCPBA

KCN

CH3 CH3MgBr

H 3O +

CH3 H2SO4 heat

OH

O Na  (c) CH 3CH 2C ≡≡ CH → CH 3CH 2C ≡≡C – Na + → CH 3CH 2  C ≡≡C  CH 2CH 2OH heat H 2O

Na liq. NH 3 H 3O + Product

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516

CH3

H H

990. HO

HOOC

COOH CH3

O

A

B

CHO

CHO

OHC CH3

OH

D

C

CHO H2/Pt ∆ E l l

OH

+

H

991. OH

O—Ts –H2O

–H+

OH +

l l

O

O

H

992. (a) (b)

993. (a)

PBr 3 (SN – 2) → C 2 H 5SNa(SN–2) : TsCl → NaCN H2SO4

OH

heat

Two inversion will finally lead to retention.

B2H6

OH

TsCl

OTs

H2O2/NaOH Me3N

Br2

(b)

NaNH2

CCl4

994. (a)

(b)

Br +

ONa

ONa

+ CH3CH2Br (1° halide)

Product

Br

C

Product

Product

C–Na+

Product

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Solutions

O–

Br

–Br–

(c)

995. CH3OH

PBr3

mg ether

Product

CH3MgBr

CH3CH2CHO

PCC CH2Cl2

PCC/CH2Cl2

OH

CH3CH2CH2OH

O OH

—MgBr Mg Et2O

OH

Br PBr3

O

PBr3

996.

H3O+

Mg/Et2O

OH

Product

PCC/CH2Cl2

MgBr

OH

997. (a) HO

NH2

(b) Br

(c) CH3O

OH

(f)

Cl

Br OH (e)

(d)

S OH

(g) D

(h) OH

—CH2CH2OH

HO

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518

O 998. (a)

O

Br

OCH3 (b)

(c) Br

OH

N3

Ph

O

OEt (d)

D

(e)

D

(f) O HO

OH

OH

999. In acid catalyzed hydrolysis, protonation activate the reaction and stability of carbocation intermediate decides the reactivity: H O

HO

O

–H+

+ H+

OH2 3°-carbocation

fast

OH2

l l

+

H OH

O+

O + H+

Slow

OH2

+

l l

OH2

1°-carbocation

In alkaline medium, the first step is nucleophilic attack of hydroxide ion,which preferably occur at less hindered carbon: O

O

less hindrance is offered from adjacent carbon.

more hindrance is offered from adjacent carbon O–

O –

CH2

1000. HO

N +

+

l l

N l l

N HO

O

O

–

HO– HO

N –N 2

O

O

H 2O

OH

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Solutions

OH

OH

O

l l

O –H2O* * OH2

H+

OH

–H+

O

O

+

OH HO

H

+

O

OH

O

H+

Ph

Ph

Ph

Ph

O * OH

1001.

+

+

O+—H –H H2O *

l l

OH *

O* Ph

Ph

Ph

1002. (a) OTs A CN

O O

O H

H A

D

O

O

O H

O

C

B CHO

O

(b)

CN

H D

C

B

1003.(a) mCPBA → H + /MeOH, (b) mCPBA → NaN 3 /H 2O, (c) (CH 3 ) 3 SBr. –H2O

1004. H+

HO l l

O

O H2N—NH2 l l

+

H—NH

H—N +

O

NH2

NH2

O –H+

H 2O

O

H—N

N

N O–

l l

NH2

O

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520

NaCN

TsCl

1005. (a)

CH3Li

H 3O +

Ph3P

PPh3

(CH3)2C

O

CH2

Product

OH H

LiAlH4

(b)

PBr3

Product

BuLi

COOH O

OH

Cl

CN NaOH H 2O

(c)

C—OH H 3O +

l l

2I H+/heat

CN

C—OH O

I

OH +

1006.

+ CH3COOH

O—H N

N—H

N

–

OAc

Product

OH

H 3O +

+ N2

–

OAc

N—H

H

O 1007.

+

H+

OH

O +

H H +

–H+

OH

Product

Join @unacademyplusdiscounts 521

Solutions

OH CH3

1008. (a) H

OH

OH and its enantiomer. (b) H

H

Ph

OH (c) CH3

CH3

CH3

H

Ph

OCOCH3

D

Br

OCH3

(d)

(e) OH CH3

1009. (a)

Mg Et2O

PBr3

—OH

Oxirane H 3O +

CH2CH2OH

O Cr2O7

(b)

OH

O

PBr3

OH

OMe

H+

Br

(1)

OMe

O

2

MeO– MeOH

O

Br

(2)

O (1)

Product

C2H5ONa

MeOH

OH

HO

C2H5OH

O

2–

H+

C2H5ONa

SOCl2

H 3O + heat

COOMe

COOMe

O

O

O HO–/heat aldol

O 1010. Br

O

CHO Br

MgBr

A

B

O

O

C

O OH

O

O

O

H

H D

E F

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522

H

OH Ph

(b)

1011. (a)

H (Racemic)

OH

Ph

CH3

O OH

(c)

(d)

H

CH3

CH3

(Phenyl attack from back side, i.e., from H-side to steric reason).

O–

CH3

OCH3 –

O

O 1012. (a)

OCH3

H 2O

KH

Product

–

OCH3

OCH3 +

OH (b) CH3

CH2

H+ –H2O

H3CO

+

H 2O +

OCH3

O +

CH3 l l

+

H 2O

CH3OH2 + O

ALDEHYDES AND KETONES NMgBr     H 3O + NaCN  → C 2 H 5 —C —C 2 H 5 → 1013. 3-pentanone.  Mg C 2 H 5 Br → C 2 H 5 MgBr  ether  aq

C 2 H 5 Br → C 2 H 5CN

1014.(a) Br — ( CH 2 — )5 Br

aq

Mg

NaCN (1.0 equivalent)

Ether

)5 CN → → Br —( CH 2 — NMgBr

O H3O+

Join @unacademyplusdiscounts 523

Solutions

or Br — ( CH 2 — )5 Br

aq

Sn HCl

dilute

→ → NaCN (Excess)

NaOH (Aldol)

CHO CHO

H2CrO4

Heat

CHO OH

CH3 (b) Br

O

CH2Br

NBS CCl4

aq

C

NaCN

Br

Br

N

Mg Ether

H 3O +

O

C2 H 5 1015.

C==C

H

C2 H 5

and

CH 3

C2 H 5

C==C

H

C2 H 5

A

OH  CH 3CH 2CH 2OH CH 3CH 2 CCH 3 D  H

B

C

F

O  1016.(a) Cl 3CCHO (b) CH 3CCl 2 —C—CH 3

G

(c) C 6 H 5CHO NO2

O —C—

CH 3CH 2CHO

O  CH 3CH 2 CCH 3

OH OH   CH 3CH 2 —C—COOH CH 3CH 2 —C—COOH   H CH 3

E

(d) O2N—

CH 3

—NO2

(e)

O

—C—

NO2

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O H

H

1017. O

CHO

B

A

OH OH

1018. (a)

CH2OH D

F

COOH

E

C

O O

A

O

B

C

OH

CHO

(b)

CHO

OH

A

B

C

O

O (c) O A

B

—COO– + O2N—

1019. (a) CH3CH—COO– (b) H2N— OH

(c)

CHO

COO–

CHO

CH2OH B

A Ph 1020. (a)

CHO (b) A

B

OH

—CH2OH

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Solutions

O

OH

OH

O

O

O

1021. (a)

(b)

CHO O

1022. (a)

O

OH

N

O

C

H+

NH

Beckmann’s rearrangement

(Oxime) A

O

(Caprolactum)

B

(b)

Ph H3C

C==N

Ph

C==N

H3C

OH

C

O

OH

CH3—C—NHPh

CH3COO –

E

F

D

O PhCOO –

Ph—C—NHCH3 G

H

H

I

H

C==N

N

CH3NH2

N

O

C==N OH

(c)

H—C—NH—

N

HCOO– M

L

K

J

O C—NH2

N

N

COO– + NH3 P

N

O

O 1023. Ph—C—CH A

Ph

CH3 C2 H 5

C2H5—CH CH3

O CH 3   Ph—C —NH—CH—C 2 H 5 D

C==N

OH

Ph C2H5—CH CH3

B

C==N

OH

C

C2 H 5 CH 3 O    C 2 H 5 —CH—C —NHPh CH 3 —CH— NH 2 E

F

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O

O

HOOC

Cl 1024.

OH–

H

–

OH–

COOH

– not

COO–

1025. (a)

OH

TsCl

NaCN

CN

SOCl2 Mg

MgCl

Ether

H3O+

O Mg

(b) C2H5Br

C2H5MgBr H3O+

aq KOH C2H5OH

PCC

CH3CHO CH

O

C—CHO

CH2CH—CHO CH3

C

O

O

Br

CH3—C

CH

CH– + EtO—C O

O

CH3

B

A

1027.

OH

CHO

1026.

H2CrO4

C—CH3

– C

EtO–

COOEt

COOEt Br

Br

C—CH3

Br Product

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Solutions

O Cl O

1028. A

D

C

B

OH O O    1029.(a) Ph—CH== CH—C—CH 3 + CH 3 —C—CH 2 —C—CH 3  CH 3 O O   (b) C 6 H 5CH== CH—C—C 6 H 5 + C 6 H 5 —C—CH== C—C 6 H 5  CH 3 O

(c) C 6 H 5CH== CH—NO 2

(d) OH

O H

(e)

(f) O

O

O

O 1030. (a)

O CH3

C6H5CHO

OH–

O (b)

CH

Aldol

O OH–

C6H5CHO Aldol

C6H5CH

CHC6H5

E

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NO 2  C6H5CHO → C 6 H 5CH==C—CH 3

OH−

(c) CH 3CH 2 NO 2 → CH 3CH—NO 2

H 2O

NH 2 NH 2   BH3 LiAlH4 CH3COOH → C 6 H 5 —CH==C—CH 3 → → C 6 H 5CH 2 CH—CH 3 O O

C OH–

(d)

CH3

Aldol

NaBH4

Product

O

O

O

(e)

CHC6H5

C6H5CHO

OH–

Aldol

CHC6H5

N2H4 OH–/Heat

CH 3 CH 3   1031.CH 3 —C== CH—CH —CH 3 A

CH 3 CH 3   CH 3 —CH —COOH CH 3 —C—COOH B  Br

CH 3  CH 3 —C—COOH  OH

C

D

CH 3 CH 3 CH 3    1032.CH 3 —CH—C ≡≡CH CH 3 —CH—C ≡≡C—CH 2CH 2CH 3 CH 3 —CH— C—CH 2CH 2CH 2CH 3 A B  O C

CH 3  CH 3 —CH—COOH CH 3CH 2CH 2COOH D

E

Cl CHO

1033. A

C Cl B

D

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Solutions

CH 3OH   1034.CH 3 —C— CH —CH 2CH 3  CH 3

CH 3  CH 3 —C— C—CH 2CH 3   CH 3O B

A

C6H5

1035.C 6 H 5CHO

H

A

OH C==N

C6H5 H

B

C 6 H 5COOH F

C==N

OH

O O   C 6 H 5 —C—NH 2 H—C—NHC 6 H 5 D

E

C

C 6 H 5 NH 2 G

O

O

O

O

CHO

COOH NNHPh

1036.

O A

C

B OH O

COOH

CHO

O F

D=E

CHO

MgBr

G

H

J

CH 3 OH CH 3 H CH 3      1037.CH 3CH 2 —C—CH 2OH CH 3CH 2 —CH—CHO CH 3CH 2 CH— CH —C—CH 2CH 3 B   CHO CH 3 C

A

CH 3 CH 3   CH 3CH 2 —C== CH—C—CH 2CH 3  CHO

CHO  CH 3 —C—CH 2CH 3  CHO

CH 3CH 2 CCH 3  O

D

E

F

1038.(a) O 3 / Zn-H 2O followed by Aldol condensation. (b) O 3 / Zn-H 2O followed by Aldol condensation. (c) X = KMnO 4 / NaOH (dilute) Y = HIO 4 , Z = dilute NaOH (Aldol)

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CH3 (d) X = CH3OCH2Cl + Ph3P

BuLi

C—

: Y = CH3OCH

—OMe :

CH3 Z = H—C—CH—

—OH

O 1039. (a)

–H +

+ H+ O

O OH

•O •

+

H

R

O

OH H+

(b)

OR

+

OH

OH –H+

OH O

•• OH

•• O ••

H

+

–H2O

+

–H+

OH2

O—H –H+

O

O

OH

O LiAlH4

1040. (a)

OR

H

OH

O

+

O

—CH—CH3

H2SO4 Heat

CHCH3 O3/Zn-H2O

H 3O +

CH3

CH3MgBr

Heat

OH

H 2O

HBr Peroxide aq KCN

Br

H 3O +

COOH

O+CH3CHO

Product

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Solutions H O+

PhMgBr

MnO

3 2 (b) PhCHO → → Ph —CH —Ph → Ph—C—Ph → Ph 3P == CH2   O OH

Ph

CH==CH 2 .

Ph O (c)

CHCH2CH2MgBr

+ CH2

H 3O +

H2SO4 Heat HBr(excess) Peroxide

Br Br2

Na

Heat

Heat

Br Br

OH Alc. KOH

OH H2CrO4

B 2H 6 H2O2/OH–

O

–

O

–

CH (CH2)9O

O (CH2)9 –

O

C

O

CH O

A

C

O C

C

O B

OH

O

O (CH2)9 O–

CH

1042. (a) O

(CH2)9O

C—CH(CH2)5—CH3

C–MgBr

H2O

Product.

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H

H C

O

O

C CH— ( CH2— )5 CH3

D

OH —CH2CHO

—CH2CH2OH

—MgBr

(b) E

F

G O

O

O

O

O

(c) H

OH

OH I

J

O

O 1043. (a)

—CH O

CH—C—CH3

(b) Ph—CH

CH—C—CH

CH—Ph

1041.Via aldol condensation : (c) Ph—CH 2 —CH==C— Ph  CHO OH O O —–    H O Mg PCC 2 1044.(a) PhBr → PhMgBr → → Ph—CH 2 —CH —CH 3 → PhCH 2 —C—CH 3 ether

O H 2O Pd-CaCO3 (b) CH 3CH 2C≡≡ CNa + ——— → CH 3CH 2C ≡≡CCH 2CH 2OH → Product. H2

Mg

O ——

H O

PBr

2 3 (c) PhBr  –→ PhMgBr → → PhCH 2CH 2OH → PhCH 2CH 2 Br

ether

O   Mg H3 CH3CH2CN (d) PhBr → PhMgBr → → Ph—C —CH 2CH 3 O+

ether

N H /OH–

2 4 (e) Product from (d) → PhCH 2CH 2CH 3

Heat

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Solutions

Na liq. NH3

Na PhBr (f) H—C ≡≡ C—H → Ph—C ≡≡ C—Ph → → Heat

H

Ph C

Ph

C

O RCO3H

H

H

Ph

Ph H

CH 3 O   1045. CH 3CH 2CH 2 C—O—CH—CH 2CH 3 CH 3CH 2 CHCH 3 CH 3CH 2CH 2CH 2OH A C  OH B

O  CH 3 —C—CH 2CH 3 D

O  CH 3CH 2 —C== CH—C —CH 2CH 3  CH 3

CH 3CH 2 C== CH—CH 2CH 2CH 3  CH 3

E

F

O OH   Mcpba CH 3CH 2 CH— CH —CH 2CH 2CH 3 CH 3CH 2 CH— C—CH 2CH 2CH 3 → A Bayer villiger oxidation   CH 3 CH 3 G

H

O OH

O

1046.

O

OCH3 A

OCH3

B

C

O COOH O E

O E

dil. NaOH

Heat

reflux

O

O

I D

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1047. (a) Ph—C —CH2CH2CH2OTs O

CH2Ph

Ph—C—CH2CH2CH2—N O O

O

CH2Ph

B

A

CH 2 Ph

Ph—C—CH 2CH 2CH 2 —N  O

CH 2 Ph

C

OTs

OH

O

(b) A

C

B

D

(c) OH

A

O

B

Ph —CH—Ph

(d)

—C—Ph

OH

A

B

O

—C

NNHPh

C C H MgBr

2 5 (e) H—C ≡≡CNa C 6 H 5CH 2C ≡≡CH → C 2 H 6 + C 6 H 5CH 2C ≡≡CMgBr

E

D

NHNH2 (f)

KOH EG

OH O   H 3O + 1048. (a) Ph—C—OMe + MeMgBr(excess) → Ph— C—CH 3  CH 3 O O   LiAlH4 H 3O + PCC (b) Ph—C—OMe  –→ Ph—C—OH → PhCH 2OH → PhCHO

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Solutions

OH CH 3 +   –H2O H 3O (c) Ph—C —OMe → Ph—C—CH 3 → Ph—C—CH 3    O OH Br Unstable

O

O CHO

CH

HOCH2CH2OH

1049. (a)

CH KCN

O

O

H+

CH2Br

CH2Br

CH2CN

OH

O CHO

O

O

CH2CH2OH

(Hemiacetal)

CH2CHO

H

Et

O (b)

CH

LiAlH4

H+

O

H+

O

+OH

O

Et

+

Sn/HCl

2

•• ••

H2O

O OH OH

O –H+

OH Me-Shift

H+

1050. (a)

–H+

–H2O +

OH

OHOH

•• H—O ••

+

O

H+

(b)

+

O

•• H—O ••

O ••

CH

+

HO

+

CH3 (Enol)

CH2

O

O

+

CH +

CH3 (Keto)

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–H+

CH3

H+

+

O

O

CH3

H–-Shift

+

O

–H+

O

O

CH2OH

(c)

O

CH2+

O

H+

+

O

+

–H2O

–H+ O

COOMe

COOMe

O CH –

1051. (a) O

– CH

H2O

+

[O]

O

O

O O

COOMe

COOMe COOH COOH OH O

H2O

O

COOMe

O

OH COOH COOH O

OH

COOMe COOMe OH O

∆

COOMe

OH

OH COOMe COOMe O

O

–

CH—COOMe

O CH—COOMe –

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Solutions

OH H+

O

(c)

+ HN—Me ••

–H2O

H

–H2O

CH

•• CH—N—OH +

N—Me

HO

Me OH

Product

–H+

+

O—H N Me

O 1052. (a)

H

CH—CH2CH2—C—CH3

H+

O

O

O

HOCH2CH2OH

O

O CH—CH2CH2—C

CH3CH2CH2CH

CH—(CH2)2—CH3

PPh3

O CH3 H3O+

H

N2H4

H2/Ni

OH–/Heat

Product

O H

C

C

Br

H

CH—CH—CH3

Br2

(b)

Br

1 equivalent

H

(c) PhCH2OH

C

C

SOCl2

CH3

KOH

C

Na/liq NH3

Heat

C—CH3

H

Mg Ether

O

PhCH2MgCl

PhCH2CH2CH2OMgCl

H 2O

Product

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Cl H3O+

(d)

+

HBr

–H+

Heat

Peroxide

Br (CH3)2CuLi

Br Br2

(e)

C2H5ONa

Heat

HBr

X

O

Peroxide

Br

OH

H2CrO4

aq KOH X Mg Ether

H+

Product

OH Br

OH 1053. CH—CH

CH2

C—CH2CH3 CH3

CH3 A

B

HO OH CH2

O

CH2 CH CH3 D

MgBr

X

E

OH C

H3C C

CHCH3

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Solutions

O

O C

CH

C

CH

CH3

1054.

CH3 O B

A

D

C

(b) RLi → CH 3I → Product.

1055. (a) N 2 H 4 / NaOH/ Heat (c)

OH

O H HCN

(d)

H 3O +

OH COOH

COOCH3

CH3OH H+/Heat

NH2 Product O

COOCH3

Hydrolyse

O

O

O

NH3

O–

O (e) H

–

+ CH3–

H+ –H2O

O

(b)

1056. (a) O

O

Ph

SOCl2

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CH 3 O— BF3   Ph—CH—CH 2 —C—CH 3

+−

1057. Ph—CH== O + BF3 → Ph—CH== OBF3 → ===

+

CH 3 O   → Ph—C —CH 2 —CH—CH 3 OH  H2CrO4 PhC ≡≡ CNa H3O+ 1058. CH 3CHO → → Ph—CH —CH 3 → → → H 3O +

PhMgBr

OH OH H    Na Ph—C—– C == C —Ph Ph—C—C ≡≡ C—Ph → liq. NH3    CH 3 CH 3 H 1059. (a)

+ CH2 +

O

O

O–

CH2

+

H

OCH3 (b)

O

O—CH2 –

H

HO

HO O X

1060. (a)

O + CH2—COOEt Br

(b)

O + PhCH

PPh3

O (c)

Zn Reformatsky reaction

H2O

CH3

Product. O

O + Cl2

∆

CH—COOEt

CH3—C

Cl

aq KCN

CN

RCO3H

Product.

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Solutions

(d)

CH2—OH

O+

H+

Product.

CH2—OH O

OH

(e)

H3

+ PhMgBr

O+

H+

Ph—C—CH3

CH2

Ph—C

∆

CH3

CH3

O (f)

Aldol

+ OH– (dilute)

H+ ∆

Product.

O

O–MgCl

1061.

OH

H 3O +

MgCl

Reverse

O MgCl + +

N

N

OH

Ag2O

1062. (a) A

+ N2

O

O O

CH3OH

O

O (b) H2C

C

(Ketene)

B

O O C

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542

O (c)

O– +

N—N –

H 2O

O

N–

N

LiAlH4

NH

NH D +

H+

(d)

+

O

HO

HO Me-Shift

+

–H+

HO

HO OH

H+

(e)

Enolization

(3 molecule)

OH

HO OH

+

HO

HO –3H2O

(f)

+

O

CH2

CH2

+

+

HO HO

O

H 2O –H+

•• HO— O O–

H

CH2

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Solutions O

O CH3

CH3 CH2N2

1063.

CH3CH2ONa

H 3O +

∆

OH

alternatively CH3OH/HCl

OCH3 CH3CH2Br

CH3I

NaH

OCH2CH3

NaH

OMe

OH

OMe

OEt OEt O

O

O

O

O

COOH

O 1064. A

Alternative

OH

B

O

O

O

CH2CH2OH

D

C

D

—CH—CH2OH COOH

A

B

O

COOH

O

1065. H

COOH HOOC

A

O B

C

H

O OH O

OH D

O

E

O

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OH

l

1066.

C

N

l

C

N

OH

O

l

l

A C

B

O C

NHCH3

N

NH2

H

O D

E

F

COOH

G

H 1067. (a)

CH3

Ph H

C

N

l

Ph

l

C

N

l

(b)

H

l

OCH3 OCH3

CH3

O N (c)

H

(e)

(d)

O O O

O

(f) NNH2 X

OH +

O

O Y

1068. (a) N 2 H 4 /NaOH/heat, H 3O + , Ph 3 P==CH 2 , (b) N 2 H 4 /NaOH/heat, H 3O + , HNCH 3 /H 2 –Pd/C (c) Ph 3 P==CH 2 , H 2 N  OCH 3 , (d) CH 3 Li/HCl/H 2O, Ph 3 P==CH 2 . 1069. III < II < I.

Z

OH

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Solutions

1070.The nucleophilic addition of cyanide ion will occur at carbonyl carbon. Therefore, the stable conformation at α-carbon will be: O OH H O H OH OH H 2 – CN – H NC H ==R 1 R As shown above, attack from side 1 will produce B and attack from side 2 will produce C. Since, attack of CN – from side 2 will suffer less steric hindrance, C will be formed in greater amount than B. H H 1071. (a)

CH3CH2MgBr

PCC

OH

O

CH3

CH3

Planar arrangement H

H +

CH2CH3 CH3

OH

OH

CH3

Major product (nucleophilic addition of CH3CH2– occur from side opposite to —CH3 group)

–

H

H3O

H Ph (attack occur from less-hindered side)

CH2I

1072.

Ph3P BuLi

Ph H

+

Ph

C6H5CH2CHO

Products

CH2CH3

(Minor product)

(b) In the most stable conform shown below: O H3C OH CH3 H H3C H CH3

H3O+

H

OH + CH3 H CH3 (Major)

Ph H CH3 OH CH3 (Minor)

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OH

OH

O HCN

(b)

CN

H

H3O+

COOH

H2O H2SO4

O

O

(c)

LiAlH4

H+

H

COOH

Heat

O

TsCl

NaCN

H OH

OH

O

O

(1) CH3Li

CN O

O CH3Li excess

K2Cr2O7

O

H

Product

O

O (d)

(2) HCl/H2O

H2SO4

O N2H4 OH–/heat

O

1073.A will proceed more efficiently. The electronegative chlorine, via inductive effect, increases the electrophilicity of the carbonyl carbon, making it more susceptible to attack by cyanide ion. O O EG H+

1074. (a)

(CH3)2S

HCl/H2O

EG H+

O

Product

O

O

O (b)

(CH3)2NH

O3

O

O3 DMS

N 2H 4

O

NaOH/∆ HCl H 2O

N 2H 4

Product

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Solutions

O (c)

H

O

O CH3Li H3

H3 C

O+

O

H+ H 2O

O CH3

OH CHO

O

O

NaBH4

O

O

H+

CH3

CH3I

NaH

Product

l l

O

O–

OH

O

OH

+

O

OH2

H+

1075. l l

HO—N—H

HO—N +

H H HO

O

N

–H2O

O

N

O O

N

O –

N

+

HO O

Cl–

●●

O

N

Cl– H

OCH3

OCH3 1076. (a)

CH3

CH3 +

CH3

CH3 (Major)

H OH

H OH (c)

+ (Major)

O (d)

N

H

HO

H

H

+

H+

O

HO

H

H

l l

N

+

N

H

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548 O

O

HOCH2CH2OH

1077.

O

O

O3

O

H

(CH3)2S

H+

O H C2H5NH2

O

O

H2/Pd

O 1078.

O

OPh

CH3O

CH3COCl

N

O

OPh

l l

CH3O

CH3O O+

O l l

+

O–Ph HO

H

H+

H O –H+

Product

O F

HCN

CN

CH3O

OH

l l

O+

H

O

CH3 HO

1079. (a) F

HO H2/Pd

F

NH2

Pd/H2

O

HCN

CHO O

EG H+

B2H6

O O

H3

O+

H 2O

Product

Pd H2

O

NH3 ∆

PCC

NaOH/H2O2

O

H

H2N

Product

OH

SnCl2 HCl

O (c)

CH3

OPh

O

(b)

Product

H

O O

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Solutions

1080.The bond angle required at carbonyl carbon is 120°. In the case of cyclohexanone, the six membered ring can accommodate this bond angle easily and thus, the driving force towards hemiacetal formation, where the bond angle is 109° is small. On the other hand, in cyclopropanone, the system is highly strained, and hemiacetal formation enables some of the strain to be alleviated by allowing the carbonyl carbon to be sp 3 hybridized. H

H+

O + CH3OH

OCH3 ring is less strained than cyclopropanone

1081. (a) CH 3COCH 2 Br, (b) CH 3CHO, (c) CH 3COCF3 , (d) CH 3CHO. 1082. A : CH 3ONa/CH 3OH, B : (CH 3 ) 2 NH, C : CH 3CH==CH 2 MgBr; H 3O + , D : CH 3OH/H 2SO 4 . CH3 O O PPh3

CH3CH

(CH3)2CuLi

1083. (a)

H2O (Conjugate addition)

CH3

CH3 CH3 O mCPBA

CH3 O O

O

O PBr3

EG H+

(b)

Mg ether

O

O

H 2O

OH

OH

OH

PCC

H3

O+

Product

H 2O

Cl

Cl H

1084.

H OH trans

OH–

H

H

SN2

H O H – [—O– and Cl are anti, substitution is preferred]

O

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550 Cl H OH

Tautomerism –OH

O

OH [H and Cl are anti, elimination is preferred]

H cis

+ H O O   1085. HO  CH 2CH 2CH 2  C  H + H + → HOCH 2CH 2CH 2  C  H →

H O+

OH

O –H+

+

OH2

O

OH

H+

CH3OH

Product

–H+

CN 1086. (a)

(b)

NH2

O Ph

O CHCH2

(c)

N + CH3CH2—CH

cis-trans – CH3O H

O

C—CHO

(e)

(d) O

CH3

O

O–

O –

1087. H A

H

H H

O

O

CH3OH

+ H more stable

H H less stable

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Solutions

O

O–

– + C

1088. (a)

H 2O

CCH3

H2/Pd

CCH3

C

Product

BaSO4

OH H2SO4 heat

(b)

mCPBA

CH2N2

(c) O

NaCN H 2O

O

O

O

CH3MgBr

H 3O +

Product

Product

OH (d)

CH3MgBr

H 2O

NaH

CH3CH2I

ether

1089. (a)

Product

OCH3

CH3ONa

mCPBA

O

CH3OH

OCH3

PCC

OH Ph3P

C

CH3 CH3

Product

O OH

O

(b)

H

CH3CH2MgBr

PCC

H3O+ H2O

H N

N

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552

OH  1090. CH 3  CH 2  CH  CHO A

CH 3  CH==CH  CHO B

CH 3  CH  CH 2CHO  Cl C

CH 3  CH  CH  CHO   Cl Cl D

CH 3CH  CH 2  CH 2OH  Cl

CH 3 CH  CH 2CH 2OH  OH

E

F O  CH 3  C  CH 3

NaNH 2 H 3O + 1091. (a) CH 3CH 2  C== CH 2 → CH 3CH 2C ≡≡CNa → → H 2O NH 3  Br KMnO4

CH 3  CH  CH 3   OH

OH  PBr3 KOH CH 3CH 2C ≡≡C  C  CH 3 → → Product EtOH  CH 3

CH 3 O  PBr3 KOH mCPBA (b) CH 3  C  OH → → →  EtOH  CH 3

HBr

HNO2

→ → Product

CH 3 CH 3   Hg(OAc)2 NaBH4 H2CrO4 (c) CH 3CH 2CH  CH==CH 2 → → → CH 3  CH 2  CH  C  CH 3 H 2O  O H 3O +

→ → Product HC ≡≡ CMgBr

Br2

(i) NaNH2 /NH3

CH 2 ==CH 2 → CH 2  CH 2  (ii) H3O+   (iii) Mg/Et 2O Br Br CH 2CH 3 CH 2CH 3   SOCl 2 B2H6 H2CrO4 (d) CH 3CH 2  C==CH 2 → → CH 3CH 2  CH  COOH → H2O2 /NaOH

PhOH

Cl 2

AlCl 3

AlCl 3

→ → Product

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Solutions

O

O O    s CH3O– CH2 == CH  C  CH3 1092. C 6 H 5  CH 2  C  CH 2C 6 H 5 → C 6 H 5  CH 2  C  CH  C 6 H 5  O O     CH3OH CH3 O C 6 H 5 —CH—C—CH 2 — C 6 H 5 ← C 6 H 5 —CH—C—CH 2 —C 6 H 5  O O–   CH 2 —CH 2 —C—CH 3 CH 2 —CH==C—CH 3 −

O  C 6 H 5  CH  C  CH  C 6 H 5 s  CH 2  CH 2  C  CH 3  O O

Ph

Ph

Ph

O –

Ph

O

CH3OH

CH3

CH3

Ph

CH3OH

H

CH3O–

O

Ph

OH

CH3

O CH3 1093. (a) H3C—

O (b)

—C—SCH2C6H5

—CH O

CH3

CH—C—CH3

O CHC6H5 (c)

O

O Br

1094. (a)

(CH3)3CO–

O –

Br

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O O

O O

Ph

Ph—C—C—Ph

Ph

O

(b) Ph

Ph

KOH

Ph

Ph s

O

O Ph

OH–

s EtOH

Ph

Ph

H Ph Ph O–

O Ph

s

Ph

EtOH –H2O

Ph Ph

O

Ph

O

O

O Ph

Ph

Ph

Ph

–H2O

H HO

Ph

Ph

Ph

Ph

1095.Difference in reactivity towards hydrolysis reaction lies in the stability of carbocation intermediate as: H

H O

O

H+

H

O+

O

+

H O

+

H

H

H2C

+

O

O

OH

OH

Allylic group provide extrastability due to resonance

H

H O

O

H+

O

+

O

H H O

+

not stabilized by allylic resonance

OH

OH

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Solutions

O H 1096. H3C

H+

H

CH3—C

–H+

+

OH

CH2

CH—OH +

–

CH2—CH—O—H

Product

H—O—–CH—CH2– +

1097. II < III < I. OH

H+

1098. O

O

O

O+

O

l l

–H+

l l

OH2

+

H

OH O

HO

H2O

OH OH +

OH

l l

–H+

Product

O 1099.

H

Br

EG H+

Mg

O

H2O

CH3CHO

H

CH3

H

Ether

O

O

OH

l l

OH

H3O+

Product CH3

O 1100. I

I – H + OCH3

O–

–I–

O

OCH3

A

OCH3 H3O+

HO H

O

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556

H2O

1101. Compound + EtO– –

O

O–

CH3

O

O CH3

H

CH3

H

CH3

50%

Planar (achiral)

H 3O +

MeMgBr

+

50%

PCC CH2Cl 2

1102. (a) PhCHO → → → Product

(b)

CH3

OH

mCPBA

OH

O

(c)

O

Mg

H 2O

O

OH

O

O

1103.

H

OH

C

B

E

D

O

F

O

O OH

Product

Br

O

A

H 3O +

CH3CHO

Et2O

H+

Br

Product

OH

O OH

CH3I

NaH

O

H+

O

O

OH 1104. B

A

CH 3 CH 3   1105. CH 3  C   C== C  CH 3   CH 3 CH  CH 3  CH 3 A

D

C

O  CH 3  C  CH 3 B

CH 3 O CH 3    CH 3  C  C  CH  CH 3  CH 3 C

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Solutions

CH 3 CH 3   CH 3  C  CH CH  CH 3   CH 3 OH

CH 3 CH 3   CH 3  C  CH==C  CH 3  CH 3

D

E

Br  B : C 6 H 5CHCH 3 ;

1106. A : C 6 H 5CH 2CH 3 ;

OH  E : C 6 H 5 CH CH 3 ;

D : C 6 H 5CHO; 1107. ClCH 2 

C : C 6 H 5CH==CH 2 ;

H

(CH 3 ) 2 CH—

1108. (a)

Ph O

EG

CHO

Ph3P

CH

H+

O

O  F : C 6 H 5  C  CH 3 .

CH2

H 3O + H 2O

Product

O

O

O

O (b)

(i) H+

H2 N

H2/Pd

H (ii) H2/Pd HO

NH

N

H+

H

1109. (a)

OH

(b) (Racemic)

(CH3)2HC

O

H

(c)

H (d)

CH3

(e) HO

H

Ph

CD2OH

CH3

CH2Ph

OH

OH

(f) Ph—CH—C(CH3)2 (g)

H

OH

CH3

OH H O 1110. (a)

N H

N

(b)

NH—C—CH3

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558

(c) PhCH 2 MgBr/Et 2O, H 3O + , PCC/CH 2Cl 2

(d) O

O

CH3 (b)

Fe/HCl

H+ H2O

H3O+

(CH3)2S

H+

O

MgBr

O3

EG

1111. (a)

Product

CH3

KI

NaNO2

NH

(e)

CrO2Cl2 Ac2O

HCl/0°C

I

NO2 CHO

CH3MgBr

H3O+

CH2Cl2

H2O

PCC

Product

I

(c)

1112.

NH2

OH

O3 (CH3)2S

TsCl

H+ BH3

OTs

Product

(CH3)3COK (CH3)3COH

B2H6 H2SO4

NaOH/H2O2

heat

Major product (not required)

PCC CH2Cl2

Product

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Solutions

OH

1113. (a)

,

OH

OH

O

OH

,

,

, OH

OH

O

O

OH

O

OH , OH

OH O

O (b) (i)

D

D

D

D

O Br

(ii)

O

D

(iii) O

O

O CHO 1114. (a)

OMe

(b) OH

(c) Ph—CH2—CH

NHCH3

O

OH CHO

(e)

(d)

OH O

O

D

CHO

1115. (a) D

D OH

(d)

O CHO

(b)

Br

(c)

CHO CH OH

OH H

C—Ph

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560

1116.A has enolizable hydrogen bonded to chiral carbon, therefore, it acquire planarity at chiral carbon when converted to its enol form. Due to planarity at chiral carbon in the enol form, it forms both enantiomers in equal amount when converted back to keto form. No such enolization is possible with B since it has no α – H at chiral carbon and hence, its optical activity is retained even in aqueous alkaline solution.

EtOH

EtO–

A H

+ O–

O Planar structure

CH2—H

1117.

O

H

Racemic mixture

O

O

O

H

O

– O

– CH2

O O

– + OH

O H2O ∆

O

O

1118. (a)

OH

(b)

(c)

CH—COOH O

O

O

(d)

(e)

OH

(f)

Br H (g) Cl

1119.

H O

O Cl

(h)

N H

OH PCC

HO–

CH2

∆ (Aldol)

CH2Cl2

H

O O

PPh3

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Solutions

O O 1120. (a)

(b) H

O

O

D

D

CHO

H (c)

O

(d)

OH

CN

(e)

N

(f)

(g) N

H

+

O

OH

O ll 1121. (a) HO

H

H

+ H+

–H+

O

l l

O

H H

+

OH2

+

O CH3 l l

H+

O OH

C

C

H

l l

(c)

H H+

CH3

–H+

O

Product

NO2

+

O

O

—NO2

H2N—O—

H NO2

OH

Ph—CH—N—O—

—NO2

–H+

+

NO2

–H2O

+

H

H

l l

Ph—CH—N—O—

—NO2

—CH—N—O—

H

H+

NO2

OH

NO2 —NO2

Ph—HC

N—O—

H

—NO2 A

NO2 – PhCN + O—

—NO2

– Ph—C

N—O—

NO2

—NO2

CH3O–

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562

1122. High proportion of enol in the equilibrium mixture of compound II can be rationalized by two factors: (a) It is conjugated system. (b) It forms a six membered cyclic ring via intermolecular H-bonding as: O

H

O H Intermolecular H-bonding.

O

O – O N

OH OH–

1123.

C

OMe

N H

– O

N

N–

H N

O

H 2O

O

OMe

N O

H N

OMe OH

O3

1124. (a)

(CH3)2S

NaOH heat

OH LiAlH4

Product (b)

O RCO3H

O

PCC CH2Cl2

Product

OH

O

O

(c)

OMe

N

O –H2O

H

OH H3O+

COOH CH3OH H+

Product

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Solutions

anti to t-butyl, shift

OH HNO2

1125. (a)

H

OH

OH

H+ +

OH

NH2

OH2 l l

Product

–H+

H–-Shift

OH

OH

OH H+

HNO2

NH2

OH2

OH

Product

O

+

–H+

+

l l

CH—OH

O

O

–

MeO–

(b)

OH

+

H 2O

Product

O OH

O –

Ph O

Ph Ph

O

O –

MeO–

OH

Ph

O H 2O

O –

O Ph

Ph

O Ph

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564

1126. (a)

O

H+

OH

OH

H 2O

+

OH

–H+

H+ –H2O

OH

OH –H+

O O

CH3 OH (i) CH3MgBr

(b)

+

H+ –H2O

(ii) H2O

O

HO–

O

O

CH2–H

HO–

O–

–

O

O

O H

H2O OH

Br

Br

1127. O

OH

O

SN2

LDA

O

+ Br– O

O– Br

Br LDA

O

CH3

+ Br– O

CH2–

O

+

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Solutions

1128. OH B

A

C

O

NOH

O E

F

D

G

CH 3 C 2 H 5   1129. CH 3  CH C==CHCH 3 A

CH3 O   CH 3  CH  C  CH 2CH 3 D

CH 3 C 2 H 5   CH 3  CH  C  C 2 H 5  OH

CH 3 C 2 H 5   CH 3  CH  C  CH  CH 3   H OH

B

C

CH 3  CH 3CH  CH  CH 2CH 3  OH

CH 3  CH 3  CH  CH==CH  CH 3 F

E

CH 3  CH 3  C== CH  CH 2CH 3 G

CARBOXYLIC ACID AND ITS DERIVATIVES 1130.Extraction of a solution of carboxylic acid and phenol in CCl 4 with aqueous solution of NaHCO 3 will bring carboxylic acid in aqueous phase as RCOONa leaving phenol in organic phase. 1131. A : 2-methylmalonic acid, B : Propanoic acid. OH

O

Br2

1132. (a)

–H+

O Br

O

• OH •

O

+

Br +

(b)

CH3COOH

+ H+

–H+ +

OCOCH3

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566 OH

(c)

+

H+ –H2O

NHCOCH3

NHCOCH3 •• NHCOCH 3 –H+

+

Product

O O 1133. (a) PhCONHCH3

O

COOH

(b)

O

(c) CH3

O

OH

O

O

(d)

OCH3 COOH dilute H2SO4

HBr NaCN 1134. (a) → → →

HBr Peroxide

dilute H2SO4

NaCN (b) → → →

O Cl

SOCl2

COOH

O

(c)

I Ag2O Br2/Heat

1135. (a) CH 3OH + Na 2CO 3

Br Mg

H3O+

ether

COOEt (b) COOH

(d) CH 3CH 2CH(COOEt) 2

I

(c) Ph—CH—COOEt  COPh

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Solutions

O —CH2NH2

(ii) H3C—

—COOH

(e) (i) H3C—

(iii) H3C—

—C—Ph

O O   1136. (a) HOOC— ( CH 2 — ) 4 COOH → C 2 H 5O—C — ( CH 2 — )4 C— OC 2 H 5 + C2H5OH H

O COOEt

EtO– EtOH

I O

O I

LDA

–

COOEt Cl

P

COOEt

aqueous NaCN

COOEt COOEt

H O+

4 3 (b) CH 3COOH → ICH 2COOH → → HOOC—CH 2 —COOH

I2

H 2O

C 2 H 5OH/ H + / Heat tEOOC

COOEt

SOCl 2 NaCN 1137. (a) PhCH 2OH → PhCH 2Cl → PhCH 2CN

CH 3  H3 SOCl 2 CH3MgCl (b) PhCOOH → → PhCOCl → Ph—C—CH 3 Excess  OH O+

EtONa

(c) EtOOC

CH2

CH—CH2Br

COOEt COOEt

COOEt

H 3O + H 2 O C2H5OH

COOH

SOCl2

COOH

Heat

COOH

COOEt

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568

O

1138. (a)

OH

Cold, dilute

Cl

Cl O

Cl

OH

Cl2

aqueous

hν

O

O

NaOH

O

O

H2CrO4

CH3 (c)

O

alkaline/KMnO4

OH

(b)

O

RCO3H

CH3

CH3

CH3Cl

H2SO4

CH3Cl

AlCl3

Heat

AlCl3

CH3

SO3H

O

SO3H

COOH

P 2O 5

O

O

Heat

KMnO4/H3O+

dilute

NaOH/Heat

H2SO4

COOH O O COOH (d)

P 2O 5

O

Heat

COOH

O

OH

O H+

COOH O

NH2 1139. (a) 6-hepten-1-ol

PCl5

NH3

NH2 O3 H2O2

COOH

Heat

NH O

NaBH4

NaCN (b) PhCH 2 Br → → PhCH 2CH 2 NH 2

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Solutions H O+

H2SO4

HBr

3 NaCN (c) PhCHO + CH 3 MgBr → Ph CH CH 3 → PhCH==CH 2 → → Heat Peroxide  OH PhCH 2CH 2CN

O 1140. (a)

P4/I2

COOH

H 3O

Na

COOH

COOH

∆

I

COOH

P 2O 5

O

Heat

O

O (b) C2H5O—C—(CH2— )4 COOC2H5

O COOEt

E t ONa

H 3O +

E t OH

He a t

O L DA

CH3

O HCOOH

1141. (a)

Heat

NH2

H—C—NH—CH—CH2CH3

C2H5MgBr

(b) CH3COOEt

C H 3I

H 3O +

Excess

OH

NH2 (c)

CH3COOH

NHCOCH3

Heat

OMgCl Br (d)

Mg Et2O

HCOCl

H3O+

Product.

CH3

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570

O COOH

COOH O

1142.

OH

COOH C

A

O D

B

OH

O

O

1143. O

A

B

COOH COOH

C

O

COOH

D

E

CH3

CH3 1144. CH3—CH—CH2COOH

CH3CH—CH—COOH

A

Br B O

COOH

O H

E

D

C

H F

CH 3 CH 3 CH 3    1145. CH 3CH 2 CHC H 2COOH CH 3CH 2 CH—CH—COOH CH 3CH 2 C== CHCOOH A C  Br B

CH 3  CH 3CH 2 —C== CH 2 D

O O 1146. NH A

B

O

OH

NH2 HOOC

HOOC

C D

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Solutions

O CH 3 O    1147. CH 3CH 2 —C —CH— C—OC 3 H 7

O  CH 3CH 2 —C —CH—CH 3  COOH

A

CH 3CH 2CH 2OH C

B

OH  CH 3CH 2CHO CH 3CH 2 CH —CH—CH 3 D  CHO E

O

COOCH3

COOH O

1148.

COOH

COOH O

C

B

A

1149. C2H5OOC—CH—COOC2H5

HOOC

COOC2H5 A O

COOH B

O O

O COOEt COOEt

EtO

C

OEt D

E

CH3 COOC2H5

1150.

COOH

A

CH3 B

O

OH

1151. A

B

C2H5OH C

D

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572

O 1152. (a) MeO—

—NHCOCH3

CH2COOH

(b) Ph—C—NHOH

(c) C O

COOH

O

C

CH3

O

OH

O (d)

(e) O

Y

X

Cl

OH 1153. (a)

OCH3

OCH3

O

A

CN  (b) Ph—C—CF3  OH C

1154. TsO—CH 2C ≡≡CH A

B

O

CN  Ph—C—CF3  OCH 3

COOH  Ph—C—CF3  OCH 3

D

E

TsO—CH 2C ≡≡CLi TsOCH 2C ≡≡ C—CH 2CH 2Cl B

NC—CH 2C ≡≡CCH 2CH 2CN D

C

HOOC —CH 2 —C ≡≡C —CH 2CH 2COOH E

O–

O– +

1155.

OCOCH3

O

O

Y

X

+ CH2

O A

CH3COCl

CH3COCl

O

O

O

Cl

O Cl

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Solutions

H 1156. (a) H

H

NaCN

CH2Cl

H

H

H 3O +

H

CH2CN

CH2COOH LiAlH4

NaBr

Product.

H

TsCl

H OH CH3

H (b) H

H

(CH3—CH—CH2)2AlH

H

COOEt

CHO

O (c)

COOH

H 3O +

O

LiAlH4

OH

OH

OH

O

O (d)

H3O+

O

NH3

OH

OH

Heat

Heat

COOH (e)

BaSO4

CONH2

CONH2

H 3O +

Pd

H2SO4

Conc. H2SO4

Et

COOH OH

Et

Heat

O

O

CH2OCOPh COOH

LiAlH4

PhCOCl

O

1157. CD3

CD3

CD2OH LiAlD4 H2O

CD2OCOPh PhCOCl

O

CD3

O

CD3

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574

CH 3  1158. Ph—CH —CH 2COOH Ph—CH—CH 2CH 3 A  COOH

CH 3  PhCH 2 CH—COOH

PhCH 2CH 2CH 3

C

D

B

PhCH(CH 3 ) 2 E

O 1159. A

COOH

COOEt

COOH

COOEt

COOEt

C

B

O

O COOEt

D

E O

O O   1160. (a) Ph—CH 2 —C—CH—C—OCH 3  Ph

(b)

18

1161. (a) HOOC—CH 2CH 2CH 2 —OH (b) H 2 N—CH 2COO − (c) HOOC—CH 2CH 2CH 2CONH 2 O

O

COONa

COOCH3 1162. (a)

O

CH2CH2CH3 Y

X

Z

(b) CH 3 —( CH 2 — )10 COOH O 1163. (a)

O

O – CH2

+ LDA

CH2O

H 2O

OH PCC

O

EtONa

(b)

—Br

H 3O +

O COOH O Heat

O H

O

O

O

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Solutions

O O   C2H5ONa Cl 2 (c) Ph—C—CH 2CH 3 → Ph—C—CH CH 3 → heat C2H5OH  Cl O O    H+ HBr Ph—C —CH== CH 2 → Ph—C—CH 2CH 2 Br → CH2 —CH2   OH OH

O Ag2O

Product O

Mg

Ph—C—CH2CH2CH2CH2OH

H2O

O

LDA

LDA

OMe

CH3I

O

O

O

O OC2H5

(e) C H O 2 5

O

Ph—C—CH2CH2Br

ether

H3O+

O

O

(d)

O

O

COOEt

EtONa

LDA

O O

O Heat

COOEt

H 3O +

CH3CH2CH2Br

O O    N 2H 4 PhC ≡≡ CNa Ph —C —C ≡≡C— Ph  1164. (a) PhCOOH → Ph— C —Cl → OH– /Heat SOCl 2

Pd BaSO4 /H2

Product ← Ph—CH 2 —C ≡≡C—Ph O (b)

Ph—C—OMe +

OMgBr MgBr

Ph—C—OMe

O H 3O + Heat

Ph—C— LiAlH4

Ph—CH

H2SO4 Heat

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576

(c)

PhCH2CH2CH2Br

C2H5ONa

COOCH3

Ph

COOCH3 +

O

OH

OH H

1165. (a)

H •• OH ••

H+

OH

O

+

O COOEt

Product

O

HO

(b)

–H+

O—H

EtO

O

O

O

HCl

O

O

HO O +

O 1166.

O

O EtO–

EtO

OEt

O– EtOOC

(EtOOC)2CH–

COOEt O H 3O +

HOOC

Heat

O

O EtO–

1167. (a)

OC2H5

OEt

CH3CH2COOC2H5 s CH3—CH—C—OEt

O

s

H 3O +

COOEt

Heat

O

O

O

O

O

CH3CHCOOEt

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Solutions

OEt

(b)

CH3MgBr

O

O

OEt

(c)

Ph

PhMgBr

O

O

O

O

[from (a)]

O

O

O CH3MgBr

(d)

HCN

CN

OC2H5

H3O+

OH B2H6

COOH

OEt

(e)

COOH

OH

PhNHNH2

CH3COOH

H2SO4 Heat

Product.

O O [from (a)]

O

O

EtO–

1168. (a)

OEt

OEt O

(b)

O

OH

OH

O

O

CH2—CH2

O

OEt CH2

H+

OEt

O

O CHMgBr

O

O

O H3

O+

O

O

N 2H 4 OH–/Heat

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578

H

O

O 1170. K2Cr2O7 + H2SO4

H 2O

l l

OH2

COOH

HO—Cr—OH

O—Cr—OH

heat

O

O

H

H

OH

OH

O H2CrO4

OH

O

H2O l l

C

CH

OCr—OH

+

O

O—H

C

C OH

OH +

C—O—CH3

l l

HOCH3

H+

OH

OH H –H+

+

OH

OH2 C

+

O

l l

H

OCH3

C—O—H –H2O

OCH3

C—OH H+

OCH3

H 2O

O

OCH3

O 1171. (a)

O Br

NaN3

H2/Pd/C HCl

Product

O

H 2O

NH2

CH3COCl

N

HCl

N—H

H2 Ni

Join @unacademyplusdiscounts 579

Solutions CHO (b)

CHO

CHO NaNO2

CuCN

HCl/0°C HCN

N2+ Cl–

NH2

CN CN

Product

OH

H2/Pd

CN PBr3

Mg

1172. CH 3OH → CH 3 Br → CH 3 MgBr ether

OH O

O CH3MgBr

H3O+

HO OH

excess

O

K2Cr2O7 H2SO4

O C 1173. (a)

–

OLi+

O

OH

C + CH3Li

Product

C

O–Li+ – CH3Li+

CH3 OH

Product

–H2O

C CH3

OH

H3O+

O–Li+

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580

H H  O+

(b) R —C ≡≡ N •• + H

O  H 2O OH2 R —C ≡≡ N—H → R —C== N—H  ••

+

H

H + OH O   H 2O R —C— NH 2 R —C — NH 2 − H+  OH H

H +

H+

O  Tautomerism R —C—NH 2 ←→ R —C== NH  OH

+

OH  + R —C— NH 3  • OH • O  S

O  R —C—OH + •• Cl

OH  + R —C== O—H + NH 3

O  R —C—OH + NH +4

O H Cl O H Cl    −    → R —C—O —S—O → R —C —O—S== O  + +  Cl Cl – Cl

O   Cl – + SO 2 + R —C—Cl ←→

Cl

R C –

O

O  O—S —Cl

Cl O   Cl − ← R —C—O—S== O

O− O  +  − + H 2O (c) R —C—N—N ≡≡ N ←→ R —C== N—N ≡≡ N → R —N ==C==O + N 2 

H 2O

H 2O + CO 2 + R —NH 2 ←

O  C

R N  H

OH ←

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Solutions

Cl–

H

H CH3

N l l

CH3

N +

Cl

1174. O

CH3O

Cl

P

Cl

O

P

CH3O

Cl

O

Cl

O +

CH3

N

+ N

H3CO

CH3O

H Cl–

Product O

LiAlH4

CH3O 1175. (a)

OH

CH3O

NaH

O

OH CH3CH2I

COOH (b)

1177. (a)

COOH

OH

LiAlH4

H3O+

Product

1176. (a) II > III > IV > I,

Product O

NaCN

(b) I > II > III > IV,

SOCl2

Mg ether

H3O+

OH

TsCl

(c) III > II > I. PBr3

LiAlH4

OH

H 2O

OCH3

H2SO4 heat, CH3OH

O

Br H 3O +

NaCN

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582

(b)

H3O+

O

COOH OH

O

H2SO4

CH3

∆

HBr ROOR

CH2

COOH

O O

aq. NaOH

H2SO4

COONa

∆

1178. (a) A will hydrolyse first, (d) Rate will decrease.

(b) Rate will increase,

OH

(c) Rate will decrease,

O O–

O Ph Ph

OCH3

1179.

Ph OCH3

– H

– OCH3

H3O+

Ph

Ph

CH3O– + Ph

O

O–

O

O

OCH3

CH3

Ph

OCH3

OCH3 O

Ph

OCH3

CH3OH

∆H = ( BE ) Reactants − ( BE ) Products = 91 + 99 − (85 + 102) = 3 kcal O–

O

O–

O

O

O

+

* O

1180. H3C

CF3

CH3

* O

l l

H

O CH3

H

CF3 + OCH3 O

CH3OH

CH3

+ CH3OH2

* O

CF3 OCH3

O

O +

CF3—C—OCH3 + CH3—C—O*– Necleophile attack at more electrophilic. C

CH3OH2

* CH3—C—OH

Join @unacademyplusdiscounts 583

Solutions

O

OH

OH O H3C

1181. H3C

H3C

COOH OH

O A

H

B

O

O CH3 O CH3

H3C

COOH

CH3—C—C—–C—OC2H5 X CH3

CH3 E

COOH D

1182. A : C 6 H 5COOCH 2CH 3 ; B : CH 3CH 2OH;

C : CH 3CHO;

1183. C 6 H 5CH==CH  COOH C 6 H 5CHBrCHBrCOOH A

C 6 H 5COCOCONH 2

E

F

HO

O

O

C

C 6 H 5COCOCOCl

D

D : C 6 H 5COOH.

C 6 H 5CH(OH)CH(OH)COOH

B

C 6 H 5COCOCOOH

O

C

O

H3C

–

–O 1184. (a)

COOH

O

O

COOH

– HO H 2O

O

O

HO

N

H TsO

COOH

HO

O–

–

TsCl

(b)

O

OH

N

H TsO

Product

N

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584

O

N—OTs s BuLi

(c)

O

O

O

BuLi

N

C

– C—N

O

OTs

O

O H2O

s

N O

Product

C O O

O

O

2–

1185. (a) OEt

+ CO3

OEt

s

H

CH2CH2Br Br

O

O

O

O

CO32–

Product –

Br

OEt

H

OEt CH2CH2Br

Join @unacademyplusdiscounts 585

Solutions

COOEt

COOEt

OEt COOEt

CH2CH2—COOEt

s

Na

(b)

– O

COOEt

COOEt

COOEt

EtOOC

COOEt

O

O

COOEt

COOEt Na/∆

s COOEt

EtOOC

EtOOC

OEt

O

O

O

COOEt

COOEt

EtOOC

EtOOC

OEt

O –

O

EtOOC

EtOOC

O t-BuOK

1186. N R

O –

N

CH3

O

EtO

O

N

CH2–

R

R AcO

O

O

O

HO Ac2O

O

Tautomerize

N

N

N

R

R

R

AcO MeMgBr

HO

OH

H2O

Me N R

OH

NaOH H2O

Me N R

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586

O

CH3

OH

–H2O

N

R

R

–

COOEt

COOEt

COOEt

COOEt

Tautomerize

Me

N

EtO–

1187. (a)

O

O –

OEt

COOEt

EtO– + O

(b) (i)

COOEt

EG H+

O

COOEt

O

CH3MgBr

O

O O

O (i) N2H4 (ii) HO–/heat (iii) H3O+

O

Br (ii)

COOEt

H2SO4

Br2/H+

heat

H 2O

O

O

O Br EG H+

O

O X

2X + Na

O

Heat

O

O

O

H2SO4 H2O

Product.

Join @unacademyplusdiscounts 587

Solutions

H2SO4

(iii)

COOEt

KMnO4

Heat

H 2O

COOH

H+/heat

O

O CH3Li excess

NaBH4

TsCl

O

NaCN

OH

CN H2SO4 H 2O

COOH

R

O (iv)

N

COOEt RNHNH2 H+

R

N

+

H COOEt

N

1188.

EG H+

O

EtO–

N

H O

N 2H 4 HO–/heat

H 2O

O

COOCH3

N+

PCC

LiAlH4

O

OEt O –

R Product

N—H

COOCH3 Product

H 3O + H 2O

O O

1189. (a)

—Br KCN

—CN

H3O+ H2O

—COOH

Br2/P

Br

(HVZ)

COOH

Join @unacademyplusdiscounts Problems in Chemistry

588 H O+

LiAlH4

3 (b) CH 3CH  CH 2CN → → Product H 2O  C6 H 5

1190.

O

O

H+

H+/H2O

H

COOH

OH

O

H—O HO

HO O

–H2O –H+

OH

HO

O

CH2MgBr

CH2MgBr

O

O

OH

+

O

H

– BrMgCH2

O+ H 2O

OH

O

1191. (a)

OH

l l

OMgBr –

O

CH2MgBr

BrMgO OMgBr

BrMg—O

O

H 3O +

OH

HO

S– (b) S

O l l

NH2 NH2

1192. (a)

(b) CH 3I (excess); Ag 2O/heat,

HS

O– S

NH2 +

O

N H

+

H

N H

O

Join @unacademyplusdiscounts 589

Solutions

(c) CH 3ONa; CH 3CH 2CH 2 Br; NaOH/H 2O; heat, (d) Mg/Et 2O; CO 2 /H 3O + ; CH 3OH/H + . O  LiAlH4 1193. PhBr → → PhCOOH → → Ph  C  NH 2 → SOCl 2

Mg

CO2

Et 2O

H 3O +

NH3

O Product

H+

Br

O  Br HOOC  CH 2  C  CH 3

1194.

C

A

HOOC  CH 2  CH  CH 3  OH

B

D

COOH 1195 CH  COOCH 3  CH  COOCH  C 6 H 5  CH 3

H C 6 H 5  CH  CH 3  OH

C==C COOH

H B

C

A

COOCH 3

COOH H H

H H

Br Br

Br Br COOCH(CH 3 )C 6 H 5

COOH

(E and F )

D

1196. All the three reaction involves SN–1 reaction of tertiary halide as: CH3 A : CH3O—

—C—Br

CH3 ∆

CH3O—

CH3

—C+ + Br– CH3CN

CH3 Here CH3O— from para position stabilize the carbocation increasing reactivity.

CH3O—

CH3 +

—C—N

C—CH3

CH3 CH3 CH3O—

+

—C—N CH3

CH3 C—CH3 OH 2 l l

–H+

CH3O—

—C—N CH3

C—CH3 OH

tautomerize

Product

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590

The above mechanism reveal that electron donating group will increase the reactivity and electron withdrawing group will decrease the reactivity. Therefore, reaction A will be the fastest and reaction B will be the slowest. OH

OH O

1197.

LiAlH4

OH

NH4Cl

H– TsCl

OEt H O

O

O

OH–

OTs

l

OH

O

O

l

HO

+

N (b)

1198. (a)

OH

(c)

OH

O O

O

O O

CH3

I N

(e)

(d)

N—CH3

(f)

H

O O O

O

C

C

OH

1199. H2N

OH

–H+

l

Ph

l

O

Ph

HN

l l

O

O

–H+

Cl O O

O

O

O

Ph

–H+ Ph

Ph O N

H—N

O

+

Ph

O HN Ph

Ph

l l

O

O

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Solutions

Cl

O

SEt 1200. (a)

OMe

OEt

(c)

(b) no reaction

CCl3 O COOH

(d)

OLi (f)

(e)

+

OEt

OHC

O

1201. (a) Ag 2O/H 2O, (b) {(CH 3 ) 3 CO}3 Al, (c) CH 3 MgCl; PCC, (d) NaBH 4 . O–

O OEt

1202. (a)

C

O

C

N

N–Li+

+ CH3Li

NaOEt

1203.

Product H 2O

Product

O

O

OEt O –

O

(b)

O

OEt

CH3MgI

Br

Br

NaOEt

EtO

EtO

Diethyl malonate

O EtO Product

O 1204. (a)

NH3

SOCl2

H 3O + ∆

O

O (b)

OEt

O OEt

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592

O

O

O (d)

(c)

COOEt O

O

O Br

CH3

1205. A

CN

CN

C

B

OH

OH

D

O OC2H5

E

O

O

O OC2H5 C2H5OK

O

C 2H 5I

OC2H5

C2H5OH

C2H5

F

O H+ ∆ 1-Phenyl-1-butanone

AMINES SOCl

H

KCN 2 2 1206. (a) R —OH → R —Cl → aqueous R —CN → RCH 2 NH 2 Ni

(b)

—OH

H2CrO4

==O

NH2OH

==N

OH Na Hg

O  LiAlH4 (c) C 6 H 5 NHCH 3 → C 6 H 5 —N— C —CH 3 → C 6 H 5 —N—CH 2CH 3   CH 3 CH 3 CH3COCl

—NH2

Join @unacademyplusdiscounts 593

Solutions NaNO

LiAlH

CuCN

CH COCl

4 2 3 (d) C 6 H 5 NH 2 → C 6 H 5 N 2Cl → → C 6 H 5CH 2 NH 2 →

HCl 0°C

LiAlH

4 → C 6 H 5CH 2 NHCH 2CH 3

CH 3 CH 3 CH 3    1207. CH 3CH 2 —CH—CH 2 NH 2 , CH 3CH 2 CH—CH 2OH, CH 3CH 2 CH—CHO, CH 3CH 2 CH(CH 3 ) 2 B

A

D

C

—CH3

1208. H2N—

O

O

1209.

OH COOH

COOH

—CN A

C

B

O

O

OH NH2

D

E

CH3 N—CH3

1210. +

+

N CH3 I–

N CH3 OH– CH3

CH3

C

B

A

CH3

CH3

+

N—CH3 OH– CH3

+

N—CH3 I– CH3

F

E

D

N(CH3)2HBr

N(CH3)2

N(CH3)3I

N(CH3)3OH

N(CH3)2HBr

N(CH3)2

N(CH3)3I

N(CH3)3OH

A

B

C

D

1211. (a) E

Join @unacademyplusdiscounts Problems in Chemistry

594 NOH (b)

NH2 Ph

Ph

N(CH3)3I Ph

Ph

Ph

Ph

A

CH3

C

B

CH3 N – Br

Ph Ph D

+

H

1212.

+ N(CH3)3

H A

H 3C

N B

CH3

O

N+

H 3C

C

CH3

CH3

O

O

H H 1213.

N3

H2 Pb/CaCO3

+

NH2

O

N

O

H

H H

–

+

N

N

1214.Hinsberg’s method : Add benzene sulphonyl chloride to the mixture of butylamine and dibutylamine and filter off the precipitate containing sulphonamides. Now dissolve the precipitate in aqueous KOH solution. Sulphonamide of butylamine will come into solution as their potassium salt leaving sulphonamide of di-butylamine insoluble. Filter the solution and treat both residue and filtrate separately with dilute acid.  H O    CH3 CH2 CH2 CH2 —N— S —C6 H5     O CH3 CH2 CH2 CH2 NH2  C6 H5 SO2 Cl  Insoluble  → (CH3 CH2 CH2 CH2 ) 2 NH  O   (CH3 CH2 CH2 CH2 ) 2 N— S —C6 H5     O

↓ KOH

K+ −

(CH 3CH 2CH 2CH 2 ) 2 N—SO 2C 6 H 5 + CH 3CH 2CH 2CH 2 —N — SO 2C 6 H 5 Insoluble

Soluble

dilute H+

dilute H+ +

(CH 3CH 2CH 2CH 2 ) 2 NH 2

+

CH 3CH 2CH 2CH 2 NH 3

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Solutions H

H

NO

N

N

N

CH3 CH3 + – N I

1215. NH2

NH2

OH

B

A

– N+ I H 3C CH3 CH3

D

C

O

H CH2CH2COOEt

1216. (a)

(b)

N—

D

N

A

N

CH2CH2COOEt

C

B

(c) H2N—CH2CH2CH2CH2—NH2(E)

O O    NH3 CHCl 3 Br2 H3 1217. (a) R —CN → R —C —OH → R —C—NH 2 → R —NH 2 → RNC O+

Heat

H 2O

H O+

NaOH

KOH

HNO

TsCl 2 3 KCN (b) R — NC → → R —NH 2 → R —OH → R —CN H 2O

O  1218. (i) PhCH 2 NH—C—CH==CH 2 (ii) PhCH 2COCH 2Cl (iii) ClCOOEt (iv) BrCH 2CH 2COOH 1219. (a) (CH 3CH 2CH 2CH 2 ) 3 B

CH 3CH 2CH 2CH 2 NH 2

A

(b) PhNHCOOH C

(c) C 2 H 5 NHCN F

B

PhNH 2

PhNHCONHPh E

D

C 2 H 5 NHCH 2 NH 2 G

O− O O  +  − 1220. H 3C—C + CH 2 —N 2 → CH 3 —C—CH 2 —N 2 → CH 3 —C  CH 2 + N 2    CH 2Cl CH 2Cl CH 2Cl( B ) O– O  +  Also, ClCH 2 —C—CH 2 —N2 → ClCH 2 —C—CH 2CH 3 + N 2 A  CH 3

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596

CH 3 CH 3   CH 3 —C—–N + —CH 2CH 3OH –   CH 3 CH 3

CH 3  1221. CH 3 —C— NHCH 2CH 3  CH 3

CH 3  CH 3 —C== CH 2 C

B

A

CH 3CH 2 N(CH 3 ) 2

(CH 3 ) 3 COH.

D

E

NO 2 1222. (a) MeC ——CH 2OH CH 2OH

(b) RCH 2 NMe 2

OH  CH2 == CO (c) CH 2 == C—CN → CH 2 == C—CN  OCOCH 3 Br

−

2 1223. (CH 3 ) 2 CH—C— NH 2 + OH – → (CH 3 ) 2 CH—C— NH → (CH 3 ) 2 CH—C— NHBr –Br –    O O O –

OH−

••

–Br −

→ (CH 3 ) 2 CH—C— N—Br → (CH 3 ) 2 CH—C—N •• → (CH 3 ) 2 CH—N ==C==O ••   O O H O

2 → (CH 3 ) 2 CHNH 2 + CO 2

O

O

O 1224. (a)

O + NH3 O

N—H

KOH

PhCH2Br

N—CH2Ph O

O

– + KBr OH PhCH2NH2

H2O

KMnO

NH Heat

LiAlH

PO

4 3 2 5 4 (b) PhCH 2 Br → PhCOOH → PhCONH 2 → PhCN → PhCH 2 NH 2 . –

OH /Heat

O 1225. (a)

+ Cl2

Heat

O Cl

aqueous KOH

O

PCC

O H

H2N—CH2CH2—NH2 ∆

N N

H2 Ni

N N H

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Solutions

H

O

O (b)

NH

O

+ NH3(excess) [form (a)]

NH

O

N

H2 Ni

N H

AROMATIC COMPOUNDS 1226. C2H5—

—CH—CH3 C2H5—

—C—CH3 C2H5—

OH

O

A

C

O —COOH

O

— [ C—

D

O

Cl

B

HOOC—

—C—CH3

]n —C—OCH2CH2O — E COOH

O 1227. (a)

—CH O

Cl

CH—C—CH3

(b) O2N

Br

Br OH

(c)

COOH COOH

(d) S

O

(e)

OCH3 (f)

NO2

(g) X : HNO 3 / H 2SO 4 followed by reduction with N 2 H 4 /OH – / Heat. Y : Reduction with N 2 H 4 /OH – / Heat followed by nitration.

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598

Ph –H+ + OH2

1228.

+

–H+

Product

+

Br NBS

1229.

aq. KOH

CCl4

OH

H+ +

+

CH3 CH3

Ph

Ph

CH3

Ph

+

C—CH3

PhCH

–H+

CH3 CHCH3

—C

1230. CH3O—

—C—CH3

CH3O—

O

B

A Cl

O

C

O —O—C—CH3

—C—CH3

CH3O— D

O

E O

1231. (a)

—C—CH3

HO—

PhCH2COCl AlCl3

O Ph

Ph

Cl2 AlCl3

Cl Ph

N 2H 4 OH–/Heat

Cl

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Solutions

OH

OH CHO

CH

+

H 3O

(b) CH3MgBr

CH3

CH—CH3

CH3Cl AlCl3 PCC

(c)

H2SO4

KMnO4

Heat

NaOH

COOH

H3O+

HO3S SO3H

(d)

+ CH3CH2CH2Cl

AlCl3

HNO3 H2SO4

NO2

1232. H3C—CH

CH2OH

H3C—CH

OC2H5

CH2I

I B

A

H3C—CH

CH2OH

CH3

CHO O

OH

D

C Cl

CH3

CHO O E

Product

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600

OH 1233. (a) PhCH2CH2CH2Br

C2H5ONa C2H5OH

CH2

PhCH2CH

(CH3COO)2Hg

NaBH4

H2O

PhCH2—CH—CH3

OH Product

Pd/BaSO4 H2

C

Cl (b)

H3O+

Ph—CH2—C—CH3

CH3C

CNa

PhCH2—C—CH3

C—CH3

PCC

O

Cl O

HNO3

Mg ether

H2SO4

O2N—

—CH2CH2OH

NBS CCl4

Product.

NO2 O

(c)

+ Cl2

AlCl3

O

—C—Cl

Cl—

AlCl3

AlCl

Mg ether

—C— Cl Oxirane

PCC

3 (d) C 6 H 6 + Cl 2 → C 6 H 5Cl → C 6 H 5 MgCl → C 6 H 5CH 2CH 2OH → +

H 3O

O OH   C6H5 MgCl H 3O + PCC C 6 H 6CH 2 —C—C 6 H 5 ← C 6 H 5CH 2 CH —C 6 H 5 ← ← I (e)

+ I2

HNO3

I AlCl3 (CH3)3CCH2COCl

C SOCl2

(CH3)3CCH2COOH B2H6/H2O2 NaOH

H3O+/Na2CrO4

(CH3)3CCH2CH2OH

O

CH2

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Solutions

O

O 1234. (a) Ph

Ph NO2

X

O

– (b) Ph—C—CH2

Ph—C—CH2—CH2—C—Ph

CN

Ph

Ph

CHO

X

Ph

O Z

Ph

O Z

Y

AlCl3

CrO2Cl2

CH3Cl

O

—CH3

—CHO

Ac2O

Product O (b) Ph

Z

O

Ph

O

CH3COOMe C2H5ONa Heat

O Br2 AlBr3

—CH

AcOH

HOCH2CH2OH

Ph

CN

H+/Heat

H3PO4

Product

H 3O +

Ph O

O2 N HNO3

CH3COCl

H2SO4

AlCl3

S

O Ph

O

S

CH—COOMe

O

O + KCN

(c)

O

O

O

1235. (a)

O

Y

O

(c)

Ph

NaBH4

Product

CH3MgBr

O CN

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602

NO2

NH2

HNO3

1236. (a)

F

Zn-HCl

HNO2

HBF4

0°C

∆

H2SO4

CH3COCl NHOC—CH3

NHCOCH3 HNO2

H 3O +

0°C

HNO3

CH3Cl

H2SO4

AlCl3

NH2 H 3O +

NO2

F

CH3 F

F

HBF4

KMnO4

HNO2

HBF4

∆

NaOH

0°C

∆

NO2

CH3

COOH –

(b) O2N—

O

—F

O –

F

+

N

H + •• N—Et

+

Et (Positive charge at p-carbon activate the nucleophilic attack) –

O O –

+

N

F

H

Et O2N—

—N

Et O

O (c) (i)

+

O

AlCl3

O OH

CH3OH H+/Heat

O

+ HF Et

N—Et +

Product

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Solutions

O (ii)

R

+

—COCl

R

AlCl3

R′

R′

O

O

C—Cl AlCl3

(iii)

+

(iv)

CH3Cl AlCl3

HNO3

CH3—

COOC2H5

H2SO4

CH3—

—NO2

Zn-HCl

CH3—

—NH2

COOH

CH3CH2OH

Ag2O

OHC—

H+

NH2

—NH2

CrO2Cl2

H 3O +

Ac2O

NH2

O OH 1237. (a)

H2SO4

KOH

∆

∆

H3O+

OCOPh

O—C—Ph HNO3

PhCOCl

H2SO4 NO2 NO2

(b)

HNO3

Br2

H2SO4

AlBr3

N2Cl Zn-HCl

NaNO2

H 2O

HCl/0°C

boil

Br

Br O O—C—Ph PhCOCl

Br

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604

CH3 (c)

COOH

CH3Cl

HNO3

KMnO4

AlCl3

H2SO4

NaOH

PhOH

H 3O +

NO2

NO2

NHCOCH3

NO2 HNO3

(d)

Product.

H+/Heat

H 3O +

HNO3

CH3COCl

Zn-HCl

H2SO4

H2SO4

NO2 NaNO2

NaNO2

HCl/0°C

Heat

Product

NH2

NO2 HNO3

(e)

CH3COCl

Zn-HCl

H 3O +

HNO3

H2SO4

H2SO4

OH Product

aq. KCN

NaNO2

NO2

Zn-HCl

HCl/0°C

H 2O

NaNO2

boil

HCl/0°C

NO2 OH (f)

OH Conc. H2SO4

OH Br2

OH Br

Br

FeBr3

SO3H NO2

NO2 + HNO3/H2SO4

Br

Br

reflux

SO3H

1238. (a)

dilute H2SO4

Cl2

NaNO2

Zn-HCl

FeCl3

HCl/0°C

Cl

aq KI

Product

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Solutions

Cl

(b)

Cl HNO3

+ Cl2/FeCl3

Zn-HCl

H2SO4

NaNO2

aq

HCl/0°C

KI

Product

NO2 NO2

NO2 Cl2

(c)

Zn-HCl

FeCl3(excess)

NaNO2

aq

HCl/0°C

KI

Product

Cl

Cl

NHOAc

NHOAc HNO3

1239. (a)

Zn-HCl

CH3COCl

HNO3

Blocking

H2SO4

H2SO4

N2+Cl– Ph

NO2

NaNO2

NO2

NH2 Br2

Zn-HCl

Br

Br

Br

aqueous

HCl/0°C

H3PO2

Br NH2

NH2

NaOH

Br

NaNO2

Br

COOH (c)

H 3O + Reflux

HCl/0°C

NO2

(b)

SO3H

NH2 NO2

PhLi

LiCl + N2 +

NO2

Conc. H2SO4

NH2

NH2

CaO/Heat

NH2

NaNO2(excess)

H 2O

HCl-0°C

Boil

NH2 O

O

O

OH

HO

Tautomer

OH

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606 NH2

H

H CH3

N: 1240.

C MeO

NH2

+ N

CH3 _ O

O MeO

CH2CH2COOMe

H 2C

H 2O –H2O

CH2COOMe H

H NH2 + _ N + CH 3 MeO

CH3

MeO

H CH2COOMe

H

HC CH2COOMe

H

H N CH3

N

–H+

MeO

MeO

H O+

CH O

+

CH3

H CH2COOMe

CH2COOMe Mg ether

NH2

+ N

PCC

2 3 1241.(a) PhBr → → → PhCH 2OH → Product

CrO Cl

2 2 (b) PhCH 3 → Product.

Ac2O Etard’s reaction

OH (c)

+ CHCl3

H 3O +

NaOH Reimer-Tieman reaction

OH

OH Conc H2SO4

(d)

Product

H 3O +

AlCl3(excess)

Reflux

CH3COCl

Product.

SO3H

OH (e)

OH H2 Ni

O H2CrO4

Cl2

C2H5ONa

∆

C2H5OH

Product

– NH3

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Solutions

O (f) PhCHO

H3

PhMgBr

O+

PCC

Ph—C—Ph

O (g)

Product

O O

+

H 3O +

HCN

AlCl3

H 3O +

Zn(Hg) HCl

HOOC

COOH

O

HF

O

CH3

CH3 HNO3

1242. (a)

CH3 Zn-HCl

H2SO4

NaNO2

H2O

HCl/0°C

Boil

HCl HCN/AlCl3

NO2

OH CH3 H3O+

CHO OH

SO3H

SO3H HNO3

(b)

OH

OH KOH

LiAlH4

H3O+

H2SO4

NO2 OH (c)

NO2

OH Br2

Br Br

Br

H 2O

Zn-powder Heat

Br Br

Br

NH2

Join @unacademyplusdiscounts Problems in Chemistry

608 OH

OCOCH3

OCOCH3

CH3COCl

(d)

AlCl3

Protection

H 3O +

CH3CH2CH2Cl

OH Conc H2SO4

(e)

∆

OH

KOH

OCH3

OCH3

H 3O +

∆

(i) NaOH

Conc H2SO4

(ii) CH3Cl

Blocking

SO3H

OCH3 KOH aq

Product

CH3

NBS

H 3O +

CH3Cl

Reflux

AlCl3

1243.PhCH 2 Br + CHCl 3 . CH3

CH3 HNO3

Conc H2SO4

1244. (a)

CH3 NO2

H 3O +

Zn-HCl

H2SO4

Product

SO3H

CH3 (b)

CHO

CH3

CHO

CrO2Cl2

HNO3

N2H4

Ac2O

H2SO4

OH–

Product

NO2 H2O

NaNO2

Boil

HCl/0°C

CH3 (c)

Conc H2SO4

KOH Heat

H3O+

Product

NO2 Zn-HCl

NaNO2 HCl/0°C H 2O Boil

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Solutions NO2 HNO3

(d)

CuCl

NaNO2

Zn-HCl

H2SO4 (Excess)

Product

HCl/0°C

NO2

NO2 Zn-HCl

(e)

CuCN

NaNO2

Product

HCl/0°C [from (d)]

NO2

NO2

NO2

NO2 NaNO2

Na2S

(f)

Zn-HCl

KI

HCl/0°C [from (d)]

NO2

NO2

Product.

NaNO2

NH2

CuBr

KCN

NaNO2

Zn-HCl

HCl/0°C

HCl/0°C

Br

CH3 (h) Toluene

Boil

NO2

HNO3

Zn-HCl

H2SO4 (excess)

NaNO2

Br CuBr

HCl/0°C

Br Product.

NO2 HNO3 H2SO4

Product.

CH3

NO2

(i)

HCl/0°C H 2O

NO2

(g) [from (f)]

I

NH2

NaNO2

CH3Cl AlCl3

NO2 Br2(excess)

Zn-HCl

FeBr3

Br

Br

Product

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610

NH2 HNO3

(j)

NHCOCH3 Conc H2SO4

CH3COCl

Zn-HCl

H2SO4

Br2

H3O+

Protection

FeBr3

SO3 H

Br

NH2

NH2 Br

Br

CuBr

NaNO2

Br

Br

Br

Br

KOH

H3O+

fusion

HCl/0°C

OH

SO3 H

OH NH2

(k)

CH3COCl

Zn-HCl

HNO3

H3O+

HNO3

Br2

H2SO4

H2SO4

FeBr3(excess)

NO2

Br Br

Br NaNO2

CuBr

Zn-HCl

KCN

Product

CH3 Br2(excess)

Br

Br

FeBr3

H 3O + reflux H 3O +

SO3H

SO3H

NH2

I Br

Br (m)

SO3H [from (j)]

Br

NO2

NO2

CH3 Conc H2SO4

(l)

Br

NaNO2 HCl/0°C

HCl/0°C

CH3

NH2

H 3O +

NaNO2

Reflux

HCl/0°C

KI aq

Br

Br

KMnO4 NaOH

Product.

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Solutions CH3

CH3 NO2

NO2

Br2

(n)

FeBr3 [from (a)]

Br

CH3

CH3 NO2

NO2

SnCl2

(o)

CH3

NO2 [from (h)]

NaNO2

H2O

HCl/0°C

Boil

NH2

CH3

OH

CH3

CH3 NO2 NaNO2

(p)

NO2

CuBr

Zn-HCl

HCl/0°C

(r) H3C—

NaNO2 HCl/0°C

NH2 [from (o)] (q) Toluene

NO2

Br

Br HNO3

Zn-HCl

NaNO2 HCl/0°C

H2SO4

—N2+Cl–

p-cresol

CN

KCN aq

H3C—

—N2+Cl–

Phenol

Product

Product.

[from (q)]

NH2

NO2 Na2S

1245. (a)

NaNO2

H3PO2

CH3COCl

Zn-HCl

HNO3 H2SO4

HCl/0°C

NO2

NH2 H3O+

CF3CO3H

NO2

Product.

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612

CHO N2H4

(b)

Cl2

OH–/Heat

AlCl3

CrO2Cl2

—Cl

H3C—

Product.

Ac2O

NO2 Na2S

(c)

NaNO2

CuCl

Product

HCl/0°C

NO2 NO2

(d)

KCN ∆

H 3O +

NO2

NaNO2

Zn-HCl

HCl/0°C

KCN aq

Product

CN

Cl

Br —CH3

(e) H2N—

CF3CO3H

—CH3

O2N—

Br2

—CH3

O2N—

FeBr3

Br H2N— CH3

Zn-HCl

—CH3

CH3 NaNO2

(f) Br

H3PO2

HCl/0°C

Br

NH2 NO2

NO2 CHO

(g)

H+ HOCH2CH2OH

O CH O

Zn-HCl

NaNO2 HCl/0°C

KCN aq

LiAlH4

CH2NH2 H3O+ Boil

CHO

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Solutions

Zn-HCl

(h)

NHCOCH3

NHCOCH3

NO2

H 3O +

CH3Cl

CH3COCl

AlCl3

CH3 NO2 KMnO4 NaOH/Heat

COOH CH3

CH3 HNO3

(i)

CH3 Zn-HCl

CH3COCl

CH3 HNO3

H2SO4

H 3O +

H2SO4

NO2

NO2

NHCOCH3

NH2 CH3 CF3CO3H

NO2 NO2 NH2 (k)

NH2 Br2-H2O

Br

Br NaNO2

H3PO2

Br

Br

HCl/0°C

Br

Br

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614

CH3 1246. (a)

CH3

CH3Cl

HNO3

AlCl3

H2SO4

COOH NO2

NO2

KMnO4 NaOH/Heat

NO2

NO2

CONH2 NO2

NH3 ∆

NO2 NO2 (b)

HNO3

NH2 Zn-HCl

H2SO4

Br2

Br

Br

CF3CO3H

H2O

NO2

Br

Br

Br

Br NO2 (c)

NO2

HNO3

Cl2

H2SO4

FeCl3

NH2 Zn-Hg

CH3COCl

Cl

Cl NH2 Br2

H 3O +

FeBr3

Cl Br

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Solutions

Cl (d)

Cl

AlCl3

+ Cl2

NO2

HNO3

Excess

H2SO4

Cl C2H5

C2H5

Cl C2H5

C2H5

C2H5 SO3H

COOH

CN

1247. A

B

SO3H C

E

D

O COOH O COOH F G

1248. (a) H3C—

Mg

—Cl

O CH2O

H 3O +

Ether

H3C—

—CH2OH

OMe Br (b)

KNH2

OMe

NH2

CHO

(c)

+ CO + HCl

AlCl3

CH3

CHO Cl2

N2H4/OH–

AlCl3

Heat

Cl

Cl

Join @unacademyplusdiscounts Problems in Chemistry

616 NO2

NH2

HNO3

(d)

NHCOCH3

Zn-HCl

Conc H2SO4

CH3COCl

H2SO4

NH2

Na2S

(e)

H2SO4

Warm

NO2

Br Br

Br

Br2

NaNO2

CuBr

Br

Br

HCl/0°C

FeBr3

NO2

NO2

NO2

H 3O +

NH2

NH2

NO2

HNO3

NO2

NO2 (f) H3C—

—NO2

Zn-HCl

CH3 COCl

HNO3

H3C—

H2 SO4

—NHCOCH3

H3 O+

NO2 NaNO2

H3 PO2

HCl/0°C

OH (g)

CHO

H3C—

CHO

Zn

CO/HCl

HNO3

Heat

H2SO4

H2SO4

NO2 CH3

CH3 N 2H 4

Zn-HCl

OH–/Heat

NO2

NaNO2

H 2O

HCl/0°C

boil

OH

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Solutions

O O O    1249. (a) Ph—C—C 2 H 5 (b) C 2 H 5 —C—OEt (c) C 6 H 5 —C—CH 3 (d) C 6 H 5CH==CH—COOEt O COCl AlCl3

1250. (a) C6H6 +

Zn(Hg) HCl

O AlCl3

(b) C6H6 +

Zn(Hg)

Cl

HCl

O

O (c)

+

O HF

O + AlCl3 HO—

O

O

O Zn(Hg) HCl

(d)

+ C2H5Cl

(e)

+ Cl2

AlCl3

NBS CCl4

Cl

AlCl3

C2H5ONa C2H5OH

OH

O

H 3O +

Mg Ether

O

(f)

+ CH3COCl

OMe

NaBH4

AlCl3

NHCOCH3

N

OMe

Cl

Br (b)

1251. (a)

OH

Br (d)

(c) COCH3

NO2

Br

COOCH3

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618

CH3

CH3

CH3 HNO3

Conc. H2SO4

1252. (a)

CH3 NO2

H2SO4

H2SO4 (Heat)

SO3H

SO3H

NO2 (b)

NO2

dilute

NO2

HNO3

Br2

H2SO4

FeBr3

NH2 Zn-HCl

Br

CH2CH3 C2H5Cl

(c)

AlCl3

Br

CH2CH3

Br—CHCH3

HNO3

Br2

H2SO4

hν

NO2

NO2 COCH3 aq

PCC

KOH

NO2 COCH3 (d)

CH3COCl

HNO3

AlCl3

H2SO4

NO2

NH2 Zn-HCl

(e) [from (d)]

CH2CH3

COCH3 N2H4/OH– Heat

NO2

NO2

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Solutions

NO2 (f)

HNO3

Cl2

H2SO4

AlCl3

Cl

CH3

CH3 H+

—N

1253.

—N—CH3

NO2+

—N—CH3 +

+

H

H

O2 N

(meta directing)

NH3

O

O 1254. (a)

+

Product.

AlCl3

Cl

NaBH4

H2SO4 Heat

B2H6

PCC

OH–/H2O2

NO2

NO2 (b)

NO2

Na2S

HNO3 H2SO4 Heat

NH2

NO2

FeSO4 H 2O 2

CH3 (c)

CH3Cl (excess)

Product.

NO

COOH KMnO4

AlCl3

NaOH/Heat

CH3

COOH CH3

CHO

CHO

O

(d)

CO/HCl

Cl2

AlCl3

AlCl3

Zn(Hg)

Cl

HCl

Mg

Cl

Ether

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CH3 O H 3O +

Mg

SOCl2

H 3O +

Ether

OH

CH3

OH NO2 (e)

HNO3

Br2

H2SO4

FeBr3

Zn-HCl

Br CH3

CH3 (f)

NH2

Br

CH2Cl

CH3Cl

Br2

AlCl3

Cl2

FeBr3

hν

Br O

(g)

+ CH3COCl

Br

O

AlCl3

HNO3 H2SO4

NO2 CH2 CH3 (h)

+ CH3 CH2 Cl

AlCl3

CH2 CH3 HNO 3

Conc. H2SO4

H2SO4

SO3 H

Product

C2H5ONa

Cl2

H3O+

C2H5OH

hν

Heat

CH2 CH3 NO2

SO3 H

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Solutions

Br (i)

Br

FeBr3

+ Br2

Conc. H2SO4

SO3H

NO2

NH2

HNO 3

(j)

Zn-HCl

CH3COCl

H3O+

CH3Cl

H2SO4

AlCl3

CH3 N CH3I

CH3

O

O

(b)

1255. (a)

(c) X

X

N X OH

O

(d)

(e) X

X CH3

O

Br NO2

1256. (a)

(b)

Cl (c)

CH3 (e)

(d)

Br NO2

OCOPh

Cl

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622

OD

OH H 2O

1257. D

D

NO2

NO2

COOH

1258. (a)

CH3Cl

KMnO4

HNO3

AlCl3

NaOH/Heat

H2SO4 Heat

HNO 3

(b) H3 C—

Zn-HCl

H2SO4

NO2 NaNO 2

NO2 CuCN

HCl/0°C

H3 C—

—CN H 3O +

Product

CH3 (c)

AlCl3

NO2

HNO3

CH3Cl

H2SO4 (Heat)

NO2 O O (d)

AlCl3

+

Cl

OH NaBH4

(e)

+ CH3 Cl

AlCl3

CH3

COOH

KMnO 4 NaOH/Heat

HNO 3 H2SO4

COOH NH 3 Heat

NO2

Product

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Solutions

(f)

+

OH

—Cl

O

H2

AlCl3

Pt

O

O OH

1259. (a)

OH

(b)

(c)

O

1260. (a)

+ CO/HCl

CHO

AlCl3

Product

C2H5C

COCH3 CNa

H3O+

OH

C—C2H5

C

Pd-BaSO4

C—C2H5

C

PCC

OH–/Heat

Product O

O (b)

N2H4

+

O

Zn(Hg) HCl

AlCl3

HOOC

HF

HOOC O

O CH3Cl AlCl3

CH3 O

O Cl AlCl 3

(c) PhH + O

Ph—C—CH2CH3

CH3MgBr

H3O+

Product.

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624

NH2

O

OMe

OH

CH3 (b)

1261. (a)

(c)

(d) HOOC

SO3H

Cl

NO2

COOH

1262. (a)

COOH Br2

CH3Cl

KMnO4

AlCl3

NaOH/Heat

FeBr3

Br

O

+

O

O

O

(b)

SOCl2

(CH3)2CHOH

Product

AlCl3

HF

HOOC

O

O N 2H 4

Product

OH –/Heat

CH3 (c)

CH3Cl

Cl2

AlCl3 (excess)

hν

CH3 —

—CH2 Cl

Mg ether

O H 3O +

CH3

Product

NO2 O

(d)

HNO3

Cl2

H2SO4

AlCl3

Mg ether

Cl

NO2

NO2 H3O+

Ag2O

OH

CH2COOH

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Solutions

O

(e)

+ CH3 Cl

AlCl3

AlCl3

Zn(Hg)

HF

O

HCl

HOOC O O

Product

NaBH4

H+ Heat

O

H

+ + I+

1263.

Cl–

OH 1264.

I

I

+ HF

CH2

CH—CH2

+

H+

C 6H 6 –H+

Allyl alcohol

+

C 6H 6

Product

–H+

COOH OH

CH3

1265.

H3C—C CH3 Cl

1266. (a)

+ Cl2

HNO3

AlCl3

Zn-HCl

H2SO4

CHO

(b)

+ CO + HCl

AlCl3

NaNO2

CuCN

HCl/0°C

Product. CH3

CHO HNO 3

N 2H 4

H2SO4

OH –/Heat

NO2

NO2

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626

NO2 HNO 3

(c)

H2SO4

Cl2

KMnO 4

HNO 3 H2SO4

Product.

COOH

CH3

CH3Cl AlCl3

CuBr

HCl/0°C

Cl

CH3 (d)

NaNO 2

Zn-HCl

AlCl3

Sn/HCl

NaNO 2 HCl/0°C

NaOH/Heat H 2O

NO2

NO2

Boil

Product.

CH3 Sn-HCl

(e)

NaNO2

H 3O +

CuCN

Product

HCl/0°C

NO2 [from (d)]

HNO 3

(f)

NHCOCH3

NHCOCH3

NO2 Zn-HCl

NO2

HNO 3

CH3COCl

H2SO4

H2SO4 Heat

NO2 Product NO2

Heat

HCl/0°C

NH2

HNO3

(g)

HBF 4

NaNO 2

NaNO2

Na2S

H2SO4 Heat

HCl/0°C

NO2

NO2 OH

OMe LiAlH4

NH2

CH3Cl

H2O

KOH

Boil

NO2

H 3O +

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Solutions

I (h)

I

HNO 3

+ I2

HNO 3

NaNO 2

Sn-HCl

HCl/0°C

H2SO4

I

NO2

H2O Boil

OH

NO2 1267. (a)

NO2

(b) N

C6H5S NO2

NO2 NO2

NO2 NHPh

Ph N—Ph

(d)

(c) O2N

CH2OH

NO2

O2N

NO2

CH2OH

OMe

CHO

OMe

1268. (a) A OMe

(b)

OMe

O

O

B OMe

OMe

OMe

(c) HO

HO

O O

E

F

G

H

OH I

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628 CH2 OH (d)

CH2 O–

OH

OMe OMe

O–

OMe J

OMe K

OMe L

NO2 – (e) CH(COOEt)2

Br

CH(COOEt)2

M NO2 N

(f) NaH/CH 3I(excess)

LiAlH 4 P

O

OMe

Br 2 / FeBr 3 Q

OMe

OMe

OMe

NO2

NH2

N2Cl

B

C

D

1269. (a) A

(b) O2N

N

O O2N

E I

I (c)

NO2

H2 N

I

O2 N I

I

H

G Br

N2 Cl CH3 CH2 O

NH2 CH3 CH2 O

(d) CH3 CH2 O I

Br

Br

J

K

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Solutions

(e) HO

HO

N2Cl

I M

L

(f) H2N

CH3

CH3

CH3

CH3

N

CH3

O

CH3

CN

CH3

N2Cl

COOH CH3

P

Q CHO

CHO

CHO

CHO

1270. (a) D

C

B

A

Cl

N2Cl

NH2

NO2

N2HSO4

OH

(b) MeO

OMe

MeO

OMe F

E CH3 NH2

H2N

CH3

CH3 N2Cl

N2Cl

HO

OH

(c) I

H

G COOH (d)

N

N

N

N2 Cl J

COOH

K

(e) No reaction (L)

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630

CH3

N

(f) N

N

H3C

+

CH3 M

CH3

P

CH3

O Br CH3

CH3

COOH

CH3

1271. Br

CH3

CH3

A

C

B

COOH

CH3 D

O COOH

N N

COOH

Br E

G

O F OMe

OMe

OMe

1272. Br A

O

B

O

OH

C

O

O

OH

OH

1273. A

O

B

OH

C

OH

CHO OH

OH

D

E O

1274. H3C—

CH3

—C—O—CH—CH2CH3 A

OH B

O C

H3C—

—COOH D

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Solutions

COOH 1275.

Br

A

COOH C COOCOCH3

B

COOH

COOH

O 1276. OH

OMe

C

B

A

COOH

COOH

CH3

O

COOH Br

Br

OH

OH

OH D

F

Br E O C—OC2H5

CH3CH2OH

1277.

CH3CHO

C6H5COOH

C

D

B A

1278.(a) Aromatic, cyclic and fully conjugated with six π-electrons. (b) Antiaromatic, cyclic and fully conjugated with 4π-electrons. (c) Non-aromatic, cyclic but not fully conjugated. (d) Antiaromatic, cyclic and fully conjugated with 4π-electrons. 1279.Conjugate base of A invoke aromatic character in the middle ring therefore, deprotonation of benzylic H in A is favoured. On the other hand conjugate base of B does not bring any stability by aromaticity deprotonation of benzylic H is not as favoured as in case of A. CH3 H3C

H

s –H+

A

Aromatic anion

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CH3 CH3

H

– –H+

B

1280. (a)

non-aromatic anion

+ Br

Br

OCH3 OCH3

OCH3 (b)

+

+

NH2 NH2

NH2 O H

H (c)

H (d)

+ O

O

O

O NO2

NO2 (f)

(e)

Cl

CH3O NO2

O

O (g)

(h) O2N

CH3

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Solutions

CHO

CHO

(i)

(j)

O I

1281.Nitration of benzene proceeds by the following mechanism:

O–

O  N +

O  + H—O—S —OH •• O—H  O

O −

O  N +

+

OH 2

+ HSO –4

+

O== N ==O nitronium ion

– O—SO3H O + N+ O

H Slow Rate determining step

NO2 +

+

H NO2

H NO2

+

Resonance stabilized σ-complex

NO2 fast

H NO2

Energy

+

Reaction energy diagram: The reaction proceeds at same rate for both benzene and hexadeutrobenzene since C—H bond cleavage is not occurring in rate determining step.

C6H6

NO2

HNO3 H2SO4 Reaction coordinate

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1282.Sulphonation of benzene proceeds by the following mechanism:

O

O

S—O–

S—OH

+ S O

+

O

O

O

HO

O

S+ H+ –H2O

+

O

O

O H

S

S

HSO4–

C6H6

+ O

O (diphenyl sulphone)

O C CH3COCl

1283.

O CH3 Na/NH 3

AlCl3

O3

t-BuOH

(CH3)2S

Products

1284.(a) CH 3COCl/AlCl 3 , Br 2 /FeBr 3 , (b) PCC/CH 2Cl 2 , NBS/hν, (c) H 2 /Rh/Pressure, NaIO 4 , (d) CH 3COCl/AlCl 3 , HNO 3 /H 2SO 4 , (e) (CH 3 ) 2 CHCOCl/AlCl 3 , Na/NH 3 /EtOH, Zn(Hg) /HCl, (f) NBS/hν, Mg/ether, CH 2O, PCC( g ) Br 2 /FeBr 3 , HNO 3 /H 2SO 4 . 1285.The reaction will be slower with acetophenone than with benzene as the acyl group is a deactivating group in electrophilic aromatic substitution. O

O–

O–

O–

+

+

+

Acetophenone

As shown above, electron withdrawing (  R) effect of acyl group decreases the electron density from the aromatic ring making it less susceptible for attack to an electrophile (Br + in bromination reaction). Also, acyl group, by resonance, brings positive charge on ortho and para positions, hence, meta-positions are relatively richer in electrons compared to the ortho/para positions. Hence, electrophile, if attacks at all, it attacks at meta positions only. 1286.The mechanism of substitution reaction is:

N

Cl – OCH3

– N l l

OCH3 Cl

+ Cl– N

OCH3

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Solutions

On the basis of above mechanism, it can be concluded that the order of reactivity will be: A < C < B. The reason for the above order are : For the pyridinium salt, one of the resonance forms bear no charge. This will be a stable, neutral species and the activation energy is, accordingly, low relative to others A and C.

N+

OCH3

OCH3 N

Cl

Cl–

–

–

– + OCH3

CH3

Cl +

+

N

Cl

CH3

N

Cl

CH3

CH3

stable, neutral species

OCH3

+

N

OCH3

CH3

In the case of 3-chloropyridine, the negative charge on intermediate can’t be delocalized, therefore, this reaction would be expected to be slower than 4-chloropyridine where the negative charge is delocalized. –

Cl – + OCH3

Cl OCH3

+

N

N

Cl

Cl OCH3

–

N

N

–

OCH3

Cl OCH3

Cl

Cl

OCH3

–

– OCH3

N

l l

N –

N

(more stable)

H +

1287.

O

l l

+

C6H6

+

H+

O—H

OH

+

HO HSO4–

+

–

HSO4

H+ –H2O +

H

OH

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636

O

+ CH3COCl

1288.

HO

AlCl3

CN NaH CH3CH2Br

HCN O

NH

—NH2

O

O

CN

O

LiAlH4

CH3COCl

H CH3O

N

l l

1289.

O H CH3

CH3O

CH3O

C

H

CH3O

H CH3

CH3O +

NH CH3O

HN

CH3O

H

CH3

+

OH2

l l

H CH3

CH3O

+

–H+

H

1290.(a) (b) (c) (d)

Product

NH

CH3O

CH3

Antiaromatic, fully conjugated, planar with 4π-electrons. Aromatic, planar, fully conjugated, 6π-electrons (two from ring oxygen). Non-aromatic, planar fully conjugated, 7π-electrons (one from radical). Non-aromatic, not fully conjugated. Cl

Cl +

1291. (a) O2N

OH

HN

CH3O

(b) NO2 OCH3

H+

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Solutions

O (d)

(c) HO3S

O O2N

NO2 OCH3

(f)

(e)

O

COOH (b) Br

1292. (a)

—NO2

O

COOH O2N O

NO2

NO2

NO2

Br

NH2 (d)

(c)

+

+

NH2

OCH3 SO3H

NH2

NO2

(f)

(e)

OCH3

NO2

1293.The σ-complexes formed by attack at ortho, para and meta positions are as follows: CH3

CH3

HNO3

CH3

+

NO2

H2SO4 +

2°

+

NO2

H

NO2

H 2°

CH3

H 3°

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638

CH3

CH3

CH3

+

para attack

+

+

2°

H

H

NO2

CH3 meta

H

NO2

3°

NO2

CH3

CH3 +

+

NO2

attack

2°

NO2

H

NO2

H

2°

H

+

2°

2°

1294.(a) CH 3Cl/AlCl 3 , NBS/hν, CH 3ONa, (b) CH 3COCl/AlCl 3 , Fe(Hg) /HCl, Na/NH 3 /EtOH, (c) H 2SO 4 /H 2O – /heat, KMnO 4 /NaOH/heat, (d) CH 3 (CH 2 ) 2 COCl/AlCl 3 , NBS/hν, CH 3ONa, Rh/H 2 /∆. 1295.  B(OH) 2 is a meta directing group due to following resonance phenomena: – OH

HO

HO

B

– B

OH

HO

– B

OH

HO

+

– B

OH

+

+

The above resonance structures reveals that  B(OH) 2 withdraw electrons from ortho/para positions deactivating the ring for attack of electrophile. Even, if the compound is subjected to electrophilic attack, it occurs at meta position. CH3

1296. (a)

Cl2

CH3Cl

AlCl3

AlCl3

CH2Br

NaN3 SN2

NBS hν

Cl

Product

Cl Br

SO3H (b)

H2SO4/H2O

Br2

CH3COCl

disulphonation

FeBr3

AlCl3

Product

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Solutions H

–

S

OH E2

C6H5—C—O—S—

1297. C6H5—CH2—Br

+

C6H5CHO + CH3

H

O

S CH3

+

1298.

+ H+ +

–H+

Product

+

CH3

CH3

H CH3

CH3 +

1299.

–H+

+ D+ +

H

D

H

3°

CH3

D+

D D CH3

CH3 H

D –H+

+

D

D +

D

D D+

CH3 D

D CH3

D

D

D

–H+

H +

D

D

H

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640

1300.Benzoic acid. The mechanism involves addition of electrons to form a radical anion in the first step as: COOH

COOH

COOH

COOH

l l

Na

EtOH

+e l

–

–

l l

COOH

COOH –

l l

EtOH

Stabilized by electron withdrawing group

O

O

1301.

+

O

AlCl3

O Br2

OH AlCl 3

H2O

O

O

O

O

C OH

OH

Zn(Hg) HCl

CH3Li(2 mol) H2O Br

Br

O

Br

Br

NH2OH

Product

Br

NNH2

1302. (a)

(b)

(c)

(d)

Br CN

O

OH enantiomers

Cl

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Solutions

Br

Br

NO2

Br2

1303. (a)

Br

HNO3 H2SO4

CH3COCl

FeBr3

AlCl3

O

O

Br

Br

Br

N2+Cl–

OH

NO2 NaNO2

H2O/∆

H+

HCl/0°C

H3

O+

EG H+

Fe HCl

O O

O

O O

O EG H+

(b)

HNO3 H2SO4 (o/p directing)

(Alternatively) Zn(Hg)/HCl

H2SO4

H 3O +

O2N

O

Ac2O

O2N

O2N O

O2N

(b)

O

CrO2Cl2

HNO3

1304. (a)

O

O

+

(c)

(Etards reaction)

(d) N

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642

NO2 Cl Br 1305. (a)

NO2

(b) O2N NO2

OCH3

(c)

(d)

COOH

(e)

OH

Ph 1306. (a)

D

D

NH2

D

(b) OH

HO

OH

D

D

(c)

D

OH

OH

CF3

O (e)

(d)

CHO CHO

N

+ CH3OH

(f)

OCH3

1307.(a) (CH 3 ) 2 CHCOCl/AlCl 3 , Ph 3 P==CH 2 , (b) O 3 /Zn–H 2O, Zn(Hg) /HCl, (c) Br 2 /FeBr 3 , HNO 3 /H 2SO 4 , Mg/ether/oxirane/H 3O + , PCC/CH 2Cl 2 , (d) Alternative : CH 3Cl/AlCl 3 , H 2SO 4 /HNO 3 , NBS/hν, Mg/ether/CH 2O, PCC/CH 2Cl 2 . 1308.CH 3OCH== PPh 3 C 6 H 5CH==CHOCH 3 C 6 H 5CH 2CHO A

B

C

1309.Only B, D and G are formed. In order to rationalize, we need to see the sigma complex in each case. In sigma complex leaging to C, E, F, H, one of the resonance form places the positive charge on C1 , which is the carbon bearing nitro group. These products are then all disfavoured.

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Solutions

NO2

NO2

NO2

σ-complex of D

σ-complex of G

NO2 +

+

O2N

σ-complex of B

NO2

+

In none of the resonance forms of the above σ-complex it is possible to place positive charge (+) on carbon bearing  NO 2 group. Sigma complex of remaining four places positive charge on C-bearing  NO 2 in one of their resonance form: NO2

NO2 +

O2 N

NO2

O2 N

O2 N +

+

σ-complex of C

NO2 O2 N

+

not favoured resonance form

O

O

O O

1310. (a)

H

O H

EG H+

NH3

Pd/C H2

∆

NH2 H Product CF3

CF3

(b)

H3O+

O O

CF3

HNO3

Fe/HCl; NaNO2

H2SO4

HCl/0°C

NO2

C6H5N(CH3)2

N2+Cl–

Product

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644

1311.Br 2 + FeBr 3 → Br + + FeBr 4– OH

OH

OH CH3

H3C

CH3

H3C

H3C

CH3

+ Br+ +

Br SO3H

O

S l l

O

OH

Br + HBr + SO3

Br–

1312.(a) I, (b) I, (c) I, (d) I, they are all aromatic. 1313.(a) III < II < I, (b) II < I < III, (c) III < II < I. NO

NO

+ E+

1314.

meta attack

+

+

+

H E

+ NO

E+

NO

NO

l l

H

H

E

E

O

N

+N

O

+

para attack

+ E

H

H

E

H

E

σ-complex is stabilized by +R effect of —NO functional group.

1315. (a)

NBS

HNO3

+ CH3CH2Cl

H2SO4

hν

NO2 C2H5ONa C2H5OH

(b)

Cl AlCl 3

+

N 2H 4 NaOH/∆

O O

H2SO4

Product

Product

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Solutions

CH3

(c)

CH3

NO2

H2SO4

Cl2 FeCl3

AlCl3

+ CH3Cl

CH3

HNO3

Cl

Cl COOH NO2

KMnO4 NaOH/∆

Cl OH CH3

OH CN

1316. (a)

(b)

CH3 OH

(c) O

OCH3 (e) OH—

(d)

H2SO4

CH3Cl AlCl3

1317. (a)

(b)

NaOH fusion

SO3/∆

HNO3

Br2

H2SO4

FeBr3

N

N

H3O+

Product

SnCl2 HCl

O

(c)

+

Cl

AlCl3

N2H4 NaOH/∆

Br2 FeBr3

Br

NBS/hν

C2H5ONa C2H5OH

Product

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646 3

1318.

2

O

+

E+

H

attack at C-2

E

O

1

H

H +

E

O

E+ attack at C-3

E

E

H

H

E

O +

Three resonance structures

+

O

O

+

Two resonance structures, less probable.

Therefore, electrophile is most likely to attack at C–2 position.

Zn-HCl

1319. (a)

N2+Cl–

NHCOCH3

NO2

CH3COCl

H 3O +

CH3Cl

NaNO2 HCl/0°C

AlCl3

H 2O ∆

Product

O

HCl

H 3O +

Na/NH3 EtOH

(b)

O

OCH3

OCH3

CH3

CH3Cl Na/∆

Cl

Product

CH2

O PPh3

OH NBS hν

(c)

Product

Mg ether

CH3I (excess)

CH2O

H 3O +

PCC CH2Cl2

NH2

H NH3 Pd-C/H2

O

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Solutions

1320.(a) II, (b) I. 1321.(a) III < I < II, (b) I < II < III, (c) I < II < III. Cl 1322. (a)

+

AlCl3

Zn(Hg) HCl

O

HNO3

(b)

Product

CH3COCl

H3O+ H2SO4

NBS/hν

CH3CH2ONa CH3CH2OH

CHO

H2O

—

H2N—

CHO

Cl

CrO3 Ac2O

1323.

Product

AlCl3

NHCOCH3

Zn HCl

H2SO4

CH3Cl

EG

O

H+

or AlCl3

O O

O

Product

CH H 3O +

Zn

H2/Pd

(CH3)2NH

HCl

N CH3

CH3

CH3

N

O

CH3

CN

CN 1324.

O

O

CHO

CN +

E+ ortho attack +

H

H

E

E

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648 CN +

CN

H

H

E

+

E destabilized by CN

CN

CN

+

E+ meta attack

+

H

E

H

E

+

Meta attack of E avoides the electron withdrawing effect of CN group as positive charge can’t interact directly with CN group. Therefore, meta attack is favoured over ortho/para attack of electrophile.

1325. (a)

(b) N

A

H B

MeO

COOMe

OMe

COOMe

O

1326.

+

l l

NH

H

N

O– NH2

NH2 OMe

OMe COOMe N

COOMe

H+ –H2O

N NH2

NH2

H

OMe

OMe H+

COOMe

COOMe –NH3 N +

NH3

OH

+

N

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Solutions

OMe

OMe COOMe

H

COOMe

H+

–H+

N

N+ H OMe

COOMe +

+ NH4

NH 3 l l

N H CH2O

Mg

PCC

1327.(a) PhBr → → → Product Et 2O

CH2Cl 2

NBS/hν CCl 4

aq. NaOH

PCC

(b) Ph  CH 3 → → → Product or

CH2Cl 2

CrO3

Ph  CH 3 → Ph  CHO Ac2O (Etard’s reaction)

OH (c) Ph—OH

CHCl3/NaOH Reimer-Tiemann Reaction

CHO OH

OH

(d)

H3O+ Reflux

CH3COCl

H2SO4

HO— O

SO3H OCH3

OCH3 (e)

CH3

H2/Ni Heat

O

OH

H3O+ Reflux

PCC CH2Cl2

O Cl2

EtONa EtOH

heat

Cl

Product

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OH OH   H2SO4 PhMgBr H3 PCC HCN (f) Ph—CHO → → → → Ph—C— Ph Ph—C— Ph → H 2O CH2Cl 2   COOH CN O+

O

O

(g)

+

AlCl3

O

Zn(Hg) HCl

COOH O

COOH HF

O

(h)

H+

+ NHNH2

Heat

N

O

H

COOH 1328. (a)

(b)

(c) Br

(e)

(d)

(f) Br O

(g)

COOH

CN

Br

(h) SO3H

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Solutions

O

O

O 1329. (a)

AlCl3

+

Cl2

Cl

AlCl3

Cl

Product

HNO3, H2SO4

N 2H 4 OH–/heat

Cl OH Br Mg Et2O

Br2

(b)

FeBr3

(i) CH3CH2CHO

H2SO4

O+

heat

(ii) H3

O H3O+

Product

NaBr

mCPBA

Br +

NH3 1330. (a)

(b)

(c)

+ Br

NO2 Br NO2

1331.

HNO3 H2SO4

Fe HCl

Ac2O

HNO3 H2SO4

H 3O +

NaNO2 HCl/0°C

—N2+Cl–

O2N— I H2SO4 SO3

NaOH fusion

H 3O +

OH

CCl4

OH

H 3O +

NaOH

COOH (I)

Product

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O 1332. (a)

O Br2

CrO2Cl2 Ac2O (Etard’s Reaction)

MeO

FeBr3

MeO

MeO Br (only product)

Ph3P

CH2

MeO Br

O

Ph3P

(b)

CH2

Product

1 equiv.

O N (c)

(d)

C6H6

CH3COCl

OH LiAlH4

H2NOH

AlCl3 NBS hν

NaOH

PCC CH2Cl2

CH3CH2CH

PPh3

Product

Product

1333. (a) Non-aromatic, (b) Aromatic, (c) Aromatic, (d) Nonaromatic, (e) Aromatic, (f) Aromatic. 1334. Since, both rings are aromatic with the following resonance structure: ⊕

s

both the rings are now aromatic hence double bond character between the rings are very less, therefore less barrier to rotation,

1335. (a) Deactivated, (b) Deactivated, (c) Activated, (d) Activated, (e) Deactivated, (f) Activated. 1336. (a)

Cl2 AlCl3

H2SO4 HNO3

Product

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Solutions Br

O

Br2

(b)

Zn(Hg)

Cl

FeBr3

NBS/hν CCl4

HCl

AlCl3

Product

O H2SO4 SO3

Cl

(c)

AlCl3

Zn(Hg) HCl

Product O

O

1337. (a)

(b)

(c)

O2N Br

O

O O

COOH

(e)

(d)

(f)

CH3

NO2

CHO

NO2

NH2

OH

O

1338. (a)

(c)

(b) NO2

O2N

Br

O NHCH2CH3

O

N

O—C—Ph

NO2

O2N (d)

(f)

(e)

Cl NO2

CF3

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654 Cl

(h)

(g)

HO—

OCH3

1339. O2N

O

l l

NHCH3 – + O N O

O2N

O

CH3

+N + N

O–

NO2

N—CH3

O

H O– –H+

+

O2N

N

O– O–

NO2

NO2 O–

N O2N

NO2

H2O

Product

NO2

1340.(a) Non-aromatic, (b) Non-aromatic, (c) Antiaromatic, (d) Antiaromatic, (e) Antiaromatic, (f) Aromatic, (g) Aromatic, (h) Aromatic. 1341.Since, double bond is in resonance and involved in aromaticity as:

s

1342. (a) C6H6

⊕

CH3Cl (excess) AlCl3

Here, both the rings are aromatic, therefore, this resonance form is a major contributor. Due to this resonance structure the double bond between the two rings has acquired significant single bond character giving low energy barrier to rotation.

CH3—

—CH3

KMnO4 NaOH/Heat

H3O+

Product

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Solutions

NO2 HNO3

(b)

NHCOCH3

NH2 CH3COCl excess

Zn HCl

H2SO4/heat

NHCOCH3

NO2 H3O+

CF3CO3H

NaNO2 HBF4

heat

NHCOCH3

NO2

HNO3

Sn HCl

H2SO4

F

F

CHO HNO3 H2SO4

CO HCl/AlCl3

(c)

NHCOCH3

NH2

NO2

NaNO2 HCl

Fe/HCl

NO2 NHCOCH3

HNO3 H2SO4

(d)

CH3

Fe HCl

Product

NH2 NHCOCH3

H2SO4

CH3COCl

H2O heat

HNO3 H2SO4

SO3

SO3H N2+Cl–

OH NO2

NH2 NO2

H 2O boil

NHCOCH3 NO2

NaNO2

NO2

dil. HCl

HCl/0°C

OH SO3H

NH2

Sn HCl

NHCOCH3 (e)

HNO3 H2SO4

Fe/HCl

CH3COCl

NHCOCH3

H2SO4 SO3

dil. HCl

CF3CO3H

Product SO3H

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O

CHO (CH3)2C HF

CO/HCl AlCl3

(f)

CH2

O CH HNO3 H2SO4

EG/H+

O N 2H 4

O CH H 2O boil

dil. HCl

NaOH/heat

SO3H (b)

(c)

Br

O

O2N O NO2

O Cl

NO2

NO2

Cl (e)

+ CH3

(f) SO3H

CH3

Cl

O2N

CH3 NHCOCH3 (h) CH3CO

Br CH3

HCl/0°C

N

1343. (a)

(g)

NaNO2

OH

OH

(d)

Fe/HCl

COCH3

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Solutions

O

O

O EG

Cl2

Cl

1344.

H+

AlCl3

AlCl3

Cl O H3

O

O H3O+

O+

O

O

Mg Et2O

H

Cl

HO O

N2H4

PCC CH2Cl2

NaOH heat

Product

HO O

Cl

1345. (a)

CH3Cl AlCl3

O AlCl3

HNO3

Zn HCl

H2SO4

NO2

NH2

NaNO2

H2O

HCl/0°C

boil

OH CH3

(b)

COOH

CH3Cl

HNO3

KMnO4

AlCl3

H2SO4

NaOH/heat

NO2

H3O+

NH4HS

NO2

Product

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658

O

O COOH

1346.

(a) A

C

B

D

NO2

NO2

Co(OAc)2

(b)

AcOH/HBr/O2

NO2

No reaction since there is no α—H to phenyl ring.

E

NO2 F

O

Br 1347. (a)

+ Br2

FeBr3

Mg Et2O

H3O+

OH

PCC CH2Cl2

Product

O

(b)

(CH3)2CHCOCl AlCl3

HNO3

Zn

H2SO4

HCl

NH2

Product

NaNO2

CuCN

LiAlH4

HCl/0°C

CN Br

Br

(c)

Br2 FeBr3

H2SO4

HNO3

SO3

H2SO4

SO3H

NO2

dil. HCl

Fe/HCl

NaNO2 HCl/0°C

SO3

NaBF4 ∆

Product

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Solutions

NHCOCH3 HNO3

(d)

H2SO4

Fe HCl

CH3COCl

HNO3

Cl Zn/HCl

O AlCl3

H2SO4

NO2 NO2

NHCOCH3

NO2 CF3CO3H

Sn/HCl

NO2

dil. HCl

NH2

NO2

1348.(a) Non-aromatic since, it has a sp 3 carbon.

(b) Aromatic.

OCH3 NO2

F (b) CH3O—

1349. (a)

O

C

—CH3

(c)

NO2

CH3 Br

(d)

—OH

(e) F

1350. (a)

(b)

O OMe

O OMe

O

OMe

OMe

(c)

CH2Ph (Minor)

O (Major)

CH2Ph

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CH3

1351. (a)

(b)

CH3 Br

CH3Cl

H2SO4

Br2

AlCl3

SO3

FeBr3

H2SO4

NaOH

SO3

fusion

dil. HCl

HNO3 H2SO4

SO3H

SO3H

ONa

OCH3 CH3I

Br2 FeBr3

Product

Product

NHCOCH3 HNO3

(c)

Fe/HCl

H2SO4

CH3COCl

H2SO4

dil. HCl

SO3

CF3CO3H

SO3H

SO3

1352. (a)

H2SO4

H 3O +

KMnO4 NaOH

(b)

t-BuCl AlCl3

KMnO4

(c)

t-BuCl AlCl3

CH3COCl

Product

H 3O +

Product

NaOH

AlCl3

Product

OMe

(d) OMe

t-BuCl AlCl3

HNO3 H2SO4

Product

Cl NH3

(e)

H2S

NO2

NaNO2 HCl/0°C

CuCN

H2SO4

CH3OH

H2O

H+/heat

Product

Product

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Solutions

(f)

CH3Br

(g)

HNO3

NaNO2

Fe/HCl

Heat

Product

HBF4/0°C

AlCl3

H2SO4

Br2

HNO3

Cl2

NH3

NaNO2

H 2O

H2SO4

AlCl3

H 2S

HCl/0°C

boil

FeBr3

Product

O (h)

+

Cl

AlCl3

HNO3

N 2H 4

H2SO4

NaOH/heat

Product

O (i)

+

Cl

HNO3

AlCl3

Zn-HCl

H2SO4

NaNO2

H 2O

HCl/0°C

boil CH3I

Br

Br

1353. (a)

Br2

H2SO4

FeBr3

SO3

Br

O

NaOH

Product

O

Cl AlCl3

SO3H

SO3H dil HCl

Zn(Hg) HCl

Product

O Cl

(b)

Br2 (excess)

Zn(Hg)

FeBr3

AlCl3

Product

HCl

Cl

O

(c)

Cl2 AlCl3

Cl

SO3 H2SO4

dil. HCl

AlCl3 Zn(Hg) HCl

(d)

H2SO4 HNO3

Fe/HCl

O

Ac2O

HNO3 H2SO4

dil HCl

Mg

D 2O

Ether NaNO2 HCl/0°C

CuCN

Product

Product

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662

NO2 HNO3

(e)

Ac2O

Te/HCl

H2SO4

CH3I

(f)

AlCl3 Br2

(g)

Fe

(h)

H2SO4 SO3

NaOH

HNO3 H2SO4

H 2S

CH3I

Heat Ac2O

NH3

CH3I

SO3

Br2(excess)

AlCl3

H2SO4

FeBr3

CH3CH2COCl AlCl3

dil HCl

NaNO2 HCl/0°C HBF4 Heat NBS

CH3—

—OCH3

Cl2 FeCl3

dil HCl CF3CO3H

HCl(dil)

hν/CCl4

NaOH Product aqueous

Product

Product

Cl CN 1354. (a) (c)

(b)

Br

NO2

NO2 Cl

OEt

NH2 (f)

(e)

(d)

O (like Fries rearrangement)

O H A:

1355.

O

Cl

O

and AlCl 3 ; B: O

D:

Ag 2O/H 2O;

E:

O Ph

MeMgBr/H 2O; C:

Product

H 2SO 4 /H 2O;

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Solutions

NHOAc CH2CH3

1356. (a) (b)

O CH3

C(CH3)3

Cl

OH (c)

(e) O2N—

(d) Br

CH3

CH(CH3)2 NO2

Cl HNO3

1358. (a)

COOH

Cl Li/Et2O

NH3/H2S

CO2/H2O

NaNO2 HCl/0°C

H2SO4 CuCN

NO2

NO2

H3O+ H2O

Product

Cl Mg/Et2O

(b)

CO2

CH3I

H2O

AlCl3

KMnO4 NaOH/heat

OCH3

Cl HNO3

(c)

H2SO4

Product

CH3ONa CH3OH

Fe/HCl

O2N

OCH3 H2O boil

HO

F

(d)

HNO3 H2SO4

OCH3

CH3ONa CH3OH

NaNO2 HCl/0°C

Fe/HCl

NaNO2 HCl/0°C

O2N

CuCN

OCH3 NaOH H2O

H3O+ H2O

HOOC

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664

Br

O 1359. (a)

Br

O (b)

—OCH3

—NO2

OCH3 (c)

—OCH3 O2N F

Cl NH2 (c)

(b)

1360. (a) MeO

NO2

Br

SO3H

OMe

NO2

OMe OMe (d)

(e)

O

I

1362.(a) (b) (c) (d) 1363.A:

Aromatic, Not aromatic, it has a sp 3 carbon which prevent delocalization, Aromatic, Not aromatic. KMnO 4 /NaOH/heat; B: SOCl 2 ; C: Et 2 NCH 2CH 2OH; D: Na 2S. OCH3

OCH3

O O

1364. HO

O HO

A

O

H

OCH3

O H3CO

B

O

OH

O

C

OH COOH

O H3CO

D

HO

OH E

O

COOH HO

F

COOH

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Solutions

—NO2

1365. (a) Cl—

(b) O2N—

Cl

(d) O2N—

—CH3

(c) H3CO—

NO2

Cl Cl

Cl

Cl Cl

1366.

Cl2 AlCl3

N2H4/OH–

O AlCl3

C O Cl

HNO3

Cl

NH4HS

H2SO4

NO2

NH2

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CARBOHYDRATES, AMINO ACIDS AND POLYMERS 1367. (a) A reducing sugar reduces Tollen’s reagent, Fehling solution as well as Benedict’s solution while non-reducing sugar donot give positive test with any one of these reagents. (b) α-D-glucopyranose has hemiacetal structure and reduces Tollen’s reagent, Fehling solution, Benedict solution while methyl-α-D-glucopyranoside has acetal structure and are non-reducing, therefore fail in the above tests. —C—O

—C—O

O—H

O—R′

C —C

Alkyl group

C —C

R(=H or CH2OH)

α-D-glycopyranose give +ve Tollen’s test

R

If R′ = CH3, methyl α-D-glycopyranoside do not gives Tollen’s test

1368. Maltose and sucrose are the two most common disaccharide. H+ Maltose + H 2O → 2 moles of glucose (C 6 H12O 6 ) H+

Sucrose + H 2O → Glucose + Fructose 1369. Cellulose and starch are the most common polysaccharides. Both on hydrolysis produces many molecules of glucose, i.e., both are glucose polymers. 1370. Lactose is a disaccharide. Lactose is a reducing sugar and it reduces Tollen’s reagent, Fehling’s solution and Benedict solution. Lactose on hydrolysis gives D-glucose and D-galactose. 1371. Epimers are the diastereomers that differs in configuration at only one stereocenter. The above two diastereomers differ in configuration only at C-2 and this can be confirmed by their reaction with phenyl hydrazine. CH==NNHPh C==NNHPh HO H H OH H OH

D-glucose + C6H5NHNH2

C6H5NHNH2 + D-mannose.

CH2OH Same phenyl osazone formation confirm that only C-2 has different configuration in D-glucose and D-mannose.

1372.

HO H H H

CHO H OH OH OH CH2OH

H2 Ni

HO H H H

CH2OH H OH OH OH CH2OH

H2 Ni

HO HO HO H

CHO H H H OH CH2OH

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Solutions C6H5NHNH2 CH==NNHC6H5

C6H5NHNH2 CH==NNHC6H5

C==NNHC6H5

C==NNHC6H5

H H H

OH OH OH

Different osazone

HO HO H

CH2OH H HO HO H

CHO OH H H OH CH2OH

H H OH CH2OH

CH==NNHC6H5 C6H5NHNH2

HO HO H

C==NNHC6H5 H H OH

CH2OH Same osazone as obtained from B.

C

1373. Amino acid exist as dipolar ion as :  + —C— NH 3  COO – (Dipolar ion) (Zwitter ion)

Due to its existence in the form of dipolar ion, there exist a very strong intermolecular attraction which is responsible for its high melting point. 1374. (a) pH = 2 indicate highly acid solution and cationic form will predominate : +

H 2 N—CH—COO – H 2 N—CH—COOH H 3 N—CH—COOH pH = 12 pH = 2  ←  →  H 3C—CH—CH 3 CH 3 —CH—CH 3 CH 3 —CH—CH 3 (b) At pH =12, anionic form predominate. 1375. (a) In strongly acidic solution —NH 2 gets protonated and cationic form will predominate : HOOC—CH 2CH 2 —CH—COOH  NH 3 +

(b) In strongly basic solution —COOH will get deprotonated and anionic form will predominate: −

OOC—CH 2CH 2 —CH—COO –  NH 2

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668

(c) At isoelectric pH, zwitterion will predominate : HOOC—CH 2CH 2 —CH—COO – .  NH 3 +

1376. Isoelectric point is the pH at which zwitter ion has its maximum concentration. Glutamic acid is a dicarboxylic acid, its zwitter ion undergo deprotonation as: HOOC—CH 2 —CH 2 —CH —COO –  NH 3 +

−

OOC — CH 2CH 2 —CH — COO – + H +  NH 3 +

In order to suppress the above ionization so that zwitter ion predominate, higher [H + ]will have to be maintained and therefore isoelectric pH will be lower. Glutamic is a monocarboxylic acid no extra H + is required to suppress above type of ionization and hence isoelectric pH is higher. +

N H 3Cl – NH 2   1377. A : H —C— COOH B : H —C— COOCH 3   CH 2OH CH 2OH

−

+

NH 3 Cl  C : H—C— COOCH 3  CH 2Cl

NH 2  D : H—C— COOH  CH2Cl

NH2  Na(Hg) D → H—C—COOH dilute H+  CH 3 O  H3O+ /Heat 1378. CH 3 —C —H + NH 3 + HCN → CH 3 —CH—CN → CH 3 —CH—COO – H O 2   NH 2 NH 3 +

Alanine

1379. (a) A step growth polymerization results from reaction of two functional group releasing out some small molecules as H 2O, HCl, CH 3OH as biproduct. O (b) CH3O—C—

O —C—OCH3 + HO—CH2—CH2—OH

Dimethylterphthalate

Ethylene glycol

O

O Heat

—OCH2CH2O—C—

—C—

Poly(ethyleneterphthalate) Decron

+ 2CH3OH

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Solutions

1380. Teflon is a polymer of tetrafluoroethylene and synthesized as follows : O O O     Heat Ph—C—O—O—C—Ph → 2Ph—C—O • ( ≡≡RO • ) Benzoyl peroxide

Heat

RO • + CF2 == CF2 → RO—CF2 —CF2• (Initiation) ROCF2 —CF2• + nCF2 == CF2 → RO — [ CF2 —CF2 —CF ]n 2 —CF2• (Propagation) RO— [ CF2 — CF2 — ]n CF2 —CF2• + RO • → RO —[ CF2 —CF2 ]— n CF2CF2 —OR RO—[CF2 —CF2 — ]n CF2 — CF2• + ROCF2 —CF2• → RO—[CF2 —CF2 — ]n +2 OR 2RO— [ CF2 —CF2 — ]n CF2 —CF2• → RO— [ CF2 —CF2 — ]2n +2OR O O   Heat 1381. (a) n HO—C —( CH 2— ) 4 C — OH + nH 2 N—( CH 2 )— 6 NH 2 → Adipic acid

O O   —[ C —(CH 2 — ) 4 C—NH — ( CH 2 — )6 NH — ] n + H 2O Nylon 6,6

(b) Due to extensive intermolecular H-bonding among the polymer chains, nylon possess very high melting point.

MISCELLANEOUS O O

1382. (a)

O NH2

(b) H

(c) CH3CH2OH + H—C—H

PhNH OH OH (d)

(e)

(f)

CN

O Br

1383. (a)

O

OH

Cl

(c) CH3COCH3

(b) O

(d) CH3

CH3

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O

COOH N

(e)

O

(g)

—N

OH

+

(f)

CH3 H

N—

—OH (h)

H NH2

O Br

1384. A :

—C

NH2

B:

Br

C:

N

D:

O H—N

O O

1385. (A)

N O Ph3P

(B)

O

O

O

(C)

CHCH3

OH– Aldol

OH

H+ Heat

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Solutions H3C

H3C

N

COOCH3

1386.

O

H3C

N

N

COOH

C6H5

COOH O

OH

O

(C)

(B)

Cocaine (A)

O   CH2N2 PCC PCC 1387. C 6 H 5CH 2OH → C 6 H 5CHO → → C 6 H 5  C  CH 3 → (i) CH3MgBr

(benzyl alcohol)

CCl 4

∆

CCl 4

(ii) H3O+

O

CH3 X

OH

X

H 3O +

CH3(CH2)3CH2MgBr

O

OH LiAlH4

1388.

PCl3

H2O

MeO

Mg ether

MeO —OMe

OH —OMe

H3O+

(MgCl)+

O

MeO

MeO

—OMe H+/Heat

MeO (Dimestrol)

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OH

O

O 1389. A

C OH B

O O

O

O Br

D

E

F

OH Br

G

O CHO O3

+

Zn-H2O

H

O O

I

Mg

O (i) 

ether

O+

PCl

C H ONa

3 2 5 1390. (a) PhCH 2 Br → PhCH 2 MgBr → Ph—CH 2CH 2CH 2OH → 

(ii) H3

NaBH4

OCH3

(CH3COO)2Hg, CH3OH

C2H5OH

Ph—CH2 —CH

CH2

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Solutions Br

(b)

HC

C

C–Na+

CH

B 2H 6 NaOH/H2O2

OH CHO (i) CH3MgBr

PCC CCl4

Product

(ii) H3O+

OH CH2CH3 OH

(b)

1391. (a)

(c)

O

OH

OH O

CH3

Br

C2H5O

1392. A

O

OC2H5 Ph

C

B

D

COOH

E

Br

1393. (a)

OH (b)

CN

OH

H

H

H

(Enantiomeric)

Ph

OH Minor

Major

Ph (c)

+

CH3 (d)

OEt OH

CH3 (e)

O

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1394. OH H+/∆

OH

B2H6/H2O2

–H2O

NaOH

1

3

Cold, dilute, alkaline KMnO4

OH

Regioselective hydroboration followed by oxidation.

OH

2

O O

LiAlH4

EG/H+

1395.

O

OEt

O

O

O

O

O

H 2O

OEt

O

PBr3

Br

OH O Product

HCl

O PhONa

O

1396. Enol 2 is thermodynamically more stable than 3 because the alkene in this tautomer is conjugated with aromatic ring as well as with carbonyl group. The C==Cdouble bond in enol 3 is isolated and the stability afforded to 2-through resonance is not available to 3.

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Solutions O

O

H

O

O

O

H 3O +

CH3

OH CH2

H 2O 2 (thermodynamically more stable)

1 H+ H

+

O

–H3O

O

3 (less stable)

O

OH +

H 2

1397.

X

O

H H

OH–

O–

O–

most stable

Y

l l

This enol is highly disfavoured-the individual 2pz orbital of the double bond are orthogonal and consequently, the pπ-bond can’t form (Berdt’s rule)

OH–

O– most stable enolate anion because the double bond is in conjugation with aromatic ring.

O– Z

O–

OH–

Less stable

Most stable, because the double bond is tetrasubstituted and consequently thermodynamically more stable than the alternative, trisubstituted enolate anion.

1398. Here, the acidity is due to COOH groups and not due to α  H. Also, COOH is a strong electron withdrawing functional group, distance of one COOH group from other determine acidic strength. Larger the distance, weaker is the acid. Thus, the order of acid strength is: C>B>D>A

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COOH 1399. A : X

1. KOH, H2O

COCl

SOCl2 or (COCl)2 or PCl3 or PCl5

2. H3O+

CH3CH2COONa

Br

Product

O

H 3O +

EtONa

B: X +

Product (Through nucleophilic addition of enolate ion)

O N

Heat

C: X + N H

O

1400.

O CH3

CH3O–

– CH2

+ CH3O–

MeOH

O OH

O

O

O

O

O

O –

MeOH

O

O

–O

OH

OH O

Heat –H2O

O

OH

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Solutions

CHO

CH2

CH

O

1401.

OH

OH O

OH

O

CF3CO3H

O

CH3 N

C

A

CH3

D

B

H2SO4/H2O E

O O

Br OH

O

Br

OH

1403.

O

O

CN

KCN

H+/∆

O CN H+

O

O 1404. EtO

OEt

EtO–

EtO

s

+ EtO EtO

O–

O

(CH3COO)2Hg H2O Styrene

NaBH4

(not formed)

OEt

–EtO–

B

1405. (a)

OEt

s

O

O OEt

O

O

more acidic

less acidic

OEt

H

H

H

O

O

PCC CCl4

O

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OH

H

B2H6–H2O2

(b)

PCC CCl4

NaOH

Br KCN

HBr

(c)

Peroxide

O H 3O + H2SO4

O

OH

OH

LiAlH4

O 1406. (a)

(b)

—C

—C

CH

CH

HgSO4

—C—CH3

H2SO4/H2O H2/Pd/BaSO4

B2H6–H2O2 NaOH

Br2 AlCl3

Product

Product

O (c)

—C—CH3

PhMgBr

H3O+

ether

H2O

Product

(from a)

QUALITATIVE ANALYSIS 1407. A : PbCO 3 , B : PbO, C : Pb 3O 4 , D : Pb(NO 3 ) 2 , E : PbO 2 , F : PbCl 2 , G : PbI 2 2+ 1408. A : Cu(NO 3 ) 2 ; B : NO 2 , C : Cu(OH) 2 , D : CuO, E : [Cu(NH 3 ) 4 (H 2O) 2 ] F : [CuCl 4 ]2– . B : 2NO 2 + H 2O → HNO 3 + HNO 2 A : ZnCO 3 ⋅ Zn(OH) 2 , B : ZnO, C : ZnSO 4 , D : Zn(OH) 2 , E : [Zn(NH 3 ) 4 ]2+ A : Fe, B : FeSO 4 , C : Fe 2O 3 , D : Fe 3O 4 , E : FeCl 3 , F : Fe(OH) 2 , G : Fe(OH) 3 SOFCl. A : BaCO 3 , B : BaO A : CrCl 3 , B : Na 2CrO 4 , C : Na 2Cr 2O 7 , D : NaCl, E : CrO 2Cl 2 , F : (NH 4 ) 2 Cr 2O 7 G : Cr 2O 3 . 1414. A : ZnCO 3 ⋅ Zn(OH) 2 B : CO 2 , C : ZnO, X : K 2 Zn 3 [Fe(CN) 6 ]2 , D : ZnS, E : Zn(OH) 2 . 1416. A : CaCO 3 , B : CaO, C : CaC 2O 4 . 1409. 1410. 1411. 1412. 1413.

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Solutions

1417. 1418. 1419. 1420. 1421. 1422. 1423. 1424. 1425. 1426. 1427. 1428. 1429. 1430. 1431. 1432. 1433. 1434. 1435. 1436.

A: A: A: A: A: A: A: A: A: A: A: A: A: A: A: A: A: A: A: A:

Bi 2O 3 + PbCrO 4 , B : Bi(OH) 3 + Pb(OH) 2 , C : Bi(OH) 3 , D : PbO 2 , E : PbCl 2 , F : BiI 3 CuCl 2 , CrCl 3 , B : CuS, C : CrO 2– 4 , D : CrO 5 Cr 2O 3 , B : Al 2O 3 , C : PbCrO 4 , D : Al(OH) 3 Fe 2O 3 , B : MnCl 2 ⋅ 2H 2O, C : Fe(OH) 3 , D : MnS, E : CrO 2Cl 2 . Pb 3O 4 , B : Fe(CrO 4 ) 2 , C : PbO 2 , D : PbS, E : Fe 4 [Fe(CN) 6 ]3 , F : PbCl 2 . HgI 2 , B : AlI 3 , C : K 2 [HgI 4 ], D : HgO ⋅ Hg(NH 2 )I, E : Al(OH) 3 ZnO, B : Ca(NO 3 ) 2 , C : Zn(OH) 2 , D : [Zn(NH 3 ) 4 ]2+ , E : CaCO 3 . Ba(NO 3 ) 2 , B : CaCl 2 ⋅ 6H 2O, C : CaC 2O 4 , D : BaCrO 4 , E : BaCO 3 . Ag 2CrO 4 , B : Ag 2CO 3 , C : Ag 2O, D : CrO 5 . Mg, B : Mg 3 N 2 , C : Mg(OH) 2 , D : NH 3 , E : HgO ⋅ Hg(NH 2 )I. FeSO 4 ⋅ 7H 2O, B : Fe 2O 3 , C : SO 2 , D : SO 3 , E : FeCl 3 . HgCl 2 , B : Hg 2Cl 2 , C : Hg, D : HgI 2 Bi 2S 3 , B : H 2S, C : Bi(OH) 3 , D : BiO –3 , E : BiI 3 ZnCl 2 , B : MgCl 2 , C : Mg(OH) 2 , D : ZnS, E : HCl. Fe 2O 3 , B : MgSO 4 ⋅ 7H 2O, C : FeS, D : Mg(OH) 2 , E : S, F : BaSO 4 . CuBr 2 , B : ZnSO 3 , C : CuS, D : Zn(OH) 2 , E : BaSO 3 , F : SO 2 Fe 3O 4 , B : AgI, C : Fe 4 [Fe(CN) 6 ]3 Ag 2CO 3 , B : CuCl 2 , C : AgCl, D : CuS, E : [Cu(CN) 4 ]3– FeCl 3 , B : ZnSO 4 , C : Fe(OH) 3 , D : Zn(OH) 2 , E : HgO ⋅ HgSO 4 AlCl 3 , B : BaSO 4 , C : BaCrO 4 , D : Al(OH) 3 , E : CoOAl 2O 3 .

COORDINATION COMPOUNDS 1437. (i) hexaamminecobalt (III) chloride. (ii) pentaamminechlorocobalt (III) ion. (iii) tetraamminesulphatocobalt (III) nitrate. (iv) potassium pentachloronitridoosmate (VI). (v) sodium dithiosulphatoargentate (I). (vi) potassium amminedicyanodioxoperoxochromate (VI). (vii) pentaamminenitrito irridium (III) chloride. (viii) potassium tetrachloroplumbate (II). (ix) copper (II) potassium hexacyanocobaltate (III). (x) potassium hexacyanoaurate (III). (xi) lithium tetrahydrido aluminate (III). (xii) sodium tetrahydridoborate (III). (xiii) sodium hexafluoroaluminate (III). (xiv) tetraaquadichlorocobalt (III) chloride dihydrate. 1438. (i) hexaammine chromium (III) hexaisothiocyanatochromate (III). (ii) tetraammine copper (II) tetrachloroplatinate (II). (iii) tetraammine platinum (II) tetrachloroplatinate (II). (iv) hexaammine cobalt (III) tetracyanonickelate (II). (v) tetraammine dichloroplatinum (IV) tetrachloroplatinate (II). (vi) hexaammine cobalt (III) hexanitrocobaltate (II). (vii) pentaamminenitrochromium (II) hexanitrochromate (II).

680

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(viii) tetrapyridylplatinum (II) tetrachloroplatinate (II). (ix) hexaammine nickel (II) hexanitrocobaltate (III). (x) hexammine cobalt (II) diamminetetrachlorochromate (II). (xi) tetraammine dibromocobalt (III) tetrachlorozincate (II). (xii) pentaammine carbonatocobalt (III) tetrachlorocuperate (II). (xiii) diammine silver (I) hexacyanoferrate (II). (xiv) dichlorobis(ethylenediamine) chromium (III) tetrachloro palladate (II). 1439. (i) µ-amidobis (pentaammine cobalt (III)) nitrate. (ii) µ-amido-µ-sulphurdioxide bis (tetraammine cobalt (II)) nitrate. (iii) tri-µ-carbonyl bis (tricarbonyl iron (0)). (iv) µ-amido-µ-hydroxotetraammine cobalt (III) bis(ethylenediamine) cobalt (III) chloride. (v) µ-amido-µ-hydroxo dibis(ethylenediamine) cobalt (III) bromide. (vi) µ-amido-µ-nitro dibis (ethylenediamine) (cobalt (III)) chloride. 1440. (i) [Co(NH 3 ) 5 Br]SO 4 (ii) [Pt(en) 2 Cl 2 ]SO 4 (iii) [Pt(Py) 4 ][PtCl 4 ] (iv) K 3 [Fe(CN) 5 (CO)] (v) Cs[TeF5 ] (vi) [Pt(NH 3 ) 3 Br]NO 3 (vii) [Co(en) 2 Cl 2 ]Cl ⋅ H 2O (viii) [Cr(H 2O) 6 Br 2 ]Cl (ix) (NH 4 ) 3 [ZrF7 ] (x) [Ni(NH 3 ) 6 ]3 [Co(NO 2 ) 6 ]2 (xi) [Cr(en) 2 Cl 2 ]2 [PtCl 4 ] (xii) Al[AuCl 4 ]3 (xiii) Fe 4 [Fe(CN) 6 ]3 1441. A : [Co(NH 3 ) 3 (H 2O)ClBr]Br ⋅ H 2O; B : [Co(NH 3 ) 3 (H 2O) 2 Br]Cl ⋅ Br 1442. A : [Co(NH 3 ) 5 Br]SO 4 , B : [Co(NH 3 ) 5 SO 4 ]Br, Ionization isomerism. 1443. A : [Pt(NH 3 ) 4 Cl 2 ]Br B : [Pt(NH 3 ) 4 Br 2 ]Cl 2 1444. A : [Cr(H 2O) 6 ]Cl 3 , B : [Cr(H 2O) 5 Cl]Cl 2 ⋅ H 2O, C : [Cr(H 2O) 4 Cl 2 ]Cl ⋅ 2H 2O. 1445. [Ti(H 2O) 6 ]Cl 4 1446. [CoCl 6 ]4− , octahedral, sp 3 d 2 -hybridization, an outer orbital complex. 1447. In case of [Co(NH 3 ) 6 ][Cr(NO 2 ) 6 ], cobalt will be discharged at cathode while with [Co(NO 2 ) 6 ][Cr(NH 3 ) 6 ], chromium will be discharged at cathode. 1448. [Cr(NH 3 ) 6 ][Cr(NO 2 ) 6 ] will have lower conductivity due to larger hydrated radius. 1449. Conducting reagent : dimer 482, non-conducting reagent : monomer 241. 1450. [Pt(NH 3 ) 4 ][Pt(NO 2 ) 4 ] and [Pt(NH 3 ) 3 (NO 2 )][Pt(NO 2 ) 3 NH 3 ] only. 1451. A : [Cr(NH 3 ) 4 BrCl]Cl B : [Cr(NH 3 ) 4 Cl 2 ]Br 1452. A : [Cr(NH 3 ) 4 Br 2 ]Cl ⋅ H 2O B : [Cr(NH 3 ) 4 BrCl] ⋅ Br ⋅ H 2O C : [Cr(NH 3 ) 4 (H 2O)Cl]Br 2 . 1453. A : [Co(NH 3 ) 3 (H 2O) 2 SO 4 ]NO 3 ⋅ H 2O, B : [Co(NH 3 ) 3 (H 2O) 2 NO 3 ]SO 4 ⋅ H 2O C : [Co(NH 3 ) 3 (H 2O)(NO 3 )(SO 4 )] ⋅ 2H 2O D : [Co(NH 3 ) 3 (H 2O) 3 ]SO 4 ⋅ NO 3 . 1454. A : [Pt(NH 3 ) 4 Cl 2 ]Br 2 B : [Pt(NH 3 ) 4 ClBr]BrCl C : [Pt(NH 3 ) 4 Br 2 ]Cl 2 1455. A : [Co(NH 3 ) 3 (H 2O)Br 2 ]ClH 2O, B : [Co(NH 3 ) 3 (H 2O) 2 Cl]Br 2 1456. [Cr(H 2O) 5 Cl]Cl 2 ⋅ H 2O 1457. (a) Red is complement colour of green wavelength and vice-versa. Hence, ligand B is producing smaller crystal field splitting. (b) [CoA 6 ]3+ 1459. [Cr(en) 3 ]3+ 1458. [Cu(NH 3 ) 4 ]2+ 1460. Due to lack of d-d-electronic transition. − 1461. (a) CN − is a strong ligand, d 8 (Ni 2+ ) has all paired electrons in Ni(CN) 2– 4 , while Cl is a weak ligand Ni 2+ ( d 8 ) has two unpaired electrons.

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Solutions

1462. 1463.

1464. 1465. 1466. 1470.

(b) Ni has d 10 -configuration in Ni(CO) 4 and complex is tetrahedral. (c) Ni in [Ni(NH 3 ) 6 ]2+ is sp 3 d 2 -hybridized. Iodide ion being a strong reducing agent, reduces Cu 2+ to Cu + . Weak ligand field of H 2O gives five unpaired electrons in d-orbitals of Fe in Fe(H 2O) 3+ 6 and no electronic transition from one d-orbital to other d-orbital can occur. On the otherhand, CN − being a very strong ligand, gives a very large crystal field splitting and electronic transition from one d-level to other require photons of high energy and absorption occur in UV-region. K 3 [Mn(CN) 6 ] : µ = 2.82 Bm. Octahedral, inner orbital complex. K 2 [MnBr 4 ] : µ = 5.92 Bm, tetrahedral. (a), (b), (d), (e) F − is a weak ligand, forms high spin complex and CoF63− has four unpaired electrons in d-orbital. CN − is a strong ligand and all six electrons in the d-orbital of Co 3+ in Co(CN) 3– 6 are paired. The trans-isomer of [CoCl 2 (en) 2 ] possess a plane of symmetry and hence, does not show optical isomerism while cis form is devoid of any symmetry, it shows optical isomerism. Cl

NH2

H2N

Cl

NH2

NH2

Co

Co

H2N

Cl

NH2

NH2

Cl

H2N

trans [CoCl2(en)2] (has plane of symmetry)

Cis [CoCl2(en)2]

1471. Complex [ZnA2 B 2 ]2+ will be superimposable on its mirror image due to presence of plane of symmetry while [ZnABCD ]2+ is optically active. A

A 2+

Zn

2+ 2+

B

Zn

A

D

B

B Possess plane of Symmetry

C Devoid of Symmetry (Optically active)

1472. [Pt(NH 3 ) 2 Cl 2 ] is a square planar complex, therefore, show cis-trans isomerism as shown below : Cl

NH3

Cl

Pt Cl

Pt NH3

Cis

NH3

H3N

Cl Trans

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On the other hand [Zn(NH 3 ) 2 Cl 2 ] is tetrahedral and all the positions around metal are equivalent, therefore, no cis-trans isomerism is observed. 1473. In octahedral complex [CrL 6 ]3+ , Cr 3+ has 3d 3 configuration. The splitting pattern and electron filling in d-orbital of Cr 3+ can be shown as:

∆0

3d3 Cr3+

The nature of ligand will affect the ( ∆ 0 ) value but not the electronic configuration in the lower level of d-orbital. Therefore, magnetic properties, which depends on number of unpaired electrons with central metal, will not be affected by identity of ligand. 1474. “a ” and “c” are identical and both are trans isomers. “b” and “d ” are pair of enantiomers and both are cis-isomer. 1475. The complex is [CoCl 6 ](C 2 H10 N 2 ) 2 and it is octahedral. Cl Cl

4–

Cl Co2+

Cl

Cl Cl

1476. The complex has formula [Ti(H 2 NCONH 2 ) 4 I 2 ]I. 1477. (a) Cr 3+ hydrolyses in water according to the following reaction: Cr 3+ + H 2O

Cr(OH) 2+ + H +

In the above reaction, H + is produced which impart acidic nature to the solution. (b) As OH – is added, a gelatinous precipitate of Cr(OH) 3 is produced at first instant which dissolves on adding excess of OH – due to formation of complex [Cr(OH) 6 ]3– as: 3OH–

Cr 3+ + 3OH – → Cr(OH) 3↓ → [Cr(OH) 6 ]3– ( aq ) 1478. (a) tetraoxalatozirconate (IV) ion (b) diaquatetrachlorocupperate (II) ion (c) amminetrichloroplatinate (II) ion (d) tetracyanotetrahydroxomolybedate (IV) ion 1479. Electronic configuration of Fe(II) is 2+

Fe

: 3d

6

4s

0

4p

0

4d

0

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Solutions

Here Fe 2+ is readily converted into Fe 3+ (3d 5 ) due to the reason that Fe 2+ , after losing an extra electron yields a stable electronic configuration : 3d 5 where the 3d-orbital is completely half-filled. With Ni(II) (3d 8 ) and Co(II) (3d 7 ), no such stable electronic configuration is formed after removing an extra electron from the 3d-orbitals, therefore, formation of Ni(III) and Co(III) from Ni(II) and Co(II) respectively are very difficult. 1480. (a) An aqueous solution of [Ni(SO 4 )(en) 2 ]Cl 2 which produce precipitate with AgNO 3 solution and it will not give any precipitate with aqueous BaCl 2 solution. However, aqueous solution of [NiCl 2 (en) 2 ]SO 4 will form a precipitate with BaCl 2 . (b) Aqueous solution of [Ni(en) 2 I 2 ]Cl 2 will form a white precipitate with aqueous AgNO 3 solution which will be soluble in excess of ammonia solution. However, aqueous solution of [NiCl 2 (en) 2 ]I 2 will form an yellow precipitate of AgI which will be insoluble in concentrated ammonia solution. 1481. (a) In [MnCl 6 ]4– , Mn 2+ has 3d 5 configuration. Since, Cl – is a weak ligand, there will be smaller crystal-field splitting. In [Mn(CN) 6 ]4– , Mn 2+ has again 3d 5 configuration, but now CN – is a strong ligand, there will be larger crystal-field splitting as:

3d5

3d5

[MnCl6]

4–

[Mn(CN)6]4–

(b) Since, strong ligand field absorb shorter wavelength light, therefore, [Mn(CN) 6 ]4– will transmits longer wavelengths. 1482. (a) Since, the complex is emitting long wavelength (yellow) light, short wavelength visible light would have been absorbed. This is also evident from the high strength ligand field given by strong ligands-cyanide ion. (b) In [Co(CN) 6 ]3– , Co is in 3d 6 state and ligand is strong, there is no unpaired electrons as:

3d6 Strong ligand field

No unpaired electron, complex is diamagnetic

(c) If NH 3 molecules are substituted for cyanide ions, complex will absorb long wavelength visible light and the shift in absorbance will be towards red region of the spectrum. This is due to the fact that NH 3 is a weaker ligand than CN – .

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1483. In high spin complex of Mn 2+ (3d 5 ), all the electrons are occupying singly the five-d-orbitals as:

Mn2+

As evident from the above orbital energy diagram for Mn 2+ , electron transition is not allowed from lower level to higher level since, any electronic transition from low energy level to high energy level will violate “Pauli Exclusion Principle”. Therefore, complex of Mn 2+ is very faintly coloured. 1484. For a spontaneous reaction ∆G ° < 0, i. e., E ° > 0. In the above example, both the reduction reactions are spontaneous. Also reduction potential of reaction (II) is more positive, corresponding ∆G° will be more negative, reaction will be more spontaneous. Therefore, [Co(NH 3 ) 6 ]3+ is more stable and [Co(H 2O) 6 ]3+ is more reactive. 1485. (a) [Cr(H 2O) 4 Cl 2 ]Cl (b) [Cr(H 2O) 6 ]Cl 3 . 1486. (a) Coordination, (b) Ionization, (c) Linkage, (d) Hydrate. O

O

OH

OH2

OH

N

N OH2

1487.

Co3+

Co3+

OH

OH2

O O

O

O H

Trans isomer

Nitrilotriacetic acid [ ••N(CH2COOH)3 ] NH H2N

1488. (a)

NH2 Co3+

H2N

NH2 HN

OH OH2

O H Cis isomer

O

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1489. Weaker the ligand field strength, smaller will be the crystal field splitting and longer will be the wavelength of absorption. Hence : [Cr(CN) 6 ]3– < [Cr(en) 3 ]3+ < [Cr(NH 3 ) 6 ]3+ < [CrCl 6 ]3– < [Cr(H 2O) 6 ]3+ 1490. Stronger the ligand field, greater will be the crystal field splitting and hence, higher will be the frequency of light to be absorbed. Hence, order of frequency of light absorbed by these complexes is : [Cr(H 2O) 6 ]3+ < [Cr(en) 3 ]3+ < [Cr(CN) 6 ]3– 1491. (a) Stronger the ligand field, greater will be crystal field splitting, smaller will be the wavelength of light absorbed. λ :[Fe(CN) 6 ]4– < [Fe(H 2O) 6 ]2+

⇒

(b) Higher the oxidation state of central metal, greater will be the crystal field splitting. Hence, there will be greater crystal field splitting of d-orbitals in Mn(III) complex compared to that in Mn(II) complex. Hence, [Mn(CN) 6 ]4– will absorb at longer wavelength (low energy photons) than [Mn(CN) 6 ]3– . Br

H2O

1492.

(a)

H2O

Br Pt

Pt H2O

Br

Br

OH2

Cis

Trans Py

Cl (b)

Pt

(c) Zn

Cl

Pt

H3N 2+

Br

Cl

Br

H3N

Pt Py

Py

has tetrahedral arrangement of ligands around it : Cl

Zn2+ H3N

NH3 NH3

Cl H2O (d)

Cl Cl

Ni H2O

Ni OH2

OH2 Cis

OH2

H2O

H2O

OH2 Cl trans

Br

NH3

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686 O

O

Cl

O

Cl

O

O

O

(e)

O

Co

O Co

Cl

O

O

O

O

O

O Cl

O

trans O Cis (Optically active) Cl en Cl

1493. (a)

en

Co

en

Co

NO2 en NO2 (trans) Cl

(enantiomeric) en Cl

(b)

en

Co

en

Co

Cl en Cl Cl (trans)

OH2 (cis) H 3N

(c)

OH2

H 3N

Cr

(d)

OH2

Cl

H 3N Cl

NH3

NH3 Cl

Cl

Pt

NH3

H 2O Cl

en NH3

(e)

Pt Cl

OH2 Co

OH2

Cl

Cl

H 3N

Cr

H3N

Cl

Cl

Cr

Cl

Cl

en NH3

Cl

en

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1494. Co(III) = 3d 6 , weak ligand, magnetic moment = 4 × 6 = 24 BM = 4.9 BM. (b) Co 3+ : 3d 6 , magnetic moment = 0; en is a strong ligand. (c) Mn 3+ (3d 4 ) and CN – strong ligand, two unpaired electron, µ = 8 BM = 2.82 BM. (d) Mn 2+ (3d 5 ) and Cl – is a weak ligand, µ = 5 × 7 BM = 5.916 BM 1495. NiCl 2 + (CH 3 ) 3 P → Compound From elemental composition, empirical/molecular formula = [NiCl 2 ((CH 3 ) 3 P) 2 ] . Since, it has no isomeric forms, complex must be tetrahedral : Cl

Ni

(CH3)3P

Cl

P(CH3)3

bis (trimethyl phosphine) dichloronickel (II)

Magnetic moment : Ni 2+ (3d 8 ) and sp 3 hybridization indicate presence of two unpaired electrons, hence, µ = 8 = 3.82 BM. 1496. Zero magnetic moment indicate that there is no unpaired electron and complex is square planar (dsp 2 hybridized Ni). CN

Br

2–

CN

Br

Ni2+ CN

Ni2+ Br

Br

CN

Cis

NH3 Cl

1497. (a)

NH3 Cl

–

NH3

Cl Co3+

Co3+ Cl

trans

Cl

Cl

NH3

Cl Cl

trans Cl

Cl Cl

(b)

NH3

Co3+

Co3+ Cl

NH3 NH3 (Facial)

NH3

Cl

NH3

H3N Cl (Meridional)

–

2–

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1498. (a) No isomers (b) No isomers. O

O O O

O O

Co3+

(c)

O

O

3–

O

O

O

O

3– Co3+

Enantiomers

O

O

O

O

O

O O

O

O

O 8

1499. Cyanide ion is a strong ligand and in 3d ( Ni

2+

) it gives square planar geometry:

Energy

dx2–y2

dxy

dz2

dxz 2–

[Ni(CN)4 ]

O

dyz

crystal field splitting.

Energy

There is no unpaired electron, complex is diamagnetic. Chloride ion is weak ligand, in 3d 8 ( Ni 2+ ), it produces tetrahedral ligand field.

[NiCl 4 ]2– crystal field splitting.

There are two unpaired electrons and complex is paramagnetic. µ = 8 BM = 2.82 BM.

O

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1500. When a substance absorbs a particular colour when bathed in white light, the perceived colour of the reflected or transmitted light is the complementary colour. Also, wavelengths of absorbed and transmitted lights are inversely related. Cyanide ion being stronger ligand than H 2O, former absorbs light of shorter wavelength than later and wavelengths of transmitted light is longer for cyanide ion (yellow) and shorter for H 2O (blue or green). 1501. The crystal field splitting energy is equal to the energy of photon causing absorption maximum. ∆=

⇒

hc 6.625 × 10 –34 × 3 × 10 8 = = 4.23 × 10 –19 J λ 470 × 10 –9

1502. Stronger the ligand field, shorter the wavelengths of photon absorbed. (a) [Co(H 2O) 6 ]3+ (b) [FeF6 ]3– (c) [CuCl 4 ]2– 0.875 1000 1503. − ∆T f = 0.56 = i × 1.86 × × ; i = vant Hoff’s factor 233.5 25 ⇒ i = 2. Hence, formula of complex is [Co(NH 3 ) 4 Cl 2 ]Cl. 1504. Rust consists of mixture of Fe(OH) 3 and Fe 2O 3⋅nH 2O. Oxalic acid, being a bidentate, chelating, strong ligand, forms stable complex with iron, removing rust as: Fe 3+ + 3C 2O 2– → [Fe(C 2O 4 ) 3 ]3– 4 CN–

2– 1505. Cu 2+ + 2CN – → Cu(CN) 2 ↓ → excess Cu(CN) 4

Soluble

White

The complex Cu(CN) 2– 4 is highly stable and it ionizes very feebly as: 2– 2+ Cu(CN) 4 Cu + 4CN – : K d is very small and equilibrium lies predominantly to left side. If H 2S gas is bubbled at this time, no precipitate of CuS is formed due to presence of very small amount of Cu 2+ ion in the solution (less than required to exceed the solubility limit). 1506. The initial green colouration is due to presence of [CuCl 4 ]2– in concentrated solution. On dilution, solution turned blue due to conversion of [CuCl 4 ]2– into [Cu(H 2O) 6 ]2+ . 1507. K f (Formation constant) =

[Fe(H 2O) 5 NCS]2+ [Fe(H 2O) 6 ]3+ [SCN – ]

=

7.3 × 10 –5 (0.2) (10 –3 − 7.3 × 10 –5 )

= 0.393

1508. Electrical conductance of electrolyte depends on number of ions available. By comparison with the conductance of standard electrolyte, nature of unknown electrolyte can be established. In the given experiment, conductance of unknown will match as : [Co(NH 3 ) 6 ]Cl 3 = FeCl 3 , [Co(NH 3 ) 5 Cl]Cl 2 = MgCl 2 and [Co(NH 3 ) 4 Cl 2 ] Cl = NaCl. 1509. CoCl 2 at low concentration appear pink coloured due to its existence as [Co(H 2O) 6 ]Cl 2 . On adding HCl, following reaction occurs : [Co(H 2O) 6 ]Cl 2 + HCl → [Co(Cl) 6 ]4–

blue-coloured

Adding HgCl 2 , pink colour is restored due to the following reaction:

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[CoCl 6 ]4– + HgCl 2 + H 2O → [Co(H 2O) 6 ]2+ + [HgCl 4 ]2– pink

colourless

1510. The complex is square planar and exhibit cis-trans isomerism in which cis-form is polar while trans-form is non-polar. Dipole moment measurement can allow distinction between these two forms : Polar

Cl

Cl

NH3

Cl

Pt

Pt NH3

H3 N Cis

H3 N

Cl trans - Non - polar

1511. Both solutions contain hydrated iron [Fe(H 2O) 6 ] and [Fe(H 2O) 6 ]3+ ions. Higher the oxidation state of transition metal, greater will be the crystal field splitting and shorter will be the absorption wavelength. The wavelength of emitted light is inversely related to absorption wavelength. On this basis, transmitted light from [Fe(H 2O) 6 ]3+ will have longer wavelength-will appear yellow, while [Fe(H 2O) 6 ]2+ solution will be brown. This can also be explained on the basis of electronic configurations : Fe 3+ (3d 5 ) and weak ligand field, completely half-filled d-orbital. Difficult d-d-transitions. Fe 2+ (3d 6 ) and weak ligand field, less than full-filled d-orbital, easier d-d-transition. 2+

REPRESENTATIVE ELEMENTS 1512. (a) B2O 3 + 3Mg → 3MgO + 2B (b) Al 2O 3 will come in solution as Na[AlO 2 ]. (c) CO 2 + 2Na[Al(OH) 4 ] → 2Al(OH) 3 + Na 2CO 3 + H 2O H H H 1513. B B H H H H  s 1514. (a) BF3 + EtOH → Et —O → BF3 ⊕

+ s (b) BCl 3 + PhNH 2 → PhH 2 N •• → BCl 3 (c) BF3 + KF → K[BF4 ] 1515. (a) HF is a weak electrolyte, but when mixed with KF, forms a strong electrolyte KHF2 which ionizes as : KHF2 → K + + H + + 2F –

Fluoride ions makes a complex AlF4− with AlF3 , making it soluble. Adding BF3 , causes AlF3 to reprecipitate as : BF3 + AlF4– → BF4– + AlF3 .

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1516. (a) In trisilyl amine, pπ - dπ back bonding occurs making nitrogen sp 2 -hybridized. H3Si—N—SiH3 SiH3

No such back bonding is possible in trimethylamine and nitrogen is in sp 3 -state.

(b) Due to presence of vacant d-orbital with silicon, it forms a network structure as: O O==Si==O

Si==O

O==Si==O

No such bonding in possible with CO 2 since carbon does not possess vacant d-orbital. Therefore, CO 2 remains as isolated gaseous molecule. 1517. (a) 6NaOH + SiCl 4 → Na 2SiO 3 + 4NaCl + 3H 2O (b) H 3SiCl + H 2O → H 3Si(OH) → H 3Si—O—SiH 3 + H 2O (c) SiF4 + 4H 2O → Si(OH) 4 + 4HF 1518. (a) Sn + 2NaOH → Na 2SnO 2 + H 2 (b) 2SO 2 + PbO 2 → Pb + 2SO 3 (c) SiH 2Cl 2 + 2H 2O → 2HCl + SiH 2 (OH) 2 → —( SiH 2O — )n Polymer

(d) 4ClCH 2SiCl 3 + 3Li[AlH 4 ] → 4ClCH 2SiH 3 + 3Li[AlCl 4 ] 1519.

H H

Be

H H

H

Be

H

Be

H H

A part of the chain structure of polymeric BeH2

Here each Be is tetrahedrally coordinated and each hydrogen bridging two Be atoms. The Be—H—Be bond is a 3 centred 2 electron bond. 1520. (a) NaH + H 2O → NaOH + H 2 (b) 2NaN 3 + Heat → 2Na + 3N 2 (c) 4KO 2 + 2H 2O → 4KOH + 3O 2 (d) NaF + BF3 → Na[BF4 ] ∆

1521. (a) [NH 4 ]2 [BeF4 ] → BeF2 + 2NH 4 F (b) 2NaCl + BeCl 2 → 2Na 2 [BeCl 4 ] Water

(c) BeF2 ( s) → [Be(H 2O) 4 ]2+ + 2F − 1522. (a) BeCl 2 monomer is linear and Be atom has two vacant 2 p-orbitals, following diagram shows one of these vacant orbitals : Cl—Be—Cl Vacant 2-p-orbital

Cl—Be—Cl Cl—Be Cl—Be—Cl

Cl Cl

Dimer

Cl—Be

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Dimerization occurs by Cl lone pair donation into vacant 2 p-orbital of Be. Hybridization of Be in dimer is sp 2 and each Be atom still has one vacant 2 p-orbital. (b) Diethyl ether is a Lewis base and BeCl 2 is a Lewis acid, adduct formation occurs. OEt2 Be Et2O

Cl

Cl

Be is tetrahedrally coordinated and sp3-hybridized.

1523. Anhydrous CaCl 2 is hygroscopic and forms a hydrate, probably CaCl 2 ⋅ 2H 2O. In the presence of a lot of water, hydrate becomes liquid since it is deliquescent. Therefore, action as a drying agent depends on hygroscopic nature of CaCl 2 . On the otherhand, CaH 2 reacts with H 2O liberating H 2 : CaH 2 + 2H 2O → Ca(OH) 2 + 2H 2 Therefore, CaH 2 is used to remove H 2O from a solvent. 1524. NaF and NaCl are ionic species. The lattice energy and hence, melting temperature, of NaF is the greater of the two because of its smaller internuclear distance. CF4 and CCl 4 are covalent compounds, and so do not adhere to the ionic model and their melting temperature are determined by van der Waal’s force which depends on molar masses. 1525. n/ p ratio for 4 Be 7 and 6 C11 are 0.75 and 0.83, less than one, i.e., their nuclei contain fewer neutrons than protons, therefore positron emitted as : 7

→

3 Li

7

+

+1 e

11 6C

→

11 5B

+

+1 e

4 Be

0

0

n/ p ratio for 6 C14 is 1.33, greater than one, i.e., its nucleus contain large number of neutrons and therefore some neutrons are converted into protons. 0n

1

→

1 1H

+

−1 e

0

, i.e.,

14 6C

→

7N

14

+

−1 e

0

1526. He consists of two protons, two neutrons and two electrons and sum of the masses of these nucleons is 4.033 amu. Therefore, ∆m = 0.304 amu per atom. ∆E = ∆mc 2 = 2.736 × 1012 J/ mol = 2.835 × 10 7 eV. 1527. Ionic mobility and conductivity in solution depends on hydrated radius of the ion. Larger the hydrated radius, lower will be the ionic mobility and conductivity in aqueous solution. Size of hydrated ion depends on extent of hydration which consequently depends on positive charge density of cation. Smaller size and higher charge gives a greater charge density and cations are heavily hydrated. Li + is most hydrated in its group and hence, has lowest ionic mobility and conductivity. 1528. Li has most negative reduction potential suggest that ∆G° should be most negative for the reaction: Li → Li + + e

E ° = + 3.05 V, greater than E ° for Cs.

On the basis of above observation, Li should be most reactive. The actual reason for higher reactivity of Cs lies on kinetics. On descending down the group, melting point of metal (Group I) decreases and the energy produced by reaction is sufficient to melt the metal, which spread metals

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Solutions

throughout the solution allowing larger surface area for reaction. Hence, kinetics favour greater reactivity of alkali metals down the group. 1529. Dilute solution of alkali metals in liquid ammonia are dark blue due to presence of solvated metal ions and solvated electrons: M + NH 3 → M + NH 3 + e − NH 3

1530. 1531. 1532.

1533. 1534. 1535.

If the solution is allowed to stand, the colour slowly fades until it disappears owing to the formation of metal amide. At concentration above 3M, solutions are copper-bronze coloured because of formation of metal ions clusters. A : Li, B : Li 2O, C : LiH, D : H 2 , E : LiOH. Mg 2+ is more heavily hydrated than Li + due to greater charge/size ratio in the former case. In salt, Be is divalent ion Be 2+ .Since Be belongs to IInd period of periodic table the only available orbitals for coordination with H 2O are 2sand 2 p and a maximum of four lone pair of electrons can be accommodated. A : Ca, B : Ca(OH) 2 , C : CaCO 3 , D : CaO, E : CaC 2 . A : BaCO 3 , B : CO 2 , C : CaCO 3 , D : BaO. Due to inert pair effect, the stable oxidation state is +2 and not +4 while for Sn both Sn 2+ and Sn 4+ are possible and it acts as a reducing agent. Sn 2+ → Sn 4+ + 2e.

1536. In non-polar solvent, dimeric structure is retained while in aqueous medium, high hydration energy is sufficient to break the covalent dimer into [Al(H 2O) 6 ]3+ and Cl − as: Cl Cl

Al

Cl •• • • Cl

Al

Cl Cl

→

– Al(H 2O) 3+ 6 + Cl

In polar (aqueous) solvent

In non-polar solvent Low T

1537. (a) B2 H 6 + NH 3 (excess) → B2 H 6 ⋅ 2NH 3 High T

(b) B2 H 6 + NH 3 (excess) → (BN) x [ Boron nitride]. 1: 2 High T

(c) B2 H 6 + NH 3 → B3 N 3 H 6 (Borazine). 1538. (a) B3 N 3 H 6 + 3HCl → B3 N 3 H 9Cl 3 (b) B2 H 6 + 6H 2O → 2H 3 BO 3 + 6H 2 (c) B3 N 3 H 6 + 9H 2O → 3NH 3 + 3H 3 BO 3 + 3H 2 (d) B2 H 6 + 6EtOH → 2B(OEt) 3 + 6H 2 (e) B2 H 6 + 2LiH → 2Li[BH 4 ] (f) B2 H 6 + HCl → B2 H 5Cl + H 2 (g) B2 H 6 + 3Cl 2 → 2BCl 3 + 6HCl

1539. A : Al 2Cl 6 , B : Al 3+ ( aq ) + Cl − ( aq ), C : AgCl,

D : Al(OH) 3 ,

E : Li[AlH 4 ]

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1540. (a) SiO 2 + 2C → Si + CO Si + 2Cl 2 → SiCl 4 (b) SiO 2 + 2C → Si + 2CO 2200°C

Si + C → SiC 1541. (a) Al 4C 3 + 12H 2O → 3CH 4 + 4Al(OH) 3 (b) CaC 2 + 2H 2O → C 2 H 2 + Ca(OH 2 ) 1100°C

(c) CaC 2 + N 2 → CaNCN + C (Nitrolium)

O

2 (d) SiC + 2NaOH + O 2 → Na 2SiO 3 + CO 2 + H 2O

(e) Mg 2C 3 + 4H 2O → 2Mg(OH) 2 + CH 3C ≡≡CH 1542. 2C 2 H 5 MgCl + SiCl 4 → (C 2 H 5 ) 2 SiCl 2 + 2MgCl 2 C2 H 5 C2 H 5   Heat (C 2 H 5 ) 2 SiCl 2 + 2H 2O → HO—Si—OH → )n —Si ( —O—   C2 H 5 C2 H 5 1543. R 3SiCl is monofunctional and act as chain terminator, hence stop the growth of propagating polymer chain as: R R R R     ( Si—O) n—O— Si— R —Si—O ( — ) n H + HO—Si— R → —     R R R R (From R3SiCl)

(growing chain)

1544. (a)

CH3 CH3 Si O O H 3C

Si

H 3C

Si O

Chain terminated

–

(b)

CH3

–

–

CH3 _

_ _

_

_

.

.

_ _

_ .

_.

(d)

_ .

. _.

(c) _ _

_

.

.

_ _ . _

_

_

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_

_ _ _.

. _ .

_ .

_.

(e) _

_

.

. _ .

. _ _ .

_

_

H O

2 1545. (a) P4 + 3NaOH → PH 3 + 3NaH 2 PO 2

(b) CaNCN + 3H 2O → 2NH 3 + CaCO 3 (c) N 2 H 4 + 2I 2 → N 2 + 4HI (d) N 2O + 2NaNH 2 → NaN 3 + NH 3 + NaOH 1546. Superphosphate : 3[Ca 3 (PO 4 ) 2 ⋅ CaF2 ] + 7H 2SO 4 → 3Ca(H 2 PO 4 ) 2 + 7CaSO 4 + 2HF 14444244443 Superphosphate

Triple superphosphate: [Ca 3 (PO 4 ) 2 ⋅ CaF2 ] + 6H 3 PO 4 →

4Ca(H 2 PO 4 ) 2

+ 2HF

(Triple superphosphate)

1547. (a) N 2O 5 + NaCl → NaNO 3 + NO 2Cl (b) N 2O 5 + Na → NaNO 3 + NO 2 (c) N 2O 4 + H 2O → HNO 2 + HNO 3 (d) 2NO 2 + Cl 2 → 2NO 2Cl (e) 2NO 2 + HCl → 2NOCl + Cl 2 + 2H 2O 1548. In N 2 , due to smaller size of 2 p-orbital, effective pπ - pπ overlap takes place giving very high “bond-energy” to N 2 molecule. In P2 , bond energy is just half of the N—N bond energy in N 2 due to poor pπ - pπ overlap between the larger 3 p-orbitals. On the other hand in P4 , each phosphorus is tetrahedrally bonded to three other phosphorus atoms giving three σ-bond to each phosphorus, greater bond energy than P2 , hence more stable P4 is formed. P N

N

better pπ-pπ overlap

P

P

poor pπ-pπ overlap

P

P P

more stable structure

1549. In NF3 , N—F bond is strongest and it behaves like other inert molecules CF4 , etc.

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O .. P O

P O

O

1550. : P O

O

O .. P P 4O 6

O

P:

O

O

P

O

P

O

O

O

P O

O

P4O10

1551. (a) Ostwald method : In Ostwald process, NH 3 is oxidised first into NO in presence of platinum, rhodium catalyst at high temperature and pressure. NO is then further oxidised into NO 2 and absorbed in water. Platinum/Rhodium 5 atmosphere, 850°C

4NH 3 ( g ) + 5O 2 ( g ) → 4NO( g ) + 6H 2O( g ) → 2NO( g ) + O 2 ( g ) 2NO 2 ( g ) 2NO 2 ( g ) + H 2O( l) → HNO 3 + HNO 2 2HNO 2 → H 2O + NO 2 + NO 3NO 2 + H 2O → 2HNO 3 + NO 2HNO 2 + 2FeO → 2NO + H 2O + Fe 2O 3 Oxidising action 5HNO 2 + 2KMnO 4 + 3H 2SO 4 → 2K 2SO 4 + 2MnSO 4 + 3H 2O Reducing action 1552. (a) Ligand oxygen has two unpaired electrons in its molecular orbitals, giving paramagnetic character while N 2 has all paired electrons in molecular orbital, and diamagnetic. (b) NO + has bond order of three while bond order of NO is 2.5. 1553. Preparation : 2H 2SO 4 → 2H + + 2HSO –4 (b)

at anode

2HSO –4 → H 2S 2O 8 + 2e − H 2S 2O 8 + 2H 2O → 2H 2SO 4 + H 2O 2 . O

Structure :

Open book like structure.

H

(a) (b) (c) (d)

H 97°

O 94°

2KMnO 4 + 3H 2SO 4 + 5H 2O 2 → K 2SO 4 + 2MnSO 4 + 8H 2O + 5O 2 K 2Cr 2O 7 + 4H 2O 2 + H 2SO 4 → 2CrO 5 + K 2SO 4 + 5H 2O 2KI + H 2O 2 → I 2 + 2H 2O 2K 4 [Fe(CN) 6 ] + H 2O 2 + H 2SO 4 → 2K 3 [Fe(CN) 6 ] + K 2SO 4 + 2H 2O

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Solutions

O  1554. (a) HO—S —OH + IV

O O   (b) HO—S —S —OH  O

O O   (c) HO—S —S —OH + III

O  (d) HO—S —OH  S

+V and + III

O O   (e) HO—S —S — OH   O O (+V)

+IV and –II

O  (f) HO—S —O—OH  O

O O   (g) HO—S —O—O—S —OH   O O

(+VI)

(+VI)

1555. In contact process, SO 2 is oxidized by air into SO 3 in presence of V2O 5 catalyst: V2O5 1 SO 2 + O 2 → SO 3 2 SO 3 is then passed through 98% H 2SO 4 , forming H 2S 2O 7 (oleum). This solution is then diluted to obtain sulphuric acid of desired strength.

1556.

boiling water

Na 2SO 3 + S → Na 2S 2O 3 Heat in air

2Na 2S 3 + 3O 2 → 2Na 2S 2O 3 + 2S Thiosulphuric acid can’t be prepared by adding acid to the aqueous solution of thiosulphate salt, since the free acid decomposes in water into mixture of S, H 2S, SO 2 , etc. It can be made in the absence of water and at low temperature. 1557. (a) 2F2 + 2H 2O → 4HF + O 2 (b) (c) (d) (e) 1558. (a)

Cl 2 + CO → COCl 2 3Cl 2 + 8NH 3 → N 2 + 6NH 4Cl 3Cl 2 + 6NaOH → 5NaCl + NaClO 3 + 3H 2O SiO 2 + 6HF → H 2SiF6 + 2H 2O Cl 2O + 2NaOH → 2NaOCl + H 2O

(b) 4HClO 4 + P4O10 → 2Cl 2O 7 + 4HPO 3 (c) 2ClO 2 + 2NaOH → NaClO 2 + NaClO 3 + H 2O (d) KBrO 3 + F2 + 2KOH → KBrO 4 + 2KF + H 2O (e) Cl 2O 6 + 2NaOH → NaClO 3 + NaClO 4 + H 2O (f) Cl 2O 6 + HF → FClO 2 + HClO 4 1559. First, water must be absent in preparation of F2 by any method, otherwise F2 , being the strongest oxidising agent, oxidises water into O 2 converting itself into HF. Secondly, HF in anhydrous state, is only slightly ionized and is therefore a poor conductor of electricity. Therefore, anhydrous HF is difficult to electrolyse. For the purpose of producing F2 by electrolytic method, a mixture of HF and KF is electrolysed. HF first combines with KF forming a conducting K[HF2 ], enabling electrolysis easier.

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698 1560. (a) F

O 103°

F

O F Cl

111°

Cl

O—O

O==I—O—I==O

F

O

O

(b) Repulsion between the lone pair reduces the bond angle in OF2 while in Cl 2O, bond angle is greater than tetrahedral bond angle due to steric crowding of the larger halogen atoms. 1561. (a) XeF2 + 2HCl → Xe + 2HF + Cl 2 (b) XeF4 + 4KI → Xe + 4KF + 2I 2 (c) XeF4 + 2SF4 → Xe + 2SF6 (d) 2XeF2 + 2H 2O → 2Xe + 4HF + O 2 3

(e) 3XeF4 + 6H 2O → 2Xe + XeO 3 +12HF + 2 O 2 (f) XeF6 + 3H 2O → XeO 3 + 6HF XeF6 + H 2O → XeOF4 + 2HF (g) 2XeF6 + XeO 3 → 3XeOF4 (h) XeO 3 + XeOF4 → 2XeO 2 F2 1562. Fe 3+ ( aq. ) is a stronger oxidising agent than Br 2 but weaker than Cl 2 . 1563. In ClF3 , Cl has ten electrons in its valence shell, seven of its own and three shared electrons from three fluorine. Since, chlorine has vacant d-orbitals (3s 2 3 p 5 3d 0 ), it can extend its valency beyond eight, i.e., it can violate octet rule. On the other hand, in FCl 3 , the central fluorine atom has ten-valence electrons which is not possible since, fluorine has no vacant d-orbitals and it cannot violate octet rule (2s 2 2 p 5 ). 1564. Due to very small size of nitrogen, their p-orbitals makes effective overlap forming strong pi-bonds and nitrogen acquire octet configuration by forming a sigma and two pi-bonds between two nitrogen atoms as: N

N

has complete octet at both nitrogen.

On the other hand, due to large size of phosphorus atom, their p-orbitals do not overlaps efficiently to form pi-bonds. Due to this reasons, phosphorus atoms forms preferably sigma bonds and at room temperature it exist as P4 containing sigma-bonds only. If heated at very high temperature, it decomposes partially into less stable P2 molecules. 1565. Sulphur contains vacant d-orbitals, can extend its valency beyond eight while oxygen lacks vacant d-orbital, can’t violate octet rule 1566. CuF2 + H 2SO 4 → CuSO 4 + 2HF 2NaCl + H 2SO 4 → Na 2SO 4 + 2HCl HBr and HI cannot be prepared by the above method because they are oxidised into elemental bromine and iodine as: 2NaBr + 2H 2SO 4 → Na 2SO 4 + SO 2 + 2H 2O + Br 2 .

Solutions

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1567. Boric acid B(OH) 3 , only partially react with water to form H 3O + and B(OH) 4 – , it behaves as a weak monobasic acid B(OH) 4 – + H 3O +

B(OH) 3 + 2H 2O

Thus, boric acid can’t be titrated satisfactorily with NaOH as a sharp end point is not obtained. Adding cis-diol or triol (glycerol) enhances acidic strength of boric acid by forming a stable complex as: B(OH) –4 + H 3O +

B(OH) 3 + 2H 2O

    C O  C  OH –2H2O   B(OH) –4 + →      C  OH   C O   

B

 OH    OH 

…(i)

–

diol –2H2O

→

    C O        C O  

B

(X)

   O  C      O  C   

s

The formation of complex of B(OH) –4 with diols, as shown above, removes B(OH) –4 from solution, driving reaction (i) in forward direction and all B(OH) 3 reacts quantitatively with water forming B(OH) –4 required for formation of stable complex (X). This increases acid strength and boric acid becomes strong acid in presence of such organic compounds. 1568. In CCl 4 , carbon has no vacant orbital where water can attack and therefore, it is inert towards water. On the other hand, Si in SiCl 4 has vacant 3d-orbitals, available for attack by water, which makes it highly reactive with water. Cl

Cl

Si

Cl

Cl

OH2

vacant d-orbitals are the site of attack by water molecule

If sufficient energy is supplied by using superheated steam, CCl 4 hydrolyses forming phosgene gas: Superheated

CCl 4 + H 2O( g ) → COCl 2 + 2HCl 1574. Initially CaCO 3 is formed as white precipitate as: CO 2 + Ca(OH) 2 → CaCO 3 ↓ + H 2O

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On further passage of CO 2 , solid CaCO 3 comes in solution by forming soluble bicarbonate as: CaCO 3 + H 2O + CO 2 → Ca(HCO 3 ) 2 ( aq ) 1575. Phosphoric acid is proton donor and gives three proton on complete ionisation, exhibiting its tribasic character: + PO 3– 4 + 3H

O== P(OH) 3

On the other hand, boric acid is electron deficient and boron has only six shared electrons in valence shell, i. e., two electrons short of stable octet configuration. Hence, H 3 BO 3 behaves as Lewis acid. Also, it can accept only one lone pair, it is monobasic: s

⊕

(a vacant orbital) + •• NH 3 → (HO) 3 B ← NH 3

(HO) 3 B

1577. 3NaHCO 3 + NaAl(SO 4 ) 2 → 3CO 2 + 2Na 2SO 4 + Al(OH) 3 1578. Inert pair effect. As we descend down a group in the periodic table, inert pair effect becomes more and more pronaunced and oxidation number decreases by two unit. 1579. Aluminium being strongly electropositive, reacts with acid components of food. 1580. In between pH = 5 to pH = 8, Al 3+ precipitate as Al(OH) 3 . As it settles, the gelatinous Al(OH) 3 traps and removes suspended solids from the water. 1581. (a) N 2 H 4 ( aq ) + HCl → N 2 H 5Cl (b) 2NO( g ) + 32 O 2 ( g ) → N 2O 5 ( g ) (c) 2NO 2 ( g ) + H 2O( l) → HNO 2 + HNO 3 (d) 2NH 3 ( g ) + 4O 2 ( g ) → N 2O 5 + 3H 2O 200°C

(e) NH 4 NO 3 ( l) → N 2O + 2H 2O Electric spark

1582. (a) 2Al + Fe 2O 3 → Al 2O 3 + 2Fe ∆H < 0 (b) Na 2O 2 + H 2O → 2NaOH + 12 O 2 ( g ) (c) Pb 2+ + O 3 + H 2O → PbO 2 + O 2 + 2H + 1583. (a) 2KO 2 + H 2O → 2KOH + 32 O 2 (b) 3O 3 + Cl – → 3O 2 + ClO –3 1584. Borax on treatment with CaF2 in concentrated sulphuric acid solution gives BF3 as: Na 2 B4O 7 + 6CaF2 + 8H 2SO 4 → 6CaSO 4 + 4BF3 + 7H 2O + 2NaHSO 4 1585. S + O 2 → SO 2 Na + H 2O → NaOH( aq ) + 12 H 2 ( g )

…(i) …(ii)

From Eqs. (i) and (ii), 2NaOH + SO 2 → Na 2SO 3 + H 2O SO 2– 3

–

+ SO 2– 4

…(iii) +

+ H 2O + Cl 2 → 2Cl + 2H {from Eq. (iii)} Boiling elemental sulphur in alkaline solution of Na 2SO 3 gives thiosulphate as: SO 2– 3 ( aq ) + S ( s) →

S 2O 32–

(thiosulphate ion)

1586. The NF3 has little tendency to act as donor molecule. The highly electronegative F atoms attract electrons and these moments partially cancel the moment from the lone pair which reduces its

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Solutions

donor power. On the other hand, in PF3 , P  F bond length is larger compared to N  F, bond, electron withdrawing effect of fluorine does not transmit effectively onto the lone pair on phosphorus in PF3 . Hence, PF3 act as a good donor molecule. Ni[CO]4 + 4PF3 → Ni(PF3 ) 4 + 4CO 1587. A : NH 3 ; B: NO; C : NO 2 ; D : N 2O 3 ; E : HNO 2 ; F : N 2 ; G :CaNCN. 1588. In POCl 3 , there is pπ − d π back bonding between oxygen and phosphorus as: O pπ – dπ back bonding

Due to this pπ − d π back bonding, P—O bond acquire double bond character and hence, P—O bond is shorter than the expected bond-length.

P

Cl

Cl Cl

1589. To attain the pentavalent state, d-orbitals must be used. Hydrogen is not sufficiently electronegative to make the d-orbitals contract sufficiently, hence, PH 5 is not formed. 1590. HCl( g ) is prepared from NaCl and H 2SO 4 as: ∆

2NaCl + H 2SO 4 → Na 2SO 4 + 2HCl( g ) HCl( g ) does not possess sufficient reducing power required for reducing H 2SO 4 , hence, the above reaction is successful in preparing HCl( g ). Contrary to HCl, HBr and HI are oxidised by H 2SO 4 as: 2HBr + H 2SO 4 → Br 2 + SO 2 + 2H 2O 2HI + H 2SO 4 → I 2 + SO 2 + 2H 2O Hence, HBr and HI can’t be prepared by the reaction of NaBr and NaI with H 2SO 4 .

1591. (a) H  C ≡≡ P ••

F

(e)

H  P H

   H  

+

F

P

F

F

1592. (a) (b) (c) (d) (e)

  (b)  H  

P4 P4 P4 P4 P4

F

+ 4Al → 4AlP +10F2 → 4PF5 +12Na → 4Na 3 P + Se → P4Se 6 + P4Se 10 + Cl 2 → PCl 3 + PCl 5 2O

2 (f) P4 + 3O 2 → P4O 6 → P4O10

••

••

(c) H  P  P  H   H H

(d)

–

O  P

O–

O O

–

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1593. Double bond is formed by side-wise (lateral) overlaping of p-orbitals holding electrons of opposite spin. For efficient pi-bonding (lateral overlap), the p-orbitals must be close-enough in order to allow them to overlap up to some extent and this will be possible if atoms are small enough in size. Oxygen, being the smallest in its group, its p-orbitals are close enough and formation of pi-bond become possible. On the other hand, sulphur is too large in size that their p-orbitals don’t come closer enough to allow the formation of pi-bonds, hence, sulphur exist as S 8 in which all bonds are sigma covalent bonds :

S

O2-molecule showing π-bond formation

S

S-atoms are much distant and no effective π-bond are formed

S8 with all single, sigma covalent bonds

1594. As we descend down to hologen group in periodic table, molar mass increases rapidly and hence, intermolecular force of attraction (van der Waals force of attraction). Due to this increase in force of attraction, volatility decreases from top to bottom and their boiling/melting point increases. 1595. (a) Xe in XeO 2 F2 is sp 3 d hybridized and shape of the molecule is F O Xe O F

(b) Xe in XeF + is sp 3 hybridized :

•• + • • Xe  F ••

Linear

(c) Xe in XeF3+ is sp 3 hybridized :

F + Xe F F

3 2 (d) Xe in XeO 4– 6 is sp d hybridized :

O O

O Xe

O

O O

4–

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(e) Xe in XeO 3 F2 is sp 3 d hybridized : O

F O Xe O F

1596. NaOH reacted with atmospheric CO 2 forming a white layer of NaHCO 3 . 1597. White phosphorus is highly reactive and it catches fire even at room temperature. It owes reactivity from its structure in which a highly strained ring is available. P

Here the P—P—P bond angles are 60° while the required bond angle is 109°. Due to this angle strain, P4 is highly unstable and reactive.

60° P

P P

Red hot

1598. (a) 2NH 3 + 3CuO → N 2 + 3H 2O + 3Cu heat

(b) (NH 4 ) 2 Cr 2O 7 → N 2 + Cr 2O 3 + 4H 2O 1599. CuSO 4 + 2KF → K 2SO 4 + CuF2 ↓ (Insoluble, precipitate as green solid) CuSO 4 + 2KCl → K 2SO 4 + CuCl 2 ( aq ) (Soluble, remains in solution giving green colour) 1600. (a) 4H 3 PO 3 → 3H 3 PO 4 + PH 3 (b) Li 4C + 4HCl → 4LiCl + CH 4 (c) 2HI + 2HNO 2 → I 2 + 2NO + 2H 2O (d) H 2S + 2Cl 2 → 2HCl + SCl 2 1601. (a) PCl 5 < CCl 4 < SiCl 4 < BCl 3 (b) F – < O 2– < N 3– < C 4– (c) I < Br < Cl < F (d) SiO 2 < H 2O < CO 2

METALLURGY 1602. Mg is extracted from sea water exploiting the fact that Mg(OH) 2 is less soluble than Ca(OH) 2 and precipitated out on treatment of sea water with lime. Sea water  Ca(OH)2  → Mg(OH) 2 ↓ 2+ contains Mg salt The precipitate is filtered off, dried and dissolved in small volume of dilute HCl as: Mg(OH) 2 + 2HCl → MgCl 2 + 2H 2O The above solution is heated to dryness and anhydrous salt is electrolysed to obtain the pure metal as: Mg 2+ + 2e − → Mg 2Cl

–

→ Cl 2 + 2e

at cathode −

at anode.

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1603. Dolomite ore consists CaCO 3 ⋅ MgCO 3 . Extraction involves: Heat

(i) Calcination: CaCO 3 ⋅ MgCO 3 → CaO ⋅ MgO + 2CO 2 (ii) Dissolution in dilute HCl : CaO ⋅ MgO + 4HCl → CaCl 2 (aq) + MgCl 2 (aq) + 2H 2O The above solution is then treated with excess of calcined ore and CO 2 is bubbled through it. CaCl 2 ⋅ MgCl 2 + CaO ⋅ MgO + 2CO 2 → 2MgCl 2 + 2CaCO 3 ↓ Precipitated

The aqueous solution is then heated to dryness and anhydrous MgCl 2 is electrolyzed in molten state to obtain pure metal. 1604. Mg itself being one of the strongest reducing agent, reduction of Mg 2+ by chemical method is not possible. 1605. Due to smaller size and high charge, Mg 2+ is heavily hydrated in aqueous solution which is responsible for its low ionic mobility and conductivity. 1606. (a) Epsom salt MgSO 4 ⋅ 7H 2O, occurs in nature as epsomite in certain gypsum deposite. It is also obtained by hydration of its monohydrate MgSO 4 ⋅ H 2O, which exist in nature as kieserite. Kieserite is allowed to remain in contact with water, when it slowly hydrated and sets to a solid mass of crystals of heptahydrate: MgSO 4 ⋅ H 2O + 6H 2O → MgSO 4 ⋅ 7H 2O Epsom salt

Epsom salt is used in medicine as a purgative and as a catalyst in preparation of H 2SO 4 . (b) 4Mg +10HNO 3 → 4Mg(NO 3 ) 2 + 3H 2O + NH 4 NO 3 1607. The net reaction of Solvay process is: 2NaCl + CaCO 3 → Na 2CO 3 + CaCl 2 . Steps involved in the process are: (i) NH 3 + H 2O + CO 2 → NH 4 HCO 3 (ii) NH 4 HCO 3 + NaCl → NaHCO 3 + NH 4Cl (iii)

Heat

2NaHCO 3 → Na 2CO 3 + H 2O + CO 2 .

NaHCO 3 being less soluble, can preferentially be crystallized leaving NH 4Cl in solution, and this forms basis of manufacture. On the otherhand, KHCO 3 being more soluble, not separated by fractional crystallization and therefore this method fails for K 2CO 3 . 1608. Baeyer’s process : Ore is roasted first to oxidise any Fe 2+ into Fe 3+ and then digested with NaOH at 130°C. Al 2O 3 dissolves, comes in solution leaving others in residue: Al 2O 3 ⋅ 2H 2O + 2NaOH → 2NaAlO 2 + 3H 2O Filtrate is treated with some Al(OH) 3 , where all aluminate is precipitated as Al(OH) 3 . Hall’s Process : Ore is first fused with Na 2CO 3 , dissolved in water and CO 2 is bubbled. CO

2 Al 2O 3 ⋅ 2H 2O + Na 2CO 3 → 2NaAlO 2 + CO 2 + H 2O → 2Al(OH) 3 ↓

Serpek’s Process :

Used when SiO 2 is the chief impurity:

Solutions

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Al 2O 3 + N 2 + 3C → 2AlN + 3CO; AlN + H 2O → Al(OH) 3 + NH 3 . 1609. Al is extracted from bauxite ore by electrolytic reduction method. Some cryolite (Na 3 AlF6 ) and CaF2 are added to the ore in order to lower the melting temperature and increase the electrical conductivity facilitating electrolysis easier: Al 3+ + 3e − → Al at graphite cathode. Thermite welding: Thermite welding is used for commercial welding of iron. It exploit strong reducing power of Al metal. Thermite is 3 :1 mixtrue of Fe 2O 3 and Al. Fe 2O 3 + 2Al → 2Fe + Al 2O 3 ∆H < 0 The heat evolved in the above reaction is sufficient to keep iron in molten state which serve the welding purpose. 3 1610. Al + 3HCl → AlCl 3 + H 2 2 2Al + 3H 2SO 4 → Al 2 (SO 4 ) 3 + 3H 2 3 H2 2 Concentrated nitric acid, being an oxidising agent, forms a protective layer of Al 2O 3 on the surface of metal, which prevent it from further reaction. 1611. Aluminium take up hydrogen in presence of Zeigler catalyst (trialkyl aluminium) as: 3 Al + H 2 + 2Et 3 Al → 3Et 2 AlH 2 Then alkene is inserted 1st between Al—H bond as: H 2O

Al + NaOH → NaAlO 2 +

Et 2 AlH + CH 2 == CH 2 → Et 2 Al—CH 2 —CH 3 Finally, applying high pressure, alkene molecules are inserted between Al—C bond as: Et 2 Al—CH 2CH 2 + CH 2CH 3 → Et 2 Al—CH 2CH 2CH 2CH 3 n-times

→ Et 2 Al — ( CH 2 —CH 2 — ) n CH 2CH 3 Hydrolysis of this long chain growing aluminium alkyls gives straight chain hydrocarbon called polyethylene. A modified catalyst TiCl 4 (Nata catalyst) causes polymerization to occur at faster rate and even a much higher molecular weight polymer is produced. 1612. Iron is extracted from its oxide ore Haematite (Fe 2O 3 ) or Magnetite (Fe 3O 4 ) by carbon reduction method, in a blast furnace. Blast furnace is charged with ore, lime-stone and coke. The reactions occurring in blast furnace are: (i) Coke is oxidised at the bottom of furnace as: 2C + O 2 → 2CO [CO 2 is not stable at this temperature] (ii) Reduction of iron oxide at the top of the furnace: Fe 2O 3 + 3CO → 2Fe + 3CO 2 Fe 3O 4 + 4CO → 3Fe + 4CO 2

706

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Limestone decomposes giving CO 2 and CaO. CO 2 combines with coke providing CO as : CO 2 + C → 2CO. The mass impurity SiO 2 is removed as silicate (slag): CaO + SiO 2 → CaSiO 3 (slag). Iron obtained here is called cast iron. 1613. In presence of moisture and CO 2 , iron gets rusted with a brown deposition on the surface: Fe + H 2O + O 2 → Fe(OH) 3 + Fe 2O 3 ⋅ nH 2O 14444244443 Rust

Rusting of iron is prevented by : (i) Coating with more electropositive metals like Mg, Ni, Zn, Cr, etc. (ii) Galvanization : Galvanization involves coating iron with Zn metal. It gives physical protection by forming a thin, hard film of ZnCO 3 on the surface in due course of time. At the same time, it also gives cathodic protection. 1614. (a) 3Fe + 4H 2O → Fe 3O 4 + 4H 2 ( g ) (b) Fe + 2HCl → FeCl 2 + H 2 ( g ) (c) Fe + 2H 2SO 4 → FeSO 4 + SO 2 + 2H 2O 2FeSO 4 + 2H 2SO 4 → Fe 2 (SO 4 ) 3 + 2H 2O + SO 2 (d) 2Fe ( s) + 3Cl 2 ( g ) → 2FeCl 3 (e) 4Fe +10HNO 3 → 4Fe(NO 3 ) 2 + NH 4 NO 3 + 3H 2O (f) 4Fe +18HNO 3 → 4Fe(NO 3 ) 3 + 3NO + 3NO 2 + 9H 2O (g) Fe + S → FeS 1615. Mohr’s salt FeSO 4 ⋅ (NH 4 ) 2 SO 4 ⋅ 6H 2O is prepared by making saturated solution of pure ferrous sulphate and pure ammonium sulphate separately, in air free distilled water at 40°C. Mixing the above solutions, and allowing to cool, the salt is obtained as pale green crystals. Mohr’s salt has the advantage over ferrous sulphate in being not efflorescent and less readily oxidized in solid state. 1616. Anhydrous FeCl 3 is prepared by direct combination of metal with dry chlorine gas at high temperature: 2Fe ( s) + 3Cl 2 ( g ) → 2FeCl 3 ( s). Anhydrous FeCl 3 is yellow coloured solid, sublime below 300°C. When heated above 500°C, salt decomposes into FeCl 2 and Cl 2 . Salt is soluble in water and its aqueous solution is brown coloured due to presence of colloidal Fe(OH) 3 . FeCl 3 + 3H 2O Fe(OH) 3 + 3HCl. On adding excess of HCl to the aqueous solution of salt, yellow coloured is restored. Hexahydrated FeCl 3 ⋅ 6H 2O is prepared by dissolving metal in dilute HCl. The resulting solution containing FeCl 2 is treated with chlorine gas to oxidise FeCl 2 to FeCl 3 . Finally solution is evaporated to crystallization on a steam bath. 1617. Tin is extracted from cassiterite ore which contain small amount of SnO 2 .The extraction involves roasting of ore followed by smelting with coke in blast furnace: SnO 2 + 2C → Sn + 2CO

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The tin obtained here is always accompanied with some iron and tungeston as impurity which are removed by liquation. 1618. (a) Sn + 2Cl 2 → SnCl 4 (b) 4Sn +10HNO 3 → 4Sn(NO 3 ) 2 + NH 4 NO 3 + 3H 2O (c) Sn + 4HNO 3 → H 2SnO 3 + 4NO 2 + H 2O (Metastannic acid)

(d) Sn + 2H 2SO 4 → SnO 2 + 2SO 2 + 2H 2O (e) Sn + 2NaOH → Na 2SnO 2 + H 2 (stannite)

1619. Extraction of Cu from copper pyrite involves the following steps: (i) Roasting : Pyrite is converted into cuprous sulphide and iron sulphide as: 2CuFeS 2 + O 2 → Cu 2S + 2FeS + SO 2 2FeS + 3O 2 → 2FeO + 2SO 2 . (ii) Smelting : Roasted ore is mixed with coke and sand and smelted in blast furnace where FeO combines with SiO 2 forming easily fusible slag FeO + SiO 2 → FeSiO 3 (slag) (iii) Self-reduction (Bessemerization) : To the molten Cu 2S, blast of hot air is passed in a controlled manner so that Cu 2S is converted partially into Cu 2O and supply of air is cut-off after some time. Now, more roasted Cu 2S is added and heated in absence of oxygen where the following self reduction reaction occur: 2Cu 2O + Cu 2S → 6Cu + SO 2 Copper obtained by this method have blistered surface due to evolution of SO 2 , and thus named as blistered copper. 1620. Moist air containing some CO 2 attacks metal slowly forming a green basic carbonate as: 2Cu + CO 2 + O 2 + H 2O → Cu(OH) 2 ⋅ CuCO 3 (Malachite green)

1621. Copper being less electropositive, below to hydrogen in electrochemical series, can’t displace H + ion from solution. 1622. When copper is heated in oxygen atmosphere, at temperature up to 1100°C, a black cupric oxide CuO is formed. If temperature is maintained above 1100°C, red cuprous oxide Cu 2O is produced. T < 1100°C

O

2 Cu + O 2 → CuO(black) → Cu 2O(red)

T > 1100°C

1623. (a) 3Cu + 8HNO 3 → 3Cu(NO 3 ) 2 + 2NO + 4H 2O 1

(b) Cu + 2HCl + 2 O 2 → CuCl 2 + H 2O O

2 (c) 2Cu + 2H 2SO 4 → 2CuSO 4 + 2H 2O

(d) Cu + 4HNO 3 → Cu(NO 3 ) 2 + 2NO 2 + 2H 2O (e) Fe 2 (SO 4 ) 3 + Cu → 2FeSO 4 + CuSO 4 1624. Ore containing PbS is partially oxidized by heating in limited supply of air. The supply of air is cut off after some time and excess of roasted ore is added where the following self-reduction occurs: PbS + 2PbO → 3Pb + SO 2 .

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1625. (i) Pb is unaffected by dry air, but in presence of moisture, it get tarnished due to formation of Pb(OH) 2 first and finally into PbCO 3 which protect it from further reaction: CO

2 Pb + air → Pb(OH) 2 → PbCO 3 . (ii) At temperature below 400°C, Pb combine with oxygen forming litharge (PbO). If the temperature above 400°C is maintained, red Pb 3O 4 is produced.

O

Heat 2 2Pb + O 2 → 2PbO → Pb 3O 4 Litharge

450°C

Red-lead

Heat

1626. Pb + 2H 2SO 4 → PbSO 4 + SO 2 + H 2O PbSO 4 forms an insoluble layer on the metal surface, which causes reaction to stop after some time. 1627. (a) 3Pb + 8HNO 3 → 3Pb(NO 3 ) 2 + 2NO + 4H 4O (b) Pb + 4HNO 3 → Pb(NO 3 ) 2 + 2NO 2 + 2H 2O 1 (c) Pb + 2NaOH + 2 O 2 → Na 2 PbO 2 + H 2O 1

(d) Pb + 2CH 3COOH + 2 O 2 → (CH 3COO) 2 Pb + H 2O 1628. Silver is extracted from its sulphide ore by cyanide process. Ore containing Ag 2S is first leached with an aqueous solution containing NaCN, where insoluble silver sulphide comes in solution by forming soluble argentocyanide complex. Ag 2S + 4NaCN → 2Na[Ag(CN) 2 ] + Na 2S Soluble

The aqueous solution is then treated with a more electropositive metal Zn where Ag is produced : Zn + 2Na[Ag(CN) 2 ] → 2Ag + Na[Zn(CN) 4 ] 1629. AgBr act as a light sensitive material and used for making photographic film. Light rays reduces Ag + into Ag. Unreacted AgBr, where light has not fallen, is finally washed away by Na 2S 2O 3 solution: AgBr + 2Na 2S 2O 3 → Na 3 [Ag(S 2O 3 ) 2 ] + NaBr Fixer

1630. Ag reacts slowly with trace of H 2S in air, forming a black layer of Ag 2S on surface: 2Ag + H 2S → Ag 2S + H 2 Black

1631. (a) Ag + 2HNO 3 → AgNO 3 + H 2O + NO 2 (b) Ag + 2H 2SO 4 → Ag 2SO 4 + SO 2 + H 2O (c) AgBr + 2NaCN → Na[Ag(CN) 2 ] + NaBr (d) AgCl + 2NH 4OH → [Ag(NH 3 ) 2 ]Cl + 2H 2O 1632. Gold (Au), being a noble metal, exist in nature in free state. Powdered ore containing Au is leached with aqueous solution of KCN in presence of oxygen where Au comes in solution by forming soluble complex [Au(CN) 2 ]− : 4Au + 8KCN + 2H 2O + O 2 → 4K[Au(CN) 2 ] + 4KOH

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Solutions

The complex is then treated with electropositive metal Zn and Au comes out of solution in a redox-displacement reaction as: Zn + 2K[Au(CN) 2 ] → K 2 [Zn(CN) 4 ] + 2Au. 1633. Au reacts with aqua-regia forming chloro auric acid as: 2Au +11HCl + 3HNO 3 → 3NOCl + 2HAuCl 4 + 6H 2O 1634. Both group I metals and copper group metals have one ‘ s’ electron in their outer orbital, but copper group metals have ten d-electrons in the penultimate orbit, giving very poor screening effect. The poor screening gives elements of copper group very small size and high density, high ionization energies, etc. Also, in Cu-group, d-electrons are involved in metallic bonding. Thus, the melting points and enthalpy of sublimation are much higher than for alkali metals. All this add up in order to make elements of Cu-group noble. 1635. MnO 2 is first fused with KOH in presence of oxygen, where it is converted into potassium manganate K 2 MnO 4 as: 1

MnO 2 + 2KOH + 2 O 2 → K 2 MnO 4 + H 2O K 2 MnO 4 is then converted into KMnO 4 by one of the following technique: (i) Passing CO 2 ( g ) : 3K 2 MnO 4 + 2CO 2 → 2K 2CO 3 + 2KMnO 4 + MnO 2 (ii) By the action of dilute H 2SO 4 : 3K 2 MnO 4 + 2H 2SO 4 → 2K 2SO 4 + 2KMnO 4 + MnO 2 + 2H 2O (iii) By the action of ozone: 2K 2 MnO 4 + O 3 + H 2O → 2KMnO 4 + 2KOH + O 2 . 1636. Chrome-iron is first heated with Na 2CO 3 to produce Na 2CrO 4 as: 7

2FeO ⋅ Cr 2O 3 + 4Na 2CO 3 + 2 O 2 → 4Na 2CrO 4 + Fe 2O 3 + 4CO 2 Roasted ore is extracted with water where Na 2CrO 4 dissolve leaving impurities behind. Na 2CrO 4 solution is then treated with concentrated sulphuric acid where Na 2SO 4 ⋅ 10H 2O separate out first followed by crystals of Na 2Cr 2O 7 . Finally aqueous solution of Na 2Cr 2O 7 is treated with KCl where K 2Cr 2O 7 separate out due to lower solubility than Na 2Cr 2O 7 : Na 2Cr 2O 7 + 2KCl → K 2Cr 2O 7 + 2NaCl. Heat

1637. (a) 4K 2Cr 2O 7 → 4K 2CrO 4 + 2Cr 2O 3 + 3O 2 (b) K 2Cr 2O 7 + 2KOH → 2K 2CrO 4 + H 2O (c) K 2Cr 2O 7 + 4KCl + 3H 2SO 4 →

2CrO 2Cl 2 ↑

(chromyl chloride)

+ 3H 2O + 3K 2SO 4

1638. (a) 2KMnO 4 + 5K 2C 2O 4 + 8H 2SO 4 → 2MnSO 4 + 6K 2SO 4 +10CO 2 + 8H 2O (b) 2KMnO 4 + 5O 3 + 3H 2SO 4 → 2MnSO 4 + K 2SO 4 +10O 2 + 3H 2O (c) 6KMnO 4 +10FeC 2O 4 + 24H 2SO 4 → 5Fe 2 (SO 4 ) 3 + 6MnSO 4 + 20CO 2 + 3K 2SO 4 + 24H 2O (d) 2KMnO 4 +16HCl → 2MnCl 2 + 5Cl 2 + 2KCl + 8H 2O

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710

1639. (a) In gaseous state, FeCl 3 remains in dimeric state as: Cl

Fe

Cl

Cl Cl

Cl

Fe

Cl

(b) In ether, FeCl 3 remains in solvated monomeric form as: Et Et

O •• → FeCl 3

(c) In water, it forms a hexahydrated complex as: FeCl 3 + H 2O → [Fe(H 2O) 6 ]3+ + 3Cl – . 1640. Iodide ion (I − ) being a strong reducing agent, reduces Fe 3+ into Fe 2+ as Fe 3+ + I − → Fe 2+ + 2 I 2 1

1641. Freshly prepared solutions of K 2SO 4 and Al 2 (SO 4 ) 3 are mixed together where potash alum crystallizes as: K 2SO 4 ( aq ) + Al 2 (SO 4 ) 3 ( aq ) → K 2SO 4 ⋅ Al 2 (SO 4 ) 3 ⋅ 24H 2O Potash olum

Potash alum is mainly used as a mordant in dyeing.