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Copyright of this book is reserved by Gujarat Secondary and Higher Secondary Education Board, Gandhinagar. No reproduction of this book in whole or in part, or in any form is permitted without written permission of the Secretary, Gujarat Secondary and Higher Secondary Education Board, Gandhinagar.

Gujarat Secondary and Higher Secondary Education Board, Gandhinagar

QUESTION BANK CHEMISTRY

Price

: ` 60.00

Published by : Secretary Gujarat Secondary and Higher Secondary Education Board, Gandhinagar

I

Contribution 1

Dr. Hasmukh Adhiya (IAS)

Principal Secretary , Education Department Gandhinagar

2

Shri R. R. Varsani (IAS)

Chairman , G.S&H.S.E. Bord, Gandhinagar

3

Shri H. K. Patel (G.A.S)

Dy. Chairman, G.S&H.S.E. Bord, Gandhinagar

4

Shri M. I. Joshi (G.E.S)

Secretary , G.S&H.S.E. Bord, Gandhinagar

Coordination 1

Shri B. K. Patel

O.S.D., G.S&H.S.E. Bord, Gandhinagar

2

Shri D. A.Vankar

Assistant Secretary (Retd.), G.S&H.S.E. Bord, Gandhinagar

5

Shri M. P. Parmar

Assistant Secretary, G.S&H.S.E. Bord, Gandhinagar

Expert Teachers 1.

Shri C. I. Patel (Convenor)

Shri Vidyanagar High School, Ahmedabad

2.

Shri S. B. Gor (Co-Convenor) Ghyanda Girls High School, Ahmedabad

3.

Shri A. I. Patel

4.

Shri V. R. Patel

5.

Shri B. R. Patel

Muktjeevan Vidhyalaya, Ahmedabad

6.

Shri K. K. Purohit

M. K. Higher Sec. School, Ahmedabad

7.

Shri M. B. Patel

New Vidhyavihar for Girls, Ahmedabad

8.

Shri B. A. Nayak

Swaminarayan High School, Ahmedabad

9.

Shri H. M. Patel

Navchetan High School, Ahmedabad

10. Shri S. B. Suthar

R.P.T.P. Science School, Vallabh Vidhyanagar

11. Shri R. N. Patel

R.P.T.P. Science School, Vallabh Vidhyanagar

12. Shri N. N. Shah

Best High School, Ahmedabad

13. Shri J. Y. Mehta 14. Shri I. B. Amlani 15. Smt. M. N. Shethiya 16. Smt. H. N. Nayak 17. Smt. P. S. Thakar

R.P.T.P. Science School, Vallabh Vidhyanagar

18. Shri G. S. Patel 19. Shri M. L. Sharma 20. Shri H. K. Patel

II

P R E FA C E Uptil now , the Students had to appear in various entrance examinations for engineering and medical courses after std-12. The burden of examinations on the side of the students was increasing day-by-day. For alleviating this difficulty faced by the students, from the current year, the Ministry of Human Resource Development , Government of India, has Introduced a system of examination covering whole country. For entrance to engineering colleges, JEE(Main) and JEE(Advanced) examinations will be held by the CBSE. The Government of Gujarat has except the new system and has decided to follow the examinations to be held by the CBSE. Necessary information pertaining to the proposed JEE (Main) and JEE(Advanced) examination is available on CBSE website www.cbse.nic.in and it is requested that the parents and students may visit this website and obtain latest information – guidance and prepare for the proposed examination accordingly. The detailed information about the syllabus of the proposed examination, method of entrances in the examination /centers/ places/cities of the examinations etc. is available on the said website. You are requested to go through the same carefully. The information booklet in Gujarati for JEE( Main) examination booklet has been brought out by the Board for Students and the beneficieries and a copy of this has been already sent to all the schools of the state. You are requested to take full advantage of the same also However, it is very essential to visit the above CBSE website from time to time for the latest information – guidance . An humble effort has been made by the Gujarat secondary and Higher Secondary Education Boards, Gandhinagar for JEE and NEET examinations considering the demands of the students and parents , a question bank has been prepared by the expert teachers of the science stream in the state. The MCQ type Objective questions in this Question Bank will provide best guidance to the students and we hope that it will be helpful for the JEE and NEET examinations. It may please be noted that this “Question Bank” is only for the guidance of the Students and it is not a necessary to believe that questions given in it will be asked in the examinations. This Question Bank is only for the guidance and practice of the Students. We hope that this Question Bank will be useful and guiding for the Students appearing in JEE and NEET entrance examinations. We have taken all the care to make this Question Bank error free, however, if any error or omission is found, you are requested to refer to the text – books.

Date: 02/ 01/ 2013

M.I. Joshi Secretary

III

R.R. Varsani (IAS) Chairman

INDEX UNIT 15 P - BLOCK ELEMENTS

1

UNIT 16 D & F BLOCK ELEMENTS

48

UNIT 17 CO-ORDINATION COMPOUNDS

63

UNIT 18 ENVIROMENTAL CHEMISTRY

76

UNIT 19 PURIFICIATION AND CHARACTERISATION OF ORGANIC COMPOUNDS

82

UNIT 20 BASIC PRINCIPLES OF ORGANIC CHEMISTRY

94

UNIT 21 HYDROCARBONS

128

UNIT 22 ORGANIC COMPOUNDS CONTAINING HALOGENS

154

UNIT 23 ORGANIC COMPOUNDS CONTAINING OXYGEN

173

UNIT 24 ORGANIC COMPOUNDS CONTAINING NITROGEN

190

UNIT 25 POLYMER

209

UNIT 26 BIOMOLECULES

220

UNIT 27 CHEMISTRY IN EVERY DAY LIFE

234

UNIT 28 CHEMISTRY PRACTICAL

241

IV

UNIT : 15 P - BLOCK ELEMENTS Important Points The elements (except He atom) having outermost valence shell electronic configuration ns2 np1 to ns2 np6 are called p-block elements. Since p-orbital can accommodate six maximum electrons, six groups 13 to 18 are there in p-block elements. We shall study group 13 and 14 i.e. Boron and carbon group elements in this unit. Generally in a row or period eletrongativity, ionisation enthalpy and oxidising power are increasing as the atomic number increases while in group it decreases as the atomic number increases. Generally, in group covalent radius, van der Waals radius and metallic character increases as the atomic number increases. It is a characteristic of the p-block elements that metal, non-metal and metalloid are included in the same group. The group 13 includes elements, Boron, Aluminium, Gallium, Indium and Thalium. Aluminium is the third most abundant element found in earth's crust. The important ores of aluminium are bauxite and cryolite. Variation in some properties of group 13 elements are like atomic radii and ionic radii, ionisation enthalphy, metallic character, electronegativity, melting point and boiling point, density, character as reducing and nature of compound are observed in elements of boron group i.e. Group 13 elements having electronic configuration ns2 np1. Hence, they possess +3 oxidation state and their stability goes on decreasing as the atomic number increases. These elements also possess +1 oxidation state and the stability goes on increasing as the atomic number increases. The chemical reactivity of group 13 elements are as given below. Group 13 elements do not combine directly with hydrogen but they combine indirectly to form hydride compounds. Boron forms a number of hydrides having molecular formula B n Hn+4 and Bn Hn+6 which are known as boranes. The other elements of this group form polymeric hydrides. Group 13 elements form MX3 type trihalides, where X = F, Cl, Br and I is unknown. AlCl3 exists in dimer form. The group 13 elements form oxides and hydroxides having fomula M2 O3 and M(OH)3 respectively. As the atomic number of elements goes on increasing, the acidic character of oxide and hydroxide goes on dcereasing. Elements of these group form octahedral complexes. Aluminium sulphate reacts and forms double salts with sulphate of NH +4 and alkali metal ions having fomula M2SO4 A2(SO4)3.24H2O or MAl(SO4)2.12H2O where M = Na+, K+, Rb+ and NH +4 . The first element (Boron) of group 13 shows anomalous behaviour. Boron is chemically less reactive and almost inert with normal chemical reagents at normal temperature but reacts with strong oxidising agents and with some typical reagents like non-metal, acid alkali and metals. Some important compounds of boron are borax powder, boric acid and boron hydride. The properties and uses of aluminium are as given in the text. Aluminium reacts with acid and base so it is amphoteric in nature. The group 14 includes elements carbon, silicon, germanium, tin and lead. The first element 1

of this group is carbon, so it is also known as carbon group elements. Carbon shows catenation property due to its small size, high electronegativity and very high carbon-carbon bond energy. Due to typical characteristic of catenation it forms number of compounds which are studied in organic chemistry. It also forms compounds with metals and nonmetals. Group 14 elements have variation in properties like atomic radii, ionisation enthalpy electropositive character, electronegativity, melting points and boiling points density, catenation and allotropy. The oxidation state of group 14 elements are +2 and +4. The trends in chemical reactivity are as given in the text. The carbon shows anomalous behaviour. The crystalline allotropes of carbon are diamond, graphite and fullerenes. The physical properties and chemical properties are as given in the text. Some important compounds of carbon are halides of carbon, carbon disulphide, carbide compounds, carbon monoxide and carbon dioxide, and the method of preparation and uses are as given in the text. The important compounds of silicon are silicon hydrides having general formula Si n H2n+2. The –1 value of bond enthalpy for silicon-silicon is 297 kJmol hence catenation character is observed in silicon and it forms limited hydride compounds, having formula Si n H2n+2 where n = 1 to 8, and these compounds are also known as silanes. The stability of silane compounds are less compared to hydrides of carbon and hence reducing power is more. Silicon also forms silicon dioxide known as silica and more than 22 allotropic structures are known in which some are crystalline and some are amorphous. Silica is acidic and hence it dissolves in liquid alkali or alkaline carbonate to form silicate compounds. Silicon reacts with only F2 and form SiF4 while reacts with Cl2 and form SiCl4. The hydrolysis of SiCl4 gives silicic acid and the mechanism for hydrolysis are in two steps. The silicones are synthetic materials containing Si  O  Si bond linkage. These compounds are polymeric substances containing R2SiO reapeating unit. The general formula is (R2 SiO)n, where R is methyl or phenyl group. The empirical formula is R2SiO which is similar to that of organic compound, ketone, so it is called silicone. The prepartion, properties and uses are as given in the text. Approximately 95% of earth's crust consists of silicates and silica compounds, containing independent SiO 44 having tetrahedral structure. The types of silicates depending upon the number of corners (0, 1, 2, 3 and 4) of the SiO 44 tetrahedron are shared with other tetrahedrons and based on that they are classified as given in the text. In a three dimensional structure of SiO2, its Si4+ partially substituted by Al3+ gives aluminosilicate are called feldspar and zeolites. In zeolites the SiO 44 and AlO54 . tetrahedron joined together in simple way to form three dimensional network. The uses of silicates are as a molecular sieves and shape selectives catalyst. One important catalyst of silicate is ZSM-5 used in petrochemical industry which converts alcohol directly into gasoline.

2

The elements of groups 13 to 18 in the periodic table are known as p-block elementes. The general electronic configuration of these elements is ns2np1–6. We have studied about the elements of groups 15, 16 ,17 and 18 in this unit. General introduction of elements of groups 15, 16, 17, 18 group 15

group 16

group 17

group 18

Nitrogen

Chalogens or

Halogen Group

Noble Gas

Identification

group

Oxygen group

Electronic configuration

ns2np3

ns2np4

ns2np5

ns2np6

N

O

F

Ne

–3, to +5

–2, –1, +1, +2

–1

–

P, As

S, Se, Te

Cl, Br, I

Xe

–3, +3, +5

–2, +2, +4, +6

–1, +1, +3, +5, +7

+2, +4, +6, +8

Sb, Bi +3, +5

Po +2, +4

– –

– –

Common name/

Group

of valence shell

Oxidation state

18 15

16

17

He

2p

N

O

F

Ne

3p

P

S

Cl

Ar

4p

As

Se

Br

Kr

5p

Sb

Te

I

Xe

6p

Bi

Po

At

Rn

Atomic radius, metallic character

Electronegativity, Ionisation enthalpy

Atomic radius, metallic character

The periodicity in properties of elements of groups 15,16,17,18. The first element of group 15, differs in many aspects from the other elements in the group. The reason for which is its smaller size, the capacity of formation of p–p triple bond between, nitrogen atoms and the nonavailability of d–orbitals. As we go down from above in the group, variations are found in properties. Dinitrogen (N2) can be prepared in the laboratory as well as on commercial level. The oxides of nitrogen element are N2O, NO, N2O3, N2O4 and N2O5 which possess resonance forms. Ammonia and nitric acid are compounds of nitrogen. Phosphorus element exists as P 4 molecule. It has many allotropes. It forms hydrides, halides and oxoacid compounds.

Polonium element of group 16 is radioactive. Oxygen forms metal oxides with metals. Ozone is a strong oxidising agent. Sulphur element possessess different allotropes. Out of these  and  allotropes are very important. Sulphur element combines with oxygen and forms oxide compounds like SO2 and SO3. Out of the different oxoacids of sulphur, sulphuric acid is very important. It is called 'King of Chemicals.' Astatine element of group 17 is radioactaive. As these elements require one electron to have stable electronic configuration, they are very reactive. As a result of this, the elements of this group are not available in free state, but are availalbe in the combined state as negative ions. The elements of this group form oxides, hydrogen halides, interhalogen compounds and oxoacid compounds. Radon element of group 18 is radioactive. As the octet structure is complete in all the elements of this group they are chemically inert. Xenon element of this group, under specific reaction conditions combine with fluorine and oxygen elements and form fluoride and oxide compounds.

3

M.C.Q. Boron Family (1)

(2)

Boron form covalent Compound due to (a) Higher ionisation enthalpy (c) Small size In diborane the Two H-B-H angles are nearly

(b) Lower ionisation enthalpy (d) Both (a) and (c)

(a) 60 , 120 (b) 95 ,120 (c) 95 ,150 (d) 120 ,180 (3) The stability of +1 oxidation state increases in the sequence (a) Al III > I > II

(b) I > II > IV > III (c) II > I > IV > III (d) IV > I > III > II

(123) n- Heptane Aromatization [Al 2 O3 + Cr2O3 ] (a) Benzene

?

(b) Toluene

(c) Hept-2-ene

(d) Hept-1-ene

(124) Benzene is obtained by heating phenol with X then X is (a) Zn dust

(b) Soda lime

(c) Sodium hydroxide(d) My / ether

(b)

(c)

(125) Which is aromatic (a)

(d) -

+

139

-

(126) In presence of light and heat toluene chlorinated and react with aq. NaOH to give (a) O - Cresol

(b) P - Cresol

(c) Mixture of (A) and (B)

(d) Benzoic acid

(127) Sulphonation of compound A followed by fusion with NaOH gives mixture of O- cresol & pcresol. compound A is (a) Benzene (c) phenol CH3

(b) Toluene (d) Benzene sulphonic acid CH3

CH COCl [ Anhy. Alcl3 ]

3 ¾¾¾¾¾¾¾ ®

(128)

(a) Kolbe`s reaction

+ HCl This reaction is known as COCH3

(b) Sandmayer reaction

(c) Diels-(Alder) reaction

(d) Fridel-craft`s Acylation

(129) In the following reaction, the product R is hot iron 3 2O ®R ® Q ¾¾¾ CaC2 ¾H¾ ¾ ® P ¾¾¾ AlCl3 tube CH Cl

(a) Benzene

(b) Toluene

(c) Chorobenzene

(d) Xylene

(130) Indentify the product (E) in the following sequence of reaction CH3 Br2 / HCl KM n O 4 NaNO 2 / HCl ¾¾ ® A ¾Sn¾¾ ® B ¾¾¾¾ 3 PO2 ¾ ® C ¾H¾ E ¾® D ¾¾¾® KOH 0 -5 C

NO2

COOH

(a)

(b)

CH3

COOH

COOH Br

Br

(c)

Br

(d)

NO2

NO2

Assertion and Reason type Questions Direction:- In each of the following Questions Read the Assertion(A) and Reason (R) carefully. Choose Correct option as under and darken trhe bubbles in OMR. (A)

If both A and R are true and R is the Correct explanation of A

(B)

If both A and R are true but R is not the Correct explanation of A

(C)

If A is true but R is false

(D)

If both A and R are false

(E)

If A is false but R is ture

(131) A : Pyrrole is an aromatic heterocyclic compound R : It has a cyclic, delocalised 6p electrons (132) A : CH4 does not react with Cl2 in dark R : Chlorination of CH4 takes place in Sunlight

140

(133) A : Neopentane forms only one monosubstitated Compound R : Neopentane has high bond energy (134) A : Addition of HBr to propene in presence of peroxide gives 1- bromo propane. R : The reaction occurs by Free radical mechanism (135) A : Styrene on reaction with HBr gives 2-bromo-2-phenylethane R :Benzyl radical is more stable than alkyl radicals (136) A : Butane-1-ol on dehydration Conc. H2 SO4 gives mainly But-1-en R :Dehydration occurs through Carbocation intermediate (137) A :Melting point of neopentane is higher than that of iso-pentane R : Neopentane Contains a quaternary carbon (138) A : Acetylene react with NaNH 2 to give Sodium acetylide and ammonia R : SP-hybridized carbon of acetylene are considerably electronegative (139) A : Addition of H 2 O to acetylene occurs in presence of dil. H2 SO4 and HgSO4 to give acetaldehyde

R : It is an example of electrophilic addition reaction (140) A : Addition of HI to vinyl chloride produces 1-chloro1-iodoethane R :HI adds to vinyl chloride againt markownikov`s Rule Matrix Match type Questions Direction :- Match the entries of column I with entries of column II. Each entry of column I may have one or more correct matching from column II. if the correct matches are A ® P, S , B ® r , C ® p , q D ® s then the correctly bubbled 4 ´ 4 matrix should be as below.

p

q

r

s

A B C D

141

(141) Match the reactions in column I with approp type of reaction as given in column II column I CH3

(A)

CH3

CH3

Cl2 [Fe Cl3]

column II

Cl

(p) Substitation Reaction

+ Cl

( i) Alc ×K O H ( ii) H 2/ N i

¾ ¾ ¾ ¾ ¾ ¾ ¾¾ C®H3

(B) C H 3 C H 2 C l (C)

C H3 -

(q) Addition Reaction

( i) H SO / H gSO ( ii) [ AlC l ]3

(r) Elimination Reaction

4 H C = C H ¾ ¾ ¾ ¾2 ¾ 4¾ ¾ ¾ ¾¾ ® CCHH O3

CH = CH2

(D)

(i) C H Cl/ [ AlCl ] ¾ ¾ ¾2¾ 5¾ ¾ ¾ ¾ 3¾ ¾®

(s) Rearrangement

(ii) ZnO / 903 K

(142)

column I

column II

(A)-CHO

(P) O/P Director

(B)-OH

(Q) Activating Benzene Ring

(C)-NH 2

(R) m-Director

(D)-CI

(S) Deactivating Benzene Ring

(143)

column I

column II

(A) Benzene

(p) Wurtz reaction of C2 H5Cl

(B) Ethene

(q) Evolves H2 when heated with sodium metal

(C) Ethyne

(r) Dehydration of ethanol

(D) Butane

(s) Electrophilic substitation

Comprehension type Questions Direction :- Question numbers 144 and 145 are based on the following paragraph. Each question has 4 options (A),(B),(C) , (D) out of which ONLY ONE is correct. choose the correct option. Paragraph :- Cyclohexene on ozonolysis following by reaction with Zinc dust and water gives aldehyde(P). Compound (p) on further treatment with aqueous KOH yields compound(Q). (144) The structure of compound (P) is

(a)

CHO

(b)

CHO

H C=O C=O H

O

CHO

(c)

CHO

(d)

O

(145) The structure of compound (Q) is OH

(a)

CHO

(b)

(c)

CHO

142

COOH

(d)

Answer key 1C 2D 3A 4D 5C 6B 7C 8B 9C 10 C 11 A 12 B 13 B 14 B 15 C 16 B 17 A 18 B 19 A 20 B 21 C 22 A 23 B 24 B 25 B 26 C 27 C 28 B 29 B 30 D 31 B

32 A 33 D 34 B 35 B 36 C 37 A 38 D 39 B 40 B 41 D 42 B 43 C 44 C 45 C 46 B 47 C 48 A 49 B 50 C 51 A 52 D 53 C 54 B 55 C 56 B 57 A 58 B 59 B 60 C 61 A 62 D

63 C 64 B 65 C 66 C 67 C 68 A 69 B 70 D 71 C 72 B 73 C 74 C 75 C 76 C 77 B 78 B 79 C 80 A 81 C 82 D 83 C 84 B 85 A 86 A 87 B 88 C 89 A 90 D 91 C 92 A 93 C

94 B 95 C 96 D 97 C 98 A 99 C 100 A 101B 102 D 103 B 104 C 105 D 106 C 107 A 108 B 109 D 110 B 111 A 112 C 113 D 114 A 115 B 116 D 117 D 118 B 119 C 120 B 121 C 122 A 123 B 124 A

143

125 C 126 D 127 B 128 D 129 B 130 B 131 A 132 B 133 C 134 A 135 D 136 E 137 B 138 A 139 B 140 C 141 A. (p), B. (r ), C. (q), D. (p) 142 A. (r ), B. (p), C. (p, q), D. (p) 143 A. (s), B. (r ), C. (q), D. (p) 144 B 145 A

Hints (1)

Wurtz Reaction :- 2 CH3CH2 Cl + 2 Na ¾Ether ¾ ¾® CH3 - CH2 - CH2 - CH3 + 2 NaCl

(2)

Alkyne Cn H2n - 2 Þ C4 H6

(3)

CH3CH 2COONa + NaOH •

(4)

¾ ¾ ¾cao ¾ ¾ ¾® soda - lime

CH3 - CH3 + Na 2 CO3

Sodalime Decarboxylation of Sodium Salt of Carboxylic acid gives alkane containig one carbon less than the carboxylic acid

Electrolysis 2 CH3CH2 COONa + + 2 H2 O ¾¾¾¾ ® CH3CH2 CH2 CH3 + 2 CO2 + H2 + 2 Naot

•

Electrolysis of aqueous solution of Sodium Salt of Carboxylic acid gives Alkane Containing even number of carbon (Kolbe`s synthesis)

(5)

This preparation is known as Kolbe`s synthesis

(6)

1 Boiling a Molecular mass a number of branches

(7)

According to Que-6

(8)

2O CH3COCl + CH3 MgCl ¾H¾ ¾ ® CH3COCH3 + MgCl 2

(9)

This reaction is carried out by Free Radical intermidiate.

(This reaction gives Ketone)

(10) R - X + 2 Na + R - X ¾Ether ¾ ¾® R - R + R - R + R - R (11) C2 H5 O H + CH 3 Mg Br ® CH4 + Mg

Br Oc2 H5

(12) LPG is a mixture of mainly propane and butane electrolysis (13) 2CH3COONa + + 2H2O ¾¾ ¾® CH3 - CH 3 + 2CO2 + 2 NaoH + H2

(14) Que-6 1

2

CH3

(15) CH3 - CH2 C 3

4

5

CH2 - CH3

3, 3 – dimethyl pentane

CH3

(16) The relative stability of conformation isomers of Alkane like Ethane, propane staggered (Anti) > Skew or Gauche > partial eclipsed > Full eclipsed (17) The relative stability of conformation isomer of cyclohexane is Chair > twist boat >boat >half Chair (18)

Alkane Butane pentane Hexane Heptane structural isomers

2

3

(19) 2R - X + Zn ® R - R + ZnX 2

5

9

(Frankland`s Reaction)

Pd /c (20) C2 H 5Cl + H 2 ¾¾¾ ® C2 H 6 + HCl

(21) when Higher hydrocabon is heated it decomposes to Lower hydrocarbon (alkane, alkene). This reaction is known as Cracking or Pyrolysis

144

(22) Methane is known as mars gas (23) because 2  carbonium ion is more stsble than 1 (24)

CH3 - CH 2 - CH 2 CH - CH 3 CH 3

isopropyl group *

*

hv CH 3 - CH - CH 2 - CH 3 + Cl 2 ¾¾¾ ® CH 2 - CH - CH 2CH 3 + CH 3 - CH - CH - CH 3 (25) CH 3 Cl CH 3 Cl CH 3 (R + S)

(R + S)

Star (*) indicate chiral (assymetric) carbon. Here in both product R and S configurations are possible so that total 4 chiral compound are possible. (26) C3 H8 + 502 ® 3CO2 + 4 H2 O Zn - Cu CH3 CH CH3 ¾HCl ¾® CH3 - CH - CH 3 ¾¾¾¾® C CH 3CH 2CH 3 + HCl 2 H 5 OH (27) 2[ H ] OH Cl

(28)

( CH 3CH 2 )2 LiCu + CH 3CI ® CH 3CH 2CH 3 + CH 3CH 2Cu + LiCl Lithium diethyl copper This is known as Core-House reaction. It is suitable method for preparation of odd carbon Alkyne 1

¾® (29) 2CH3 CH CH3 + 2 Na ¾ ¾ Br ether

iso-propyl bromide CH3

3

4

2,3- dimethyl butane CH3 Br

+ Br2 ®

(30)

2

CH3 - CH - CH3 + 2NaBr CH3 - CH - CH3

+ HBr

Methyl cyclo pentane

1- bromo-1-methyl cyclopentane

(31) Alkene shows mainly electrophilic addition reactions. (32) CH2 = CH2 + Br2 ¾CCl ¾ ¾4 ® Br - CH2 - CH2 - Br b

µ

.KoH (33) C H C H CI ¾Alc ¾¾ ® CH2 = CH2 + KCI + H2 O 3 2

( b - Elimination reaction) b

(34) CH3 CH2

µ

b

Conc.H SO

2 4® CH - CH3 ¾¾¾¾¾¾ 160° c OH

CH3 - CH = CH - CH3 + CH3 - CH2 - CH = CH2

Dehydration of butane-1-ol or 2-ol gives trans-but-2-en as the major product. According to Saytzeff rule But-2-ene is more Substituted alkene and therefore it will be major. KMnO4 / KOH ¾® CH 2-- CH 2 (Baeyer`s test for unsaturations) (35) CH2 = CH2 ¾¾¾¾ OH OH

145

(36) CH3 - CH = CH2 + HCI ® CH3 - CH - CH3 (Markownikoff’s Rule) Cl (37) CH3CH = CH2 + HBr ¾Peroxide ¾¾® CH3 - CH2 - CH2 - Br (Antimarkownikoff `s Rule) (38) C6 H5CH2 - CH = CH2 + HBr ¾Peroxide ¾¾® C6 H5 - CH2 CH2 Br (Anti Markownikoff) (39) The disappearace of the purple colour of KMnO 4 in its reaction with an alkene is the test for unsaturation (double bond). is known as Baeyer`s Test. dil × H SO CH2 = C - CH 2CH3 + H2 O ¾¾¾¾¾¾¾® CH 3 Hg + 2 (40) CH3 2

4

CH 3 C CH 2CH3 OH

R - CH CH - R dil × aq. KMnO 4 ® (41) R - CH = CH - R ¾¾¾¾¾¾¾¾¾ Room temp. OH OH

Conc. KMnO 4 (42) R - CH = CH - R ¾¾¾¾¾¾¾¾ ® Heat

(Markownikoff)

glycol

R – COOH + R - COOH

(43) Ozonolysis CH2 = CH2 + O3 ®

(44)

CH2

Alk . KmnO4 ¾¾¾¾¾¾¾ ®

CH2

Zn 2HCHO ¾¾¾¾ HO ® 2

OH

OH

Acidic (45) CH 2 = CH2 + 2 [ 0] ¾¾¾¾¾ KMnO4 ® HCHO + HCHO

(46)

CH 2 - CH 2 + Zn ¾ ¾D ® CH2 = CH2 + ZnBr2 Br Br

(47) CH3 - CH = CH - CH2 - CH3 + HBr ® CH3 - CH2 - CH - CH2 CH3 + CH3 CH - CH 2 CH2CH3 Br Br

(48) According to Que-43 (49)

CH2 = CH2 + HOCI ® CH2 - CH2 OH Cl

.KOH (50) CH2 = CH2 ¾HBr ¾® CH3 CH2 Br ¾aq¾ ¾® CH3 CH2OH

CHI3 Iodoform

(51) According to Que-35 +

H+ = CH - CH 3 (52) CH3 - CH - CH CH2 ¾¾® CH3 - CH -(2 CH3 CH3 °)

1, 2 ¾¾¾¾¾¾¾ Hydride shift ®

+

Br CH3 CH - CH2 CH 3 Br CH C CH2 – CH3 ¾ ¾ ® 3 CH3 (3°) CH3

product will be major because 3 - Carbonium is more stable than 2  146

(53) According to Markownikoffs Rules Peroxide (54) CH3 CH = CH 2 + HCl ¾ ¾¾® CH3 - CH - CH 3 Cl

This reaction follows markownikoff`s rule. Because peroxide effect or Anti markownikoff`s rule is applicable only to HBr. It is not for HCl and HI. (55) Ozonolysis of alkene gives two carbonyl group (C=0) for cleavage of each > C = C < s (i ) O3 CH3CHO + CHO + HCHO CH3 - CH = CH - CH = CH2 ¾¾¾¾¾¾¾ (ii ) Zn / H 2O ® CHO (56) 3R - CH = CH 2 +

1 B2 H 6 2

H 2O2 ( R - CH 2 - CH2 ) 3 B ¾¾¾¾¾ OH - ®

– N-2, OH – HCl 3 R - CH2 - CH 2 + (57) CH3 - CH = CH2 + N OCI ® CH3 - CH - CH2 Cl NO

According to markownikoff addition of NOCl (58)

2 CI 2 CH2 CH3 ¾SO ¾¾ ®

(59) CH3

.koh CH2 CH2 CI ¾Alc ¾¾ ®

CH3 KOH C CH2 - Br ¾alc ¾.¾ ® CH3 CH3

CH = CH 2 + KCl + H 2O

CH3 + Methyl C CH2 ¾¾¾¾¾ ® Shifting ¾ CH3

1 carbonium (less stable) CH3

CH3 CH3 E limination of C CH2 - CH3 ¾¾¾¾¾¾¾¾ ¾ ® b - Hydrogen

+

C = CH - CH3 CH 3

3 carbonium (morestable) CH2 CH (60) | CH

+

CH2 200° c ® || ¾¾¾¾ CH2

(Diels-Alder Reaction)

CH2

(61)

SP 2

SP 2

SP

SP

Þ CH2 = CH - C º CH

Hybridization is considered by no. of s - bonds.

(62) CH2 = CH2 + Br2 ® BrCH2 - CH2 - Br (Brown)

(Colour less)

Disappearance of brown colour is the test for unsaturation (63) 2CHCI 3 + 6 Ag ® HC º CH + 6 AgCl (64) CH3 CH Cl2 + 2 alc. KOH (65)

HC º CH + 2 KCl + 2 H2O

Liq . NH 3 ¾® CH3 – C º CH + 2 NaBr + 2 H2O CH3 CH – Cl 2 – Br + 2 NaNH 2 ¾ ¾¾ Br

147

(66)

D CH2 - CH2 + 2 KOH ( alc ) ¾ ¾ ® HC º CH + 2 KBr + 2 H2 O

Br

Br

(Vicinal)

(67) H – C – COOK ||

OR||

H – C – COOK Potasium Maleate (cis) (68) HC º CH

Acetylene H – C – COOK ¾¾¾¾¾ + 2H O ® 2

CH |||

2

+ 2CO 2 +2NaOH + H2

KOOC – C – H

CH

Potasium Fumrate (trans)

At anode

Na / liq.NH 3 3CH2 Br HC º C Na + ¾CH ¾¾ ¾® CH3 - CH2 - C º CH + NaBr 196 K Sodium Acetylide

(69) According to Que-68 Fe (70) 3CH º CH ¾¾® C2 H6 OR 

Benzene

Nil ( N )2 (71) 4CH º CH ¾¾¾¾ ¾ ® ( HC = CH ) 4 OR

Cyclooctutetruene

H 2 SO4 / Hg (72) CH º C – CH 2 – CH 3 + H 2 O ¾¾¾¾¾¾ ® CH2 = O – CH2 - CH3 OH +2

Tax tomerise ® CH –C– CH - CH ¾¾¾¾¾¾ 2 3 3 O

(73) CH º CH ¾HBr ¾® CH2 = CH - Br ¾HBr ¾® CH3 - CHBr2 Ethylidene bromide (74) CH º CH ¾HCN ¾ ¾® CH2 = CH - CN HOCI HOCI (75) CH º CH ¾ ¾¾® CI - CH = CH - OH ¾ ¾¾® H2 O Cl - CH - CH - OH ¾-¾ ¾® CHCI 2

Cl

At Cathode

OH

(76) CH3 - CH 2 - C º CH

CH = O KMnO4 / KOH

CH3CH2 COOH + CO2 + H2O

(terminal alkyne) Oxidation of terminal alkyne gives acid and CO 2 KMnO4 (77) CH3 - C º C - CH3 ¾¾¾¾ KOH ® CH3COOH + CH3COOH

(nonterminal alkyne) Oxidation of nonterminal alkyne gives mixture of two acids (78) CH º CH + CH3CH2 MgBr ® CH3CH3 + CH º CMgBr (79) According to Que-76

148

(80) HC º CH + O3 ® CH

CH

H–C–C–H || || O O

Zn/ H 2 O

¾ ¾¾®

Glyoxal (81) CH3 - C º C - CH3 + O3 ® CH 3 CH

3

– C – C – CH

/ H2 O ¾Zn¾¾ ® CH3 – C – C – CH3 || || O O

Buta -2-3-dione (Dimethyl Glyoxal) .KOH (82) CH3 CH2C º CH ¾Alc ¾¾ ® CH 3 - C º C - CH 3

Terminal alkyne undergo iso merisation to give non-terminal alkyne (83) CH3 - C º C - CH3 ® CH3 - CH2 - C º CH Non-terminal alkyne undergo iso merisation to give terminal alkyne Br 3 PO . 4 (84) CH3COOH ¾LiAlH ¾¾4 ® CH3CH2 OH ¾H¾¾ ® CH2 = CH 2 ¾¾2 ® . KOH Br - CH2 - CH2 - Br ¾alc¾¾ ® CH º CH + 2 KBr + 2 H2 O

(85) Alkyne with Lindlar`s Catalysist ( Pd / BaSO4 ) gives Cis- alkene while with Na / Liq. NH 3 (Birch Redaction) gives trans-alkene pd (86) CH º CH + 2 H2 ¾Ni¾/¾ ® CH3 - CH3

(87) According to Que-85 (88)

Mg 2 C3 + 4 H2 O ® CH3 - C º CH + 2 Mg ( OH ) 2

(89) Hydrogen attached to triplebonded carbon is acidic (90) with Ag ( NH 3 ) 2

+

1- Alkyne will give reaction while 2- Alkyne will not give reaction

SO3 H

D + H2 O (g ) ¾ ¾ ®

(92)

+ H2 SO4

(93) 6 p electrons (94) According to Hukel’s Rule Compound to be aromatic it should have ( 4 n + 2 ) p e s i.e It should have 2,6,10,14 p e electrons. But cyclo octa tetraene has 8 p e so it is non-aromatics OH

(95) phenol

is aromatic but not aromatic Hydrocarbon

+

(96)

It has 4 p e there for not aromatic COCH3

(97)

CH 3COCl ¾¾¾¾¾ [ Al Cl ] ® 3

Acetophenon.

(98) Nitration, Sulphonation, Halogenation, Alkylation and Acylation of Benzene are Electro philic Substitation Reactions. +

(99) NO2 is electrophile 149

(100) Br + is electrophile (101) Beacause double bonds of Benzene breaks on addition of H2 Cl

Cl

Cl

hv +3Cl 2 ¾¾ ® Cl

(102)

Benzenehexachloride(BHC)

Cl

Cl Cl

(103)

+ HCI

+ CI 2 ¾ ¾¾® [ FeCI3 ]

(104) m - Directors Þ - NO2 , - SO3 H , -COOH , -CN , CHO, -COR (105) O–P Directors Þ - CH3 , - OH , - OR , - NH 2 , - X (Halogen) NO2

conc.HNO3 ¾¾¾¾¾ [ H SO ] ®

(106)

CH3

CH3

CH3

2

4

+

Because – CH3 is O/P Director NO2

CH3

CH3

Cl2 ¾¾¾¾ [ FeCl ] ®

(107)

Cl

3

CH3

CH3 OH

(108)

Zn ¾¾¾  ®

O-Cresol

+ Zno

Toluene CH2 Cl

CH3

Cl2 ¾¾¾ uv ®

(109)

Because - NO2 m - Directors

Benzylchloride COOH

CH3

KMnO4 / H ¾¾¾¾¾ ® 3[O ] ¾ +

(110)

CHO

CH3

CrO2 Cl2 ¾¾¾¾ ® 2[O ] ¾

(111)

+ H2 O

+ H2 O CH3

CH3

O2N

NO2

H 2 SO 4 +3 HNO3 ¾¾¾¾ 111° c ®

(112)

2,4,6-Trinitrotoluene(TNT) NO2

CH3

CH3

O2N

®

(113)

COOH

NO2

KMnO 4 ® ¾¾¾¾¾ H+

NO2

NO2

NO2

NO2

NO2

Soda lim e ¾¾¾¾¾ ® 

NO2

NO2

1,3,5-Trinitrbenzene(TNB) CH3

(114)

CH3 Br

+ CH3 Br + 2 Na ¾ ¾ ¾® ether

CH3

+ 2 NaBr

O-Xylene This is known as wurtz- Fitting Reaction 150

(115) Upon Oxidation of o, m, p-Xylenes forms corresponding dicarboxylic COOH COOH

COOH COOH

COOH

Phthalic acid

Isophthalic acid

Terthalic acid

CH2 = CH 3

CH = CH2

(116)

COOH

H2 ¾Ni¾+¾ ®

Styrene

Ethylbenzene CH2 = CH 3

(117)

CH = CH2

[O ] ¾¾¾¾ 600° C ®

AlCl 3 ] + CH2 = CH2 ¾[¾¾ ®,

+ H2

(119) According to Que-55 (120) - NO2 is m- Director (121) -COOC2 H 5 m - Director (122) Bond length µ 1/ Bondorder there for bond length in

CH3

CH3

CH2 [ Al O + Cr O ]

CH2 | (123) CH2

2 3 2 3 ® CH3 ¾¾¾¾¾¾¾¾ + 4H2  [o ] | Toluene CH2

CH2 n-Heptane OH

(124)

Zn ® ¾¾¾ 

+ Zno

(125) p e s in , , , are 4,4,6,4 there for according to Hukelis Rule (c) is Aromatic CH3

CCl3

3 Cl / hv

2 ¾¾¾¾¾ Heat ®

(126)

COOH

3 NaOH ® ¾¾¾¾¾

CH3

(127)

Sulphonation

Compound A (Toluene)

CH3

CH3

CH3

¾¾¾¾¾¾¾ ®

+ 3NaCl + H 2O

SO3H

NaOH ® ¾¾¾¾ fusion

+ So3H

151

CH3

OH

+ OH

H2 O (129) CaC2 ¾2¾ ¾® HC = CH + Ca ( OH ) 2

CH3

CH Cl

hot iron ® 3CH = CH ¾¾¾¾¾ tube

CH3 Br

Br2 ¾¾ ®

Br

/ HCI ¾Sn ¾¾ ®

NO2

NO2

NaNO / HCl

2 ¾¾¾¾¾¾ ¾ ® 0.5° c

NO2

CH3

COOH

CH3

H PO

Br

+ HCI

3

CH3

CH3

(130)

3 ¾¾¾¾ [ AlCl ] ®

3 2 ¾¾¾¾® 2[ H ]

Br

Br

KMnO

4 ¾¾¾¾ ® KOH ¾

[O ]

N = N – Cl

(131)

Pyrole is aromatic beacuse it is cyclic, and contains ..

N | H

6 p e s ( 2 ´ 2 p -bonds + lonepair of N) (132) Correct explanation :- Chlorination of CH 4 is free radical reaction and free radicals are obtained in sunlight.

CH3 | (133) CH3 – C – CH3 | CH3

Correct R:- In neopentane four identical CH3 groups are attached to

4  - carbon there for only one monosubstitated product is possible (134) R is Correct explanation of A (135) Correct A:- gives -1- bromo-1- phenyl ethane Correct R:- Benzyl cation is more stable than alkyl cation (136) Correct A:- Gives But-2-ene as a major product (137) Correct explanation :- Neopentane being symmetrical packs more closely in the crystal lattice than isopentane (138) R is the correct explanation of A (139) Correct explanation :- Hg +2 ion being an electrophile attacks on p e of triplebond and then nucleophilic attack of H 2 O occurs. (140) HI adds to vinyl chloride according to markownikoffis Rule. (142) O – P directors except halogen activates benzene ring while m-director deactivates. O

3 (144) ¾¾¾¾¾ Zn / H O ® 2

CHO CHO

aq. KOH

¾¾¾¾¾ ®

CHO

152

Unit-22 - Organic Compounds Containing Halogens MCQ 1.

Haloarenes contain halogen atom(s) with carbon atom(s) possessing hybridization. (A)

2.

SP 3

SP

(C)

SP 2

(D)

(C)

chloramphenicol (D)

B&C

Who is very effective for typhoid ? (A)

chloroquine

3.

4.

(B) (B)

chloroform

A

chlorapetite

B

(i)

Goiter

(p)

Anaesthetic

(ii)

Malaria

(q)

Potential blood substitutes

(iii)

Typhoid fever

(r)

chloroquine

(iv)

haloethane

(s)

chlor amphenical

(v)

fluori nated compounds

(t)

Thyroxine

(A)

i-t, ii-q, iii- p, iv-r, v-s

(B)

i-t, ii-s, iii- p, iv-q, v-r

(C)

i-t, ii-q, iii- s, iv-r, v-p

(D)

i-t, ii-r, iii- s, iv-p, v-q

Which of the following is vinylic halides. X

(A) 5.

(B)

X

(C)

A

X

B

CI (i)

CH3

CH2 Cl

(ii)

(p)

2, chloro toluene

(q)

1, Benzyl chloride

(r)

1, chloro toluene

(s)

chlorophenyl methane

(t)

ortho chloro toluene

(A)

i - t, ii - s

(B)

i - p, ii - s

(C)

i - r, ii - q

(D)

i - r, ii - q

153

(D) A & C

6.

How many structural isomers are formed from C5H11Br. (A)

7.

5

(B)

H

10

(C)

CH3

CH3 H H Br

H H

(B)

H

H H 3C

CH3 H CH3 BR

(D)

8

(D)

H

H H H BR CH3 H H Br

Which is sandmeyer's reaction from the following. N3+ X

(A)

–

OH

+

(C) (D)

– N+ 2 X

OH + HCl + N

– N+2 X

273–278K

+ H 2O

(B)

2

N=N

H

H

(D)

+ KI

+ N2

OH + HCl

? X ; X = .............

+ HBr

H

(A)

I

Above all H

9.

(C)

Correct structural formula for 3-Bromo 2-methyl But-l-ene (A)

8.

6

Br H

HH

H

(B)

H

(C)

Br

HH

HH

H H

None of these H CH 2–C = CH 2

10.

+ HBr Peroxide

Y. For Y =

CH–CH – CH 3 2

(A)

CH 2 – CH 2CH 2 – Br

(B)

Br CH 2 – C – CH 3

(C)

?

CH – CH 2 – CH 3

(D)

Br

154

Br

11.

A

B

(CH )

CO

(i)

2 CH 3 Br + NaI ¾¾3¾ ¾® CH 3 I + NaBr

(ii)

2CH 3 Br+Hg 2 F2 ¾¾ ® 2CH 3F+Hg 2 Br2

x : Finkel stein

12.

14.

15.

16.

O

3 2 CH 3CH 2Cl+NaI ¾¾¾¾ ® C2 H5 I+NaCl

(iv)

2C 2 H 5 Br + COF2 ¾ ¾® 2C 2 H 5 F + COBr2

(v)

COF2 + 2CH 3 Br ¾ ¾® 2CH 3 F + COBr2

(A)

1-y, 2-x, 3-x, 4-y, 5-x

(B)

1-x, 2-y, 3-y, 4-x, 5-y

(C)

1-x, 2-y, 3-x, 4-y, 5-y

(D)

1-x, 2-y, 3-y, 4-x, 5-x

y : Swart reaction

R - OH + PX 3 ¾ ¾® R - X + Z. For Z = ..................?

(A) 13.

( CH )

(iii)

X 2 PO 4

(B)

x + HCl Aniline =Y

HPO3 X 2

– N+ 2X

(C)

H 2 PO3 X

(D)

H 3 PO 3

For x and y Respectively as

(A)

NH 3 , 298 - 300K

(B)

NaNO3 , 273 - 278K

(C)

NaNO 2 , 273 - 278K

(D)

NaNO, 298 K

Arrange each set of compounds in order of increasing boiling points. (A)

1-chloropropane < Isopropyl chloride < 1-chlorobutane

(B)

1-chlorobutane < 1-chloropropane < isopropyl chloride

(C)

1-chlorobutane > 1-chloropropane > isopropyl chloride

(D)

iso propyl chloride > 1-chlorobutane > 1-chloro propane

For nucleophilic substitution bimolecure SN2 reaction give the correct order of reactivity. (A)

20 halide < 30 halide < 10 halide

(B)

30 halide < 20 halide < 10 halide

(C)

30 halide < 20 halide > 10 halide

(D)

10 halide 20 halide > 10 halide (D) 20 halide 30 halide < 10 halide Predict the order of reactivity of the following compounds in SN1 reaction. (A) Br

(B) Br

(C) Br

20.

(D) None of these Give the major & minor part for x and y

CH 3CH 2 – CH 2 – CH – CH

OH 3

OH

Br

21.

22.

–

–

CH 3 – CH 2– CH 2– CH = CH x - pent 1 - ene CH 3 – CH 2– CH = CH – CH y - pent 2 - ene

2

3

(A) x = 81% ; y = 19% (B) x = 19% ; y = 81% (C) x = 50% ; y = 50% (D) x = 80% ; y = 20% Predict the order of reactivity of the following compounds in SN2 reactions. (A)

C6 H 5CH 2 Br > C6 H 5CH . ( CH 3 ) .Br > C6 H 5C ( CH 3 )( C6 H 5 ) Br > C6 H 5CH ( C6 H 5 ) Br

(B)

C6 H 5CH 2 Br > C6 H 5CH . ( CH 3 ) .Br > C6 H 5CH ( C6 H 5 ) Br > C6 H 5C CH 3 .

(C)

C 6 H 5CH 2 Br < C 6 H 5CH (CH 3 )Br < C 6 H 5CH (C 6 H 5 )Br < C 6 H 5C(CH 3 )(C 6 H 5 )Br

(D) b & C Predict the true R or S for the following structres r

p

(i) s

q r

(ii) s

p

s

(iii) q

r p

q

P = I; Q = Br, S = Cl; r = H (A) i - S ii - R, iii - R (C) i - S ii - R, iii - S

(B) (D) 156

i - R ii - R, iii - S i - S ii - S, iii - R

( C6 H 5 ) Br

23.

C2H4Cl2 contains how many isomers. (A)

24.

25.

26.

27.

29.

30.

31.

(B)

5

(C)

8

(D)

2

Reactivity order for HX in Lucas test. (A)

HBr > HCl < HI

(B)

HCl < HBr < HI

(C)

HBr > HCl > HI

(D)

HCl > HBr > HI

Chloroform + O 2 light X + HCl, for X = .......... ?

(A)

CH 2 OCl

(B)

CH 2 Cl 2

(C)

COCl

(D)

COCl 2

Alkoh olic ¾¾ ® A KOH

CH 3CH 2 CH2 I

NaNH 2 ¾¾ ® C For ; C - NH 3 - NaBr

Br 2 ¾¾ ® B Alkoholic KOH

=

(A)

Alkenol

(B)

Alkene

(C)

Alkaiene

(D)

Alkele amine

?

For given compounds gives proper reactivity order for SN1 reaction. (x)

28.

0

Me

Me

(y)

Br

Me

Br

(z)

Br

Me

(A)

Z>X>Y

(B)

X >Y> Z

(C)

Y>Z>X

(D)

X>Z>Y

Which compound gives inversion isomer during SN2 reaction. (A)

(C 2 H 5 )2

(C)

CH 3Cl

CHCl

(B)

(CH 3 )3 CCl

(D)

(CH 3 )3

C.Cl

CH 3CH 2Cl + Ag CN

Ethenol x + Agcl for x = ................? 

(A)

CH 3CH 2 CN

(B)

CH 3CH 2 N = C

(C)

CH 3CH 2 N = C

(D)

a&b

Give the monomar of PVC. (A)

Propine

(B)

I-chloro ethene

(C)

ethene

(D)

1-chloro propane

Predict non chiral compound. (A)

2, chloropanten

(B)

1-chloropanten

(C)

1-chloro 2-methyl panten

(D)

3, chloro 2-methyl panten

157

32.

10 amine react with ethenolic KOH in presence of CHCl3 gives...... (A) iso saynide (B) Aldehide (C) cynide (D) Alcohol Cl

Cl

Cl

33.

Cl

34.

(A) Diphenyl dichlorop propare (C) 1, 1, dichloro diphenyl ethyl chloride Predict the true R/S for the following structures (A) i - S, ii - R, iii- R, iv - S (C) i - R, ii - R, iii- R, iv - S

Cl

H

Give IUPAC Name

H

H 2N

37.

i - R, ii - S, iii- R, iv - R i - R, ii - S, iii- S, iv - S

Br

(A) R and R (B) (C) S and S (D) 2 Predict the order of reactivity in R-X for SN reaction. (A) R-F >>R-Cl > R-Br > R-I (B) (C) R-F < R-Br < R-F < R-Cl (D) Freons which is use in industry

R and S S and R R-I > R-F > R-Cl > R-Br R-F R3CX > R2CHX (D) R2CH2X < RH2X > R3CX

158

40.

CH

3

(1) 41.

42.

Br + Nu– ® CH3Nu + Br– for this reaction arrange given Nu– in increasing order of reactivities. (2)

Pho -

CH 3O -

(3)

OH -

(A) 2 > 4 > 3 > 1 (B) 4 > 3 > 1 > 2 (C) Predict the decreasing order for neucleo philicity.

(4)

AcO-

4>3>2>1

(D)

1 > 2 >3 > 4

(A)

I - < Cl - < Br -

(B)

Cl - < Bi - < I -

(C)

I - < Br - < Cl -

(D)

Br - > Cl - > F -

Which is the D configration of alanine NH 2

(A)

(B)

CH 3

H

(C)

H CH 3

COOH

COOH H 2N

NH 2

COOH

H

CH 3

NH 2

H

COOH

43.

CH 3

(D)

A

B

1.

CH 3CH 2 MgCl + H 2 O ® Ethane + Mg(OH )Cl

p. elimination reaction

2.

2CH 3CH 2 Br + 2 Na ® CH 3 (CH 2 )2 - CH 3 + 2 Nax

q. substitution reaction

3.

KOH + Ethanol CH 3CH 2 ¾¾¾¾¾® Ethene D | Cl

r. courtz reaction

– + + OH

4. (A) (C)

Tertiary but ylalcohol

1-r, 2-q, 3-s, 4-p 1-s, 2-p, 3-r, 4-q

(B) (D)

s. grignard reaction 1-s, 2-r, 3-q, 4-p 1-s, 2-r, 3-p, 4-q

44. A (1) (2)

Cl

2 Br

B + 2NaCl

+ 2Na Ether

CH3 + 2NaCl

+ 2Na + CH 3 Br Ether

(p) Fridal craft reaction (q) Fitting reaction

Cl

(3)

C6H5Cl + HC3 – C – Anhyd.AlCl 3

+ 2NaCl

(r) wurtz fitting reaction

O O

(A)

1-r, 2-q, 3-s

(B)

(s) Alkylation reaction.

1-r, 2-q, 3-s 159

(C)

1-a, 2-s, 3-r

(D)

1-q, 2-r, 3-a

45.

Na / Ether ® R' X ¾¾¾¾¾

. Identify R1

CH 3 (A)

(C)

46.

CH 3 - C - CH 3 |

CH 3 | (B) CH 3 - C | CH 3

CH 3 | CH 3 - C - CH 2| CH 3

CH 3 | (D) CH 3 - C | CH 3

|

Mg

H 2O

B.

Mg Br

Mg Br

(A)

(B) Mg Br

Mg Br Br

(C)

48.

CH 3 | C | CH 3

Identify A & B respectively Br + Mg dry.Ether A.

47.

CH 3 | C - CH 2 | CH 3

(D)

For recemisation. (A) 50 : 50 mixture & opticall active. (B) 75 : 25 mixture & optically inactive (C) 50 : 50 mixture & optically inactive (D) 25 : 75 mixture & optically active Classify the configuration of product of a symmetric carbon atom. Consider the replacement of a group X by Y in the following reaction.

H

C 2H 5

H 3C

Y

C 2H 5 x

H CH 3

C 2H 5

CH 3

y

A+B

H

A

Process for A, B and A+B as under (A) A-retention of configuration A+B racemisation, B-Invention of configuration (B) A-Inversion of configuration, A+B retention of configuration, B-recemisation. (C) A+B - Inversion of configuration, B-retention of config n; A-recemisation. (D) None of them

160

49.

Predict the conect order for following reaction the approaching nucleophilies. P = Nu :

C–X

q Nu :

H H

50.

51.

54.

55.

Nu :

H C–X

; S

(A)

NaNO2 + 2 HCl C6 H 5 NH 2 ¾¾ ® C6 H 5 N 2Cl + NaCl + H 2O 273 - 278 K

(B)

C6 H 5 N 2Cl

Send Mayur reaction ¾¾ ® C6 H 5 Cl + N 2 DCuCl2 + HCl

(C)

C6 H 5 N 2Cl

Send Mayur ¾¾ ® C6 H 5 Br + N 2 + HCl DCuCl2 + HBr

Nu :

C–X

KI

D C 6 H 5 N 2 Cl ¾¾® C 6 H 5 I + KCl + N 2

¾® Chlorobenzne + HCl + SO 2 ; where X = .......? Phenol + X ¾

(A) 53.

R

(A) Q < P < R < S (B) Q > P > R > S (C) P > Q > R > S (D) Q < P < R < S Predict the decreasing order of density for same Alkyl or cryle group. (A) R / Ar - Br > Ar / R Cl > R / Ar - I > R /Ar > F (B) R / Ar I > R / Ar – Br > R / Ar Cl > R / Ar – F (C) R / Ar - I < R / Ar – Cl < R / Ar – Br > R / Ar – F (D) R / Ar – F > R / Ar – Cl < R / Ar – Br < R / Ar – I Predict the improper chemical reaction

(D) 52.

H H – C –X ; H

PCl 3

(B)

SOCl 3

(C)

PCl 5

(D)

SOCl 2

Predict the no. of isomers for possible genimal dihalide in (A) 0 (B) 4 (C) 3 (D) 9 for vinyl halide compound which is true ? (A) chloro ethen (B) 1-chloro 3-phenyl propene (C) 1–Bromo cyclohexa-1enc (D) All of above Which is correct for aelaillic halide compounds. (A) 3, Bromo, 2-Methy, Buyt 1-ene (B) 4-Bromo, 3-Methyl Buyt-2-ene (C) 3, Bromo 2-Methyl Propene (D) 1-Bromo Buyt -2-ene.

161

56.

A

B

1. 2. 3. 4.

Benzail Bromaid Ethilidin bromaid Finail Bromaid Alail Bromaid

M. N. O. P.

CH3CH2Br2 – H2C = CHCH2Br

(A) (C)

1-M, 2-N, 3-O, 4-P 1-P, 2-N, 3-O, 4-M

(B) (D)

1-P, 2-N, 3-M, 4-O 1-N, 2-O, 3-M, 4-P

Br

CH 2Br

Cl

57.

Give IUPAC Name

58.

(A) 2-Phenyl 2-chloropropane (B) (C) isopropyl chlorobenzene (D) Which alkylhalide known as Neopentyl chloride ?

1-chloro, 1-Methyl ethyl Benzene 1 dimethyl, 1, chlorobenzene H

Cl

(A)

Cl

(B)

H

H H

59.

60.

61.

H

H

H

(C) H

H

(D) Cl

Cl

H

R - H + PX5 ® R - X + B + HX B menas = ...................? (A)

H 3 PO 3

(B)

HPOX 3

(C)

POX 3

(D)

H 3 PO 4

C 2 H 5 I ¾AgNO ¾¾2 ® X. ..............?

(A)

C2H5 - O - N = O

(B)

C2H5 - N = O

(C)

C 2H 5 – O – N

O Ag

(D)

C 2H 5 – O – N

O O

Which compound known as kiral molecule & giving optical isomers ? (A) 2 – chloropropane (B) 2 – Methyl butane (C) 2 – Bromo butane (D) 2, 2 dichloro butane

162

62.

C 2 H 5OH + NaCl + H 2SO 4 ® A + NaHSO 4 + H 2 O A + 2Na B + 2NaCl ? For A & B respectively dry Ether

63.

64.

65.

66.

(A)

Ethyl Bromide, ethane

(B)

Methyl chloride ; ethane

(C)

Ethyl Bromide butane

(D)

Ethyl chloride butane

X Y Benzene ¾¾® Chlorobezene ¾¾® For X & Y respectively..

(A)

Cl 2 , [FeCl3 ] ; Cl 2 [FeCl3 ]

(B)

Cl 2 , [dry AlCl3 ] ; [dry AlCl3 ]

(C)

Cl 2 , [FeCl3 ] ; [dry AlCl3 ] C 2 H 5 Cl

(D)

Cl 2 , [FeCl3 ] ; [dry AlCl3 ] CH 3 Cl

Silver nitrate not giving the Precipates with chloro form because of ............. (A)

Chloroform is organic compound.

(B)

Chloroform is non ionised in water

(C)

AgNO3 indouble in chloroform

(D)

Chloroiform is insoluble in water

Which compound on oxidation gives Banzoic Acid ? (A)

Chlorobenzene

(B)

ChloroTolune

(C)

Benzyl Chloride

(D)

Chloro phenol.

(C)

a&b

Which is Gemexine compond ? (A)

67.

68.

DDT

(B)

BHC

(D)

None of these

Chloroform on slow oxydation with air gives (A)

Formile Chloride

(B)

Formic Acid

(C)

Cl 3C - COOH

(D)

COCl 2

(B)

CH 3CHO

(D)

Benzophenone

For which compound iodoform test not possible ? (A)

C 2 H 5OH COCH 3

(C) 69.

70.

71.

Reaction of CCl4 with AgNO3 soultion gives... (A)

White PPT of Agcl

(B)

No reaction take place

(C)

NO2 gas produce

(D)

CCl4 dissolved in AgNO3 Solution

What is DDT ? (A)

Insecti side

(B)

Bleaching agent

(C)

Weedkiller

(D)

Oxydiging agent

Predict the condition for conversion of Benzyl bromide form Tolluene. (A)

Bl2 / FeBr3

(B)

HBr

(C)

Bl2 / Light

(D)

Br2

163

72.

73.

74.

75.

Which compound having stranger C-X ? (A)

CH 3 - F,

(B)

CH 3Cl

(C)

CH 3 I

(D)

CH 3 . Br

For C–Cl of chlorobenzene campair with C-Cl of ethyl chloride (A) Larger bond length & weak bond. (B) Shorter bond lenth & weak bond (C) Larger bond length & strenger bond (D) Shorter bond length & strenger bond Predict the correct order of bond energy for C-X (where x = cl, Br, I) (A) C–I > C–Br > C – Cl (B) C–Cl > C – Br > C – I (C) C–I > C – Cl > C – Br (D) C – Br > C – I > C – Cl [ FeBr3 ] Ethylbenzene + Br2 ¾¾¾ ® ............ ?

CH 2CH 2Br

CH 2CH 3

(A)

Br

(B) CH 2CH 3

CH 2CH 3

(C)

(D) Br

76.

77.

78.

79.

80.

Br

Which compound having zero dipolmoment. (A) cis 1-2, dichloro ethelene (B) 1, 1, dichloro ethelene (C) trans 1-2, dischroethelen (D) none of these Which compound formed when ethenolic KOH react with normal propile bromide ? (A) Propane (B) Propanol (C) Propene (D) Propine Benzen + n–propail chloride ¾any ¾AlCl ¾ ¾3 ® .......... ? (A) propile benzene (B) (C) 3-propile 1-chloro benzne (D) For Lucast test............

n-propile benzene none of these

(A)

conc HCl +dry Zncl2

(B)

conc HNO3 + dry Zncl2

(C)

conc HNO3 + ZnCl2( aq )

(D)

HCl + ZnCl 2( aq )

(B) (D)

1,1 dichlorobutane none of these

Geonetrical isomerssm is not shown in by (A) 1, 2 dichlorobutane (C) 1, 3 dichlorobutane

164

81.

Which type of product produce. Circal Centre in the reaction CH 3CHO + HCN ® CH 3 - CH 3 - CH (OH ) - CN

82.

(A) Dextro rotatory (B) Mesoisomer (C) Levo-rotatory (D) Recemic mixture On the base of inductive effect predict the increasing stability order + CH

2

(1)

83.

+

+

(2)

(3)

CH 3 C CH 3

(A) I > II > III (B) I < II < III (C) III > I > II (D) II = III < I According to -I effect predict in creasing order of stability. + CH 2

(1)

+ CH 2

+ CH 2

X

(2)

(3)

X

X

84.

(A) III > I > II (B) II < I < III (C) I < II < III (D) III < II > I Which among the following compunds will be most recetive for SN1 reaction ? (A)

Cl

(B)

Cl

Cl CH3

(C)

(D)

85.

Phosgene react with excess of benzene in presence of anhyd Alcl3 to give. (A) benzoylchlorde (B) benzophenone (C) benzoicacid (D) benzbinacol

86.

The correct order of reactivity of the following compound ith H 2 O is ......... (i)

87.

(ii)

(iii)

Br Br (A) IV > III > I > II (C) I > IV > III > II The reactivity of the compounds............ (I) MeBr (III) MeCl (A) I > II > III > IV (C) IV > III > I > II

Cl CH3

(iv) (B) (D)

Br I > II > IV > III I > IV > II > III

(II) (IV) (B) (D)

PhCH2Br P - Meo C6H4Br decreases as IV > II > I > III II > I > III > IV

Br

165

Br2 / Fe CH2CH = CH2

88.

Br2 / Water (A)

Br

Br

(C)

y

; x & y respectively

CH2CH = CH2,

CH2 – CH2Br,

(B)

x

Br

CH2 – CH – CH2 – Br | Br CH2 – CH – CH2 | | Br Br

CH2 – CH = CH2,

CH2 – CH2 – Br Br

89.

90.

91.

92.

93.

(D) None of these Geometrical isometrism is not shown by ? (A) 1, 4 dichloro, 2-pentene (C) 1,2 dichloro, 1-pentene

(B) (D)

1,1 dichloro 1-pentene 1,3 dichloro 2-pentene

The no. of structural & configurational isomers of a bromo compound C5 H 9 Br obtained by the addition of HBr on 2-pentyne respectively are............ (A) 2, 4 (B) 4, 2 (C) 1, 2 (D) 2, 1 HBr Hint : CH3 - CH2 º C - CH3 ¾¾¾® CH3 CH2 - C = C - CH3 ; CH3 CH2 - C = C- CH 3 | | | | H Br Br H Thus, each one of the these will existas (I) as a pair of geometrical isomer. Structural isomers = 2 & stereosomers = 4. How many enantiomer pairs are obtained by monochlorination of 2, 3-dimethyl butane ? (A) two (B) three (C) four (D) one Which intermidiat increase the SN2 reaction ? (A) carboniyan ion (B) activated complex ion (C) Freeredical (D) carbenion ion

3CCl 4 + 2SbF3

¾SbCls ¾ ¾® high Pr essure

A + 2SbCl 3 ; A = .....................

(A)

CH 2 Cl 2

(B)

CCl 3 F

(C)

CCl 2 F2

(D)

CClF3

166

94.

95.

96.

97.

98.

99.

Which compounds are used as a Antiseptic ? (A)

CCl 4

(B)

CHI 3

(C)

CHCl 3

(D)

CCl 2 F2

CH3 | H – C – D Which is the configuration of this compound. | Br

(A) S(B) R (C) L (D) D Reaction of chlorobenzene with Mg in presence of dry other will gives compound A. Compound A reacts with aqueous halogen acid give which product. (A) Phenol (B) Phenyl Ether (C) Phenyl ketone (D) Benzene In sulfonation of chlorobenzene gives ................ (A) Benzen sulfonic acid (B) m-chloro benzene sulfonic acid (C) O & P Chlorobenzen sulfonic acid (D) O-chlorobenzen sulfonic acid C2H5I

¾Ethenol ¾¾® KOH

Br2 A ¾¾® B ; Where B = ................

(A)

CH 3 - CHBr2

(B)

CH2Br – CH2Br

(C)

CH.Br = CH.Br

(D)

CH 3CH 2 Br

Which is the true drecreasing order for (A)

CH 3Cl ; CH 3 - Br, CH 3 - F

(B)

CH 3 Br, CH 3 - F, CH 3 - Cl

(C)

CH 3 - CI ; CH 3 - F, CH 3 - Br

(D)

CH 3 - Br, CH 3 - Cl, CH 3 .F

100. Which is not a sendmayer reactant ? (A)

Cu 2 (CN )2 + KCN

(B)

Cu 2 Cl 2 + HCl

(C)

Cu 2 I 2 + KI

(D)

Cu 2 Br2 + HBr

CH3 101. H3C

CH2 OH gives which observation in Lucas test ? CH3

(A) (C)

reaction not take place Oilly point on layer

(B) (D)

167

give colour layer milky solution

102. Which compound having zero dipole moment ? (A) Cis 2-Butene (C) 2-Methyl 1-Propene 103. In which compound ciral centre is absent ?

(B) (D)

Tras 2-Butene 1-Butene

(A)

DCH2CH2CH2Cl

(B)

CH 3CH.D.CH 2 Cl

(C)

CH 3CH.Cl.CH 2 D

(D)

CH 3CH 2 CH.D.Cl

104. Which compound is optically active ? (A) 2,2 dimethyl pentene (B) 3-Methyl pentene (C) Butene (D) 2-Methyl Penten 105. How d and l isomer are differ from eachother ? (A) On the basis of reactivity with unoptical reactant (B) On the basis of melting point (C) On the basis of solubility of unoptical reactant (D) On the basis of optical rotation of polarised right. 106. Main product of reaction between tertiary butyle chloride and sodium ethoxide (A) 1-Butene (B) 2-butene (C) 2-Methyl Prop. 1-ene (D) 2 Etoxy 2-Methyl propane 107. Which compound gives only one monochloro product ? (A) 2-Methyl penten (B) 3-Methyl pentene (C) Neopentene (D) 2, 3 dimethyl pentene 108.

(CH 3 )3 COCl having IUPAC name ? (A) (C)

3-chlorobutene tertiary butile chloride

(B) (D)

2-chloro, 2-Methyl propane n-bytyl chloride

109. C3 H 8 + Cl 2 ¾light ¾® ¾ C3 H 7 Cl + HCl is which type of rxn. (A) Addition (B) Substitution (C) elimination (D) Re-arrangement 110. 1, Chloro butane react with Alcoholic KOH produce............ (A) 1-Butenol (B) 1-Butene (C) 2-Butene (D) 2-butenol 111. Which compound contains Antiseptic property ? (A) Trifloro methane (B) Triiodo methane (C) Tetra chloromethane (D) dichloro methane 112. In presence of which compd. Benzene react with I2 to form Iodobenzene ? (A) HI (B) SO 2 (C) H2O (D) HNO 3

168

113. Give correct reaction for methyl floride (A)

CH 4 + HF ®

(B)

CH 3OH + HF ®

(C)

CH4 + F2 ®

(D)

CH 3 Br + AgF ®

114. Which compound on liquification SN1 reaction is not possible ? (A)

CH 2 = CHCH 2 Cl

(B)

C6H5Cl

(C)

C6H5CH(CH3)Cl

(D)

C 6 H 5CH 2 Cl

115. Which compound show optical isomerism ? H

(A)

H

HO

(B)

CO2H H

(C)

CO2H OH

CO2H

(D)

CO2H H

Cl

116. Onheating Chloroform with silver powder gives ........... (A) Ethane (B) Ethyne (C) Methane (D) Eethene 117. Which compound gives more reactivities in SN1 reaction of (A)

C 6 H 5CH 2 Br

(B)

C 6 H 5C(CH 3 ) (C 6 H 5 ) Br

(C)

C 6 H 5 CH (C 6 H 5 ) Br

(D)

C 6 H 5CH (CH 3 ) Br

118. CH 2 = CH 2 + HCl ® X + NaI ¾Acetone ¾¾® Y + NaCl x and y Respectively (A) (C)

Iodo ethane, chloroethane Chloro ethane, iodo ethane

Me CH2Br

+

119.

Anhy.AlCL3

(B) (D)

1, 2 dichloro ethane, Iodo form. Broimoethane, iodoethane

C;C=?

H

(A)

CH3

Br

(B)

CH3

(D)

None of these

H

(C) CH3

169

120. The number of isomers for the compound with colecular formula C2BrClFI is ... (A) 3 (B) 4 (C) 6 (D) 5 121. In the presence of peroxide, hydrogen chloride and hydrogen iodide do not give anti Markovnikov addition to alke nes because. (A) both are hightly tonic (B) one is oxidising and other is reducing (C) (D)

one of the steps is exothermic in both cases all the steps are exothermic in both the cases

122. An SN2 reaction at an a symmetric carbon of a compound always gives. (A) a product with the optical rotation (B) a single stereomer (C) a mixture of diastereomers (D) a product with opposite optical rotation 123. A solution of (+) 2-chloro-2 phenyl ethane in toluene recemises slowly in the presence of small amount of SbCl5 due to the formation of (A) carbanion (B) carbenec (C) carbocation (D) Free redical 124. The final product formed by distilling ethyl alcohol with excess of Cl2 and Ca(OH)2 is (A)

CH 3CHO

(B)

CCl3CHO

(C)

CHCl 3

(D)

(CH 3 )2 O

125. The reaction of CH3CH = CH

OH with HBr gives.

(A)

CH3CH.Br.CH2

OH

(B)

CH3CH2CH.Br

(C)

CH3CH2CH.Br

OH

(D)

CH3 – CH.Br.CH2

(B)

Hydrohelognation reaction

126. A new C-C bond formation is possible in (A) Cannizzaro reaction

Br Br

(C) Clemmensen reduction (D) Reimer tiemann reaction 127. The product of reaction of alcoholic silver nitrate with ethyl bromide is (A) Ethylene (B) Ethyl nitrite (C) Nitro ethane (D) EthylAlcohol 128. Which of the following will have zero dipolemoment ? (A) 1,1 dichloro ethelene (B)

cis-1-2 dichloro ethylene

(C) trans. 1,2-dichloro ethelene (D) none of these 129. Chlorination of Tolnene in the presence of light and heat followed by treatment with aqueous NaOH gives. (A) (C)

O-cresol Ben zoic Acid

(B) (D) 170

P-cresol 2, 4 dihydroxy toluene

130. 2-Phenyl 2-chloro propane on treat ment with alcok gives mainly. (A) 2-phenyl propene (B) 3-phenylpropane (C) 1-phenyl propan 2-ol (D) 1-phenyl propane 3-ol 131. Only two isomeric monochloro derivaties are possible for (A) n-hexane (B) 2, 4 dimethyl pentane (C) benzene (D) 2, methyl propane 132. The C-H bond distance is longest in (A)

C2H 2

(B)

C2H 4

(C)

C 2 H 2 Br

(D)

C2H6

133. Chloroform on treatment with alcoholic KOH + aniline gives (A) phenyl isocyanide (B) Phenol (C) Cyanobenzene (D) None of these 134. The molecular formula of a saturated compound is C2H4Br. The formula permits the existance of two. (A) Functional isomers (B) Position isomers (C) optical isomers (D) Cis trans isomers

Answer Key 1. c 2. c 3. d 4. c 5. b 6. d 7. c 8. b 9. c 10. c 11. c 12. d 13. c 14. c 15. b 16. a 17. b

18. c 19. a 20. b 21. b 22. b 23. d 24. b 25. d 26. c 27. b 28. c 29. c 30. b 31. c 32. a 33. b 34. c

35. c 36. d 37. d 38. a 39. b 40. c 41. b 42. c 43. d 44. d 45. a 46. d 47. c 48. a 49. b 50. b 51. b

52. d 53. c 54. d 55. a 56. b 57. b 58. a 59. c 60. d 61. c 62. d 63. d 64. b 65. c 66. b 67. d 68. d

69. b 70. a 71. a 72. a 73. d 74. b 75. d 76. b 77. c 78. a 79. a 80. b 81. d 82. b 83. c 84. c 85. b 171

86. c 87. d 88. a 89. b 90. a 91. d 92. b 93. c 94. b 95. a 96. d 97. c 98. b 99. c 100. c 101. a 102. b

103. a 104. b 105. d 106. c 107. c 108. b 109. c 110. b 111. b 112. d 113. d 114. b 115. b 116. b 117. b 118. c 119. b

120. c 121. c 122. b 123. c 124. c 125. c 126. d 127. c 128. c 129. c 130. a 131. d 132. d 133. a 134. b

Unit-23 - Organic Compounds Containing Oxygen Important Points A Organic compounds containing oxygen - I 1. ®

2. ®

®

3. ®

®

4. ®

5. ® ®

Preparation of alcohols: Monohydric alcohols are prepared by the hydrolysis of alkyl halides with aqueous alkali, hydration of alkenes, hydrolysis of ester, reduction of (alde hydes, ketones, acids and acid derivatives). Grignard reagents is also used to prepare monohydric alcohols. Physical properties of alcohols: The boiling points of alcohols are much higher than comperatively same molecular masses of alkanes, ethers and alkyl halides. This is due to intermo lecular Hbond. For isomeric alcohols, the boiling points are in the order 10 > 20 > 30. Due to the formation of H-bond between alcohol and H2O modecules, alcohol with lower number of carbons are soluble in water. Chemical properties of alcohols: Alcohols exhibit three types of reactions, (i) Reaction in which O-H bond cleaves (ii)Reaction in which C-O bond cleaves (iii)Reaction in which whole molecule of alcohol participate. Victor-Meyer’s test and Lucas reagent are used to distinduish 10 , 20 and 30 alcohols.Oxidation reactions are also used to distiguished between 10, 20 and 30 alcohols. Preparation of phenol: Phenol is prepare from cumene, diazonium salt, benzene and coal tar. Physical properties of phonol: Phenols have higher boiling point than the corresponding hydrocarbon and aryl halides. This is due to the presence of intermolecular hydrogen bonding. Phenols are more acidic than alcohols because phenoxide ion is stabilised by resonance. The presence of electon withdrawing group like NO2, increases the acidic strength of phenol and electron donating group like R, decreases the acidic strength of phenol. 172

6. ®

7. ®

Chemical properties of phenol: Reaction of phenols are mainly of two types, (i) Reaction involving OH group (ii) Reaction involving phenyl group Preparation of ethers: Ethers are mainly prepared by Williamson’s synthesis which involces the heating of alkyl halides with sodium or potassium alkoxides or phenoxides. Dehydration of alcohol at 140 c also gives ether.. Chemical properties of ethers: (i) Reaction involving cleavage of C-O bond with dilute acid or with HX (ii) Electrophillic substitution reactions occuring in aromatic ring 0

8. ®

B: Organic compounds containing oxygen - II 1.

Carbonyl compounds:

®

+ – Organic compounds containing carbon-oxygen double bond ( C = O) O are C called C carbonyl =O group or carboxy group compounds. In aldehydes, the carbonyl group is attached to one hydrogen atom and one alkyl (or aryl or hydrogen atom) group, while in ketones it is attached to one alkyl and one arly group or to two alkyl (or aryl) groups, which may be same or diffrent. If carbonyl group is attached to one hydroxyl group, the compounds are know as carboxylic acids. In carboxylic acid compounds, if the hydrogen of hydroxyl group is substituted by alkyl or aryl group the compounds are known as esters, but if it is substitued by acyl group, the compounds are known as acid anhydrides. If the carbonyl group is attached to chlorine and to amino group the compounds are known as acid chlorides and amides respectively. The general formula of these compounds are expressed as

Aldehyde

Ketone Carboxylic acid

Anhydride

Acid Chloride Amide

2.

Structure and nature of carbonyl group:

®

Carbonyl carbon atom is sp 2 hybridised and form three s -bonds and one p bond. All the three

®

s-bond lie in same plane having angle 1200. The p- bond lies both above and below the C-O s bond. Thus the carbonyl carbon, oxygen atom and two atoms which are directly bonded to the carbonyl carbon lie in one plane, and is confirmed by electron diffraction and spectroscopic studies. Due to higher electronegativity of oxygen atom relative to carbon atom the carbonyl group is polarized and carbonyl carbon becomes electrophile (Lewis acid) and oxygen becomes nucleophile (Lewis base). Carbonyl group is polar in nature and has dipole moments.Aldehydes and ketones 173

have dipole moments 2.3-2.8 D. The resonance structures are as shown below :

C= O C

+ – C =O

3.

Physical properties of aldehydes and ketones:

®

The polar carbonyl groups have dipole-dipole interaction between opposite ends of the C = O O C group dipoles and hence due to weak intermolecular attraction the melting points and boiling points of aldehydes and ketones are higher than corresponding non-polar compounds. The order of boiling points is carboxylic acid > alcohol > isomeric ketone > isomeric aldehyde > ether > hydrocarbon. Due to hydrogen bonding with water molecules the aldehydes and ketones upto three carbon are soluble in water. The aromatic aldehydes and ketones due to presence of larger hydrocarbon parts (like benzene ring etc.), are insoluble in water. All adehydes and ketones are fairly soluble in organic solvents like benzene, ether, alcohols, chloroform etc. Chemical properties of aldehydes and ketones: Due to presence of hydrogen atom, the carbonyl group of aldehyde is much more reactive than ketone. Aldehydes and ketones undergo nucleophilic addition reaction because the carbonyl carbon atom is slightly positively charged. In nucleophilic addition the first step is reversible and also slow, so it is a rate determining step. The second step is reversible. Due to steric effect and inductive effect the aldehydes are more reactive than ketones. Most of the aldehydes and aliphatic methyl ketones, due to less steric hindrance are more reactive.

®

4. ®

®

Aldehydes and ketones react with NaHSO3 and give bisulphite addition product which are usu ally crystalline solids. On hydrolysis they give original aldehydes and ketones, so this reaction is useful for separation and purification of aldehydes and ketones. Addition of HCN and Grignard reagent to the aldehyde and ketone which give a -hydroxy carboxylic acid and 10 , 20 ,30 alcohols respectively.. Addition of alcohol to aldehyde give hemiacetal and further acetal, while ketone give the same product.

®

Nucleophilic addition reaction of aldehydes and ketones with NH 3 and its derivatives ( H 2 N - Z ) are catalysed by acids. Aldehydes and ketones on reduction give 10 and 20 alcohols respectively.. Aldehydes and ketones can be reduced to hydrocarbon by using different reagent like WolffKishner reduction, Clemmenses reduction, red phosphorus with HI. Ketones on reduction with magnesium amalgam and water give the product pinacol. Oxidation of aldehydes : Tollens’ test, Fehling’s test and Benedict’s test give the product carboxylic acid. Fehling’s test and Benedict’s test are not given by aromatic aldehydes. Oxidation of ketones by strong oxidizing agents like con. HNO3, KMnO 4/ H 2SO 4, K2Cr2O7/H2SO4 give mixture of carboxylic acids. 174

+ – C =O

Oxidation of aldehydes and ketones containing CH 3CO -group give iodoform test.

®

5.

6. ®

Aldol condensation and cross aldol condensation are the reactions given by aldehydes and ketones having a -hydrogen atom or atoms using dilute alkali as catalyst. Cannizzaro reaction is given by aldehydes and ketones which do not have an a -hydrogen atom by using con. NaOH or 50 % NaOH. Electrophilic substitution reactions of aromatic aldehydes and ketones are nitration, sulphonation and halogenation. Preparation of caroxylic acids : Carboxylic acids are prepared from : • Primary alcohol and aldehyde • Alkyl benzene and alkenes • Nitriles and amides. • Grignard reagents • Acid halide (chloride) and anhydrides • Esters Acidic nature of carboxylic acids : Carboxylic acids are stronger acids than phenol and alcohols. For convenience the strength of an acid is generally indicated by its pK a value rather than its K a value.

pKa = –log Ka

7. ®

®

Factors affecting strength of acids are • effect of electron-donating group • effect of electron withdrawing group • attachment of phenyl or vinyl group directly to carbonyl group. Physical and chemical properties of carboxylic acids : Carboxylic acid in aqueous solution form intermolecular hydrogen bonding with water molecules. Carboxylic acids are cyclic dimer in vapour phase or in aprotic solvents. The reactions of carboxylic acid are • Reactions involving cleavage of O-H bond. • Reactions involving cleavage of C-OH bond • Reaction involving -COOH group Substitution reaction in hydrocarbon part of carboxylic acid are halogenation and ring substitution as bromination, nitration and sulphonation. Carboxylic acids are used in different fields.

175

MCQ 1.

Numbers of isomeric alcohols of molecular formula C5 H12O are (A) 5

2.

3.

4.

5.

(B) 8

(C) 6

(D) 9

Which of the following will produce only one product on reduction with LiAlH4 ? (A) CH3COOCH2CH3

(B) CH3CH2OCOCH2CH3

(C) CH3COOCH3

(D) CH3CH2OCOCH2CH2CH3

CH 3 CH 2 CH 2 OH Can be converted to CH 3 CH 2 CH 2 COOH by the following sequence of steps:

(A) PBr3, KCN, H2/{Ni}

(B) HCN, PBr3, H3O+

(C) PBr3, AgCN, H3O+

(D) PBr3, KCN, H3O+

LiAlH 4 H 2 C = CH - COOH ¾¾¾ ® X . What is "X" ?

(A) CH 3CH 2 COOH

(B) CH 3CH 2 CH 2 OH

(C) H 2C = CH - CH 2OH

(D) CH 3CH 2CHO |

P+I H O Mg HCHO ® A ¾¾¾ ® B ¾¾¾ ® C ¾¾¾ ®D, In the following Sequence of reactions, CH 3CH 2OH ¾¾¾ ether 2

2

the compound D is:

6.

(A) propanal

(B) n-butyl alcohol

(C) butanal

(D) n-propyl alcoho

Acid catalysed hydration of alkenes except ethene leads to the formation of (A) primary alcohal (B) mixture of primary and secondary alcohols (C) secondary or tertiary alcohol (D) mixture of secondary and tertiary alcohols

7.

8.

During dehydration of alcohol to alkenes by heating with Conc H2SO4, the initiation step is: (A) elimination of water

(B) formation of an ester

(C) formaton of carbocation

(D) protonation of alcohol molecule

which of the following compounds will give positive iodoform test ? (I) 3- methyl propan-2-ol (III) I - methyl cyclopentanal (II) I - phenyl propan-1-ol (A) I and III

(Iv) 3- phenyl propan-2-ol. (B) I and IV

(C) II and III 176

(D) II and IV

9.

Which of the following alcohal on heating with conc H2SO4 gives product, which show geometrical isomerism ? (A) 2,4- dimethyl pentan-3-ol

B) 2- methyl butan -2-ol

(C) butan-2-ol (D) all of the above 10. Propan -l-ol and propan-2-ol can be distingaished by (A) oxidation with alkaline KMnO4 followed by reation with Fehing Solution (B) oxidation with acidic dichromate followed by reation with Fehling Solution (C) oxidation by heating with copper followed by reation with Fehling Solution (D) oxidation with conc H 2SO 4 followed by reaction with Fehling Solution 11.

Which one of the follwing will most readily be dehydrated in acidic condition ? O

(A)

OH

(B)

O

O

(C)

O

(D) OH

OH

12.

The best reagent to convert pent - 3 - en-2-ol into pent-3en-2-one is (A) acidic permanganate

(B) acidic dichromate

(C) chromic anhydride in glacial acetic acid (D) pyridinium chlorochromate 13.

CH 3CH 2 OH can not be prepare by which of the following reaction ?

(A) Hydrolysis of ethyl acetate (B) Hydroboration of ethene followed by oxidation in basic medium (C) Reaction of ethyl chloride with alcoholic potassium hydroxide (D) Reaction of ethylacetate 14.

The most suitable reagent for the conversion of primary alcohol into aldehyde with the same number of carban is (A) acidified K2Cr2O7 (B) alkaline KMnO4 (C) acidified KMnO4

15.

Na P + Br OH ¾¾¾ ® x ¾¾¾ ® y What is y ? 2

(A) 16.

(D) pyridinium chlorochromate

(B)

(C)

(D)

An organic compound "X" on treatment with PDC in CH 2Cl2 gives compound "Y". Compound "Y", reacts with I2 and alkali to form yellow precipitate. The compound "X" is (A) acetone

(B) ethanal

(C) ethanol 177

(D) acetic acid.

17.

How many optically active stereoisomers are possidle for butan -2,3 -diol ? (A) 1

18.

(B) 2

20.

(D) 4

The correct order of boiling points is for n- Butyl alcohol

tert - Butyl alcohol

(I)

(III)

iso - Butyl alcohol

Sec- Butyl alcohol

(II)

(IV)

(A) I > II >IV > III (C) II > III > I > IV 19.

(C) 3

(B) III > IV > II > I (D) IV > III >II > I

The boiling point of glycerol is more than propanol because of (A) hydrogen bonding

(B) hybridization

(C) arrangment of molecules

(D) size of molecule

An organic compound X is oxidised by using acidified K 2Cr2 O7 . The product obtained reacts with phenyl hydrazine but does not give silver mirror test. The possible structure of X is

21.

(A) CH 3 CH 2 OH

(B) CH 3 CO CH 3

(C) ( CH 3 )2 CH OH

(D) CH 3 CHO

( CH3 )3 CMgCl on reaction with D2O produces (A) ( CD )3 CD

22.

24.

25.

(D) ( CH 3 )3 CD

Lucas test is associated with (A) alcohols

23.

(B) ( CD3 )3 CH (C) ( CH 3 )3 COD (B) phenols

(C) aldehydes

(D) carboxylic acids

______________alcohol reacts immediately with anhydrous ZnCl2 + HCl and gives insoluble chloride (A) Methanol

(B) Ethanol

(C) Isopropyl alcohol

(D) 2 - Methyl propan - 2-ol,

Glycerol is more viscous than ethanol due to (A) many hydrogen bonds per molecule

(B) high boiling point

(C) high molecular weight

(D) Fajan's rule

4.6 gram ethanol when reacts with sodium metal___________is formed.

(A) 11.2 litre H 2 at STP

(B) 1.12 litre H 2 at STP

(C) 1.12 litre O 2 at STP

(D) 11.2 litre H 2 at STP 178

26.

(i ) B H

2 6 P ¬¾¾¾¾¾¾¾ (ii ) H O / OH -¾ 2

(A) Both ƒk™u 27.

H O CH2 ........ ¾¾¾ ® Q P and Q respecctively are 3

2

CH2OH

(ƒ Bk™ )u

CH3 OH

(C)

+

CH2OH

and

C CH H32 H3 (D) CHCOH 2OHand OH and

CH2OH

C2 H 5OH and C6 H 5OH can be distinguished by

(A) Br2 + H 2 O

(B) I2 + NaOH

(C) FeCl3

(D) Both (B) and (C)

28.

In the Lucas test of alcohols, appearnce of cloudiness is due to the formation of (A) aldehyde (B) alkyl chloride (C) acid chloride (D) ketone 29. When ethyl alcohol is treated with cl2 we get (A) CH 3 CH 2 Cl

(B) CH 2 Cl CH 2 OH

(C) CH Cl2 CH 2 OH

(D) CCl3 CHO

30.

When a compound (Molecular formula C3H8O ) is treated with acidic sodium dichromate we get compound "X". When"X" is treated with methyl magnesium bromide followed by hydrolysis compound "y" is formed. The compound y is (A) isopropyl alcohol (B) tertiary butyl alcohol (C) iso butyl alcohol (D) methyl ethyl ketone 31. Idenfify P,Q and R in the following reactions, (i) OH

(ii)

P ¾¾® D

COOH OH

OH

Zn ¾¾¾ ®Q D

C2 H 5 I (iii) R ¾¾¾¾ ®

OC 2H 5

(A) Sodium oxide benzene, Sodium phenoxide (B) Sodalime, benzene, potassium phenoxide (C) Zn, cyclohexanone, Sodium ethoxide (D) Sodium, cyclohexanone, potassium benzoate 32. Salicyladehyde and o-nitrophenol are less soluble in water because, (A) - CHO and - NO2 groups are not polar. (B) they are aromatic compounds. (C) intra molecular H-bond is present (D) their molecular weights are high. 179

33.

The final product of the following reaction is / are OH

CHCl3 50%KOH ¾¾¾¾ ® × ¾¾¾¾¾® KOH OH

CH

(A)

OH

(B) OH

OH

CH2 OH +

(C)

OMe

34.

OH

Given I =

C

COOK

, III = OH

OH

The decreasing order of the acidic character is (A) I >II >III (B) III > II >I (C) II > III >I 35.

COOK

CH2 OH +

(D)

NO2

, II = OH

COOK

O

(D) II >I >III

An organic compound `X' With molecular formula, C7 H8O is insoluble in aqueous NaHCO3 but

dissolves in NaOH . When treated with bromine water, `X' rapidly gives `Y' ( C7 H5OBr3 ) The compounds `X' and `Y' respectively, are (A) benzyl alcohol and 2,4,6 - tribromo-3-methoxy benzene (B) benzyl alcohol and 2,4,6 - tribromo-3-methyl phenol (C) o-cresol and 3,4,5 -tribromo-2- methyl phenol (D) m-cresol and 2,4,6 -tribromo -3- methyl phenol 36. Willamson's Synthesis is used for the preparation of (A) acid (B) ester (C) ether (D) alcohol 37. P - cresol reacts with chloroform in alkaline medium to give the compound A, which adds hydrogen cyanide to form the compound B. The latter on acidic hydrolysis gives chiral carboxylic acid. The structure of the carboxylic acid is CH3

CH 3

(A) OH

CH2COOH

(B) OH CH3

CH3

(C)

CH(OH)COOH

CH2 COOH

(D) OH

OH

180

CH(OH)COOH

38.

Which of the following has maximum acidic strength ? OH

OH

(A)

(B)

(C)

(D)

NO2

NO2

39.

OH

OH

CH 3

The Structure of the compound that gives a tribromo derivative on treatment with bromive water is CH3

(A)

(B)

OH CH3

(C)

CH2 OH

CH3

OH

(D) OH

40.

(I) Benzene 1,2 - diol (II) Benzene 1,3 -diol (III) Benzene 1,4 -diol (IV) Phenol The incresing order of boiling points of above mentioned compounds is (A) I < II