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PREFACE Dear readers, I feel deeply indebted to all of you for such a tremendous response to earlier editions of this book. The revised edition of this meticulously crafted book has been written based on the latest syllabus of both engineering and medical entrance examinations. Some major highlights of this edition are : • Introduction of a new chapter entitled ‘Physical Properties of Organic Compounds’. • Addition of new challenging problems in each chapter for JEE Advanced aspirants. • Chapter on ‘Isomerism’ has been thoroughly revised for better understanding of the topic. • Concept strengthening Matrix questions have been added to each chapter keeping in consideration the pattern of entrance examinations. • Sincere efforts to remove all errors. Almost all suggestions received from learned fellow teachers, as well as, many students have been paid due attention and incorporated to the best possible extent. I am confident and sure this book will help students reinforce their fundamentals of Organic Chemistry. Suggestions from readers for further improvement of the book are welcome.
Vaibhav Trivedi Msc-IIT-R, NET [email protected]
Dedicated to my Parents
CONTENTS Chapter 1
THE LIVING WORLD 1-8
Chapter 2
NOMENCLATURE OF ORGANIC COMPOUNDS 9-18
Chapter 3
ISOMERISM
Chapter 4
REACTION MECHANISM (General Organic Chemistry)
Chapter 5
ACID & BASE
151-168
Chapter 6
HYDROCARBON (Alkane, Alkene & Alkyne)
169-202
Chapter 7
ALKYL HALIDE & GRIGNARD’S REAGENT
203-230
Chapter 8
ALCOHOL, ETHER, EPOXIDE & GRIGNARD’S REAGENT
231-264
Chapter 9
ALDEHYDE & KETONE
265-294
Chapter 10
CARBOXYLIC ACID & ITS DERIVATIVES
295-318
Chapter 11
AMINES
319-340
Chapter 12
BIOMOLECULES
341-354
Chapter 13
AROMATIC CHEMISTRY
355-394
Chapter 14
PRACTICAL ORGANIC CHEMISTRY
395-400
Chapter 15
PHYSICAL PROPERTIES OF ORGANIC COMPOUNDS
401-404
19-50 51-150
Main Features 1. Hybridization refers to the phenomenon of mixing of two or more than two atomic orbital of same energy (or nearly same energy) to produce two or more than two orbital of same energy, identical shape & size. Lone pair of electron (lp) + sigma bond pair of electron (s bp) + (–ve) charge = Hybridization for eg. s bp = 3 lp = 1 (–ve) charge = 0 sum = 3 + 1 + 0 = i.e. sp3 (3+1= 4)
sigma bp = 3, lp = 0, –ve charge = 0 Sum = 3 + 0 + 0 = 3 (sp2)
sigma bp = 3, –ve charge = 1, lp = 0 sum = 3 + 1 + 0 = 4 (sp3)
2. ‘N’ can acquire four states.
•
(When four bonds are present then ‘N’ has +ve charge)
•
(When three bonds are present then lone pair of e– is present on Nitrogen)
•
(When two bonds are present on ‘N’ then there is a –ve charge along with lp on ‘N’
For eg. H — C
π σ π
N Here N has a lp hence hybridization of ‘N’ is sp (one lp + 1bp)
3. The lone pair of electron in conjugation with double bond does not consider in hybridization. and
4. A carbon attached to one (or no) carbon atom is called a primary (1º) similarly if a ‘C’ is attached to two, three & four ‘C’ atoms it is referred as secondary (2º) ,tertiary (3º), & quaternary (4º) carbon respectively. 5. The hydrogen attached to 1º, 2º & 3º ‘C’ atoms is called 1º, 2º & 3º hydrogen respectively.
2
Problems in Organic Chemistry
6. Starting from the simplest member of the molecular formula of the compounds of a particular class arranged in an increasing number of ‘C’ atom then a series is obtained in which the member differ from the next member by ‘CH2’ group. Such a series is called homologues series. 7. Mol. Wt. of homolog ∝ boiling point ∝
1 solubility in water
Objective Questions (−)
1. Which type of hybridization is absent in CH 2 = C = CH − CH = CH (a) sp2 (b) sp (c) sp3
(d) all are present
2. How many 3° H are present in
(a) 1
(b) 2
(c) 3
3. Sequence of hybridization in CH3—CH = CH — CN (a) sp3, sp2, sp, sp2, sp2
(b) sp3, sp2, sp2, sp2, sp2
(d) sp3, sp2, sp2, sp, sp2
(c) sp , sp , sp , sp, sp 3
2
4.
which statement is false? (a) Four carbon atoms are sp2 hybridized (c) It contains four secondary – H – atoms
5.
(a) It contains two 2ºC atoms (c) It contains 81- ‘H’ atoms
(b) It contains two 2º ‘D’ atoms (d) The ‘C’ joined with Cl is sp2 hybridized
which is correct? (a) It contains four sp2 hybridized C atoms (b) It contains nine sp3 hybridized C atoms (c) It contains two 3º – H – atoms & two sp2 hybridized ‘C’ (d) It contains ten- H-atoms & two sp2 hybridized ‘C’ atoms.
7.
(b) One nitrogen is sp2 & other is sp3 hybridized (d) One nitrogen is 2º & other is 3º
which statement is correct for this compound?
6.
2
In which species bold ‘C’ atom is not sp2 hybridized (a) I (b) II
(c) III
8. In which case state of hybridization is changing.
(d) 4
(a)
(c)
(b)
(d) (a) & (c)
(d) I & III
3
The Living World
9. In which case state of hybridization is changing. (+)
(−)
−H (b) CH3 — CH3 CH3 — CH 2
− H( + )
(−)
− H( + )
−H (a) CH3 — CH3 CH3 — CH 2
(c) = CH 2
CH = CH 2 2
(+ )
(−)
(–)
CH ≡ CH CH (d) CH ≡ C
10. In which case state of hybridization on each atom is retained. (a) CH 2
(+ )
= CH 2 + H
(+ )
→ CH3 — CH 2 (b)
( −) (+ ) 1 liq. (c) → H 2 + CH ≡ C Na CH ≡ CH + Na NH3 2
(d) (b) & (c)
11. CH3CH 2CH 2OH CH3CHO CH3CH 2OH HCHO I II III IV Which statement is correct? (a) I, II, III & IV are homolog (c) B.P. of I will be lesser than III 12.
(b) I & II, III & IV are homolog (d) Solubility of (IV) in water will be greater than II
which statement is correct?
(a) (b) (c) (d)
Two sp hybridized ‘C’ atoms & six - 2º - H atoms are present in it Two 3ºC - atoms & two sp2 hybridized ‘C’ atoms are present in it. N is sp3 & two - 2º - C atoms are present in it. Two 3º – H – atoms, Eight – 2º – H – atoms & four – sp3 – hybridized ‘C’ atoms are present in it.
13.
State of hybridization of bold element is changing as:-
sp3 → sp 2 → sp 2 (a)
3 3 3 (b) sp → sp → sp
sp3 → sp3 → sp 2 (c)
3 2 2 (d) sp → sp → sp
14. (a) x > y > z
(c) z > x > y
(d) y > x = z
(c) sp3
(d) None
15.
correct order of bond length is : (b) y > x > z
is called nitrene. Hybridization on ‘N’ is : (a) sp
(b) sp2
16. Which statement is incorrect about :
(a) (b) (c) (d)
Oxygen gives one of its electro pair to the empty p-orbital of 2nd carbon atom Each carbon is sp2 hybridized Each carbon has one e– in unhybridized pz orbital Except 2nd C - atom each carbon has one e– in unhybridized pz orbital
17. Which type of bond is absent in benzyne?
(a) sp2 – sp2 s bond
(b) sp2 – s s bond
(c) sp2 – sp2 p bond
(d) sp2 – p s bond
4
Problems in Organic Chemistry
Passage - I Enzymes present in bacteria can convert p - amino benzoic acid in to folic acid. N
H2N
N
O H CH2CH2COOH
N
N
CH2 – NH
C–N–C H
OH C O
(Folic acid)
OH
Answer the question from 18 to 26 18. Which statement is in favour of ‘N’? (a) All are sp2 hybridized (c) Four are sp2 hybridized & three are sp3 hybridized
(b) All are sp3 hybridized (d) Four are sp3 hybridized and three are sp2 hybridized
19. How many ‘C’ atoms are sp2 hybridized in the folic acid? (a) 14 (b) 21
(c) 12
(d) 15
20. How many ‘C’ are sp3 hybridized? (a) 4 (b) 3
(c) 5
(d) 6
21. How many p bonds are present in it? (a) 11 (b) 10
(c) 6
(d) 12
22. Number of lone pair of electron possessed by nitrogen is / are : (a) 3 (b) 5 (c) 7
(d) None
23. How many H – atoms are present in folic acid? (a) 19 (b) 23
(c) 18
(d) 17
(c) 4
(d) 3
24. How many ‘C = O’ bonds are present in folic acid? (a) 5 (b) 6
25. How many ‘C’ atoms (which are not the part of ring) are sp hybridized? (a) 4 (b) 3 (c) 2
(d) None
26. Ratio of N — H bond to C — H bond is : (a) 1 : 3 (b) 4 : 7
(d) 2 : 9
2
(c) 2 : 5
Passage - II H2 / Ni can reduce almost all types of multiple bonds
= CH 2
2 → CH CH CH NH CH — C ≡ N 3 2 2 2
H Ni
O OH || | H2 CH3 — C— CH3 → CH3 — CH — CH3
; While PCC can convert alcohol (RCH2OH) in to aldehyde (R – CHO)
Consider the following compound.
Answer the question from 27 to 36
Ni
5
The Living World
27. In compound A ‘N’ does not exhibit :
(a) sp hybridization
(b) sp2 hybridization
(c) sp3 hybridization
(d) ‘N’ exhibits sp, sp2 & sp3 hybridization
28. In (A) carbon exhibits :
(a) sp hybridization
(b) sp2 hybridization
(c) sp3 hybridization
(d) All of these
29. In compound (C) :
(a) All ‘N’ are sp3 hybridized
(b) Two nitrogen’s are sp3 hybridized and one is sp2 hybridized
(c) All nitrogen’s are sp2 hybridized
(d) One nitrogen is sp3 hybridized & two are sp2 hybridized
30. In compound (C)
(a) All ‘C’ are sp3 hybridized
(b) 9 ‘C’ are sp2 hybridized & two are sp3 hybridized
(c) 9 ‘C’ are sp3 hybridized & two are sp2 hybridized
(d) One ‘C’ is sp hybridized while rests are sp3 hybridized
31. In compound (C) total ‘N — H’ bonds is :
(a) 2
(b) 3
(c) 4
(d) 1
(c) 1 : 4
(d) 3 : 17
32. In compound (C) ratio of N — H to C — H bonds is :
(a) 4 : 17
(b) 4 : 15
33. How many carbon atoms of (B) are sp2 hybridized which are not the part of ring?
(a) 1
(b) 2
(c) 3
(d) None
(c) 1 : 3
(d) 3 : 3
(c) 5
(d) 6
34. In compound (B) ratio of ‘C = O’ to C = C bonds are :
(a) 2 : 3
(b) 1 : 1
35. Total double bonds present in compound (B) are :
(a) 4
(b) 3
36. In compound (B)
(a) One ‘C’ is sp hybridized 5 ‘C’ are sp2 hybridized and rest ‘C’ are sp3 hybridized
(b) On ‘C’ is sp hybridized 7 ‘C’ are sp2 hybridized and rest all are sp3 hybridized
(c) One ‘C’ is sp hybridized 8 ‘C’ are sp2 hybridized and rest are sp3 hybridized
(d) One ‘C’ is sp hybridized and rest all are sp3 hybridized
Passage - III Succinoyl sulphathiazole was used as a chemotherapeutic agent beginning in 1942. Its structure is given below.
Answer the question from 37 to 40 37. Select the correct statement for the above compound:
(a) Both sulphur atoms are sp2 hybridized
(b) Both sulphur atoms are sp3 hybridized
(c) One sulphur atom is sp2 hybridized & other is sp3 hybridized.
(d) One sulphur is sp3 hybridized and other is sp hybridized
38. Which statement is correct about nitrogen? (a) All N atoms are sp2 hybridized (c) All N atoms are sp2 hybridized.
(b) All N atoms are sp3 hybridized (d) One N is sp3 hybridized while rest two N atoms are sp2 hybridized
6
Problems in Organic Chemistry
39. Total p bonds present in this compound are: (a) 6 (b) 7
(c) 8
(d) 9
40. Which statement is true about pentagonal ring? (a) Both sulphur and Nitrogen are sp hybridized. (b) Both sulphur and nitrogen are sp2 hybridized (c) Sulphur is sp3 hybridized however nitrogen is sp2 hybridized (d) N is sp2 hybridized but sulphur is sp hybridized.
Answer Key 1. (c)
2. (a)
3. (c)
4. (b)
5. (c)
6. (c)
7. (a)
8. (c)
9. (b)
10. (c)
11. (d)
12. (b)
13. (c)
14. (c)
15. (c)
16. (c)
17. (d)
18. (a)
19. (d)
20. (a)
21. (a)
22. (c)
23. (a)
24. (d)
25. (b)
26. (a)
27. (c)
28. (d)
29. (a)
30. (a)
31. (c)
32. (a)
33. (b)
34. (b)
35. (c)
36. (b)
37. (c)
38. (a)
39. (b)
40. (b)
SOLUTION 1. (c) CH2 = C = CH – CH = CH(–) ↑ ↑ ↑ ↑ ↑ sp2 sp sp2 sp2 sp2
2. (a)
3. (c)
4. (b)
Each ‘C’ atom present in ring is sp2 hybridized. Nitrogen containing positive charge is tertiary while other nitrogen is secondary. 5. (c)
Each ‘C’ atom is sp3 hybridized 6. (c) Carbon of CO group is sp2 hybridized rest all carbon atoms are sp3 hybridized
7
The Living World
7. (a) Here negatively charged carbon is sp hybridized. 8. (c) Nitrogen in CH3CH2CN is sp hybridized while it is sp2 hybridized in 9. (b) Carbon is sp3 hybridized in ethane while it is sp2 hybridized in CH3CH2(+) 10. (c)
11. (d) Homolog should have same functional groups. Boiling points of homolog ∝ Molecular weight Solubility of homolog in water ∝ 1 / molecular weight 12. (b)
Except double bonded carbon atoms rest all ‘C’ are sp3 hybridized. Double bonded carbon atoms are sp2 hybridized. 13. (c)
14. (c) Hybridization % s character sp3 25 sp2 33.33 sp 50 % s character ∝ electronegativity ∝
1 / Bond length
Bond length of y will be lesser than z & x because due to resonance this bond will acquire partial double bond character. 15. (c)
16. (c) Empty 'p' orbital
2 17. (d) sp — sigma bond
H H sp2—sp2π bond H
H sp2 — sp 2 sigma bond
8
Problems in Organic Chemistry
Passage – I sp2
sp2
NH2
> > > > >
O O H
sp2
N
N
CH2—CH—C—OH 2 C—N—C
N sp2
sp2
N OH
CH—NH 2
H C—OH
sp2
sp2
O
All Carbon atoms present in ring are sp2 hybridized. All Carbon atoms of C = O bond are sp2 hybridized Rest all carbon atoms are sp3 hybridized Total N – H bonds = 4 , Total C– H bonds = 12 Each N atom possesses lone pair of electrons.
Passage – II sp3 OH H CH—C N
3 sp H H sp
CH2OH
3
sp2
sp3
sp2 O H C—C
sp (A)
CN
2
sp
N sp2
N
sp2 CH2OH sp
N H
sp2
> In compound C total numbers of N — H and C — H bonds are 4 and 17 respectively. > In compound B Total number of C = O bonds = 2
Passage – III
O
O
sp2
sp2
HO —C—CH—CH 2—CHN 2 sp2 sp3
sp2
O N
sp2 sp2
sp3
S—NH sp2 sp2
sp2
O 3
sp
2
S
sp
sp 2
CH2NH2
sp3
(C)
Subjective Questions (A) Write the I.U.P.A.C names of following figures. (1)
(2)
(3)
(4)
(5)
(6)
(8)
(7) OH
(9)
O OH
OH
OH
(B) Draw the structures of the following. 1. 1, 3 – Dicyclobutyl propane 3. Cyclo butyl pentane 5. Cyclo hexan -1 D, 2 D – diol 7. 4, 7 – Epoxy- 4 hydroxy decanoic acid 9. 4 – Ethyl – 6 nitro oct – 5 en – 3 - one
(10)
2. 4. 6. 8. 10.
4– tert- butyl phenol N- methyl butanamide 7, 7 – Dimetyl bicyclo (2, 2, 1) heptane 3– chloro – 4 hydroxy pentanoic anhydride Bicyclo (2 ,1 0) pentane
10
Problems in Organic Chemistry
11. 13. 15. 17. 19.
3 – Bromo cyclo hexane carbaldehyde Cyclo butane methanoyl chloride 2 – Hydroxy ethanoic -3- bromo propanoic anhydride. 1,3.5 trimethylene cyclohexane. 2, 3 – Di carboxy cyclo pentanone.
12. 14. 16. 18. 20.
2 – Aza cyclo pentanone. 3 – (4-nitro phenyl) cyclo hexane carboxylic acid. (2, 9), (4,7) – Di epoxy decan – 1, 10 – dioic acid. Bis (2, 3 - trichloro methyl) cyclo hexane carbaldehyde. Spiro (3, 4) octane.
(C) Write the I.U.P.A.C. names of the following
(1)
(2)
(3)
(4)
(5) (CH3)3C(CH2)4COOH
(7) H3 CC[(p – Br – C6H4)]2COOH
(9)
(6) CH3CH2CH(p – Cl – C6H4) CHBrCH2CH3 CH 2 — CH — CH—CH2COCl (8) | | | COCl COCl COCl
(10)
(11)
(12)
(13)
(14)
O N
CH3 (15)
(17)
(16) CH3 – O – CH2 – CH2 – CH2OH
(18)
(19) CH3 — CH OH CH Cl CO O CO CH OH CH3
(20) H — CO NCl2
(21) HCOOCHCHOHCH3
(22)
11
Nomenclature of Organic Compounds
(23)
(25)
(24)
—— H
(26) CH3— CO — CH = CH — C — CH — CH = CH — CONH2
O (27) CH3 — CH = C H —
(29)
OH
— (CH = CH)2 — COOH (28) (CH2 = CH)4C
(30)
Multiple Choice Questions 1.
I.U.P.A.C. name of this compound is:(a) Decanoic anhydride (c) 2, 9 Di carboxy oxa cyclo nonane
2.
(b) 2, 9 Epoxy decan 1, 10 dioic acid (d) 2, 3 Di Carboxy -2-oxa cyclo nonane
I.U.P.A.C name of this compound. (a) N, N – Di methyl – 3 – chloro butanamide (c) 3 – Chloro – N, N – di methyl butanamide
(b) 4 – Chloro – N, N – di methyl pentan – 2 – one (d) 2 – Chloro – 4 – di methyl amino butanone
3. Which alphabet has I.U.P.A.C name similar to the alphabet K? (a) W (b) X (c) Y
(d) Z
4. Me2CCl CH2 CH = CH – CH = CH – COOH I.U.P.A.C name of this is: (a) 7-Chloro-7-methyloctan -2,4-dienoic acid. (b) 7-Chloro-7,7-dimethyl-2,-3-heptadienoic acid. (c) 2-Chloro-2-methyl-4,5-octadiene-7-oic acid. (d) 7-Chloro-7-methyl oct-2-3-dien-1-oic acid.
5.
I.U.P.A.C. name of this compound is:(a) Oxa – cyclo – 3 hexanone (c) 3 – Oxa cyclo hexanone
6. I.U.P.A.C. name of
(b) Oxa – cyclo – 3 – pentanone (d) 3 – Oxa cyclo pentanone is:
(a) 1, 2, 3 Trimethyl bicyclo(2, 1, 0) Pentane (c) 1, 2, 5 Trimethyl bicyclo(2, 1, 1) Pentane
(b) 1, 2, 5 Trimethyl bicyclo(2, 1, 0) Pentane (d) 1, 2, 3 Trimethyl bicyclo(2, 1, 1) Pentane
12
Problems in Organic Chemistry
7. The structure of 11’, 31’ 1” Tercyclobutane is:-
(a)
(c)
8.
(b)
(d)
I.U.P.A.C name of it is:-
(a) 2 - Carboxy ethyl but - 3 - en -1 - oate (c) 3 - Ethoxy carbonyl - buten - 4 - oic acid
(b) 3 – Ethoxy Carbonyl -but -3- enoic acid (d) None
9. I.U.P.A.C name of
(a) Hex - 5 - ene - 1 - yne (c) Hex – 1-en -5 yne
10.
(b) Hex -1 -yne - 5 - ene (d) Hex - 5 yne -1 - ene
I.U.P.A.C name of this compound (a) 2 – Chloro – 5 – bromo - hexane. (c) 2 – Bromo – 5 – chloro hexane
(b) 5 – Bromo – 2 – chloro hexane (d) 5 – Chloro – 2 – chloro hexane
O 11. I.U.P.A.C. name of
O
O (a) Propanoic anhydride. (c) Propanoic butanoic anhydride.
(b) Butanoic propanoic anhydride. (d) 4 – Oxa octan –3, 5 – dione
12. I.U.P.A.C. name of
(a) Methyl -chloro butanoate (c) 4 – methyl butanoate.
(b) 1 – Chloro pentyl ethanoate (d) Methyl -2 – chloro butanoate.
13. Suggesst the I.U.P.A.C name of CH3 – CHClCON Cl Br (a) 2,N-Di chloro - N- bromo propanamine. (b) 2-Chloro-N - Bromo -N - chloro propanamide. (c) N -Bromo – 2,N- di chloro propanamine (d) N -Bromo –N- chloro 2- chloro propanamine 14. I.U.P.A.C name of I 15.
Br (a) 4-Bromo-3-iodo cyclohexa-1,3-diene (c) 1- Bromo-2-iodo-1,3-cyclohexadiene
(b) 1- Bromo-2-iodo cyclohexa-1,3-diene (d) (b) & (c) both are correct
I.U.P.A.C. name of it is:(a) 6-(2-Chloro cyclo hexane)-5-chloro hexane (c) 3-Chloro-1-(2-chloro cyclo hexyl) hexane
(b) 5-Chloro-6-(2-chloro cyclo hexyl) hexane (d) 2-Chloro-1-(2-chloro cyclo hexyl) hexane
13
Nomenclature of Organic Compounds
16. I.U.P.A.C name of (a) 1 – Bromo propan – 1, 2 – di one (c) 2 – Oxo – 3 – propanoyl bromide
(b) 3 – Bromo propan – 2, 3 – di one (d) 2 – Oxo propanoyl bromide
17. I.U.P.A.C. name of
(a) 2, N- Di chloro butanamide (c) 2-Chloro-N-chloro butanamide
(b) N,2-Di chloro butanamide (d) N-Chloro-2-chloro butanamide
18. I.U.P.A.C name of CH3—CH—C—CH—COOH | || | OH O CONH2
19.
(a) 2-Carboxy hydroxy 3-oxo pentanamide (c) 2-Amide-4-hydroxy -3-oxopentanoic acid
I.U.P.A.C. name of it is:(a) 2-Oxo cyclo pentane carboxylic acid (c) 1-Carboxy-2-cyclo pentanone.
20. I.U.P.A.C. name of
(b) 4-Carobamoyl-2-hydroxyl-3-oxo pentanoic acid (d) 2-Carbamoyl-4-hydroxyl-3-oxo pentanoic acid
(a) Bicyclo (7,3,3) nonane (c) Bicyclo (7,5,1) nonane
(b) 1-Oxo cyclo pentan-2-carboxylic acid. (d) None
is:
(b) Bicyclo (7,1,1) nonane (d) Bicyclo (5,1,1) nonane
(b) Cyclo hexyl methanone (d) Cyclo hexylidene methanone
21. I.U.P.A.C name of
(a) Cyclo hexanone (c) Oxy cyclo hexane
22. Which is not the name of CH3OH? Methanol Methyl alcohol (I) (II) (a) III & IV (c) IV 23.
Carbinol (III)
Wood alcohol (IV)
(b) III (d) I, II, III & IV all are correct
CH2 — CH = CH2 I.U.P.A.C. name of this is:(a) allyl cyclo butane (c) 2 – Propenyl cyclobutane
(b) Propenyl cyclobutane (d) 3-Cyclo butyl propene
24. I.U.P.A.C. name of
(a) Chloro – di methoxy methane (c) Methoxy chloro ethoxy methane
(b) Dimethoxy chloro methane (d) Chloro – methoxy, methoxy methane
14
Problems in Organic Chemistry
25. Probable structure for the compound 2 – Bromo ethyl – 2 – chloro propanoate is: (a)
(b)
(d)
(c)
CH2Br
O Cl
O 26. Probable structure of the compound 3 – Chloro penta - 2, 4 – dienoic anhydride is:
(a) (CH2 = CH —
= CHCO)2O
(c) CH2 = CH — CH = CH COO CO CH =
(b) (CH2 =
— CH = CHCO)2O
— CH = CH2 (d) (a) & (c) both are correct
27. The I.U.P.A.C name of compound CH3CH2—CH—CH—CHO is :|| | O CN
(a) 2 – Cyano – 3 – oxo – pentanal (c) 2 – Cyano – 1, 3, pentandione.
(b) 2 – Formyl - 3 – oxo pentane nitrile (d) 1, 3 – Di oxo – 2 – cyano pentane
28. CH3CO CH (CH3CO) COOCH3 has I.U.P.A.C name: (a) Methyl – 2, 2 di acetyl ethanoate (c) Methyl – 2 – acetyl – 3 – oxo butanoate
(b) 2, 2 acetyl – 1 – methoxy ethanone (d) None
Bu | 29. I.U.P.A.C name of Pr—C—Ac | Et
(a) 3 – Ethyl – 3 – propyl 2 – heptanone (c) 4 – Acetyl – 4 – ethyl nonane
(b) 3 – Acetyl – 3 – propyl heptanone (d) None
30. Structure of Spiro (3, 4) octane is: (a)
31.
32.
(b)
(c)
(d)
I.U.P.A.C. name of this compound is:-
(a) 1, 4, 7, 10 – Tetraoxa dodecane (c) 12 – crown – 4
(b) 1, 4, 7, 10 – Tetraoxa cyclodecane (d) 4 – crown - 12
I.U.P.A.C. name of this compound is:- (a) 2 – cyano – 4 – chloro formyl nitrobenzene (c) 3 – cyano – 4 – nitro benzoyl chloride
(b) 4 – chloroformyl – 2 – cyano nitrobenzene (d) 3 – chloro formyl – 2 – nitro benzo nitrite
15
Nomenclature of Organic Compounds
33. Which is not correctly matched? (Compound) (IUPAC name)
(a)
(b)
(c)
(d)
Ethane
2 – (1 – methyl butyl) cyclo butanol
OH Phenol
34. I.U.P.A.C name of H2N — N
2 – cyclobutyl pentane
CO2H
(a) 4-Hydrazono cyclo hexanoic acid (c) 4-Hydrazono cyclo hexane-1-carboxylic acid
(b) 4 – Hydrazono benzoic acid (d) none of these
35. Which of the following groups is always taken as a substituent in the I.U.P.A.C nomenclature? (a) — NH2 (b) — CHO (c) — NO2 (d) — CN 36. Which prefix is not suitable for –CH=CH–CH3 ? (a) allyl (b) Propenyl
(c) 2 – propenyl
(d) (a) & (c)
37. I.U.P.A.C name of
(a) 2,2/-Di carbaldehyde-6,6/-di nitro bi phenyl (c) 6,6/-Di nitro bi phenyl-2,2/-di carbaldehyde
(b) Di nitro biphenyl di carbaldehyde (d) None
38. The compound
(a) 1, 2 – Bicyclo (2, 2, 2) hexandiol (c) Bicyclo(2, 2, 2)hexan – 2, 3 – diol
39. The compound 2 – cyano penta 2 – 4 – dienal contains. (a) 4p bonds (b) 3p bonds
(b) Bicyclo (2, 2, 2) hexan 1, 2 diol (d) Bicyclo(2, 2, 2) – 5, 6 – hexandiol (c) 5p bonds
(d) 6p bonds
40. I.U.P.A.C name of
(a) 2 – oxo – 5 ethoxy carbonyl cyclohexane carboxylic acid (b) 5 – Ethoxy carbonyl – 2- keto cyclohexane carboxylic acid (c) Ethyl – 2- carboxy – 3 – oxo – cyclohexyl methanoate (d) 3 – Ethoxy carbonyl – 2- carboxy cyclo hexanone
16
Problems in Organic Chemistry
41. Cinnamic acid can be identified as: (a) Phenyl butenoic acid (c) 2 – Phenyl -2 – Propenoic acid
(b) 3 – Phenyl prop – 2 – en -1 – oic acid (d) 3 – Phenyl propanoic acid
42. I.U.P.A.C name of crotonaldehyde is: (a) Butanal (b) Butynal
(c) But – 2 – enal
(d) 2 – Butynal
43. I.U.P.A.C name of
(a) 2 – Chloro spiro ( 2, 5) octane (c) 1 – Chloro spiro (2, 5) octane
44. The compound
(b) 2 – Chloro spiro (5, 2) octane (d) 1 – Chloro spiro (5, 2) octane can be named as:-
(a) 1 – Oxo – 2 – carbethoxy cyclo pentan (c) 2 – Methoxy carbonyl cyclo pentanone
(b) 2 – Methoxy carbonyl cyclo pentanone (d) None
45. HCOOCHBrCH2Cl can be named as: (a) 1– Bromo – 2 – chloro propanoic acid (c) 1 – Bromo – 2 chloro ethyl methanoate
(b) 2 – Bromo – 3 – chloro propanoic acid (d) 1 – Bromo – 2 chloro ethyl ethanoate
46. 47. 48. 49.
50.
Can be named as:(a) Hept–6–en–2, 4–diyne (c) Hepta–2, 4 diyn–6–ene
(b) Hepta – 3, 5 – dien–1– yne (d) Hept–1–en–3, 5–diyne
can be named as:(a) Butyl cyclo hexane
(b) Pentyl cyclo hexane
(c) Cyclohexyl butane
(d) Cyclohexyl pentane
(c) Cyclohexyl octane
(d) Octyl cyclohexane
can be named as:(a) Cyclohexyl heptane
(b) Heptyl cyclohexane
can be named as:(a) 4 – Cyclohexyl butanoic acid (c) Carboxy ethyl cyclohexane
(b) 3 – Cyclohexyl propanoic acid (d) Carboxy propyl cyclo hexane
can be named as:-
(a) 2-hydroxy butyl pentane (c) 2-butyl cyclobutanol
(b) 2-hydroxy butyl hexane (d) 2-pentyl cyclobutanol
17
Nomenclature of Organic Compounds
Answer Sheet 1. (b)
2. (c)
3. (b)
4. (a)
5. (c)
6. (b)
7. (c)
8. (b)
9. (c)
10. (c)
11. (b)
12. (b)
13. (b)
14. (b)
15. (d)
16. (d)
17. (c)
18. (d)
19. (d)
20. (d)
21. (d)
22. (d)
23. (d)
24. (a)
25. (d)
26. (a)
27. (b)
28. (c)
29. (a)
30. (d)
31. (c)
32. (c)
33. (b)
34. (c)
35. (c)
36. (d)
37. (c)
38. (c)
39. (c)
40. (d)
41. (b)
42. (c)
43. (d)
44. (c)
45. (c)
46. (d)
47. (a)
48. (a)
49. (b)
50. (d)
SOLUTIONS (A) Write the I.U.P.A.C names of following structures (1) 2-Cyclo butyl-5-cyclopropyl nonane (2) 7, 8-Di cyclo butyl cyclododecan-1,2-dione (3) 1-Cyclobutyl-3-[1,1-di methyl ethyl]cyclotetradecane (4) 2, 5 Dicyclobutyl 1-methyl bicyclo (2,1,0) pentane (5) 8, 9 Dicyclobutyl- 4, 13-di methyl hexadecane (6) 2,3-Di methyl bicyclo (2,2,0)hexane (7) 7, 14-epoxy cyclotrideca –8, 12-dien-1, 2, 4, 6 - tetraol (8) 3, 3-Di (2-hydroxy ethyl) pentan - 1, 5-diol (9) spiro (3, 5) nonan - 1, 3 - di thiol (10) 2, 4 Difluoro - 2, 4 di methoxy pentane (B) Draw structures of following (1)
(2) (3)
(4) CH3(CH2)2CONHCH3 (5)
(7)
(10)
(6)
(8) (CH3CHOHCHClCH2CO)2O (9)
(11) (12)
(13)
(14)
(16)
(17)
(19)
(20)
(15) HO-CH2COOCOCH2CH2Br
(18)
18
Problems in Organic Chemistry
(C) Write the I.U.P.A.C. names of the following
(1) (3, 6), (4, 5)-Di epoxy octan-1, 8-di carboxylic acid
(2) 2, 3, 5-Trimethyl bicyclo (2, 1, 0) pentane
(3) 2-[4-cyano phenyl] cyclohexanone
(4) N-Ethyl N-propyl butanamine
(5) 6,6-Dimethyl heptanoic acid
(6) 3-Bromo-2-(4-chloro phenyl) heptane
(7) 2, 2-di – [4-bromo phenyl] propanoic acid
(8) Butan-1, 2, 3, 4-tetra carbonyl chloride
(9) 3-cyclopropyl buta-1, 3-dien-1, 4-di carboxylic acid
(10) 11’, 2’1’’, 2’’1’’’-Quarter cyclopropane
(11) 2-Chloroformyl-3-methyl butanoic acid
(12) 2-Aza cyclohexanone
(13) Thia – cyclohexan-2-ol
(14) N-methyl aza-cyclohexanone
(15) 1-aza-4-oxa cyclohexane
(16) 3-Methoxy propanol
(17) 1-[1-methyl ethyl]-3-cyclopropyl cyclohexane
(19) 2-Chloro-3-hydroxy butanoic-2-hydroxy propanoic anhydride
(20) N, N-Di chloro methanamide
(21) 2-hydroxy propyl methanoate
(22) Propane-1, 2, 3-tri nitrile
(23) 12-crown-4
(24) Hexa-1, 3, 5-triyne
(25) Methane
(26) 4-Hydroxy-5, 8-di oxo-nona-2, 6-dien-1-amide
(27) 6-bromo-nona-2, 4, 7-trien-1-oic acid
(28) 3, 3-Di ethylene penta-1, 4-diene
(29) 1, 2-Dicyclobutyl ethane
(30) 1-methyl butyl cyclo hexane
(18) 3, 3-Di (4-hydroxy phenyl methyl) but-3-en-1-oic acid
Main Features 1. If two compounds have same molecular formula but different structural formulae or physical or chemical properties then they are called isomers and the phenomenon of their existence is called isomerism. Isomerism is of two types (i) Structural isomerism (ii) Stereo isomerism 2. Structural isomerism: In it the two isomers have different arrangement of atoms with in the molecule. Structural isomerism is of the following types Chain isomerism (or skeleton isomerism) These isomers have same positions of multiple bond & functional group but differ from one another in the length of parent ‘C’ chain CH3 — CH2 — CH2 — CH2 — OH
n – Butyl alcohol (4C)
Isobutyl alcohol
(3C)
Position isomerism: These isomers have same length of ‘C’ chain but differ from one another in the position of multiple bonds, branch or functional group. CH2 CH — CH2 — CH2 — CH3 and CH3 — CH CH — CH2 — CH3
3. Functional isomerism: These isomers have different functional groups i.e. isomers belong to different homologues series. > CH3CH2CHO & CH3COCH3 & CH2 = CH — CH2OH (Aldehyde) (Ketone) (Alcohol) > CH3CH2CN & CH3CH2NC CH3CH2OH & CH3OCH3 > CH3COOH & HCOOCH3 > Ring chain isomerism is also an example of functional isomerism. For example n-butene & cylobutane, butyne & cyclo butene Metamerism: It is due to the difference in the nature of alkyl groups attached to the same functional group. CH3 — O — CH2CH2CH3 & CH3CH2OCH2CH3 Methyl propyl ether Di ethyl ether Tautomerism: This type of isomerism has two functional isomers present in dynamic equilibrium.
% enol content ∝ 1/ Temperature ∝ conjugation % enol content in decreasing order is as follows.
∝
H – bonding
∝
aromatic character
20
Problems in Organic Chemistry
> PhCOCH2COCH3 > MeCOCH2COMe > MeCOCH2COOC2H5 > MeCOCH2COO Et > MeCOCH2COOMe > MeCOCH3
4. Stereo isomerism: When isomers have the same structural formula but differ in relative arrangement of atoms or groups in space within the molecule, these are known as stereoisomer and phenomenon of their existence is called stereoisomerism. Stereoisomerism is of two types. (a) Geometrical isomerism: Here the isomers differ in spatial arrangements of atoms or groups around the double bonded carbon atoms.
> Calculation of geometrical isomers: Alkene R1(CH = CH)n R2 Geometrical isomers (1) When n = 1 or R1 = R2 2n
(2) When R1 = R2,
n = even
2
n –1
+2
n –1 2 ( n + 1)
–1
2n – 1 + 2 2 (3) When R1 = R2, n = odd > Geometrical isomers possess different physical properties. Their chemical properties are similar but not identical. > Due to good packing of trans isomers, it has higher melting point than cis isomer (b) Optical Isomerism (Enantiomerism): > The isomer which does not rotate the plane of ppl is called optically inactive and which rotates the plane of ppl is called optically active. The isomer which rotate the ppl towards left is known as laevo (l) while which rotates the ppl to right is known as dextro (d).d & l isomers are non super imposable mirror images of each other & are known as enantiomers The optical isomers which do not have mirror image relationship are called diastereomers > Equimolar mixture of d & l isomers of a same compound is called dl or (+) or racemic mixture. > Calculation of optical isomers: (i) Compounds which have unsymmetrical molecule with one or more chiral centres. In such compounds if n is the number of chiral carbons, then Optically active forms = 2n, Enantiomeric pair = 2n / 2 Optically inactive form = 0
(ii) If it is symmetrical & n = even then optically active forms = 2n–1 enantiomeric pair = n –1
1 n –1 (2 ) 2
Optically inactive forms = 2 2 (iii) If is symmetrical & n = odd then Optically active forms = 2n – 1 – 2
n –1 2
= a (say)
Enantiomeric pair = a/2, Optically inactive form = LEVEL – I
n –1 2 2
Subjective Questions 1. Draw the Fischer projections of the following.
(a)
(b) (c)
21
Isomerism
(d)
(e)
(f)
CH3
OH
CO (i) (h) H H NH2 NH
(g)
D
( j)
2. In Q.no.1 which is / are meso forms? 3. Assign the names R or S to the following.
(a)
(b)
(c)
Cl (d)
(e)
(f) O
O CH2I
(g)
(h)
(i)
Cl (j) H
CH2OD (k)
(l)
D
(m)
(n)
(o)
22
Problems in Organic Chemistry
4. Assign the names E & Z to the following.
(a)
(b)
(c)
Cl—CH2 (d)
(e)
C CH2COCH3 (f) C Br NHCOCH 3
HOH2C (h) C OHC
(g)
( j)
HOH2C
CH2CN C
(i)
CH2CH2NH2
CH2CN C
C
OHC
CH2CH2NH2
Multiple Choice Questions 1. Which among the following can show both the stereoisomerisms i.e. optical & geometrical?
(a)
(b)
(c)
(d)
2. What is the relation between the following three structures?
(a) 1 & 2 are enantiomers (c) 2 & 3 are structural isomers
(b) 2 & 3 are enantiomers (d) 1 & 2 are homomers
3. I.U.P.A.C name of
(a) 2 – Chloro cyclohexan-1-ol (c) 1-Chloro cyclohexane-2-ol
4. D – form of the compound (a)
(b) 1 D-Chloro cyclohexane-2D-ol (d) 2D-Chloro cyclohexan-1L-ol
PhCHOHCH3 can be written as:-
(b)
(c)
(d)
CH2CCl3
CH3
23
Isomerism
5. Optical rotation of the following isomer is + 36°
Which isomer will have the optical rotation equal to-36°?
(a)
(b)
(c)
(d)
6. Consider the following compounds:-
(1) CH3NO2 (2) (3) (4)
The compounds which can also exist in its enol form is / are:-
(a) 1
(b) none
(c) all
(d) 1 & 3
7. How many geometrical isomers are possible for CH3 (CH = CH)3 CH3
(a) 4
(b) 5
(c) 6
(d) 8
8. How may geometrical isomers are possible for CH3(CH = CH)3C2H5
(a) 4
(b) 5
(c) 6
(d) 8
9. An optically active compound is placed in a polarimeter tube of length 5 dm & rotates ppl by 60°. If 40 gm compound is present in 200 ml of solution then calculate specific rotation & angle of rotation of optically active compound if above solution is diluted up to 2 lit.
(a) 60° and 12°
(b) 60° and 6°
(c) 30° and 12°
(d) 30° and 6°
10. In which isomer plane of symmetry is present Br H
Cl
(a)
(b) (c)
(d) OH
H OH
H
Cl
H
COBr
11.
F
COCH3
This is:-
Br
(a) R, Z
(b) S, Z
(c) S, E
(d) R, E
H
24
Problems in Organic Chemistry
12. The structures shown below can be related as:
(a) Position isomers
(b) Diastereomers
(c) Enantiomers
(d) Homomers
13. Identify the meso isomer
(a)
14.
(b)
and (a) Chain isomers
(b) Positional isomers
(c)
(d)
are:(c) Functional isomers
(d) All of these
15. Which is highly stable?
(a)
(b)
(c)
(d)
16. The number of cis–trans isomers possible for the following compound are:
(a) 8
(b) 4
(c) 2
17.
(a) I & II are diastereomers (c) I, II & III, I are enantiomers
(b) I & III are diastereomers (d) II & III are enantiomers
(d) 16
25
Isomerism
18. If (+)-2-chloro butane had specific rotation 10.6° when pure. The amount of (–) – 2 – chloro butane in the mixture of (+) & (–)-2-chloro butane which has rotation of 7.2° would be. (a) 49.66% (b) 50% (c) 18.5% (d) 83.9% 19. Which is optically active?
(a)
(b)
(c)
(d)
20. Identify the species which is optically active.
(a)
(b)
(c)
(d) NH2Cl
21. Which among the following compound has (L) configuration?
CHO H
HO (a) H
(b)
(c)
(d)
OH CH2OH
22. The specific rotation of pure enantiomer is +36°. Its observed rotation if it is isolated form a reaction with 75% racemisation & 25% retention. (a) +36° (b) +9° (c) +27° (d) Zero
23.
AcCH2COOEt, AcCH2CO2Me, PhCOCH2Ac
(I)
(II)
(III)
Decreasing order of % enol content in the given compounds is:(a) I > II > III > IV (b) I > III > II > IV (c) I > IV > II > III
(IV) (d) I > II > IV > III
24. How many steroisomers are possible for the following?
(a) 8
(b) 6
(c) 2
(d) 4
25. Total number of cyclic structural as well as stereo isomers possible for a compound with the formula C5H10 is: (a) 5 (b) 7 (c) 10 (d) 6 26. Arrange the following compounds in decreasing order of heat of hydrogenation. (I)
(a) I > II > III > IV
(II) (b) IV > III > II > I
(III) (c) IV > I > III > II
(IV) (d) IV > III > I > II
26
Problems in Organic Chemistry
27. If stereochemistry about the double bond in the compound shown below is cis, the number of enantiomers possible for this compound would be: (a) 2
CH2BrCH – CH = CH CHBr – CH3 (b) 4
28. The structures given below are
(a) Enantiomers
(c) 6
(d) 8
and
(b) Conformational enantiomers (c) Rotamers
(d) Geometrical isomers
29. How many optically active isomers are possible for CH3(CHCl)3CH3 (a) 2 (b) 3 (c) 8
(d) 6
30. How many meso forms are possible for CH3(CHCl)4C2H5 (a) 2 (b) 3
(d) Zero
(c) 1
31. Which statement is wrong for the given compound?
(a) I.U.P.A.C. name of it is 5-Bromo-hex-3-en-2-one (c) It has Z configuration
(b) It has (S) configuration (d) All are correct
32. In which solvent % enol content of CH3 CO CH2 COCH3 will be least: (a) Benzene (b) H2O (c) D2O
(d) Hexane
Passage - 1 Multiple bonds do not possess free rotation. The isomerism generated due to restricted rotation is called geometrical isomerism. The isomer in which similar groups are on the same side of multiple bonds is called cis & isomer in which similar groups are on the opposite side of double bond is called Trans isomer. In case of C = N & N = N Multiple bond the designation cis & Trans are replaced by Syn & Anti respectively. Answer the questions from 33 to 37 33. Which is Syn ethyl phenyl ketone oxime?
CH3CH2
OH C
(a)
(b)
(c)
(d)
N Ph
34. Which does not contain centre of symmetry?
(a)
(b)
(c)
(d) (b) and (c)
27
Isomerism
35. Which will not exhibit geometrical isomerism?
CH3 (a) C Et
CHCH3 (b)
(c) Ph2N2 (d)
36. Which is cis isomer?
(a)
(b)
(c)
(d) All of these
37. Which is more stable:-
(a) Cl
Cl
Cl (b)
Cl
Cl
(c)
Cl
(d)
Cl Cl
Passage II Some times isomer does not contain chiral carbon atom but it shows optical isomerism. It is due to the presence of chiral plane. Biphenyls and cumulenes exhibit this type of optical isomerism. Cumulenes are those unsaturated hydrocarbons which contains consecutive double bonds. If cumulene contains odd no. of double bond it is always optically inactive and in case of even number of double bonds it shows optical isomerism only when it is asymmetrically substituted. Answer the questions from 38 to 40. 38. Which is optically active:
(a) CH3CHClCH3
(c)
(b)
(d)
28
Problems in Organic Chemistry
39. Which is resolvable? (a) (P — CH3C6H4) CH = C = CHCH3
(b) CHCl = C = CH2
(c)
(d) Threo tartaric acid
40. Which does not contain plane of symmetry:-
(a)
(c)
(b)
(d)
Passage III Configuration is that representation of a molecule in which relative positions of atoms or groups of atoms are shown. Configuration does not change on rotation but changes only when bond breaks & forms in new direction. If two exchanges are made in an isomer then configuration of isomer does not change(Retention). Answer the questions from 41 to 43.
D
CH3 41. (I)
(II)
(III) H
NH
Which statement is correct? (a) I & II are same (c) II & III are same
OH
H CO (IV) NH2
(b) I & IV are same (d) III & I are same
42. The configuration of
is similar to:-
Cl Cl
(a)
Cl
Cl
Cl
(b)
(c)
(d) Cl
Cl
Cl
43. Which statement is false for isomers? (a) They have same molecular formulae (c) They have same empirical formulae
(b) They have same vapour density (d) All are correct
29
Isomerism
44. Which pair of isomerism is not possible together? (a) Functional and position (c) Metamerism & functional
(b) Ring chain & functional (d) Chain & functional
45. Decreasing order of % enol content of the following compounds would be:
(1) (2) (3) (4) (a) 2 > 3 > 4 > 1 (b) 2 > 1 > 3 > 4 (c) 4 > 1 > 3 > 2 (d) 2 > 1 > 4 > 3 46. Which is not erythro?
(a) Meso tartaric acid
(b)
(c)
(d)
Ph Me ∆ → A & B. Which of the following statement is true? CO2H Me
H 47. HO2C H Ph
(a) A & B are structural isomers. (c) A & B are geometrical isomers
48.
(b) A & B are enantiomers (d) A & B are diastereomers
aq ¾¾¾ ® NaOH
+
This reaction is leading to % racemisation (a) 40% (b) 60%
49. (1)
(c) 80%
(2)
Resolvable compound is / are:-
(a) Only 2
(3)
(b) 1, 2 & 3
(c) Only 4
(d) 20%
CH3CH2CH2COONa (4)
(d) 2 & 4
50. Largest % enol content will be found in:O
(a)
O (b) (c) (d)
30
Problems in Organic Chemistry
51. The wrong statement concerning the structure A, B & C is: O Ph Et
Ph
Et
Ph
OH
CH3
OCH3
(a) (b) (c) (a) A & C are functional isomers (b) B & C can show geometrical isomerism (c) A & B are tautomers (d) Degree of unsaturations of A, B & C are not same 52. The priority sequence for the group CH3 , C14 H3 , D & H will be in the order
(a) 3 > 2 > 1 > 4
(1)
(3)
(2)
(4)
(b) 3 > 4 > 1 > 2
(c) 2 > 3 > 1 > 4
53. Alternating axis of symmetry is not present in:
(d) 2 > 1 > 3 > 4
D
Cl
Cl
NO2
H
CH3
(a)
(b) H
H (c)
(d)
D F
Br
CH3
Cl
Cl
H H
54. Maximum number of atoms present in one plane of the following isomers is: H
H C=C=C=C
HS
(a) 8
(b) 6
D
(c) 9
(d) 4
55. Optical isomerism will not be exhibited by: O Br
(a)
O
Br
(b)
(c)
P OMe OMe
D
(d)
P OMe Me
56. Which pair does not represent metamerism?
Me
HO
Cl
(a) Ph—OCO
& Ph—COO
(b)
Me
&
Cl OH
(c) CH3CH2NH2 & CH3NHCH3
(d)
&
57. How many structural isomers are found to be optically active obtained by the mono chlorination of methyl cyclo butane: (a) 1 (b) 2 (c) 3 (d) 4 58. Which will have least stable enol form?
O (a)
O
O
O
NO2 O (b) (c)
O NO2 (d) I
I
31
Isomerism 1
H
4
59. H
O Which H will involve in keto - enol tautomerism? H
O
3
H 2
(a) Only 1
(c) Both
CMe2CMe2H
Cl 60.
C
(d) none
HO
C
C
CH3
F
(b) 2 & 4
C CH2CH2CH3
ClO
Configurations of these two compounds are:(a) Z, Z (b) E, E
(c) Z, E
(d) E, Z
61. How many optically active isomers are possible for 1,3,5-tri methyl cyclohexane. (a) 2 (b) 4 (c) 6
O
NHCOMe
(d) none
OH
O 62. Configuration of
and
are:-
H
Br
(a) RR
(b) RS
(c) SS
(d) SR
63. Which will show enantiomerism? (a) O
Me Me (c)
NH (b)
Me
Br (d)
Me
Br 64. If in an organic compound two COOH groups are present on same C atom then such a compound looses one molecule of carbon dioxide on heating.
HOOC 3
HOOC
NMe2 1
In compound X configuration of C1 is:(a) R (c) may be R or S
2
Heating ¾¾¾¾ ®X
COOH (b) S (d) R & S names can not be assigned
65. The most stable form of 2 – nitro ethanol is:-
H H (a)
H OH
NO2 NO2 H (b)
H
H
H
H OH
H
NO2 H OH (d)
H (c) H
H H
NO2
H
H OH
32
Problems in Organic Chemistry
66. The most stable form of 1, 2 – Di nitro ethane is:-
H H
NO2
(a)
H (b)
H
NO2
NO2
NO2 NO2
H
H
H H 67. The most stable form of cyclo hexane 1, 2-diol is:-
H
H
H
(c)
H (d)
H
H
NO2 H
OH OH
(b)
H
H
OH
OH
H
OH (c)
H NO2
H
(a)
NO2
H
H (d)
OH OH
H H
OH 68. The most stable form of 1, 2 di chloro cyclo hexane is:-
H
Cl Cl
(a)
(b)
H
H
Cl Cl
H
H
Cl
Cl (c)
(d)
Cl
Cl H
H H
69. The most stable form of cyclo hexan – 1, 4 – di ol is:H
OH OH (b) H
(a) H
OH OH H (c)
H OH
OH
H
OH
(d) H
OH H
33
Isomerism
70. The most stable form of 4 – Bromo cyclo hexanol is:-
OH
Br (a) Br
H
OH (b) H
H
OH (c)
Br
H OH (d)
Br 71. CH3CH2CH2CHO & CH3CH—CH3 | CHO
(a) Metamers
(b) Skeleton isomers
(c) Position isomers
(d) Homomers
72. Total number of aldehyde & ketone possible for C4H6O are: (Excluding stereo isomers) (a) 10 (b) 6 (c) 8 73.
and
O H
Ac
(d) 7
are:-
Ac
O H
(a) Enantiomers
(b) Geometrical isomers
(c) Diastereomers
(d) Position isomers
74. Which will have highest dipole moment?
D (a) D
Cl (b)
Cl
H
(c) H H
H
H
(d) H
Cl
H
75. How many stereo isomers are possible in A & B?
Br
Br (A)
(a) 2 & 2 respectively
(B)
(b) 2 & 4 respectively
(c) 0 & 2 respectively
(d) 0 & 4 respectively
76. The absolute configurations of chiral centers 1, 2 and 5 in the following molecule are:-
Cl
4 5
3 2
OH
1
CHO
(a) 1R 2R 5R
(b) 1S 2R 5S
(c) 1R 2S 5R
(d) 1S 2S 5S
77. Which will have very small enol content?
O (a)
O
O
O
O
O
NO2
(b) (c) (d)
O
34
Problems in Organic Chemistry ( +)
(–)
( +)
(–)
CH 2 = N —O 78. CH3 — N — O || | OH O (Nitromethane)
(Aci-nitromethane)
Nitro methane is stable than aci – nitromethane because (a) C = N bond is less stable than N = O bond. (b) Hybridisation of ‘C’ changes from sp3 to sp2 which increases electro negativity of ‘C’ (c) Nitromethane is more stabilized by resonance than aci – nitro methane is not. (d) Both are equally stable.
79. Maximum enol content is in:-
(a)
(b)
(c) CH3COCH2COCH3
80. Assertion -Conformers are not isolable at room temperature
Reason -These are unstable at room temperature (a) Assertion is True, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true
81. Assertion HDC = C = CHD can exhibit enantiomerism
Reason – It contains chiral carbon atom (a) Assertion is True, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true
82.
Assertion - Me3C-NO2 can exhibit position isomerism but not tautomerism Reason – Acidic hydrogen is absent in this compound (a) Assertion is True, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true
83.
Assertion- HDC=C=C=CHD can exhibit optical isomerism Reason- Entire molecule is planar (a) Assertion is True, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true
84.
Assertion- Resolution of meso isomer in to two optically active forms is not possible Reason- It contains plane of symmetry (a) Assertion is True, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true
(d) PhCOCH2COOCH3
35
Isomerism
85.
Assertion: - % enol content in CH3CSCH3 is greater than that of CH3COCH3 Reason:-S—H bond is weaker than O—H bond because of large size of sulphur than oxygen. (a) Assertion is True, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true
86. Consider the following compound (A)
Assertion: - Compound A readily undergoes tautomerism
Reason: - Enol form of A is stabilized by H — bonding
(a) (b) (c) (d)
Assertion is True, Reason is true: Reason is a correct explanation for assertion. Assertion is true Reason is true: Reason is not a correct explanation for assertion Assertion is True, Reason is False Assertion is False, Reason is true
87. Consider the following three compounds
Assertion: - % enol contents of I, II & III vary from 100 to 80 & 1.4 respectively Reason: - In this series H – Bond becomes longer and weaker (a) Assertion is True, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true
88. Assertion: -
If it is rotated along y axis its configuration does not change
Reason: - Configuration changes only when bond breaks and forms in new direction. (a) Assertion is True, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true
89.
Assertion: - A mixture of cis & trans but – 2 – ene can be resolved at room temperature Reason: - At room temp temperature cis transforms are not interconvertible (a) Assertion is True, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true
36
Problems in Organic Chemistry
O 90. Assertion: - CH3CH2O—S—CH3 &
are not position isomers
O Reason: - These are differing from each other due to the difference in the arrangement of CH2 group. (a) Assertion is True, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true
Answer Key 1. (b)
2. (d)
3. (d)
4. (b)
5. (c)
6. (a)
7. (c)
8. (d)
9. (b)
10. (b)
11. (c)
12. (b)
13. (c)
14. (b)
15. (b)
16. (a)
17. (c)
18. (a)
19. (c)
20. (b)
21. (c)
22. (b)
23. (c)
24. (c)
25. (b)
26. (d)
27. (a)
28. (b)
29. (a)
30. (d)
31. (d)
32. (b)
33. (d)
34. (c)
35. (d)
36. (d)
37. (a)
38. (d)
39. (a)
40. (c)
41. (b)
42. (c)
43. (d)
44. (c)
45. (d)
46. (d)
47. (d)
48. (c)
49. (d)
50. (d)
51. (d)
52. (d)
53. (c)
54. (a)
55. (b)
56. (b)
57. (a)
58. (d)
59. (b)
60. (c)
61. (d)
62. (a)
63. (d)
64. (d)
65. (c)
66. (a)
67. (d)
68. (c)
69. (d)
70. (a)
71. (b)
72. (d)
73. (d)
74. (a)
75. (d)
76. (d)
77. (b)
78. (c)
79. (a)
80. (c)
81. (c)
82. (a)
83. (d)
84. (b)
85. (b)
86. (c)
87. (a)
88. (d)
89. (a)
90. (a)
Multiple Choice Questions (More Than One May Correct) 1. Which of the following compounds will show geometrical isomerism? (a) Pent – 2 – ene (b) 1 – Chloro hexene (c) 3, 4 – Diethyl hex – 3 – ene (d) 1, 2- Di chloro ethene 2. Aldehydes and ketones on reaction with NH2OH form oximes which can exhibit geometrical isomerism. R1
R1 C
O + NH2OH
H2O +
R2
C
N—OH
R2
Which among the following aldehyde and ketone will show geometrical isomerism after treatment with NH2OH ?
(a)
(b) (c) (d)
3. Which among the following is not ‘E’ isomer? (a)
(b)
(c)
(d)
37
Isomerism
4. Which statement is not in favour of the following compound?
(a) It can show geometrical isomerism (c) It possesses plane of symmetry
(b) It can show optical isomerism (d) Rotation about C1 – C2 & C3 – C4 is restricted rotation
5. Select the objects which are achiral due to the presence of plane of symmetry.
I (a)
(b)
(c)
(d) Br
6. Select the objects which are achiral due to the presence of plane of symmetry as well as centre of symmetry.
(a)
(b)
(c)
(d) None of these
7. Which is not correctly matched?
(a)
&
(b)
(c)
are enantiomers
&
are position isomers
&
are functional isomers
(d)
and
8.
Products (mono chlorination)
are enantiomers
Select correct statements regarding above reaction. (a) 4 Structural isomers are produced (b) 5 Structural isomers are produced (c) Total number of isomers obtained in this reaction are 6 (d) Two structural isomers are optically active
Cl
38
Problems in Organic Chemistry
9. The compound which can exhibit geometrical as well as optical isomerism is/are:CH I (b) 2 (c)
(a)
(d) None of these
10. An optically active compound is/are:-
(–) Br (a)
(–)
NH3Br (b)
(c)
(+)
N
Et
Br
(d) None of these
(+)
CD3 Br 11. For the formula C3H6O (a) Maximum number of possible alcohols is ‘4’ (c) Maximum number of possible ether is 2
(b) Maximum number of possible Ketone is 2 (d) Maximum number of possible aldehyde is 2
12. Compound(s) whose enol form is not possible is:
(a)
(b)
(c)
(d)
13.
Select correct statements (a) I & II are identical
(b) II & III are identical
(c) II & III are diastereomers (d) All are diastereomers
14. An optically active compound containing one chiral centre is found to exist in to two forms X & Y. The select correct statement. (a) Mixture of X & Y can be separated by distillation. (b) Equimolar mixture of X & Y is found to optically inactive due to internal compensation. (c) If 40% X & 60% Y are mixed, 80% racemisation will take place. (d) Meso form of this organic compound is not possible.
Answer Key 1. (a), (b), (d)
2. (c), (d)
3. (a), (b), (d)
4. (a), (b)
6. (a), (b), (c)
7. (a), (b), (d)
8. (a), (c), (d)
9. (d)
11. (a), (c), (d)
12. (b), (c)
13. (a), (b)
14. (c), (d)
5. (b), (c), (d) 10. (b), (c)
39
Isomerism
LEVEL – II
Multiple Choice Questions 1. You have following four compounds CH3CH2CHO CH2 = CHCH2OH CH2 = CHOCH3 (1) (2) (3) (4)
Select the correct statement (a) 2 & 3 are functional isomers as well as position isomers (b) 1 & 3 are chain isomers (c) All are functional isomers (d) 3 & 4 are position isomers
2. Which is not resolvable?
(a) EtNHCH3 (b)
(c) POClBrI
(d) All are resolvable
3. Identify the compound which has the least value of heat of combustion (a)
(b)
(c)
(d)
4. You have given the following transformations. Trans –but-2-ene
Cis-but-2-ene………….…….………....(1)
Anti-n-butane
Gauche-n-butane…....….…………………(2)
R-butan-2-ol
S-butan-2-ol ....…………….………..……..(3)
S-butan-2-ol
R-butan-2-ol .....…….…..….……………....(4)
For which transformation least activation energy is needed: (a) 2 (b) 1 (c) 3
(d) 4
5. You have four compounds
(1)
(2)
(3)
Geometrical isomerism will be exhibited by:(a) 2 & 4 (b) 1, 2 & 4
(4)
(c) All
(d) None
C is isomer will be more stable than trans isomer when (a) n > 10 (b) n < 10
(c) n < 13
(d) n = 10
(A) & (B) are:(a) Rotamers
(c) Enantiomers
(d) Structural isomers
(CH2)n 6. General representation of cyclo alkene is
C H
C H
7.
(b) Geometrical isomers
40
Problems in Organic Chemistry
8. Consider the following four isomers.
(1)
C2H2
(2)
(3)
Which is correct:-
(a) 2 & 4 possess centre of symmetry
(b) 1, 2, 3 & 4 contain axis of symmetry
(c) 2 & 3 possess alternating axis of symmetry
(d) 4 contains centre of symmetry while 2 contains plane of symmetry
BF3 (4)
9. Consider the following compounds
(1)
(2)
Which option satisfy for the above two?
(a) Both can show geometrical as well as optical isomerism
(b) Both can show optical isomerism.
(c) Both are chiral but only 2nd can show geometrical isomerism
(d) Both can show geometrical isomerism.
10. Consider the following compounds.
SO3H (1)
NO2 (2)
(3)
(4)
(C2H)4
Optically active compound is / are:(a) Only 1 (b) 1 & 2
(c) 4, 2 & 3
(d) None
Matrix Match Questions Each question contains statement given in two columns which have to be matched. Statement (A, B, C , D) in column I have to be matched with statements (p, q, r, s) in column II. The answers to these questions have to be appropriately bubbled as illustrated in the following examples. If the correct matches are A – p, A – s, B – r, B – q, C – q, D – S, then the correctly bubbled 4 x 4 matrix should be as follows. A
p
q
r
s
B
p
q
r
s
C
p
q
r
s
D
p
q
r
s
41
Isomerism
11.
Column – I
Column – II
SH Br (A)
(p) R
H
(q) S
(r) E
H
Br
Br HO
H
(B) HO
Br
OH (C) D H
Br CH2I (D) 12.
(s) Z
Column – I Cl (A) No. of positional isomers of
Column – II (p) 2
(B) Total structural isomers for Pr – NH2 (q) 3 (C) Optically active isomer for dichloro cyclo hexane (r) 4 (D) Structural isomers for nitrotoluene (s) 1 13. Matrix Match Pair of isomers Relation Br
H
I
(A)
and
I
(p) Enantiomers
H Br CH3
Br
CH3 CH3
(B)
CH3
Br
H
H
and
H
Br
Br
H
(C)
and
(r) Homomers
(s) Position isomers
Br
CH3
Br
CH3 and
(D) Br
(q) Diastereomers
Br
CH3
CH3
42
Problems in Organic Chemistry
14. Column – I Column – II (A) CH3COCH2D (p) optically inactive but its enolic form can show geometrical isomerism O (B) CH3CH2CHClCCH3
(q) optically active & its enolic form can show geometrical isomerism
Et Br (C)
Br
H
H
(r) can show metamerism
Et (D) CH3CH2COCH2CH3
(s) can show position isomerism
Answer Key 1. (c)
2. (a)
3. (a)
4. (a)
5. (d)
6. (b)
7. (a)
8. (c)
9. (b)
10. (b)
Answers Matrix Match 11. 12. 13. 14.
(A) – q, r, (B) – p, q (C) – q (D) – r (A) – q (B) – r (C) – p (D) – r (A) – r, (B) – s, (C) – p, (D) – r (A) – p, (B) – q, r, s (C) – r, s, (D) – p, r, s
SOLUTION LEVEL - I 1. Draw the Fischer projections of the following. CH3 OH
H (a) H
H
Cl (c) Br I
Cl (b) H Cl
CH3
I
Br (e) I
H (g) H OH
OH Cl (f) NH2
Cl
Cl
H Br
H
I
CH3 NH (h)
H CO
H
D
I
Br
Br CH3
Br
NH2 ( j)
Cl
Br (d)
Br
NH2
CH3
H (i)
CH3
CH3
OH
Br
CH3
Br
H
H
CH3
43
Isomerism
2. Only (g) is meso isomer as it has plane of symmetry 3. (a) F(1), CONH2(2), D(3), H(4) by using double exchange phenomenon configuration is ‘R’ (b) Cl(1), ethynyl(2), ethenyl (3), ethyl(4) by using double exchange phenomenon configuration is ‘ S’
O
3
2
3
3
COBr
1
(c)
2
NH2
4
4
1
1
2
R configuration
4 3
1 Cl (d)
3 (3)
→
2
→
(2) (1)
O
1 R configuration
4
O
4
2 (4) (e) CH2I(1), CH2F(2), CH2CH2I(3), CH2CH2CCl3(4) by using double exchange phenomenon configuration is ‘R’
Cl 2O (f)
(g) (h) (i) ( j)
4 3
same as (d), S configuration
CH2I 1 Cl(1), F(2), ethyl (4) – configuration is ‘S’ OH(1), NHCOCH3(2), COCH3(3), CH3(4) – configuration is ‘R’ Same as (c) configuration is ‘R’ H(4), Cl(1), D(3), CH2OD(2) by using double exchange phenomenon configuration is ‘S’
(k)
(l) Same as (k) it is SS
CH3
3
CH3
CH3 H
4H
Br
H
Br
3
by using double exchange phenomenon it is S configuration
by using double exchange phenomenon it is S configuration
3
CH3
CH3 H
H CH3
CH3
CH3
(n) H
S, S
2
OH
Br
1 Br
Br1
Br
2
→
(m)
4H
Br1
4
2
Br CH3 by using double exchange phenomenon it is R configuration
Br
2
H R, R
4H
Br 1 CH3
3 by using double exchange phenomenon it is R configuration
44
Problems in Organic Chemistry
(o) Same as (k) (1) & (m) it is (R, S)
O
O
H
2
Br 4. (a) I
COBr 1 E H
1
2
(b)
E 2
1
1
NH
E 1D
(c) H 2
C
C 1
1 2 H NH 2
Z C
COBr1 D
C
C
C 2
C
CHO 2 H Cl
F
C2
ans-E, Z, Z
C
C CHO
C 2
C F 1
COBr
H NH2
Cl 1
Z
2 2
CONHCH3
HO
(d)
C FO
1
2
2
NHCOCH3
1
1
Z (e) CH COCH
1
C
2
3
Z
NHCOCH3
(f) Same as previous it is Z. (g) Here out of lone pair & OH, hydroxyl group will get 1st priority. It is an‘E’. (h) CHO (1), CH2OH (2), CH2CN (1), CH2CH2CN (2) it is E 2
H 1
(i) It is E
1
H2 1
(j) It is E
HO18— CH2
OH 2 C=C OCl
HO—CH2
2
1
Answers to Objective Questions CH3
CH3
H
CH3
1. (b)
Trans isomer is optically active.
H
H (cis)
CH3 H (Trans)
while cis isomer is optically inactive due to the presence of plane of sysmmetry. 2. (d) 1 & 2 are identical H D D Cl Cl
H D D
Cl
Cl H
{1}
(rotate entire molecule to 120°) 3. (d) 4. (b) Phenyl group should be kept at the bottom of vertical line
H
{2}
45
Isomerism
5. (c) Because proposed isomer & (c) are enantiomers
H
COOH H Cl
Cl H COOH
(R, R)
(S, S)
Mirror
HOOC H
Cl Cl COOH
OH
O
6. (a) CH3 — N O (+)
CH2
N
(+)
O
7. (c) 2n–1 + 2(n – 1) / 2 n = Number of double bond = 23–1 + 2(3–1) / 2 = 4 + 2 = 6 8. (d) 2n = 23 = 8 9. (b) Specific rotation is a constant quantity for a particular isomer hence it should be 60° and can be calculated as: sp. rotation =
q× 100 60°× 100 = = 60° C× 20 × 5
from q1V1 = q2V2, 60 × 0.2 = q2 × 2, q2 = 6°. 10. (b) Both the carbon atom & Boron present in the ring are sp2 hybridised. (2)
'E'
H 11. (c)
F
(1)
COBr
COBr
(3) (4)
COCH3
F (2) 'S'
(2)
Br (1)
COCH3
Br (1)
12. (b) There is no mirror image relationship between the two isomers. CO2H CO2H H OH H OH 13. (c) Fischer projection of it is:= H OH HO H CO2H CO2H 14. (b) Here length of carbon chain remains same but position of methyl group is changing. OH
H3C
OH
H3C
H
15. (b) It is stabilized by H – bonding
H 16. (a) 17. (c) I = (R, R) configurations II = (S, S) configuration III = (S, S) configuration Hence I, II & III, I are enantiomeric pairs 18. (a) suppose wt of mixture is 100 gm Let amount of (+) form be x gm then amount of (–) form will be (100 – x)gm specific rotation of (+) form = +10.6° specific rotation of (–) form = –10.6°
46
Problems in Organic Chemistry
x (+10.6°) + (100 – x) (–10.6°) = 7.2 x = 50.34 % (+) 2 chloro butane = 50.34 % (–)2 chloro butane = 100 – 50.34 = 49.66 % 19. (c) Except (c) rest all possess plane of symmetry 20. (b) It is free from all type of symmetry elements while (a) possesses centre of symmetry & (c) & (d) possess plane of symmetry. 21. (c) First chiral centre from the bottom is considered
CHO H L–configuration H OH
OH OH H CH2OH
22. (b) Racemic mixture has zero specific rotation thus, observed rotation will be due to retention i.e. 36° x 25 / 100 = 9° 23. (c) Enol form of I is highly stable as its enol form is aromatic.
O
OH
Enol forms of II , III & IV is stabilized by H- bonding. Enol form of IV is stable than that of II & III because of more resonance (due to phenyl group)
H OH | PhC CHCOCH3 PhCOCH2 COCH3
O
OR
O
PhC CH
C — COCH3
24. (c) Both the rings are perpendicular to each other. 25. (b) 26. (d) Stability ∝ 1 / Reactivity ∝ 1 / heat of hydrogenation. 27. (a) 28. (b) Since these conformations have mirror image relation 29. (a) Chiral Carbon = 3 Optically active isomers= 2n – 1 – 2n – 1/2, = 23 – 1 – 23 – 1/2, = 2 30. (d) It does not have plane of symmetry. (2)
(1)
H
COBr
(3) (4)
31. (d)
F
COCH3 (2)
Br
F (2)
COBr COCH3
Br
'E'
(1)
(1)
(4) In Fischer projection
(1) 'S'
(2) (3)
32. (b) ‘CO’ group undergoes hydrogen bonding with H2O. 33. (d) OH & ethyl are on the same side of double bond 34. (c) 35. (d) 36. (d) In chair form substituents at 1, 3 positions are cis when these are located at equatorial position. Similarly substituents at 1, 2 positions are cis when one is located at equatorial position while other at axial position
47
Isomerism
37. (a) Minimum van-der-waal repulsion
38. (d) Possesses plane of symmetry.
H
H C
C
C
39. (a) H3C
CH3
It does not contain plane of symmetry. 40. (c) It contains chiral plane 42. (c) Same as question no.17 44. (c) Factual question
41. (b) Same as question no. 17, 43. (d) Factual question 45. (d)
H D
H
CH3
Me
46. (d)
Me
CH3
H3C
D
D
H
Threo
H
47. (d) On heating one mole of carbon dioxide comes out.
Ph
Ph
Me
H
H H H
–CO 2
CO2 H
HO2C
Ph Me CO2H Me
H HO2C H
&
Me H Me
Me
H
Ph (B)
Ph
Ph
(A) Both contains plane of symmetry & hence optically inactive. Me 48. (c) H
Br Et A
Me
Me aq. NaOH
OH
H
+
HO Et
Et (40%)
B
H (60%)
C
100 g of A gives 40 g of B & 60 g of C. Now 40 g of B will undergo racemisation with 40 g of C. Hence total wt of racemic mixture will be 40 + 40 = 80 thus, 80 % racemisation will occur. 49. (d) 1st contains plane of symmetry & 3rd does not contain chiral carbon atom. 50. (d) e nol form is more stable because of extended resonance. 51. (d) Degree of unsaturation of all isomers is ‘5’. 52. (d) Higher the atomic number higher is the priority. When atomic numbers are same then higher the atomic weight higher is the priority 53. (c) 54. (a) 55. (b) It contains plane of symmetry. 56. (b) Because these are position isomers 57. (a) 58. (d) Enol form is not stabilized by H-bonding because Iodine can not form H-bond.
OH 59. (b) Bridge ‘C’ of bicyclo compound can not have double bond until ring contains 8 ‘C’ atoms possible. 60. (c) Out of cyclopropyl & propyl group former gets higher priority.
2 HO C ClO 1
1 C CH2CH2CH3 2
is not
E
48
Problems in Organic Chemistry
CH3
CH3
61. (d) CH3
CH3
CH3
CH3 plane of symmetry
plane of symmetry
2 NHCOMe O 62. (a)
O 1
by using double 3 exchange phenomenon it is R configuration
2
4
Br1 63. (d) It does not contain plane of symmetry. 64. (d) Because after heating two valencies of C1 are identical.
HOOC 3 NMe2 HOOC 1 2
Heating –CO 2
2 OH 2
1
3 R configuration
3 H4
HOOC
NMe2
H
H COOH We can not assign priorities to the valencies of C1 because 2nd & 3rd carbons are identical. Thus, R-S nomenclature is not possible. NO2 COOH
H
1
OH intra mol- H -bond
65. (c) It is stabilized by H bonding.
H
H H
66. (a) Nitro group can not form H bond with hydrogen of C-H bond thus, anti form is more stable 67. (d) Equatorial OH are trans to each other & can form H bond with each other. 68. (c) Both Cl are trans to each other.
OH
OH 69. (d)
H
Intra molecular - H - bond H
70. (a) Since intra molecular - H - bonding is not possible thus, chair form will be more stable than boat form. 71. (b) Because position of CHO group does not change & length of carbon chain is affected.
72. (d)
, 7 isomers
3 2 1 73. (d) O
4 5
5
Ac H
3 2
4
1 O
Ac H
49
Isomerism
1 D 74. (a)
µ res = µ1+ µ2
Cl 2
D
75. (d) A is optically inactive & can not show geometrical isomerism while B is optically active (one d & one l) & can show geometrical isomerism (one cis & one trans). 76. (d) See q.n.3 (c) 77. (b) Because it has zero dipole moment & its enol form is not stabilized by H-bonding 2 O µ res = µ1– µ2 = 0
1O 78. (c) Nitromethane is stabilized by resonance
O 79. (a)
OH Enol form is aromatic & stable.
H 80. (c) Conformational energy is very small so one conformer can convert itself in to other at room temperature thus, these are not isolable at room temperature. 81. (c) It contains chiral plane
H
H CHIRAL plane
C = C =
C
D
D
82. (a) Carbon attached with nitro group does not contain H atom 83. (d) It does not possess chiral plane.
H
H C = C = C =C
D
D
84. (b) It is not the mixture of two components. 85. (b) 86. (c) Enol form is stable due to aromaticity . 87. (a) As we move from I to III distance between functional groups increases thus, H-bonds of enol forms become longer & unstable. 88. (d) 89. (a) At room temperature cis & trans forms are not inter convertible (large energy is required) 90. (a) These are mesomers
More Than One May Correct 1. (a, b, d) 3, 4 diethyl hex-3- ene can not show geometrical isomerism because double bonded carbon atom contains two ethyl Et Et groups C C Et Et
50
Problems in Organic Chemistry
2. (c, d)
3. (a, b, d)
4. (a, b) since 1st & last benzene rings contain same substituents hence it can not show geometrical as well as optical isomerism. 5. (b, c, d) In (a) centre of symmetry is present
6. (a, b, c)
7. (a, b, d) In (a) both the isomers are optically inactive and same. In (b) both the isomers are chain isomers. In (d) both isomers are identical. 8. (a, c, d) 9. (d (a) & (c) can not show geometrical isomerism while (b) can not show optical isomerism. 10. (b, c) In (a) chiral ‘C’ atom is absent 11. a, c, d 12. b, c 13. a, b
H 3. (a) 4. (a) 5. (d)
..
..
14. c, d d & l forms can not be separated by distillation. LEVEL - II 1. (c) All have same molecular formula but different functional groups. Me N 2. (a) Because of amine inversion. Et N
Me Et H
less van-der-waal forces of repulsion, more stable alkane, least value of heat of combustion Conformational energy is very small. So one conformer can convert itself in to other There is no geometrical isomerism up to seven carbon atoms. (Cyclo heptene)
(CH2)n 6. (b) H
C
C
n= 6 to 9 cis is more stable than trans n= 10 cis & trans are equally stable H n > 10 trans is more stable than cis
7. (a) Rotation about single bond is free. 8. (c) See alternating axis of symmetry in your text book. 9. (b) In both the cases rings are perpendicular to each other thus, due to absence of any symmetry element both are optically active. 10. (b) In both the cases phenyl groups are present on different plane.
4
Reaction Mechanism (General Organic Chemistry) Main Features
1. Inductive effect It is defined as the displacement of sigma electrons from less to more electro negative atom. It is of two types viz. +I & -I effects (–)
+I groups: -
O— > R3C — > R2CH — > RCH2 — > CH3 ss
s–
s+
C—C—C—C
+I group
X
(+ )
-I groups: -
NMe3 — > NO2 > CN > SO3H > CHO > CO > COOH > COCl > COOR >
CONH2 > F > Cl > Br > I > OH > OR > NH2 > C6H5 > H s+s+ s+
C—C—C—C
X
s–
—I group
Due to – I effect acidic strengths of carboxylic acids & alcohols increase F——CH 2 — COOH > Cl Cl Br < CH2—COOH because –I groups help in the dissociation of OH bond Due to +I effect acidic strength decreases because +I groups create problem or obstruction in the dissociation of OH bond Due to +I effect basic strengths of amines increases while by – I effect basic strength decrease
CH3 CH3 > NH > CH3 > NH2 > NH3
Order of basic strength (CH3)2NH > CH3NH2 > (CH3)3 N > NH3 For amines other than methyl amines R2NH > R3N > RNH2 > NH3
2. Resonance (Mesomeric effect): It occurs in following five systems.
(i) C
(v)
C—C
C
C—C
C
(ii)
(+)
C
C—C
(iii)
(–)
C
C—C
(iv)
C
C—C
(lone pair)
(Free radical)
Resonance µ ∝ Resonance energy ∝ Resonating structures ∝ Stability
3. Electromeric effect: - It is defined as the displacement of p electrons between two atoms in the presence of attacking reagent. 4. Aromaticity: - Those species which are planar, cyclic & contains (4n + 2) p delocalized electrons are called aromatic (4n + 2) p e– rule ( Huckel rule ) n = 0 2 p e–
52
Problems in Organic Chemistry
6 p e–
n = 1
n = 2 10 p e– 6 p e– , Cyclic & planar n = 3 14 p e– (Aromatic) Aromaticity ∝ Resonance energy ∝ Resonance.
5. Antiaromaticity: - The species thus called antiaromatic when shows planar cyclic and contains 4np delocalized electrons 4p e– , cyclic, planar, and hence antiaromatic. 6. Electrophile (E+):- Those species are called electrophile in which central atom has incomplete octet or central atom can receive electrons in its empty d orbital.
For eg. F+, NO2+, AlCl3, ZnCl2, PCl5, PCl3 ————— etc.
7. Nucleophile (Nu-):- These are the species in which central atom has complete octet and these can donate electrons to elelctrophiles in chemical reaction.
•• ••
••
••
••
••
••
For e.g. F ,CN ,H2 O, NH3, R — O — R & R — S — Hetc. (–)
(–)
8. Amphiboles: - These are the species which can favour the attack of (E+) or N u– on itself. If C forms multiple bond with another electronegative atom then that species behaves as amphibole.
H | For eq., CH3—C ≡ N, CH3 — C
H | O, CH3 — C
O || O, CH3 — C — OH........etc.
Free Radicalsubstitution Reaction (F.S.R.) 1. It involves three steps, chain initiation, chain propagation & chain termination. In very first step generation of free radical takes place. 2. In halogenations of alkane numbers of products depend upon the variety of hydrogen’s present in the alkane.
mono chlorination h
CH3CH 2 CH3 Cl2 CH3CH 2 CH 2 Cl Cl3 —C HCH3 |
Cl 3. The following is the decreasing order of case of abstraction of different kinds of hydrogens
Benzylic Hydrogen > allylic hydrogen > 3° — H > 2° — H > 1° — H > CH3 — H > vinylic —H
Benzylic carbon CH—CH=CH—CH 2 3
Allylic carbon
Vinylic carbon 4. Attack of free radical on C—H bond is rate determining step R — H + X* ——→ R* + H — X 5. The relative rates of formation of alkyl radicals by chlorine radical: 3° > 2° > (5.0) (3.8) Bromine radical –
1° (1.0)
1° (1.0)
3° (1600)
>
2° (82)
>
6. For allylic substitution NBS (N – bromo succinamide), NCS (N – chloro succinamide) can be used. CH3 — CH = CH 2 → Br — CH 2 — CH = CH 2 NBS hν
53
Reaction Mechanism (General Organic Chemistry)
Electrophilic Substitution Reaction (E.S.R) It is the characteristic property of aromatic compounds. General mechanism of electrophilic substitution reaction is given below.
E (+)
(+)
E
E
Slow
+ E
E
~ = (+)
(+)
(+)
(Wheland complex)
H
E
E (+)
Fast
H
(+)
+
1. Types of E.S.R. Reaction 1. Halogenation 2. Nitration 3. Sulphonation 4. Friedel craft
Reagent X2/Fe or X2 lewis acid conc HNO3 / conc H2SO4 or Acetyl nitrate or NO2BF4 or EtNO2 Oleum or H2SO4 + SO3 RCOCl / Anh AlCl3 or (RCO)2 O/Anh AlCl3 or RX / Lewis acid
2. Directive influence: - On the basis of it groups can be classified in to four types.
+ R (N H2, O H, N HR, O R, N COR...........)
–R (NO2, SO3H, CHO, COOH…………...) +I (Alkyl groups) Tautomeric (f, Cl, Br & I) Out of these four classes+ R + I & Tautomeric groups are ortho & para directing & goups of – R class are meta directing. Groups of –R & tautomeric class are ring deactivating on the other hand groups of + R & + I class are ring activating.
3. Introduction of third group – Here two cases may arise. Case I: - When both groups already present have same agreement
NO2
SO3H OH
CH3
Case II: - When both groups already present do not have same agreement then. When both groups are ortho Para directing, the more powerful group controls the orientation. The directing power of each group is in the following order. O (–) > NH2 > NR2 > OH > OMe, NHCOMe > X (halogens)
OH
Cl Me OH
NH2
OMe
When both groups are Meta directing, it is difficult to introduce a third group. At this time drastic conditions are required for the introduction of third group. The following is the general order of directing power of each group, and it is the less powerful group which controls the orientation. Me3N(+) > NO2 > CN > SO3H > CHO > COMe > CO2H
54
Problems in Organic Chemistry
When one group is ortho Para directing and the other is Meta directing then o, p directing group controls the orientation.
OH
SO3H
Electrophilic and free Radical Addition reactions 1. It is the characterstic property of C = C & C ≡ C systems. 2. In presence of sun light radicals are generated and free radical addition reaction takes place while in dark addition takes place via ionic mechanism i.e. electrophilic addition occurs. 3. Electrophilic addition follows markownikoff rule on the other hand free radical addition follows anti markonikoff rule. Br
| CH3 — CH = CH + HBr ¾¾ ® CH3 C HCH3 (95%) (Markonikoff Product)
CH3 — CH = CH + HBr ¾¾¾¾ ® CH3CH 2 CH 2 Br (95%) Sunlight
(Anti Markonikoff Product)
4. KHARSCH EFFECT: - When unsymmetrical alkene is treated with HBr in presence of active peroxide Anti markownikoff addition takes place. CH3 — CH=CH 2 ¾¾¾¾¾ ® CH3 — CH 2 — CH 2 — Br HBr (RCO) 2 O 2
(major)
5. Stereochemistry of electrophilic addition reaction:- Always anti addition takes place on alkene and alkyne Anti addition
C is form ¾¾¾¾¾¾ ® Racemic Mixture
CH3 H
CH3 C
CH3 + H
C
in CCl4
Br 2
CH3
H
Br
Br
H
+
Br
H
H
Br
CH3
CH3
Anti addition
Trans ¾¾¾¾¾¾ ® Meso isomer
CH3 CH3
H C
C
H
+ CH3
Br2
in
H
Br
CCl 4
H
Br CH3
Elimination Reactions 1. These reactions are the reverse of addition reaction. In these reactions double and triple bonds are produced.
CH2 — CH 2— Br
+ KOH
–HBr
alc
H CH 2 = CH 2 ¾¾¾® CH3 = CH 2 Br +HBr
(Addition)
CH2
CH2 (Elimination)
55
Reaction Mechanism (General Organic Chemistry)
2. Types :- Three types E1 (Unimolecular Elimination) ( +)
(–)
R — CH 2 — C H 2 ¾¾¾¾ ® R — CH 2 — C H 2 ¾¾¾® RCH = CH 2 (–) Slow
–Br
|
OH –H 2O
Br
Rate = R [Substrate] or [alkyl halides] E2 (Bimolecular Elimination) (–) OH OH H
R—CH—CH2
–
Slow
H
+
R— CH
CH2
Br
(–)
R— CH
CH2 + H2 O + Br
–
Br
Rate = k [Substrate] [Base] E1CB (Elimination via the formation of conjugate base) (–)
OH
H
R — CH — CH2 Br
(–) OH H2 O (Slow)
(–)
R — CH — CH2
R— CH
(–)
CH2 + Br
Br (Conjugate base)
Rate = R [Substrate] [Base] 3. Two rules are followed in elimination reactions. (a) Saytzeff rule: - In 1, 2 elimination reaction highly substituted alkene in formed as a major product.
H alc KOH
CH2 — CH — CH — CH3
(Minor)
Cl
H
CH2 = CH— CH2— CH3+ CH3CH
CH—CH3
(Major)
(b) Hoffman rule: - In 1, 2 elimination reaction less substituted alkene is formed as major product.
H (–) CH2 — CH — CH — CH3 OH H
CH2
CH—CH—CH +3 CH—CH 2 3 (Minor)
NMe3
CH —CH3
(Minor)
(+)
Hoffman elimination generally occurs when leaving group is poor like N+Me3, P+Me3 & F etc.
AROMATICITY LEVEL - I
Objective Questions 1. For which compound pKa is least?
(a)
(b)
(c)
COCH3
Ac
2. Consider the following compounds
(i)
(ii)
(iii)
(d)
56
Problems in Organic Chemistry
Which statement is correct? (a) (ii) & (iii) are aromatic while (i) is non aromatic (c) All are anti aromatic
(b) (i) & (ii) are anti aromatic but (iii) is aromatic (d) All are aromatic
3. Which is aromatic?
(a)
(b)
(c)
(+) (–)
(d) All of these are aromatic
4. Which is not aromatic?
Ph
Ph C (+)
N
(a)
⊕
(b)
(c)
(d) (b) & (c)
5. Consider the following two compounds.
(A)
(B)
Which statement is correct regarding the stabilities of the above two compounds? (a) Both are equally stable (b) B is less stable than A (c) B is more stable than A (d) unpredictable
6. Out of following four compounds
Ph
Ph
Ph
Ph
Ph
(+)
N–Ph
Me— N
(–)
OMe O (i)
Aromatic compounds are:(a) i & iv
N
(–)
CH3
O (ii)
O
(iv)
(iii)
(b) i, ii, & iv
(c) ii & iv
(d) ii, iii & iv
7. Consider the following compounds:(–) (+)
C if planar (B)
(A)
Select the correct Statement (a) All are aromatic (c) (A) & (C) are aromatic (B) is anti aromatic
NMe2 NMe2
(C)
(b) All are anti aromatic (d) (C) is aromatic, (B) is anti aromatic & (A) is non aromatic
8. Consider the following dissociations (–)
(I)
C1
——→
(+)
+
C1
(II)
C1
——→
(–)
+
C1
(III)
C1
——→
(–)
+
C1
(+)
(+)
57
Reaction Mechanism (General Organic Chemistry)
Identify the correct energy level diagram regarding above dissociations. I II III
P.E
II I III
P.E
(a)
III I II
P.E
(b)
P.E
(c)
T
T
II III I
(d) T
T
9. Correct order of aromaticity of following compounds will be:
(I)
(II)
N
(III)
O
S
H (a) I > II > III (b) III > II > I 10. Consider the following compounds.
(c) I ; III > II
O
O
(d) II > I > III
O
S
O (A)
S (C)
(B)
Select the correct statement (a) All are aromatic (c) (B) & (C) are aromatic while (A) is non aromatic
S
(b) Only (B) is aromatic (d) Only (C) is aromatic
11. Which with release H2 gas on reaction with ‘K’ metal Ph
Ph
(a)
Ph
(b)
(c)
(d) both (a) and (c)
Ph
12. The compound which can convert itself into aromatic compound by intramolecular rearrangement is:O O O O NH HN (a) (b) (c) (d) All of these O O 13. Consider the following reactions:I (+)
1.
I
AgNO3
(+)
AgI ↓ +
NO3(–)
2.
AgBF 4
(–)
BF4
+
AgI
(–)
3.
EtOK
(+)
K
If rate of these reactions are R1, R2 and R3 respectively then (a) R1 > R2 > R3 (b) R3 > R1 > R2
(c) R3 > R1 = R2
(d) R2 > R3 > R1
58
Problems in Organic Chemistry
14. Identify the species which is not aromatic:-
(+)
(a)
(+)
(b)
(c)
Fe
(d)
15. Pick out the most acidic compound:-
H
H
(a)
H
H
H
(b)
H
(c)
H
H
(d)
16. Which of the following species is least stable?
NH
O
(a)
(b)
O
(c)
(d)
Passage – I Cyclic compounds with planar geometry & (4n +2) p delocalized e– are called aromatic while anti aromatic if they contain 4np delocalized e–. Consider the following compounds.
Answer the question from 17 to 22 17. Which compound is aromatic? (a) Only A (b) B, C
(c) Only C
(d) C & D
18. Compound which is antiaromatic (a) A (b) Only B
(c) B, C & D
(d) B & C
19. Which is neither aromatic nor anti aromatic? (a) B (b) D
(c) A & B
(d) Only A
20. Compound with 4np e– is/are: (a) B & F (b) A, B & C
(c) B & D
(d) B, C & D
21. Compound with all 4np delocalized electron is/are (a) B & C (b) Only B
(c) B & D
(d) None
22. Which does not have planar geometry? (a) B (b) C & D
(c) Only A
(d) A & D
59
Reaction Mechanism (General Organic Chemistry)
Passage – II Reagents LiAlH4, SeO2 & PCC can do following transformations. OH
O ||
|
LiAlH 4
® CH3 — CH — CH3 CH3 — C— CH3 ¾¾¾¾ 2 ® CH = CH — CH OH CH 2 = CH — CH3 ¾¾¾ 2 2
SeO
O
OH |
||
CH 2 = CH —CH — CH3 ¾¾¾ ® CH 2 = CH — C— CH3 PCC
Answer the question from 23 to 25 23. The compound which can be converted into aromatic from non aromatic by SeO2 & PCC is:
(a)
(b)
(c)
(d)
24. The compound which can be converted into anti aromatic by treating with SeO2 & PCC is:
(a)
(b)
(c)
(d)
25. The compound which is aromatic but becomes non aromatic when reacts with LiAlH4
O
O
O
(a)
(b)
(c)
(d)
O
Passage – III Consider the following compounds (+)
..
N
N
H
H H CH3O2C
OCO
Answer the question from 26 to 30 26. How many rings are aromatic in this compound? (a) 4 (b) 3
(c) 2
(d) 5
27. Number of delocalized electron present in pentagonal ring are: (a) 2 (b) 4 (c) 6
(d) None
28. How many rings have planar geometry (pentagonal as well as hexagonal)? (a) 6 (b) 4 (c) 3
(d) 2
60
Problems in Organic Chemistry
29. Total delocalized p electron present in the rings of compound are: (a) 18 (b) 14 (c) 12
(d) 20
30. If pentagonal ring containing nitrogen has ‘x’ delocalized p electron & hexagonal ring containing nitrogen has ‘y’ p electron then (a) x > y (b) y > x (c) x = y (d) None
Passage – IV Acid strength of organic compounds can be compared by comparing the relative stabilities of conjugated base or anion produced after the release of H+ ion
HA
H(+)
+
A(–) (conjugate base)
Higher is the stability of conjugated anion more acidic will be the acid. Acid chaacter a 1 / pKa Consider the following compounds
(I)
(II)
(III)
(IV)
Answer the questions from 31 to 33 31. Which among the following can release H2 gas on reaction with NaH?
(a) I
(b) II
(c) IV
(d) III and IV
(b) IV
(c) I
(d) III
32. Which will have least pKa?
(a) II
33. Which among the following will favour the release of hydride ion?
(a) II
(b) IV
(c) Only I
(d) Both I and III
34. Which statement is correct about the following two compounds?
CCl2
(–)
O
(A)
(B)
(a) Both are aromatic (c) A is aromatic while B is antiaromatic
CH3
(b) Both are antiaromatic (d) A is antiaromatic while B is aromatic
35. Which is aromatic?
CMe2
(a)
O
NH2
(+)
(–)
(b)
(c)
N
(d)
N H
O
61
Reaction Mechanism (General Organic Chemistry)
36. Consider the following compounds
(ii)
(i)
(iii)
If dipole moments of (i), (ii) & (iii) are x, y & z respectively then:(a) x > z > y (b) x = y = z (c) y = z > x
(d) z > x > y
37. Consider the following compounds
••N
••N—H (A)
S
N (B)
Highly Aromatic compound is (a) A (b) B
O
N (C)
(c) C
(d) all are equally aromatic
38. Consider the following reaction ••N
H
••N—H
+
(A) Which is correct about A
(a) A is aromatic (c) A is non aromatic
39. Assertion:-
(b) A is antiaromatic (d) It is nonaromatic but planar
This compound has large dipolemoment
Reason: — C = C bond between two rings on dissociation makes the entire molecule aromatic & stable (a) Assertion is True, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true
40.
Assertion: - Pyrrole is aromatic however it contains sp3 hybridised N Reason: - It is cyclic and contains six p delocalised electrons (a) Assertion is True, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true
41.
Assertion: - Cyclopropene is hydride ion donor Reason: - The carbocation formed after the removal of hydride ion from cyclopropene is stabilized by resonance (a) Assertion is True, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true OH O OH O (+)
H
(+)
42. Assertion:-
;
O
OH
(i)
(ii)
H
O
OH (iii)
(iv)
62
Problems in Organic Chemistry
Compound (i) readily isomerises to hydroquinone (ii) when treated with dilute acid but (iii) could not be isomerises to (iv) Reason: - (ii) is aromatic while (iv) is non aromatic (a) Assertion is True, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true
Ph 43. Assertion:-
Ph
H
COO–t–Bu
Ph
H
(i)
(ii)
Ph
COO–t–Bu
Compound (i), upon treatment with t – BuOK/ t – BuOD for 31 hours gave 5% D exchange at the starred hydrogen. Compound (ii), similarly treated, and gave 22% exchange in 30 min. Reason: - Presence of double bond makes (i) stable & irreactive (a) Assertion is True, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true
Matrix Match Questions Each question contains statement given in two columns which have to be matched. Statement (A, B, C, and D) in column I have to be matched with statements (p, q, r, s) in column II. The answers to these questions have to be appropriately bubbled as illustrated in the following examples. If the correct match are A – p, A – s, B – r, B – r, C – q, D – S, then the correctly bubbled 4 × 4 matrix should be as follows.
44. List – I (molecule / ion)
A
p
q
r
s
B
p
q
r
s
C
p
q
r
s
D
p
q
r
s
List – II (Properties)
(A)
Br
(p) Anti aromatic
(B)
Br
(q) Aromatic
(C)
(+)
(r) Non aromatic
Br
(s) Gives ppt of AgBr with AgNO3
(–)
(D)
63
Reaction Mechanism (General Organic Chemistry)
45. List – I
List – II
(A)
(p) Aromatic
(q) Non Aromatic
(r) Evolve H2 with NaH
(s) Anti aromatic
(+)
(B)
O
(C)
(D)
B—H
46. List – I
List – II
(A)
(p) Non planar
(q) Non Aromatic
(r) Aromatic
(s) (4n + 2)p delocalised electrons
(–)
(B)
(C)
(D)
(+)
47. Column – I
(A)
(B)
Column – II
(p) Aromatic
(q) Non aromatic due to loss of planarity
O EtO—C
NEt2
(C)
O
(D)
(r) Non Aromatic because Huckel rule is not followed.
C—OEt
Et2N
(–)
(s) Anti aromatic
64
Problems in Organic Chemistry
48. Column – I O (+) (A)
(P) aromatic
(B)
(Q) highly unstable ion
(C)
(R) highly unstable molecule
(S) highly stable ion
O
Column – II
(–)
(D)
Answer Key 1. (d)
2. (a)
3. (b)
4. (c)
5. (c)
6. (b)
7. (d)
8. (b)
9. (d)
10. (b)
11. (d)
12. (d)
13. (b)
14. (b)
15. (a)
16. (c)
17. (d)
18. (b)
19. (d)
20. (c)
21. (b)
22. (c)
23. (b)
24. (a)
25. (b)
26. (a)
27. (c)
28. (b)
29. (d)
30. (c)
31. (c)
32. (a)
33. (d)
34. (a)
35. (d)
36. (a)
37. (a)
38. (a)
39. (a)
40. (d)
41. (b)
42. (a)
43. (c)
Answers matrix match:44. (A) – r, s, 45. (A) – q, r, 46. (A) – r, s, 47. (A) – p 48. (A) – p, s
(B) (B) (B) (B) (B)
– r, s, – p, – r, s, – p – r
(C) (C) (C) (C) (C)
– p, – q, r, – r, s –s – q
(D) (D) (D) (D) (D)
–q – r, s – p, q, s – q, r –p
RESONANCE Multiple Choice Questions (–)
O O 1. How many resonating structures are possible for the given species:-
O
(–)
O
(a) 1
(b) 2
(c) 3
(d) 4
2. Which among the following is not stabilized by resonance?
(a)
(b)
(c)
(d)
65
Reaction Mechanism (General Organic Chemistry)
3. Identify the wrong statement regarding tautomerism and resonance
(a) Tautomerism usually involves making and breaking of sigma as well as pi bonds while in resonance only the electrons in pi bonds or lone pair on hetero atoms shift; the sigma framework is not disturbed.
(b) Tautomerism may involve a change in hybridization of atoms resulting in a change in the shape of molecule.
(c) The two tautomeric forms exist together though the equilibrium may shift to either side with a change in condition. The tautomer has a physical reality whereas the resonating structures are imaginary.
(d) All are correct.
4. In which case a chance of resonance is least:-
NH2
NH2
(a)
(b)
NH2
NH2
(c)
(d)
5. Which is not coplanar?
O
O
(a)
(b)
(c) HCONH2
(d)
6. In which case lone pair indicated is not involved in resonance. (a)
N
(b)
C = N
•• — CH — O— (c) CH 2 — CH3 (d)
..
..S
••
H 7. Among the following identify electron-withdrawing group (in resonance).
( +)
(a) —OCH = CH2
(+)
(b) — N = SH
(c) —NHAc
(d) —OTs
8. Consider the following species. (–)
O
F
(II)
(III)
NO2
NO2
O2N (I)
CHO
Correct order of p e– density inside the benzene ring is:(a) I > II > IV > III (b) II > III > IV > I
(c) III > II > IV > I
(IV)
(d) II > IV > III > I
9. Consider the following species.
F
(I)
OH
NH2
OCH3
(II)
(III)
(IV)
Correct order of p e– density inside the benzene ring is:(a) IV > III > II > I (b) III > IV > II > I
(c) III > II > IV > I
(d) IV > II > III > I
66
Problems in Organic Chemistry
10. Compare the relative sability of following reasonating structures (–)
O
O
O
(+)
CHO
CHO
(I)
(a) I > II > III
(–)
NH2
H
(III) (c) II > III > I
(d)
I > III > II
NH2 CH2 == NH
11.
CHO (I)
C
O
(II)
(b) III > II > I
NH2
(+)
(II)
(III)
(IV)
Among these compounds, the correct order of C – N bond lengths is:(a) IV > I > II > III (b) III > I > II > IV (c) III > II > I > IV
(d) III > I > IV > II
12. What is the oxidation state of nitrogen in both the resonating structure respecively?
2
(a) –3 & – 3
13.
(b) –3 & +3
O
+
(+)
H
(c) +3 & –3
(d) –3 & –4
Stable intermediate
CH3O
Most probable structure of this intermediate would be:O
O (+)
OH
(a)
H
(b)
CH3O
O
(+)
(c) CH3O
(+)
CH3O
14. Which of the following has longest C – O bond: (a) CH2 = CH – CH2 – OH (c) CH2 = CH – OCH3
(d) CH3O
H
(b) CH3 – CO– CH3 (d) CH2 = CH – CH = CH – OCH3
15. Which of the following has highest resonance energy?
O
O
(a)
(b)
O
O
(c)
16. Out of the following three species which of the two have equal C – O bond lengths: CH3COO(–) HCOO(–) HCOOH (I) (II) (III) (a) II, III (b) I, III (c) I, II
(b)
(d) None
(+)
67
Reaction Mechanism (General Organic Chemistry)
17. (1) CH 2
( +)
= CH — CH 2
(2) CH 2
(+ )
= CH — CH = CH — CH = CH — C H2
(+)
(3)
2
(+) CH
Correct order of resonance energy
(a) 1 > 2 > 3 > 4
(4)
(b) 4 > 3 > 2 > 1
(c) 3 > 4 > 2 >1
(d) 2 > 4 > 1 > 3
18. Consider the following reduction process
E1
E2
E3
(1)
(2)
(3)
Here E1, E2 & E3 are activation energies of reduction processes of (1), (2) & (3) respectively. Relation between these activation energies is:(a) E1 > E2 > E3 (b) E3 > E2 > E1 (c) E2 > E3 > E1 (d) E1 > E3 > E2
19. Resonance effect is not possible in:-
.. S
(a)
(b)
N – H
(c) Both (a) & (b)
.
O
N
(d) (+)
.
O
20. Out of I, II, III & IV equivalent resonating structures will be obtained in:(–) (–)
2–
—O
CO3 (I)
(a) I, II & III
(–)
RSO3
(II)
(III)
(b) I & III
(IV)
(c) II & IV
(d) I
21. You have three molecules as indicated below, where l1, l2 & l3 are bond lengths. CH2
l2
CH — O — CH3
Relation between l1, l2 & l3 would be (a) l1 > l2 > I3 (b) l3 > l2 > l1
(c) l2 > l3 > l1
CH2
l3
.
CH — S — CH3
..
l1
..
CH2
CH — NH — CH3
(d) l3 > l1 > l2
22. Consider the following ions.
O
(–) O
(–) O (I)
Correct order of stability is:(a) III > II > I
O(–)
(b) I > II > III
(II)
(c) I > III > II
(III)
(d) I > II = III
68
Problems in Organic Chemistry
23. In whch case first resonating structure is more contributing than second resonating structure. (–)
O
(b) CH2
(+)
CH — O — CH2
(+)
and
CH3 — C
and
CH2
NH2 (+)
CH — O
CH2
.
(a) CH3CONH2
..
(+)
Br (+)
(c)
O
(d)
Br
and
(–)
O
and (–)
24. Relation between bond dissociation energies of bonds a, b, c & d is:c
a CH
CH
(a) a > b > c > d
d CH3 — CH
b
(b) d > b > a > c
(c) a > d > b > c
CH2
(d) a > d > c > b
25. Correct order of resonance stabilization of given anions is: PhSO3(–) AcO(–) PhO(–) CH2 = CH—CH = CHO(–) (1) (2) (3) (4) (a) 3 > 1 > 4 > 2 (b) 1 > 3 > 4 > 2 (c) 1 > 2 > 3 > 4 (d) 3 > 1 > 2 > 4 26. Compare the bond length of C — N bonds in following species NH2 O
N O
(a) I > II > II
NH2
(I)
(II)
(b) I > III > II
(III)
(c) III > I > II
(d) II > III > I
27. In the given molecule lone pair present on hetero atoms are numbered as 1, 2, 3 & 4
1N
N
O 3
S 4
2
Me
Correct order for these lone pairs in order of their participation in resonance will be:(a) 2 > 1 > 4 > 3 (b) 4 > 3 > 2 > 1 (c) 4 > 2 > 1 > 3
(d) 2 > 1 > 3 > 4
28. Correct order of stability of resonating structures given below is:CH 2
= CH — Cl
(+ )
(I)
(a)
I > II > III
(b)
III > II > I
(+ )
(–)
CH 2 — CH = Cl
CH 2 — CH = Cl
(II)
(III)
(–)
(c)
I > III > II
(d)
II > I > III
69
Reaction Mechanism (General Organic Chemistry)
Passage - I Consider the following two compounds A & B 1 2
CNH—NH
4
3
5
6 NH
O
O (A)
(B)
Answer the questions from 29 to 32 29. In compound A correct order of electron density inside the benzene ring will be: (a) 1 > 2 > 3 (b) 2 > 3 > 1 (c) 2 > 1 > 3
(d) 3 > 2 > 1
30. In compound B correct order of electron density inside the benzene ring will be: (a) 4 > 5 > 6 (b) 6 > 4 > 5 (c) 5 > 6 > 4
(d) 5 > 4 > 6
31. The benzene ring with least electron density is: (a) 1 (b) 2
(d) 3
(c) 4
32. In compound B correct order of electron density inside the benzene ring when NH is replaced by S will be: (a) 4 > 5 > 6 (b) 6 > 4 > 5 (c) 5 > 6 > 4 (d) 5 > 4 > 6
Passage - II When any species undergoes resonance it becomes stable. Higher the number of resonating structures higher is the stability of the species. Resonance also affects the bond length. Answer the questions from 33 to 36 33. Which has least dipole moment? (a) Flurobenzene (b) Chlorobenzene
(c) Bromobenzene
(d) Iodobenzene
34. In which case bond length of C = O bond is least? (a) Carboxylate ion (b) formate ion
(c) Methyl acetate
(d) Acetamide
35. In which case bond length of C — O bond is maximum. (a) Carboxylate ion (b) Phenoxide ion
(c) p-nitro phenoxide ion
(d) 2,4 di nitro phenoxide ion
36. Pyrrole undergoes resonance as:(–)
(+)
(+) (–)
N H
N H
N H
(I)
(II)
(III)
Correct order of stability of these resonating structures is:(a) I > II > III (b) III > II > I
(c) I > III > II
(d) III > I > II
Answer Key 1. (d)
2. (c)
3. (d)
4. (c)
5. (d)
6. (b)
7. (b)
8. (b)
9. (c)
10. (a)
11. (c)
12. (a)
13. (a)
14. (a)
15. (b)
16. (c)
17. (c)
18. (c)
19. (d)
20. (b)
21. (b)
22. (c)
23. (c)
24. (c)
25. (c)
26. (b)
27. (d)
28. (c)
29. (b)
30. (c)
31. (a)
32. (d)
33. (a)
34. (c)
35. (b)
36. (c)
70
Problems in Organic Chemistry
REACTION INTERMEDIATES AND ATTACKING REAGENTS (Carbocation, carboanion carbene, Nitrene, Benzyne, Electrophile, Nucleophile)
Multiple Choice Questions 1. Hybridization of negatively charged carbon in given species are respectively:O CH3 CH3
(–)
(–)
CH
(a) sp3, sp3, sp3
CH2
(–)
CH — CH2
(b) sp3, sp2, sp3
(c) sp3, sp2, sp2
(d) sp2, sp2, sp2
2. Which is better attacking reagent (strong base) among the following? (a) CH3O(–) (b) HCOO(–) (c) PhO(–)
(d) CH2 = CH—O(–)
3. Which among the following will have least electrophilicity? (a) BF3 (b) BBr3
(d) BI3
(c) BCl3
4. Identify the carbocation in which rearrangement is not possible.
CH3
CH3
(a)
(+)
(+)
(+)
(b)
(c)
(+)
(d)
CH3
5. Highly stable carbocation is:(+)CH
(+)CH
(+)CH
2
2
2
NO2
(a)
(b)
(c)
(+)CH
2
NH2
(d)
NH2
NO2 6. Highly stable carbocation is:(+)CH2
(+)CH 2
(a)
(b)
(+)CH 2
(+)CH 2
OCH3
(c)
NH2
(d)
NO2
7. Identify the option in which Ist intermediate is found to be more stable than 2nd CH3 — N — CH3
(a)
CH3 — CH — CH3
, CH3 — C — CH3 (+)
(+)
(b)
(+)
CH3 — C — CH3 (+)
(+)
(c)
(+) , CH — CH — CH 3 3
(d) CF3(–) , CCI3(–)
,
71
Reaction Mechanism (General Organic Chemistry)
8. Triplet carbene has (a) Tetrahedral geometry
(b) Bent shape
(c) Linear geometry
(d) Trigonal planar shape
(c)
(d) None
9. Which carbocation is aromatic? (+)
NH
(a)
(b)
(+)
N H
••
10. Cis–but–2–ene + C H2 (triplet) ——→ [X], [X] will be:
(a)
H3C
CH3 C—C H H
(b)
H3C
H C—C
H
H3C
(c)
CH3
(+ )
C—C H
CH3 H
(d) both b & c
••
11.
Which statement is incorrect about CH3 and CH 2 (triplet) (a) Both serve as electrophile (b) In both the cases central carbon atom possesses 6e- in outermost shell (c) In the both cases carbon is sp2 hybridized
(d) CH3+ is triangular planar however CH 2 is linear
••
12. Pick out highly unstable carbonium ion:
(+ )
(+ )
(+ )
(+ )
(a) CH3
(b) (F3C) 3 C
(c) F — C — F
(d) CH3 CH 2
|
F
13. Highly stable carbocation is:
( +)
(a) CH 2
= CH — CH 2
(+ )
(b) CH 2 = CH — CH —OCH3
|
OCH3
(c) CH 2
( +)
= CH — CH — OCH3
(d) CH 2
|
(+ )
= CH —CH — OCH3 |
OCH3
NMe3
( +)
(–) CH2
(–) CH2
(–) CH2 NO2
14.
OH
NH2 (2)
(1)
(–) CH2
Correct order of stability is:(a) 2 > 4 > 1 > 3
(4)
(3)
(b) 2 > 4 > 3 > 1
(c) 2 > 3 > 4 > 1
(d) 2 > 3 > 1 > 4
(c) 2 > 3 > 1
(d) 2 > 1 > 3
15. Correct order of stability of following carbanions will be:(–)
F—C—F
Cl
F
|
(–)
Cl—C—Cl (1)
(a) 3 > 2 < 1
(–)
|
(2)
(b) 3 > 1 > 2
(3)
72
Problems in Organic Chemistry
16. Consider the following:-
H
H
H
H
(1)
Out of these which will form highly stable carbocation by the removal of H (a) 1 (b) 2 (c) 3
(3)
CH 2
= CH — CH 2
In which case positive charge density is least on bold C? (a) 3 (b) 2 (–)
(4) CH—C—H 3
(d) 4 () 2C ( + ) — CH3 (4)
CH3 (3)
(2)
(c) 1
(d) 4
(–)
(–)
18. 3 C
||
( +)
( +)
17. 3C( +) (1)
(–)
CH 2 = CH—CH 2
(CH 3 ) CH 2
2 C — CH3
(2)
(3)
(4)
(1)
O
H
(2)
H
In which case negative charge density is maximum on C? (a) 1 (b) 2
(c) 4
(d) 3
19. Consider the following radicals •
•
•
(1)
(2)
(3)
Correct order of stabilities of these radicals is:(a) 4 > 2 > 1 > 3 (b) 4 > 2 > 3 > 1
(4)
(c) 4 > 3 > 2 > 1
(d) 2 > 4 > 3 > 1
20. Which statement is incorrect about the following:••
(+ )
••
CH3 — N, CCl,2 CH3 •• (a) All can serve as electrophile (b) In all cases reactive centre possesses 6e- in its outermost shell
••
••
(c) CH3 N and C Cl2 both have two forms singlet and triplet •• (d) All are short lived species (i)hv (ii) Cyclohexene
21. RN3 ¾¾¾¾¾¾® Product. Identify the product of this reaction?
R R
(a)
N—R
(b)
R
R
R
(c)
(d)
22. Consider the following carbocation:(+)
(+) (+)
(1)
(2)
Correct order of their stability will be: (a) 3 > 4 > 2 > 1 (b) 3 >2 > 4 > 1
(3)
(c) 4 > 3 > 2 > 1
(+)
(4)
(d) 3 > 4 > 1 > 2
73
Reaction Mechanism (General Organic Chemistry)
23. Consider the following carbanion: (–)
(–)
COCH3
(1) CH 2COCH 3
(2)
Correct order of their stability is:(a) 3 > 2 > 4 > 1 (b) 3 > 2 > 1 > 4
(–)
COCH 3
(3)
(4) CH 2
(c) 2 > 3 > 4 > 1
24. More stable carbocation is: (a) p – Chloro benzyl carbocation (c) p – Methoxy benzyl carbocation
(–)
= CH — C H 2
(d) 2 > 3 > 1 > 4
(b) p – Nitro benzyl carbocation (d) Benzyl carbocation
25. The most reactive carbanion among the following is:(–)
(a) CH 2 CH2
(c) p — OMe — C6 H 4 — CH 2
(–)
(b) CH 2 (–)
(–)
(d) p — NO2 — C 6H 4 — CH 2
26. Carbocation undergoes rearrangement to get stability. In a chemical reaction following carbocation is generated as intermediate. CH3 (+)
C2H5 — C — C
O
H
Select the correct statement regarding above carbocation
(a) It is stabilized by hydride shift towards electrophilic carbon
(b) It is quite stable and can not show rearrangement (c) It is stabilized by methyl shift towards electrophilic carbon (d) It is stabilized by ethyl shift towards electrophilic carbon
27. Rank the following free radicals in increasing order of their stabilities CH2 (1)
(a) 4 < 3 < 2 < 1
(2)
(b) 4 < 1 < 2 < 3
(3)
(c) 4 < 2 < 3 < 1
(4)
(d) 3 < 4 < 2 < 1
28. H2 gas will be liberated in
(a)
+ K metal (1 mole)
(b)
+ NaH
(c)
+ K metal (2 mole)
(d) Both (b) & (c)
29. Identify the species which is not an example of electrophile:
(+ )
(a) (C 2 H5 )4 N
(+ )
(b) H3 O
(+ )
(c) Cl
••
(d) C Cl2
74
Problems in Organic Chemistry
30. Which of the following is most stable carbanion intermediate?
(a)
(b)
(d)
(–)
(–)
(–)
(c)
(–)
— C — CH3 C2H5
Passage - I When H+ attacks on OH group of an alcohol carbocation is formed:H ( +)
R — OH ¾¾¾® R (+) + H 2 O Rate of formation of carbocation depends on the stability of carbocation. Higher the stability of carbocation easily it can be formed. Answer the questions from 31 to 34. 31. In which case attack of H+ on OH is more appreciable.
F (a) F— C—OH F
OH (b) CH3OH
F (c) F
F F
(d) (F3C)3COH
32. Which of the following will give highly stable carbocation on reaction with H (+) ?
(a) Ph3COH
(b)
COH 3
Ph
|
(c) CH 2 = C — OH
OH
|
(d) CH3 — C — CH = CH 2 |
Ph
33. Which will produce highly unstable carbocation on reaction with H+?
OH
OH
(a) (CF3)3COH
OH (1) Ph OH (2)
34.
OH (3)
(b) CF3OH
(c)
(d)
75
Reaction Mechanism (General Organic Chemistry)
Correct order of ease of removal of OH by H+ is:(a) 3 >2 > 1 (b) 1 > 2 > 3
(c) 3 > 1 > 2
(d) 1 > 3 > 2
Passage - II Carbocation, carbanion & free radicals are stabilized by resonance. Except carbanion, carbocation & free radical exhibit hyperconjugation however carbanion shows reverse hyperconjugation. Stability of free radical should be checked by resonance and I-effect only. Answers the question from 35 to 38 35. Consider the following carbanion (–) (1)
(–)
(–)
MeOCH = CH — CH 2
CH 2 = CH — CH 2
CH 3
CH 2F
(3)
(4)
(5)
(2)
Correct order of their stability is:(a) 2 > 3 > 1 > 5 > 4 (b) 3 > 1 > 2 > 5 > 4
(–)
(–)
Ph 3 C
(c) 3 >2 > 1 > 5 > 4
(d) 3 > 5 > 2 > 1 > 4
36. Most stable carbocation is:(+)
(+)
(a)
(+)
(+)
(b)
(c)
(d)
37. In which case chances of resonance is least?
(–)
(–)
(–)
(–)
CHCF3
CHCF3
CHCF3
CHCF3
(a)
(b)
(c)
(d)
38. Which one of following will form with greater ease?
(a)
(c)
(b)
• (d) CH 2 — — CHCH 2 — C H 2
Passage - (III) Answer the question from 39 to 42
+ 2Na
(A)
+ Na
(B)
(X) + (Z)
[X] 39. Which is aromatic? (a) A & Z
(b) B & Z
(c) Only Z
(d) None
76
Problems in Organic Chemistry
40. Which is radical anion? (a) A
(b) B
(c) Z
(d) None
41. Which is correct about A, B & Z? (a) A & Z are same (b) A & B are same
(c) A ≠ B ≠ Z
(d) All are wrong
42. Which is anti aromatic? (a) X
(c) B
(d) None
(b) A
43. Which intermediate is optically active?
(–)
(+ )
•
(a) MeCHF
(b) CF2 I
(c) CHFBr
(d) All are optically inactive
44. If resonance effect is not considered then least stable carbocation will be:(+ )
( +)
( +)
(a) 3 C
(b) F3CCH 2
(c) CH 2
(+ )
= CH — CH 2
(d) CH3
45. Consider the following carbocations (+)
(+)
(+)
CH2—CH 2 (1)
(2)
Examples of classical carbocation is /are:(a) 1, 2 (b) 2, 3
(3)
(c) only 3
(d) 1, 2 & 3
46. Four containers A, B, C & D contain FCH2Br, F3CBr, vinyl bromide and allyl bromide respectively now equal amount Ag+ ions are added in each container and white precipitates are formed in three containers. The container in which precipitation does not occur is: (a) A (b) B (c) C (d) D 47. Which will undergo disproportionation to give alkene?
( )
(–)
(a) CH3 — CH 2 (+)
(+)
CH2
CH2
OCH3
••
•
(b) CH3 — CH 2
(c) CH3 — CH — CH3
(d) CH3 C H 2
(+)
(+)
CH2
CH2
OCH3
O
(3)
(4)
48. (1)
(2)
OCH3
These ions will follow the stability order:(a) 1 > 3 > 4 > 2 (b) 1 > 3 > 2 > 4
(–)
(c) 4 > 3 > 1 > 2
(d) 4 > 1 > 3 > 2
49. Which of the following is most unlikely intermediate? (+)
(a)
(b)
(c)
(d)
50. How many rearrangements will occur frequently in the given carbocation?
Ac Ac
(a) 1
(b) 2
(+)
(c) 3
(d) None
(+)
77
Reaction Mechanism (General Organic Chemistry)
Matrix Match Questions Each question contains statement given in two columns which have to be matched. Statement (A, B, C, and D) in column I have to be matched with statements (p, q, r, s) in column II. The answers to these questions have to be appropriately bubbled as illustrated in the following examples. If the correct match are A – p, A – s, B – r, B – r, B – q, C – q, D – S, then the correctly bubbled 4 × 4 matrix should be as follows. A
p
q
r
s
B
p
q
r
s
C
p
q
r
s
D
p
q
r
s
51. List – I (Reaction intermediate) CH3( ) •• CH2 CH3(–)
(A)
(B)
(C)
(D) CCl2
List – II (Available empty p – orbital / s)
(p) 0
(q) 1
(r) 2
(s) 3
52. List – I (Molecule / intermediate)
List – II (Pattern of bond present)
(A)
(p) sp – sp2 s bond
(B)
(q) sp3 – sp3 s bond
(r) sp2 – sp2 p bond
(s) sp2 – sp2 s bond
S
(C)
(D)
N2
(+)
53. Column – I Column – II (+)
(A)
CH2
(B)
CH2
(C)
(D) CH3CH2(–)
(p) sp2 hybridised carbon
(q) sp3 hybridised carbon
(r) Most stable intermediate
(s) Intermediate is not stabilised by resonance
(–)
CH2 — CH2
54. Column – I Column – II N D N → (A) (p) Carbocation is produced
(B)
H( + )
OH
→
(q) H2O is produced
78
Problems in Organic Chemistry D (C) CH2CO → (+) (–) CH 3 D (D) → C2H5 C == N == N — SO2Ph
(r) N2 is produced (s) SO2 is produced (t) Carbene is produced
Answer Key 1. (c)
2. (a)
3. (a)
4. (a)
5. (c)
6. (a)
7. (a)
8. (c)
9. (a)
10. (d)
11. (c)
12. (b)
13. (b)
14. (d)
15. (b)
16. (b)
17. (c)
18. (d)
19. (b)
20. (c)
21. (a)
22. (a)
23. (d)
24. (c)
25. (a)
26. (b)
27. (a)
28. (d)
29. (a)
30. (b)
31. (a)
32. (b)
33. (a)
34. (d)
35. (b)
36. (a)
37. (c)
38. (c)
39. (a)
40. (b)
41. (a)
42. (d)
43. (d)
44. (a)
45. (b)
46. (c)
47. (c)
48. (c)
49. (d)
50. (c)
Answers matrix match 51. (A) – q, (B) – q, r, (C) – p (D)-q 53. (A) – p, (B) – p, r (C) – q, s (D) - q, s
52. (A) – s, (B) – r, s, (C) – q, (D) – p, s 54. (A) – r, t (B) – p, q (C) – t, (D) – r, t
Nucleophilic Substitution Reactions Multiple Choice Questions 1. How many transition states will be observed in following transformation? Br +
(a) 2
NaNH2 in water —— NaBr
(b) 3
+
(c) 1
NH2
(d) 4
2. The reaction Ph3CBr + EtOH ——→ Ph3COEt + HBr will involve. (a) One transition state (b) 2 - transition states (c) 3 - transition state (d) four transition state 3.
Which Statement is correct about aliphatic nucleophilic substitution reactions? (a) In SN2 racemisation takes place. (b) In SN1 complete racemisation occurs (c) In SN1 partial racemisation occurs and isomer with inverted configuration forms more. (d) In SN1 partial racemisation occurs and isomer having configuration similar to that of substrate forms more.
4. Pick out the correct order of rate of SN1 for the following compounds
Cl
Ph
Cl
CH—Cl Ph
(I)
(a) I > II> III >IV (c) I > III > II > IV
(II)
(III) (b) III > II > I > IV (d) III > I > II > IV
Me2CHCl
(IV)
79
Reaction Mechanism (General Organic Chemistry)
5. In which case racemisation takes place. C2H 5
C2 H5
(–)
|
(a) CH 3 — CH—CH 2Br
(c)
(–)
|
OH
→ 1
OH
(b) CD3 — CH—CH 2 Br → 1
SN
SN
(–)
OH
→ CH2Br 1
SN
(d) All of these
6. A species is said to be a good nucleophile if (a) It has large size and high electro negativity. (c) It has large size and less electronegativity.
(b) It has small size and high electro negativity (d) It has small size & less electro negative.
7. Correct order of rate of SN2 is:
(I)
Cl
(a) II > III > IV > I (c) II > IV > III > I
(II)
Cl
(III)
Cl
(b) III > II > IV > I (d) I > III > II > IV
(II)
(IV)
Cl
8. Correct order of rate of SN1
(I)
Cl
Cl
(III) Ph2CHCl
(IV)
Cl
(a) III > IV > I > II (c) III > IV > II > I
CH3 9. CH3 CH3
C — O — CH3 +
(b) I > III > IV > III (d) III > I > IV > II
HI (cold and dil) —— products
Select correct statement(s) (a) This reaction occurs via SN2 path way (c) In this reaction 3°-iodide is produced
(b) In it primary iodide is produced (d) Reaction can occur by SN1 as well as SN2 path way
10. Which is correct?
H
Et
(a)
D
Pr
(–)
C2H5O
Recemisation
Br
(b)
Et
Me
C2H5OH
Walden inversion
Br
Et
(c) Me
Br
C2H5OH
Racemisation
(d) All are correct
Ph 11. CD2 = CH — CH2Br is subjected to nucleophilic substitution reaction by EtO(–) in EtOH. Which of the following statement will be most appropriate? (a) Both SN1 & SN2 give two products. (b) Both SN1 and SN2 give only one product. (c) SN1 gives two products but SN2 gives one product. (d) SN1 gives one product but SN2 gives two products.
80
Problems in Organic Chemistry
12. Suggest the very suitable solvent for the reaction given below CH2Br NaOH —————→
(a) H2O
(b) C2H5OH
CH2OH
(c) HCONMe2
(d) C6H6
13. In which case chances of SN2 are maximum
(a)
Br + C 2 H5 O
(c)
Br + C2 H5O in DMF
(–)
(–)
COOH
(b)
Br + C2H5O(–)(in C6H6)
(d)
Br + C 2 H5 O in DMSO
(–)
COOH (1) aq. NaOH
14. D
Br → D +
OH
(2) H
C2H5
(in H2 O)
C2H5
This nucleophilic substitution reaction occurs via the formation of:-
O
D
COOH
(a)
(+)
D δ– (b) OH
COOH δ–
Cl
D
(c)
Et
Et
Et
COOH s– Cls– HO
(d) O
Et D
15. If the compound given below is subjected to SN1 reaction by aq. NaOH then order of ease of removal of bromine will be:-
Br3 Br1 HN Br2
(a) 1 > 2 > 3
(b) 2 > 3 > 1
I
I 16. (I)
(c) 3 > 2 > 1
O
(II)
(d) 2 > 1 > 3
I
O
(III)
I
(IV)
O
SN1
Ease of reaction among these compounds will be in the order as:(a) I > III > II > IV (b) IV > III > II > I (c) III > II > I > IV
17. Which among the following reaction satisfies the potential energy diagram given below?
change in free energy
reaction coordinates
(d) III > IV > II > I
81
Reaction Mechanism (General Organic Chemistry)
(a)
Br
+
(b)
Br
+
aq NaNH2
NaBr
+
NH2
H Br
+
NH2
——
NH3
CH3
CH3
(c) CH3 — CH — CH2 — Br
+
——→
NH3
HBr
+
CH3—C—NH2 CH3
(d)
+ Br
H2 O
OH
——
OH aq.
→ (A), (A) & (B) are:Na 2CO3
aq.
18. (B) ← NaOH
Br OH
(a)
OH &
OH
OH
(b)
OH
(c)
&
& CO3Na CO3Na
OH
OH
O
(d)
OH
OH OH &
Br
OH
CH3 C H OH
19.
2 5 → A (major), Product A is:-
CH3 CH3
OEt
|
|
(a) CH3 — C— CH 2OC2H5 |
(b) CH3 — C— CH 2 — CH3 |
CH3
CH3
CH3
|
(c) CH3 — C
CH 2
(d) None
20. Which one of the following will be hydrolysed with maximum rate?
(a) C6H5Cl
Cl
(b)
(c)
CCl
(d) C6H5CH2Cl
3
21.
Br
+
NaOH
solvent
—————
OH
For which solvent rate of SN2 will be maximum (a) 100% H2O (b) 75% H2O + 25% acetone (c) 100% acetone
(d) 21% H2O + 75% acetone
82
Problems in Organic Chemistry
Me Br
Br 22.
H H
H Me OH
OH
(–)
OH A is:in DMSO
(a)
H
(b)
H
H OH
H
H
Me OH
H
(c)
H
Me H
H OH
(d) All of these
OH
23. Which will yield white precipitate with AgNO3?
Cl
(a)
CH2Cl
(b)
(c) CH2 = CH — Cl
(d)
Cl
24. Which path is better for the preparation of ROH? (–)
R—F
OH I
(a) Path I
(–)
(Path I) ROH (–) OH RI ROH (Path II)
(b) Path II
(c) Both
(d) None
25. Consider the solvolysis (SN1) of the following halides. Cl Ph
(I)
(II)
(III)
Ph Ph
CH—Cl
(IV)
Ph CHI Ph
Cl
Correct order of rate of solvolysis willl be:(a) II > III > I > IV (c) IV > III > I > II
(b) III > IV > II > I (d) IV > III > II > I
H |
26. In the given reaction CH3 — C — Br |
[X] will be:NaI Acetone
C2 H5 H
(a)
I
(b) CH3CH2 — CH2 — CH2 — I
(d) Mixture of (a) & (c)
Et H
(c) I Et
27. Arrange the following in decreasing order of nucleophilicity in ethanol medium: (–) (–) (–) F Cl Br (1) (2) (3) (a) 1 > 2 > 3 > 4 (b) 3 > 2 > 1 > 4 (c) 4 > 3 > 2 > 1
(–) I (4) (d) 2 > 3 > 4 > 1
28. Arrange the following groups in order of decreasing leaving groups ability for SN reaction. (–)
(I) CH3 SO 3 (a) I > II > III > IV
(–)
(II) CF3SO3 (b) IV > I > II > II
(–)
(III) Tert BuO (c) III > II > I > IV
(–)
(IV) I (d) II > I > IV > III
83
Reaction Mechanism (General Organic Chemistry)
29. Which among the following will undergo hydrolysis rapidly?
Br
Br
(I)
(a) I
Br
Br
(II)
(III)
(b) II
(IV)
(c) III
(d) IV
30. Arrange the following in decreasing order of SN2:-
Br
Br
(1)
(2)
(a) 1 > 4 > 2 > 3
31.
Br
Br
(3)
(b) 1 > 4 > 3 > 2
(4)
(c) 1 > 2 > 4 > 3
(d) 1 > 3 > 4 > 2
OH
CH2Br
(–)
OH ————— in water
Which among the following is correct potential energy diagram for the above reaction?
(a) G
(b)
G
reaction progress
reaction progress
(c)
G
(d)
G
reaction progress
reaction progress (–)
(–)
(–)
32. Arrange the nucleophiles O H, Se H, and S H in order of their nucleophilicity in di methyl formamide (DMF) medium
(–)
(–)
(–)
(a) OH SH SeH
(–)
(–)
(–)
(b) SeH OH SH
(–)
(–)
(–)
(c) SeH SH OH
(–)
(–)
(–)
(d) SH OH SeH
84
Problems in Organic Chemistry
33. Consider the following reaction. HOH CH3 — CH2 — S — CD2 — CH2 — Br → [X], [X] will be: (a) CH3CH2 — SCD2CH2OH (b) CH3CH2SCH2CD2OH (c) Both (a) & (b) are correct (d) None
34. Alcohols on reaction with HX produce alkyl halides like:D ROH + HX ¾¾ ® RX + H2O
How many carbocations will form as intermediates in following reaction? OH
Ph + HBr
(a) 1
(b) 2
——→ major product
(c) 3
(d) 4
35. Consider the following reaction D Me3CBr ¾¾ ® Me3C — OEt ………….. I
Me3CBr
C2 H5OH → in 25% H 2O
C H OH
2 5 Me3CBr → Me3C — OEt …………..II in 25% H O 2
Me3C — OEt …………..III
Correct order of rate of SN1 will be:(a) III > II > I (b) II > III > I
36.
Which is correct about the nueleophilicity of halide ion? (a) In DMSO order of nueleophilicity is F(–) > Cl(–) > Br(–) > I(–) while in water it is I(–) > Br(–) > Cl(–) > F(–) (b) In DMSO order of nueleophilicity is I(–) > Br(–) > Cl(–) > F(–) while in water it is F(–) > Cl(–) > Br(–) > I(–) (c) Order of nueleophilicity is same as F(–) > Cl(–) > Br(–) > I(–) in both water & DMSO (d) Order of nueleophilicity is same I(–) > Br(–) > Cl(–) > F(–) in both water & DMSO
(c) III > I > II
(d) I > II > III
37. Consider the following reactions:-
CH3 Me2 COH CI aq CH → ………….. II | NaOH 3 Ph Ph
NaOH Ph 3CBr → Ph 3COH ………….. I (aq)
Ph Br
Ph OH
NaOH
NaOH
CH3—CH2Br → CH3CH2OH………….. III → in DMF in DMF Out of these four reactions in which case rate of reaction will increase by the addition of NaOH (a) I (b) II (c) III (d) IV
………….. IV
38. Which among the following will undergo SN1 rapidly? O
(a)
(b)
Cl
O
(c)
O
Cl O
Cl (d) H—N
O
Cl
39. Consider the following reactions: (I) CH3CH2Br + NaCN ——→ CH3CH2NC + NaBr (III) CH3CH2Br + AgNO2 ——→ CH3CH2ONO + AgBr Which reaction is / are not correct? (a) III & I (b) III, II & I
(II) CH3CH2Br + AgCN ——→ CH3CH2CN + AgBr (IV) CH3CH2Br + NaNO2 ——→ CH3CH2NO2 + NaBr (c) III
(d) I, II, III & IV
85
Reaction Mechanism (General Organic Chemistry)
40. What is [X] in the following reaction:HOH EtSCH 2 CH — Cl [X] |
CH3
Me
(–)
|
(a) EtSCH 2 CH—Cl
(b) EtS—CH—CHO 2 H
|
OH (+) (c) EtS — CHMe
CH3
CH2
CH3
CH3 + NaOH
41.
——→
H
Et
(d) both (a) & (b)
OH
Et H
Br
The most efficient solvent for above reaction is:-
(a) H2O
O
O
O
O
(b)
(c) H2O + EtOH (4 : 1 ratio) (d) H2O + EtOH (3 : 1 ratio)
42. The reation R — X + NaOH ——→ ROH satisfies following Ea diagram
The Ea diagram (3) belongs to (a) R – F (b) R – Cl CH OH
3 43. (A) ← (–)
(a)
H Me HO
OMe
OH
2
& H Me
Me (c) OMe & H
CH OH 4
Me
HO
(d) R – I
3 → (B), (A) & (B) are:H SO
H Me
CH 2O
(c) R – Br
HO
Me
H HO MeO
(b)
Me
MeO H Me & H MeO
(d)
OH
MeO & Me MeO
OH
HO
CH3 Me
H MeO
OH
86
Problems in Organic Chemistry
44. For the following reaction:CH3 — C H — CH3 KOH KCl CH3 — CH — CH3 |
(aq)
|
OH
Cl
100%
Rate law can be written as:-
(a) K1 [alkyl halide] [KOH] + K2 [alkyl halide]
(b) K2 [alkyl halide]
(c) K1 [alkyl halide] [KOH]
(d) K1 [alkyl halide] [KOH] — K2 [alkyl halide]
45. CD2 = CH — CH2 — Br is subjected to SN1 and SN2 reactions separately which of the following statement is correct.
(a) Both SN1 and SN2 give two products
(b) Both SN1 and SN2 give only one product
(c) SN1 gives two products but SN2 gives only one product (d) SN1 gives one product but SN2 gives two products
46. Which among the following chloro derivative of benzene would undergo hydrolysis most rapidly with aq. NaOH ?
NO2
NO2
(a) F — Cl
(b) p — Me2N—C6H4Cl
(c) O2 N
Cl (d) O2 N NO2
47. Match the following Substrate
(A)
(B)
Rate of solvolysis by ethanol
Br
(p) 10–10
(q) 10–3
(r) 1
Br
(C)
Br
CH3
(a) A ——→ p, ——→ B ——→ q. C ——→ r (c) A ——→ p, B ——→ r, C ——→ q
(b) A ——→ r, B ——→ q, C ——→ p (d) A ——→ r, B ——→ p, C ——→ q
48. Which sequence of nueleophilicity is correct?
(a) nBuO(–)> SecBuO(–)> tert BuO(–)
(c) SecBuO(–)> tertBuO(–)> nBuo(–)
(b) tert BuO(–) > SecBuO(–)> nBuO(–) (d) SecBuO(–)> n-BuO(–)> tertBuO(–)
49. The best leaving group among the following is:
(a) CH3COO(–)
(b) CH3O(–)
(c) CH3SO3(–)
(d) CH ≡ C(–)
50. Among the given compounds choose the two that yield same product in SN1 reaction
Br
Br
Br Br (2)
(1)
(a) (1) & (2)
(b) (1) & (3)
(3)
(c) (1), (2) & (3)
(4)
(d) (2) & (4)
Cl
87
Reaction Mechanism (General Organic Chemistry)
Et
O
+ HI (cold and dilute) ——→ [X]
51.
Select correct statement regarding the product [X] (a) It is 3° alcohol (b) It is 2° alcohol I
(c) It is 1° iodide
CH3
52.
+ NaOH (in crown ether) —— [X],
the product [X] is
H
H
CH3
H
(a) Optically active
(b)
H
HO CH3
HO
(c)
Note:
(d) It is optically active
H
H
& can not show geometrical isomerism
(d) None is correct
For more questions on N.S.R see hydrocarbon (chapter - 6) & alkyl halide (chapter - &)
Answer Key 1. (b)
2. (c)
3. (c)
4. (a)
5. (b)
6. (c)
7. (b)
8. (d)
9. (c)
10. (c)
11. (c)
12. (c)
13. (d)
14. (d)
15. (d)
16. (a)
17. (b)
18. (c)
19. (b)
20. (c)
21. (c)
22. (c)
23. (a)
24. (b)
25. (d)
26. (c)
27. (c)
28. (d)
29. (a)
30. (a)
31. (d)
32. (a)
33. (c)
34. (a)
35. (d)
36. (a)
37. (c)
38. (a)
39. (d)
40. (d)
41. (b)
42. (d)
43. (b)
44. (a)
45. (c)
46. (c)
47. (b)
48. (b)
49. (c)
50. (a)
51. (d)
52. (b)
Free Radical Substitution Reaction (FSR) Multiple Choice Questions
1.
+ Br•——→ HBr + X • X radical would be:-
CH2
(a)
(b)
(c)
(d)
88
Problems in Organic Chemistry
2. Maximum numbers of products which can be obtained by the mono chlorination of methyl cyclo butane are (Excluding stereoisomers)? (a) 2 (b) 3 (c) 4 (d) 5 Mono Chlorination
→ 3. (CH3)2CH — CCl2 — CH (CH3)2 [X] (major), [X] is: (a) (CH3 ) 2 CHCCl 2CHCH 3 (b) (CH3)2CCl — CCl2 — CH (CH3)2 |
CH2 Cl
(c) Both a & b are in equal amount
(d) (CH3)2CCl — CCl2 — CCl(CH3)2
Cl A hn
4.
A may be:-
→ (a) NCS (c) Cl2 / hv
(b) tertiary butyl hypochlorite (d) all of these
5. You have following four compounds. CH3 Ph
CH3 CH 3
(I)
|
|
|
(II) CH3 — CH — CH — CH3 (III)
|
(IV) CH3 — C — C — Ph |
|
CH3 Ph
Select those compounds which will give two products on mono chlorination by Cl2/hv (Excluding stereoisomers) (a) II & IV (b) II & III (c) I & II (d) I, II & III
6.
Rate of abstraction of these numbered hydrogen’s will follow the order
H1
H3 H2
(a) 1 > 2 > 3
(b) 2 > 1 > 3
(c) 3 > 2 > 1
(d) 2 > 3 > 1
NBS in CCl hn
7. CH3 — CH2 — CH = CH2
4 → [X], [X] may be:
Br |
CH 2
(a) CH3 — CH — CH
(c) CH3 — CH2 — CH — CH 2 |
(b) CH3 — CH = CH — CH2Br (d) (a) & (b)
|
BrBr NOCl hn
8.
→ [A], (a)
NO
[A] is:(b)
Cl
N—OH
(c)
9. CH4 + 2CoF3 ——→ CH3F+ 2CoF2 + HF……………… (1) NBS CH3 — CH = CH2 → hn
CH 2 — CH CH 2….…………….. (2)
|
Br Propanone CH3CH2Cl + I(–) → CH3CH2I + Cl(–) …..…………….. (3) Bu SnH
2 CH3CH2Cl → CH3 — CH3 …..…………………... (4)
Which reaction is Wohl - Ziegler reaction (a) 1 (b) 2
(c) 3
(d) 4
(d)
None
89
Reaction Mechanism (General Organic Chemistry)
10. A hydrocarbon C8H18 gives only one mono chloro derivative with Cl2/hv. Hydrocarbon is:
(a)
(b)
(c)
(d)
11. A hydrocarbon A (vapour density = 43) on monochlorination gives chloro alkanes. If only one of the formed chloro alkane contains chiral carbon atom then A would be: (a) n-hexane (b) iso hexane (c) 2, 3-di methyl butane (d) 1, 1 - di methyl cyclobutane 12. Following steps are involved in halogenations of alkanes
X2 ——→ 2X*
……….…………………….. (1)
X* + RH ——→ R* + HX
……….…………………….. (2)
R* + X2 ——→ RX + X*
…………..………………..… (3)
Which is rate determining step? (a) 1 (b) 2
(c) 3
(d) none
13. Expected mono bromination products in the given reaction would be- (Excluding stereoisomers) Br hn
2→
(a) 3
(b) 5
(c) 4
(d) No Product
14. Which of the following reactions will yield 2, 2 - dibromo propane?
hn (a) CH3 — HC = CH2 + 2HBr → (c) CH3 — HC = CH2 + 2HBr ——→
hn (b) CH3CBr = CH2 + HBr → (d) both a & c
15. In which substrate two mono bromo derivatives will be formed by NBS/ hv /CCl4 (Excluding stereoisomers)
(a)
(b)
(c)
H O K 2CO3
NBS
(d)
Et
1
:
2 → →
h
+ Cl2
16.
A, A will be –
2 O
CHOHBr
(a)
(b)
C—COBr
(c)
COCH2Br
(d)
Cl
2→ 17. C10H20 one mono chloro derivative, hn
So A will be-
(a)
(b)
(c)
(d)
CH2 C2HBr
90
Problems in Organic Chemistry
18. Which of the following is not an example of free radical scavenger? (a) O2 (b) Diphenyl aniline (d) Catechol
(d) T.E.L
19. Iodination of alkane can be carried out by (a) I2 / hn (b) I2 / in CCl4
(d) I2 / AlCl3
20.
(c) I2 / HIO3 / hn
Iodination of alkane is difficult because:(a) I2 is more reactive in comparison to other halogens & hence, forms various side products (b) Propagation step is exothermic. (c) HI formed serves as reducing agent and reduces alkyl iodide in to alkane & I2. (d) All are correct. Cl
2 → (CH3)2 CClCH(CH3)2 (A) % Yield of A would be:21. (CH3)2CH CH (CH3)2 hv (a) 54.54 (b) 50 (c) 45.45
22.
Hydrocarbon
+ Cl 2
[containing 6 carbon]
KOH EtOH
h
[1 mol]
(d) cannot be predicted.
NBS hn
(P) does not react with chlorine in dark
The product P would be:-
OEt Cl
Cl
(a)
OH
(b)
(c)
(d)
Br
Br
23. Consider the following reaction:CH3 — CH — CH — CH3 |
D
Br
|
CH3
+
Br* → X + HBr
(free radical)
Identify the most probable structure of [X] D |
(a) CH3 — CH — CH — CH 2 |
|
CH3
•
(c) CH3 — C — CH — CH3 D
|
• (b) CH3 — CH — C | CH3
•
D
CH3
CH3 | • (d) CH 3 — CH — CH — CH 3
CH3 •
24. Consider the following CH3 — CD — CH — CH3 Br ——→ Free radical |
|
CD3
CD3
Identify the free radical predominantly produced in above reaction.
(a) CH3 — C — CH — CH3
•
||
(b) CH3 — CD — CH — CH3
CD3 CD3
| • CD 3
• (c) CH3 — CD — CH — CH 2
(d) CH3 — CD — C• — CH3
|
CD3
25.
|
CD3
|
CD3
|
CD3 |
CD3
Which statement is correct? (a) Chlorination of alkane is a selective reaction because chlorine free radical is highly reactive (b) Chlorination of alkane is a selective reaction because chlorine free radical is less reactive (c) Bromination of alkane is a selective reaction because bromine free radical is highly reactive (d) Bromination of alkane is a selective reaction because bromine free radical is less reactive
91
Reaction Mechanism (General Organic Chemistry) hn 26. C2H6 + Br2 → C2H5Br + HBr
………………….k1
hn C2D6 + Br2 → C2D5Br + DBr
………………….k2
hn C2H4 + Br2 → C2H4Br2
.………………….k3
hn C2D4 + Br2 → C2D4Br2
.………………….k4
If k1 , k2 , k3 & k4 are the rate constants of the given reactions then:(a) k1 = k2 (b) k3 > k4 (c) k1 > k2
(d) k3 = k4
Answer Key 1. (d)
2. (c)
3. (a)
4. (d)
5. (c)
6. (d)
7. (d)
8. (c)
9. (b)
10. (c)
11. (c)
12. (b)
13. (c)
14. (c)
15. (a)
16. (c)
17. (c)
18. (d)
19. (c)
20. (c)
21. (c)
22. (c)
23. (b)
24. (d)
25. (d)
26. (d)
Aromatic Electrophilic Substitution Reaction Multiple Choice Questions 1. Which among the following is the correct ptential energy diagram for sulphonation of benzene?
(a)
∆G
G
(b)
reaction progress
reaction progress
(c)
G
reaction progress
(d) G
reaction progress
92
Problems in Organic Chemistry
Ph 2.
OH
+
H( + )
→ X, X would be :-
HO
1 mole
Ph
Ph (a)
(b)
(c)
O
(d)
Ph
Ph
HO
Ph
O Ph
3.
OH
Ph
Ph
OH
Ph
Ph
H( + )
→ X, the product ‘X’ would be:-
OH Ph
(a)
O
O (b)
O
(c)
(d)
OH
Ph
O
Ph OH 4. In previous reaction which intermediate will not form:(+)
(–)
O
O
OH Ph
(a)
O
(b)
(c)
(+)
Ph
O
O
OH Ph
(d)
Ph
O H
5. When 3 - Nitro acetophenone is treated with Fe/Br2 it gives.
COCH3 Br
(a)
COCH3
COCH3
COCH3
Br
NO2
(b)
(c)
NO2
NO2
(d)
Br
NO2
Br 6. In which case Friedel Craft reaction by RCl / AlCl3 is not attainable. (a) Benzoic acid (b) Salol (c) Acetanilide
(d) Aniline
OH Br Water
7.
2 → [A], [A] would be:-
SO3 H OH
OH Br
Br
(a)
Br
(b)
OH
OH Br
Br
(c)
(d)
Br SO3H
SO3H
Br
SO3H
93
Reaction Mechanism (General Organic Chemistry)
8. Identify the product formed in the following transformation.
N O
2 5 → major product
O2N
(a)
(b)
(c)
(d)
NO2
NO3 NO2
Fuming HNO
3→ X (major), X would be :-
9.
NO3 NO2
SO3H NO2
(a)
SO3H
(b)
SO3H
(c)
O2N
CH3
CH3
(d)
SO3H
O2N
CH3
CH3
10. Consider the following reaction O Cl + ZnCl2 Cl → P. product P would be : N one mole
N O
(a)
4
11.
O
O
O
(b)
N
N
H
H
(c)
Cl
Cl
(d) Ph
Ph
N
N H H
3
2 At which location chances of electrophilic nitration will be maximum:N1 H (a) 2 & 5 (b) 4 & 5 (c) 1
5
(d) 1 & 2
12. 4 - Phenyl butyl chloride can be converted in to naphthalene by using: (a) Anhydrous AlCl3 & H+ (b) Anhydrous AlCl3 & Pd / D (c) Cl2 / Fe & Na / Ether (d) Anhydrous ZnCl2 & H3O+ 13. The major product formed when benzo trichloride is treated with [HNO3 / H2SO4 / heat] is : (a) o - nitro benzo trichloride. (b) m - nitro benzo trichloride. (c) p - nitro benzo trichloride. (d) 2, 4 - Di nitro benzo trichloride. 14.
Consider the following four statements (1) Benzene does not favour the attack of Nu (–) on it self. (2) On benzene ring E(+) can attacks easily because benzene ring is electron rich. (3) Benzene ring does not undergo addition reaction readily because after addition reaction benzene ring looses its aromaticity. (4) When electron with drawing groups are present on benzene ring, attack of Nu(-) on benzene nucleus becomes easy.
Out of these statements, correct statements are. (a) 1, 3 & 4 (b) 1, 2 & 4
(c) 2 & 4
(d) All are correct
94
Problems in Organic Chemistry 1
8 9
7
15. Electrophilic substitution reaction in naphthalene occurs at:-
6
2 3
10 5
16. 17.
4
(a) Position - 1 (b) Position - 2 (c) Position - 9 (d) Position - 6 Identify the group which is meta directing for nucleophilic substitution reaction in benzene:(a) SO3H (b) CN (c) isopropyl (d) COOTs Identify the reagents which can convert benzene in to phenyl cyclohexane:OH Cl (+) CH2OH & H (c) (a) & H3PO4 (b) & AlCl3 (d) All of these
18. In which electrophilic substitution reaction isotope effect is observed significantly. (a) Nitration (b) Sulphonation (c) Friedel craft reaction 19. In which case o/p ratio will be maximum when subjected to Cl2 / Fe:Cl I F
(d) Halogenation
(d)
(a)
(b)
(c)
Br
20. Which of the following compound will undergo friedel craft reaction with slower rate: (a) C6H6 (b) C6D6 *
(c)
(d) the rate is same in all of these.
14
{*=C } H( + ) C6 H 6
H (1 mol) Ni / D
2 21. Iso pentyne → → [X], [X] would be:-
CH3
|
(a) — CH2 — CH 2 — CH — CH 3
(b) — CH — CH(CH3)2 |
CH3
(c) —CMe 2
(d) — CH2 — CH — Et
|
|
Et
CH3
22. You have following quaternary ammonium ions (+)
(+)
ArNMe3 (I)
23.
(+)
ArCH2NMe3 (II)
Arrange the folowing in order of % of meta substitution (a) I > II > III (b) II > I > III G
ArCH2CH2NMe3 (III)
(c) III > II > I
(d) I > III > II
FeCl3 + BrCl → [X] (Major)
What should be [X]if G is sec-butyl? G G Br (a) (b)
G
G
Cl
(c)
Cl
(d)
Br
95
Reaction Mechanism (General Organic Chemistry) (i) H ( + ) (ii) Ph − H
24.
→ [X], [X] may be:(a)
(b)
Φ
(c)
(d) Both (a) & (c)
(i) FeCl
3→ + (5 - chloro pent - 2 - ene) [A], [A] will be:(ii) H ( + )
25.
(iii) Pd / D
(a)
(b)
(c)
(d)
26. Sulphonation of benzene differs from rest of electrophilic aromatic substitution reactions because: (i) In sulphonation neutral electrophile attacks on benzene nucleus. (ii) It is reversible. (iii) It requires drastic conditions like high pressure and temperature. (iv) It requires inert atmosphere.
The correct statement is / are:(a) 1, 3 (b) 1, 2, 4
(c) 1, 2
(d) 1, 2, 3 & 4
27. Correct order of electron density inside benzene ring in the following compound is:-
HOOC
A
O
CO
O
B
C
ONH2
NO2
(a) A > B > C
(b) B > C > A
(CH2 )4 — CH2Cl
28.
(c) C > A > B
Anh AlCl3 HBF4 NO 2 → [X] → [Y] would be:-
(CH2)4CH2NO2
(a)
(c)
O4Cl
O2N
(b)
(d)
O2N
O2N
29. For the conversion of benzene in to naphthalene which reagent is not needed?
(a) Succinic anhydride / AlCl3 / H2O
(b) H+
(c) Zn / Hg + HCl, Pd / ∆
(d) All are needed.
30. Select the correct statement:
(a) Chlorination of nitro benzene is easier than nitration of chloro benzene.
(b) -CHCl2 & CCl3 groups are o & p directing.
(c) In aniline and phenol friedel craft methylation is not possible by CH3Cl & Anh. AlCl3
(d) (b) & (c) are correct
(d) A > C > B
96
Problems in Organic Chemistry
31. Consider the following reaction Et |
Fe/Cl 2 C6H6 + Me3CCH2Cl Ph — CMe 2 , [W] Major
NH
NH EtONO
C S
C S
2 →
[X] major
NO2
COR
COR
CH3
CH3 [ Y] major
H S O
2 2 7→
SO3H
[ Z] major
Cl / FeCl
2 3→
Cl
The product which is least likely to form (a) [W] (b) [X]
(c) [Y]
(d) [Z]
Passage - I IPSO attack is a kind of aromatic substitution reaction in which a non hydrogen substitute in aromatic ring is substituted by other sustituent. Answer the questions from 32 to 36. 32. Which is not an example of IPSO attack:-
OH O2N
OH COOH
NO2
conc HNO conc H 2SO 4
3→
(a)
NO2 CH3 CH3 C
CH3 (+)
(+)
(b)
+ H
+ C Me3
Cl NO2
O2N
NO2
(c)
+ H2O ——→
NO2
OH
(d) All of these are the examples of IPSO attack.
NO2 + HCl
NO2
97
Reaction Mechanism (General Organic Chemistry) (i) i − BuCl / AlCl
3→ [X], [X] would be:(ii) MeCl / AlCl
33.
(iii) Fe / Br2
3
CMe3
Me
CMe3
CH2CHMe 2
(a)
(b)
(c)
Me
(d)
Me
Br Br
Br
Me
Br
34. The IPSO attack most likely to occur is: (a) ΦCΦ3 + H+ ——→ [CΦ3]+ + Φ - H (c) Φ — Me + H+ ——→ Me+ + Φ — H
(b) ΦC(CF3)3 + H+ ——→ Φ — H + C+(CF3)3 (d) All are possible
35. Para nitro phenol can show IPSO attack when it is treated with (a) nitrating mixture (b) Cl2 / Fe (c) Bromine water
(d) All of these
CH3
CH3
+ Me2C = CH2
H+ / D
36.
→
CMe3
Which of the following statement is correct for this reaction:(a) It is ArSE (IPSO attack) (b) It is an example of β - elimination. (c) It is both ArSE and β - elimination (d) None
Passage - II o / p or m - directive influence of a group can be explained by the stability of carbocation (wheland complex or σ complex) formed as intermediate. When electrophile attacks on substituted benzene different carbocation intermediates form.Their relative stabilities help us in determining the directing nature of the group present on benzene ring. Answer the questions from 37 to 40. 37. Most stable intermediate among the following is:-
NH2 CH3
NH2 (+)
E
(a)
(+)
(c)
OMe
OMe 38.
(b)
NO2
E
E
+
→
(+)
E [A]
Which statement is not correct about [A] (a) It is a sigma complex (b) +R effect of OMe helps in the stabilization of positive charge. (c) It has three resonating structures. (d) Stability of [A] indicates that OMe is o/p directing group.
(+)
(+)
(d)
E
98
Problems in Organic Chemistry
39. Most stable intermediate among the following is:-
(a)
(–)
OH NO2
Cl
OH NO2
Cl
(b)
OH NO2
Cl
(c)
(–)
F
(d)
(–)
OH NO2
(–)
40. Identify the most stable intermediate which should be formed during electrophilic substitution in Indole.
N H (+)
(a)
E
E
(b)
N
N
H
H
(+)
(+)
(c)
(+)
N
E
(d)
N
H
E
H
Passage - III Rate of aromatic electrophilic substitution depends upon the electron density inside the benzene ring. If benzene ring is substituted by electron donating groups, electron density inside the benzene ring increases thus, E(+) attacks on ring with more efficiency and hence rate of aromatic electrophilic substitution increases. In the same way rate of electrophlic substitution decreases if benzene ring is substituted by electron withdrawing group. Answer the questions from 41 to 45. 41. Rate of ArSE will be maximum in:-
NH2
NHOMe
NHAc
(a)
(b)
(c)
(d)
NPh2
2
G 3
1
G
G 42. 1
2
3
If G1, G2 & G3 are OMe, Cl & CHO respectively then correct sequence of benzene nucleus (1, 2 & 3) in order of their decreasing electron density is :-
(a) 1 > 2 > 3
(b) 2 > 3 > 1
(c) 3 > 2 > 1
43. You have three substituents G1, G2 & G3 as follows G1 = NO2,
G2
= OH,
G3
=
(d) 3 > 1 > 2
Me3C
Identify the wrong statement
(a) If benzene ring is substituted by G3 then on nitration para nitro derivative will be formed as major product.
(b) Out of p — G1— C6H4 — G3 & p — G2—C6H4 — G3 former will have higher electron density in benzene ring than later.
(c) From G3 — Ph, it is difficult to get o — G1 — C6H4 — G3 as a major product by electrophilic aromatic nitration.
(d) All are correct.
99
Reaction Mechanism (General Organic Chemistry)
44. Arrange the following compounds in decreasing order of rate of ArSE CH2Me CH3
I
II
(a) I > II > III > IV
(b) IV > III > II > I
45. Highest electron density inside the benzene ring will be in: (a) Φ — Cl (b) Φ — F 46.
(+)
Ph
•
Cu
+
(Free Radical)
N2
CHMe2
CMe3
III
IV
(c) I > III > IV > II
(d) III > IV > I > II
(c) Φ — Br
(d) Φ — I
——→ [A], [A] would be:(–)
(a)
(b)
(c)
(+)
N = NCu
(d)
47. For friedel craft alkylation rate law can be written as: Rate = K [Ar — H] [RX] [MX3]
Where MX3 is Lewis acid (Catalyst) The expected order of effectiveness of Lewis acids in friedel craft alkylation will be :(a) AlCl3 > BF3 > FeCl3 > SnCl4 (b) BF3 > AlCl3 > FeCl3 > SnCl4 (c) FeCl3 > AlCl3 > BF3 > SnCl4 (d) AlCl3 > FeCl3 > BF3 > SnCl4
48. Which among the following will produce 3 products on nitration by HNO3 + H2SO4? Et
Et
CH3 CH3 NO2
NO2 I (1)
(2)
(a) 1 & 4
CH3
SO3H SO3H (4)
(3)
(b) 2, 3 & 5
(5)
(c) Only 1
(d) 2 & 5
49. What would be the least possible molecular weight of hydrocarbon which is optically active and undergoes Friedel Craft reaction with RX in presence of anhydrous AlCl3? (a) 122 (b) 108 (c) 134 (d) 132
CMe3 (+)
N O 2 BF4(–)
50.
→ [A], [A] will be:-
Me3C
CMe3
CMe3 51.
(a)
O2N Me3C
CMe3
CMe3
(b)
O2N Me3C
CMe3
NO2 CMe3
(c)
O2N
CMe3
(d) both a & b
Which statement is not correct about benzene and pyrrole (C4H5N) (a) Both are aromatic but pyrrole ring has less electron density in comparison to benzene ring. (b) Aromatic electrophilic substitution in pyrrole is difficult in comparison to benzene. (c) In pyrrole 3rd & 4th locations are more sensitive towards ArSE. (d) Pyrrole has 6π delocalized electrons.
100
Problems in Organic Chemistry
OH H( + ) C6 H 6
52.
→ [X], [X] would be:-
(a)
O
(b)
(c)
(d)
CD2H 53. Which of the following can show both –R & +R effect depending upon situation? OTs (1)
CH == CH2
NO (2) (b) 1, 3, 4, 5
N == NH (4)
(3)
(a) 2, 3, 4 54. Most sensitive benzene nucleus towards ArSE is present in
(c) 1, 2, 3, 4
CHS (5) (d) 2, 3, 4, 5 OCH3
OCH3 (–)
N
O
(a)
(b)
(c)
(d)
CH 3
OMe CH3 2 moles of AlCl CH3Cl
3→ Product
55.
The product of this reaction would be:OMe CH3
CH3
(a)
(b)
OMe CH3
OMe CH3
CH3
(c)
(d) Both b & c
CH3 AlCl3
EtONO 2
56. Ph — H + DCl →[X] →[Y] is:- (Excess)
NO2
NO2 D
(a) C6H5NO2
D
(b)
D
D
D
(c)
NO2
(d)
D
D CH3 57.
D
D
——→ CH3CO
SO3H
During this transformation the reaction which should be carried out at last is: (a) Friedel craft acylation (b) Sulphonation (c) Friedel craft alkylation 58. Which among the following will undergo Friedel Craft reaction (F. C. R.)?
(d) Unpredictable H
N
(1)
(a) only 2
S
(2) (b) 2 & 5
(3) (c) 1, 2, 3, & 5
N
(4)
(5) (d) All will show F.C.R.
101
Reaction Mechanism (General Organic Chemistry)
59. Which among the following can be used as a solvent in friedel craft alkylation (by RX/AlX3)? (a) Φ — Me (b) Φ — CH2Cl (c) Φ — NO2 (d) Φ — OH
60.
OH
Sequence of reagents required for this conversion is:(a) H3PO4 / ∆, & LiAlH4 (c) H3PO4 / ∆ & Rh / ∆
(b) PCl5, alc. KOH, H+/∆ & Rh /∆ (d) Both (b) & (c)
61. Which is not produced as intermediate in following transformation? (+)
Ph CH2OH
H
——
(+)
(a)
(b)
(c)
(+)
(d)
(+) (+)
62. What is [A] in the given reaction:-
SO3H HO
NH2
Br water
2 → [A]
Br
SO3H
HO
HO
(a)
NH2
Br
Br
(b)
NH2 Br
Br
Br
Br
HO
Br
Br
NH2
(c)
HO
(d)
NH2
Br
Br
63. Which among the following friedel craft reactions is not possible:AlCl
3 PhCH2Cl + Ph — H → Ph2CH2 ……………..….. (1)
3 Ph2CHCl + Ph — H → Ph3CH ...………………... (2)
3 Ph3CCl + Ph — H → Ph4C .…….……………... (3)
3 p — i — Bu — C6H4 — Cl + Ph — H → p — i — Bu — C6H4 — Ph……(4) (a) 2, 3, 4 (b) 3, 4 (c) 4, 1
64.
The sequence of reactions by which benzene can be converted in to 3 - ethyl - 5 - nitro benzene sulphonic acid is:(a) Friedel craft alkylation, Nitration, Sulphonation (b) Sulphonation, nitration, Friedel craft alkylation (c) Nitration, Friedel craft alkylation, sulphonation (d) Sulphonation, Friedel craft alkylation, nitration
AlCl
AlCl
AlCl
(d) 2, 3
102
Problems in Organic Chemistry
CH2
65.
COCH3
HBr acetic acid / heat
→ [A], [A] would be :-
(a)
(b)
(c)
O
CH2OH
66.
H( + ) Heat
→
(a)
67. 2Ph — OH +
[X], [X] is:-
(b)
(c)
(a)
C
(d)
HCl
→ [A], [A] would be:-
CH3
O
(d)
OH
C
(b)
CH3
(c) HO
OH
CMe2
(d) HO
CH2COCH 2
OH
68. The reagents required for the conversion of toluene in to 2 - (4-methyl phenyl) butan - 2 - ol is: (a) But - 1 - ene & H2O (b) Butan -2-ol / H3PO4 (c) Butan - 2 - one / H3PO4 (d) both (a) & (b) 69. When benzene is treated with methyl epoxy ethane in presence of (i) AlCl3 (ii) H2O a compound X (C9H12O) is formed. X will be:OH OCH2 CH 2 CH3 OH CH2CH 3 CH2OH (a) (b) (c) (d) 70. Predict the major product of bromination of the compound given below by using Br2 / FeBr3 in the dark: NO2
CH3 NO2
NO2
(a)
NO2
Br (b) CH3
(c) Br
CH3
Br NO2
(d) Br CH3
CH3
103
Reaction Mechanism (General Organic Chemistry)
71. Arrange the following in increasing order of rate of electrophilic nitration. Cumene anisole Benzene (1) (2) (3) (a) 1 > 2 > 3 > 4 (b) 2 > 3 > 4 > 1 (c) 2 > 3 > 1 > 4
Benzoic acid (4) (d) 2 > 1 > 3 > 4
72. How many carbocations are produced during the following transformation? 2 3
(a) 1
(b) 2
(c) 3
(d) 4
73. Which among the following is the correct potential energy diagram for nitration of benzene?
(a) G
(b)
∆G
reaction progress
reaction progress
(c)
G
(d) G
reaction progress
reaction progress
74. In which case mono substituted derivative of benzene is not produced. (a) Benzene + N2O5 (b) Benzene + Oleum (c) Benzene + CH3COCl + anh. AlCl3 (d) Benzene + CH3Cl + anh. AlCl3
NO2
NO2 H OH CH3
75.
O2N
——
NHCH3
O2N
N Me
This conversion can be performed by (a) H(+) /∆ (b) AlCl3
(c) EtOH + NaOH
(d) All of these
104
Problems in Organic Chemistry (i) CH 2 N 2
→ [A], [A] would be:76. 3-Chloro propene (ii) AlCl / C H 3
6 6
CH2 —CH2 = CH2
(a)
(c)
(b)
(d)
AlCl
77.
3 + C2H5 — Cl → Ph - Et
For the better yield of ethyl benzene, benzene & ethyl chloride should be taken in ratio. (a) 1 : 15 (b) 15 : 1 (c) 1 : 1 (d) 1 : 1 & AlCl3 should be taken in excess OH
OH
78.
NO2
+
steam distillation
———————→
[X], the distillate
NO2
How many products will be obtained by the sulphonation of [X] (distillate)? (a) 1 (b) 2 (c) 3
79.
1 Mole of oxirane is mixed with 2-moles of benzene & catalytic amount of H3PO4 is then added product formed in this reaction
(a) Ph2 (CH2)2
(d) 4
would be:-
(b) PhCH2CH2OH
(c) PhCH = CH2
(d) Reaction is not possible
AlCl
3 80. CCl4 + Benzene (excess) → [X], [X] would be: (a) Φ2CCl2 (b) Φ3CCl (c) Φ4C
(d) ΦCCl3
X
81. Ph — H + I2 → Ph — I + HI X would be: (a) HNO3 (b) FeI3 O
(c) HI
(d) Fe / I2, ∆
Br FeCl3
82.
2 → Product, Product will be:-
O Br O
O
(a)
(b)
O (100%)
Br
Br
O (100%)
Br
O
O
(c)
O (100%)
(d)
O
&
Br O
O
105
Reaction Mechanism (General Organic Chemistry)
Answer Key 1. (b)
2. (b)
3. (c)
4. (c)
5. (d)
6. (d)
7. (c)
8. (c)
9. (c)
10. (c)
11. (a)
12. (b)
13. (b)
14. (d)
15. (a)
16. (c)
17. (d)
18. (b)
19. (c)
20. (d)
21. (c)
22. (a)
23. (d)
24. (b)
25. (d)
26. (c)
27. (c)
28. (c)
29. (d)
30. (c)
31. (b)
32. (d)
33. (b)
34. (a)
35. (c)
36. (c)
37. (a)
38. (c)
39. (c)
40. (b)
41. (c)
42. (a)
43. (b)
44. (a)
45. (b)
46. (c)
47. (d)
48. (d)
49. (c)
50. (c)
51. (c)
52. (d)
53. (a)
54. (a)
55. (a)
56. (b)
57. (c)
58. (c)
59. (c)
60. (d)
61. (c)
62. (c)
63. (b)
64. (b)
65. (a)
66. (c)
67. (c)
68. (c)
69. (a)
70. (a)
71. (d)
72. (c)
73. (a)
74. (d)
75. (a)
76. (d)
77. (b)
78. (b)
79. (a)
80. (b)
81. (a)
82. (d)
Electrophilic and Free Radical Addition Reactions Multiple Choice Questions 1. Which is correct potential energy diagram for following transformation?
(a)
G
(b)
G
reaction progress
(c)
G
reaction progress
reaction progress
(d)
G
reaction progress
106
Problems in Organic Chemistry
2. CH3 — CH = CH2 + Br2 → Products The product, not expected in this reaction is:in NaCl
Br
Br
(a)
CH2Cl
(b)
Cl
CH2Br
(c)
CH2Br
(d) All of these
KMnO 273K
4 → [X] 3. Cis -but- 2 – ene
CH3 Me H
Me
(a) H HO
OH
H (b)
(c) H Me
OH
H
HO
Me H
OH
(d) Both (a) & (b)
OH
CH3 4. How many carbocation are produced during the following transformation? OH
OH + dil H2SO4 ———→
(a) 4
5.
D
D C
H
=C H
(c) 6
(d) 3
is subjected to halogenation by X2 / in CCl4. In which case meso isomer will be produced?
(a) Iodination
(b) Bromination
(c) Chlorination
(d) Fluorination
H /D
6.
(b)
2 → [A], [A] will be:-
(a)
(b)
(c)
(d)
3 7. CH3 — CH = CH2 + Cl2 → CH3 — CHCl— CH2 — Cl
AlCl
Select the correct statement regarding the mechanism of this reaction. (a) This reaction occurs via radical mechanism. (b) Cyclic carbonium ion is not formed as an intermediate. (c) AlCl3 helps in the generation of Cl(+) (d) AlCl3 has no role in this reaction.
8. Consider the following compounds.
Ph2C = CH2 1
(CH3)2C = CH2 2
Correct order of rate of electrophilic addition reaction is:(a) 1 > 2 > 3 (b) 3 > 2 > 1
Ph2C = CHCF3 3 (c) 1 > 3 > 2
(d) 3 > 1 > 2
(c) Trans – 2 – butene
(d) 1, 3 – Butadiene
9. The compound with highest heat of hydrogenation is:
(a) Me2C = CH2
(b) Cis – 2 – butene
107
Reaction Mechanism (General Organic Chemistry)
10. CH2 = CH2 + H2 → CH3 — CH3 Ni D
It is an example of:(a) Free radical addition reaction (c) Nucleophilic addition reaction
(b) Eleclrophilic addition reaction (d) Molecular addition.
11. Acetylene ——→
This reaction can be performed by:-
(a) CH2 = C = O / hv
CH = CH 2
12.
(b) CH2N2 / hv
(c) CH2I2 / Zn (Cu)
(d) All of these
(c)
(d)
H( + ) EtSH
→ [A] Major, [A] will be:-
SEt
(a)
SEt
(b)
SEt
SEt
13. Consider the following reactions.
Br
(1)
→ + HBr
(3)
Peroxide + C2 H 5SH →
The reaction with wrong product is / are:(a) 2 (b) 2 & 3
SEt
hn
Cl
hn
Br
(2)
+ HCl →
(4)
+ HBr →
(c) 3
(d) 1 & 4
O (+)
H3O → [A],
14.
[A] would be:-
CH2OH
OH
(a)
(b)
CH2OH
(c)
(d)
CH3 OH
OH
OH
15. When cyclo hexene is treated with chlorine water followed by aq NaOH, another compound [X] is produced. The compound [X] can be identified as:-
(a)
(b)
OH
(c)
O
(d)
Cl
OH
(i)Cl / H O (ii)Na 2CO3
16.
2 2 → [A], [A] would be:-
Cl
OH
Cl
Cl
OH
(a)
OH
(b)
OH
(c)
O
(d)
CO3
108
Problems in Organic Chemistry
17. Which statement is correct about A & X?
CH3
CH3
|
|
dil
A Ph — CH — C CH 2 H 2SO4 [X]
(a) (b) (c) (d)
On dehydration ‘A’ gives alkene (major product) which can show stereoisomerism ‘A’ is optically active alcohol and contains 2- chiral carbon atoms. On reduction by H2 / Ni compound ‘X’ gives optically active hydrocarbon which contains one chiral carbon atom. ‘A’ is an optically inactive alcohol.
Ac
18.
CH—CH 3 + HBr ——→ [A], [A] may be:-
NC
(a)
Ac
CH2— CH3
NC
(b)
Ac CH—CHBr—CH 3 NC
Br
OH
(c)
Ac
CH—CH 2 —CH2Br
(d) CH3 C Br CH—CH—CH3 NC Br
NC
19. (O2N)2 C = CH — CH3 [A]
Which statement is not correct about [A] (a) HBr will add on it according to Markonikoff rule (c) It will favour nucleophilic addition H( + ) D 2O
20.
→ [A], [A] will be:D
(b) Addition of HBr on [A] will give antimarkonikoff product (d) Both (b) & (c)
(a)
OD
(b)
D
(c)
OD
OD
(d)
OD
Cl H 2O
21.
2 → [A], [A] is:
OH OH
Cl
Cl
(a)
(b)
Cl
(c)
(d)
Cl
Cl
Cl O
(+)
H3O
22. CH3—CH—CH—CH3 ————→
— cis-CH3—CH—CH—CH 3
+
tans-CH3—CH==CH—CH3
+
CH3
H
D2
Br2
.....................(1)
————→
.....................(2)
.....................(3)
4 ————→
.....................(4)
Ni ∆
in CCl4
————
CH3 H
+
Br2
Racemic mixture will be produced in:(a) 1 & 4 (b) 2 & 3
in CCl
(c) 1, 3, & 4
(d) 3 & 4
109
Reaction Mechanism (General Organic Chemistry)
23. In the reaction given below [X] is an organic compound & can be identified as:hn
( Ph2CH)2N2 + Br2 → [X], (a) Ph2CHNBr—NBr — CHPh2 (c) Ph2CHBr (+)
H D2O
24
A A & B respectively are :-
(+)
D H2O
(b) Ph2CHNBr2 (d) All of these
B
(a)
, OD
,
(b)
OD
OD
D
D
OD
D
(c)
,
OH
OD
OH (d)
,
D
OH
25. dil H SO Excess
2 4 —————— → [X], Major
Ph
How many chiral ‘C’ atoms are present in [X]? (a) 3 (b) 4
(c) 5
(d) 6
26. When 2 – iodo – 1 – phenyl propene is subjected to addition reaction with HI in dark, an organic compound [X] is formed which is: (a) 1, 2 – Di iodo – 1 – Phenyl propane (b) 2, 2 – Di iodo – 1 – Phenyl propane (c) 1 – Phenyl propene (d) Both (a) & (b) 27. CH 2 be:-
= CH — CH = C* H2
* 14 [A] C == C If 1 mol of [A] is treated with 1 mol HBr in dark then expected product/s will
Br
Br
|
• (a) C H 2 = CH –CH — C H3
(c) BrCH 2 CH CH — CH3 & BrCH 2 CH CH — CH 3 (d) All of these
|
= CH –CH — C• H3
*
*
Ph 28.
(b) C H 2
3 1 2
Here double bonds are numbered as 1, 2 & 3. The correct order of these double bonds towards eletrophilic attack is:(a) 1 > 2 > 3 (b) 3 > 2 > 1 (c) 1 > 3 > 2 (d) 3 > 1 > 2
Me 29.
+ HBr ——→ [A], major [A] will be:-
Ph
Br
Me (b)
(a)
Ph H
H
(c)
Ph
Ph H
H
Me
Me
Br
Br
Me
(d)
Ph Br
110
Problems in Organic Chemistry H( + )
30. CH2 = CH2 + H2O → CH3CH2OH
For this reaction rate law is:-
(a) Rate = k [CH2 = CH2] [H+]
(b) Rate = k [CH2 = CH2] [H2O] [H+]
(c) Rate = k [CH2 = CH2]
(d) Rate = k [CH2 = CH2] [H2O]
31. The intermediate of the reaction is:
Phenyl ethylene + ICl ——→ Product ( )
I |
(a) Ph — CH — CH 2
( )
I
Cl |
(b) Ph — CH — CH 2
Cl (+)
(+)
(c) Ph—CH —CH2
(d) Ph—CH—CH 2
32. Correct order of reactivity of the following halogens and interhalogens towards electrophilic addition to C = C is: (a) BrCl > Br2 > IBr > I2 (b) Br2 > I2 > BrCl > IBr (c) BrCl > IBr > Br2 > I2 (d) IBr > Br2 > BrCl > I2 33. Which of the following compounds is most reactive towards electrophilic addition reaction?
(a) CH2 = CH2
(b) (F3C)2C = CH2
NO2
(c) NC — CH = CH — CHO
(d) O2 N
CH = CH 2 NO2
34. An alkene (A) on reduction gives an optically active alkane of least molecular weight & on treatment with H+ gives 3° carbonium ion which can not show rearrangement. What should be the structure of (A) if it can show geometrical isomerism?
(a) CH3CH2CH(CH3)CH = CH — CH3
(b) CH3CH2CH(CH3)CH2 — CH = CH2
(c) CH2 = CH — CH(CH3)CH2 — CH = CH2
(d) CH3 CH CCH 2CH 2CH 3 |
CH3
35. End product of the following reaction
O + HBr (1mol) ——→ is :-
O
(a) O
O
(b) HO
OH
(c) Br
Br
Br
OH
(d) O
OH
Br
OH
36. When Cis- but-2-ene is treated with Br2 in carbon tetra chloride medium the product formed will be: (a) (2 R, 3 S) di bromo butane (b) (2 R , 3 R) di bromo butane (c) (2S, 3 S) di bromo butane (d) Mixture of (2 R, 3 R) & (2 S, 3 S) di bromo butane
Answer Key 1. (d)
2. (a)
3. (d)
4. (c)
5. (d)
6. (a)
7. (c)
8. (c)
9. (b)
10. (d)
11. (d)
12. (c)
13. (a)
14. (b)
15. (c)
16. (a)
17. (c)
18. (b)
19. (d)
20. (b)
21. (a)
22. (a)
23. (c)
24. (d)
25. (c)
26. (c)
27. (d)
28. (c)
29. (d)
30. (a)
31. (c)
32. (a)
33. (a)
34. (d)
35. (b)
36. (a)
111
Reaction Mechanism (General Organic Chemistry)
Elimination Reactions Multiple Choice Questions
EtOH + KOH
1.
→ [X], [X] will be:-
Cl
(a)
(c) Both (a) & (b)
(b)
(d) No product will form
CH3
2.
+
alc KOH ————— major product
Cl
(a)
( )(–)
Which product will not form in this reaction? (a) CH3OH (c) H2O conc H3PO4 ∆
(a)
(d)
(b) Me3N (d) No product will form in this reaction
A, the product A would be
(b)
(c)
(d)
(c)
(d)
H PO
(a)
(b)
conc H SO
2 4 6. CH3CH2CH2OH ——————→ ∆
3 4→ Major [P], [P] would be:OH D
5.
(c)
4.
Products
3. (CH 3 )4 NOH
(b)
(a) E1 pathway
CH3CH==CH2
(b) E2 pathway
+
H2O
(c) E1cb pathway
(d) Unpredicable
112
Problems in Organic Chemistry
7. Select the reaction in which chances of SN1 are maximum
(a)
I + EtNH2 —→ Product
(c)
+ C2 H5 O —→ Product
+ NH2(–) —→ Product
(b) I
(–)
(d)
I + C2H5OH —→ Product
I
Ph + alc KOH —→ X (major), [X] would be:Me
8. CH3CHI—CH Erythro
Me
Me H
(a)
H Ph
OH H
Me
(b)
Me
Ph
H
(c)
Me
H
(d)
H H
OH Ph
Me Me
Me
9. Number of possible products obtained by the acid catalysed dehydration of 3-methyl-pentan-2-ol is (do not consider stereoisomers):
(a) 1
(b) 2
(c) 3
(d) 4
OH dilute acid
10.
→ [X], [X] can be identified as:-
OH
OH
(a)
(b)
(c)
(d)
H (+) (–)
11.
Heat
→ [X] Major, [X] will be:Ph NMe3OH
Ph
(a)
(b)
Ph
(c)
(d) None of these.
Ph 12. The sp3 hybridised carbon atom can not bear two groups which interacts with each other as shown below & elimination occurs:OH C
——————→
H2O
+
C == O
——————→
HF
+
C == O
OH OH
C
H-bond
H-bond
F
(+)
OH
H3O
Predict the final product of the following reaction:- Cl
(a)
O
(b)
O
(c)
O
(d)
O
113
Reaction Mechanism (General Organic Chemistry)
13. Which alkyl halide would you expect to be least reactive in an E1 reaction? (a) CH CH —CH — CH (b) Ph — CH — CH3 2 3
|
|
Br
Br
C — CH 2 — CH3
(c) CH 2
(d) Ph — CH2 — CH2Br
|
Br (i) H / Ni / D
2 → 1, 3 – butadiene + Me3N Substrate will be:14. Substrate (ii) CH I(excess) 3
(iii) AgOH / D
(a)
Br
N
N
H
H
H
H
[X] H I
NH2
, [X] would be:-
Br H
H
D
(a)
D
(d)
D (Product)
D
(c)
N
15.
(b)
H
(b)
I
D
D H
H
(c)
H
D
(d) Both (b) & (c)
D
16.
In this reaction % of [X] & [Y] are respectively:(a) 15 % & 85% (b) 50 % & 50%
(c) 85 % & 15%
(d) 55 % & 45%
Passage - I Acid catalysed dehydration of alcohol follows E1 mechanism. In it carbocation is formed as intermediate. Rate of dehydration of alcohol depends on two factors viz stability of carbocation & acid strength of H present at adjacent position to positively charged carbon atom in intermediate. Answer the following questions from 17 to 19 17. Consider the following alcohols
OH
(1)
OH
NO2 (2)
Correct order of rate of acid catalysed dehydration will be:(a) 1 > 2> 3> 4 (b) 2> 1 > 3 > 4
OH
OH
NMe2
NO2
(3)
(c) 3 > 2 > 4 > 1
(4)
(d) 3 > 2 > 1 > 4
18. Correct order of rate of acid catalyzed dehydration if nitro group is replaced by methyl group (a) 1 > 2 = 4 > 3 (b) 2 > 1 > 3 > 4 (c) 3 > 4 > 2 > 1 (d) 2 = 4 > 3 > 1 19. Correct order of rate of acid catalyzed dehydration if dimethylamino group is replaced by methyl group (a) 1 > 2> 3> 4 (b) 2> 1 > 3 > 4 (c) 3 > 2 > 4 > 1 (d) 3 > 2 > 1 > 4
114
Problems in Organic Chemistry
20. Match the following Reaction Product Br
alc KOH
Me →
(A)
OH
H PO D
3 4→
(B)
F 21.
(p)
CH2
(q)
3°BuO(–)
→ (C) (a) A —→ p, B —→ q, C —→ r (c) A —→ r, B —→ p, C —→ q
O
OH
(r) (b) A —→ q, B —→ p, C —→ r (d) A —→ r, B —→ q, C —→ q
O
(–)
OH → H2O + D
This reaction occurs via:(a) E1 mechanism
(b) E2 mechanism
22. CH3CH2ONa +
(c) E1CB mechanism
60° C ———
(d) E2 as well as E1CB
+ OEt
Br [X]
In this reaction % of [X] & [Y] are respectively:(a) 75% & 25% (b) 25% & 75%
[Y]
(c) 50% & 50%
(d) almost 100% [X] is produced
23. Main products of reactions 1 & II are respectively:
(I) Et3CBr
CH OH
3 ———— 25°C
CH3
(II)
C—H
+
50°C ———— CH OH
NaSH
Br
3
Et CH3
CH3
(a) Et3COMe and
H
(b) Et2C==CH2 and H
Et SH CH3
(c) Et3COMe and
(d) Et2C==CH2 and
SH
H
Et SH CH3
SH
H Et
Et
Matrix Match Questions Each question contains statement given in two columns which have to be matched. Statement (A, B, C , D) in column I have to be matched with statements (p, q, r, s) in column II. The answers to these questions have to be appropriately bubbled as illustrated in the following examples. If the correct match are A - p, A - s, B - r, B - r, B - q, C - q, D - S, then the correctly bubbled 4 x 4 matrix should be as follows.
115
Reaction Mechanism (General Organic Chemistry)
A
p
q
r
s
B
p
q
r
s
C
p
q
r
s
D
p
q
r
s
24. Reaction
Graphs
(A) E1
(p) Rate
Base
(B) E2
Free (q) Energy
Temp
(C) SN1
(r) Rate
Base
(D) SN2
(s) Rate
25. If r1 & r2 are rate of reactions then match the following. substrate Column - I Column - II
(A)
H PO ∆
4 CH3 CH2 OH 3 → alkene
(r1)
H 2SO4 CD3 — CH 2 — OH → ∆
alkene
(p) Follows saytzett rule
(r2)
( )
H
CH3CH2 CH 2OH alkene
(B)
(r1)
( )
H CH3CHOHCH3 alkene (r )
(q) r1 r2
(r) r1 > r2
2
OH |
(C)
H( +) ∆
Ph — CHCH3 → alkene (r1)
( +)
H CH3CH2 CH 2OH → alkene ∆ (r ) 2
CH3 CH3
(+)
C—CH2—CH 3 OH
(D)
CH3 CH3
H
alkene (r1)
(s) r1 < r2
(+)
CH—CH—CH3 OH
H
alkene (r2)
Answer Key 1. (d)
2. (b)
3. (c)
4. (c)
5. (b)
6. (b)
7. (a)
8. (c)
9. (c)
10. (b)
11. (a)
12. (b)
13. (c)
14. (b)
15. (c)
16. (c)
17. (d)
18. (c)
19. (d)
20. (c)
21. (c)
22. (d)
23. (b)
116
Problems in Organic Chemistry
Answers Matrix Match 24. A → r, s ; B → p, s, q ; C → s, r ; D → q, s, p 25. A → q ; B → s ; C → r ; D → p, r
Multiple Choice Questions (More Than One May Correct) 1. Identify the compound whose resonating structure can show geometrical isomerism:-
O
(a) CH3CONHCH3
(b)
NH
(c) NH2CONHNH2
(d) PhCONH2
2. Identify the species where resonance is not possible:-
(–)
(–)
(–)
(+)
(a) CH 2Cl
(b)
(c)
F
(d) CH 2 CHO
3. In which case 1st resonating structure is more stable than 2nd (–)
( )
(a) CH3 — C
O
—
( )
CH3 — C O
O
O —
(b)
(–)
(+)
(–)
(c)
(–)
—
(+)
(+)
O
—
(d)
O
O
O (+)
4. Which is not correctly matched? (a) CO32– ——→ Bond order of C — O bond is 3/2 (c) ClO4(–) ——→ Bond order of Cl —O bond is 7/4
(b) CH3COO(–) ——→ Bond order of C — O bond is 4/3 (d) PO43– ——→ Bond order of P — O bond 5/4
5. In which case 1st species has more resonance energy than 2nd.
O
(a)
,
O
(b)
,
O
(c) PhCOO(–), PhO(–)
(+)
(d)
6. Correct order of leaving group character is : (a) CCl3COO(–) > CH3COO(–) > CH3O(–) > PhO(–)
O
(c) CCl3COO(–) > PhO(–) > CH2 = CH — O(–) > CH3O(–)
(+)
(b) OTs(–) > CH3SO3(–) > PhO(–) > OH(–) (d) HCOO(–) > PhO(–) > CH3COO(–) > CH3O(–)
7. Which among the following is/are aromatic?
CH3
H
O
B
(a)
NH
(b)
N H
O
CH3
N
(c)
N
(d)
N
N
H
H
O
117
Reaction Mechanism (General Organic Chemistry)
N2 8. Which statement is not in favour of
N1
H (a) It is aromatic (c) 2nd nitrogen is less basic than 1st nitrogen. gen
(b) It looses its aromaticity when 1st nitrogen reacts with H+ (d) It becomes antiaromatic when H+ ion attacks on 1st nitro-
9. The compound from which NH2(–) can not abstract H+ is/are:
(b) CH2 = CH2
(a) CH3CH3 CH3
(c) CH3OH CH3
|
(d) HCOOH
CH3
|
|
10. CH3 — CH —CH — CH3 CH3 — C — CH 2 CH3 CH3 — C CH — CH3 Reagent
|
|
Br
(Y)
OCH3 (X)
Which are true statements about reagent & products (X & Y)? (a) ‘X’ forms when CH3O(–) / CH3OH is treated with the substrate at elevated temperature. (b) ‘Y’ forms when CH3O(–) / CH3OH is treated with the substrate at elevated temperature . (c) ‘X’ forms when substrate is treated with CH3OH. (d) ‘X’ forms when substrate is treated with CH3O(–) / DMSO.
11. Select the true statements regarding the following four reaction intermediates. (+)
(I)
(II)
(–)
(III)
(IV) (–)
(+)
(a) All are aromatic (c) IV is more stable than III
(b) II is less stable than I (d) III is aromatic while II is antiaromatic
12. Which of the following compound will not give Friedel craft acylation under ordinary conditions?
SO3H
NO3
CH3
NO2
(a)
(b)
(c)
(d)
NO2 NO2 13. Under what conditions chances of SN1 are maximum: (a) When medium is polar (c) When medium is polar aproteic
(b) When base is weak (d) When substrate is tertiary
14. Under what conditions chances of E1CB are maximum: (a) When substrate contains poor leaving nucleophile. (c) When attacking base is poor
(b) When attacking base is strong (d) Substrate contains good leaving nucleophile
CH3 15. Which among the following statement is not correct about xylenes
CH3 ?
(a) Ortho xylene on mono nitration gives only one product (b) Para xylene on mono nitration gives only one product (c) Meta xylene is thermodynamically most stable than ortho and para xylenes
118
Problems in Organic Chemistry
(d) Among all xylenes para derivative has maximum melting point.
Cl product;-
16.
Cl
Product of this reaction is/are
(a)
(b)
R (–)
17. R—C ≡≡ C
+
(c)
(d)
H 50°C
H—C
———→ Products
C—Br H
R
Which statement is correct regarding this reaction? (a) R—C ≡≡ H will produce as one of the product (b) Alkene is produced as major product which can not show geometrical isomerism (c) Reaction occurs via E1CB path way (d) Reaction occurs via E2 path way
18. In which case first alkene is more stable than 2nd are
(a)
,
(c)
,
(b)
, ,
(d)
19. In which case I species is stable than 2nd
(a) S
S , (–)
O
O
(b) CF3(–), CCl3(–)
(–)
(–)
(–)
O
(c) CF3CH2(–), CCI3CH2(–)
O
(d)
, O
20. In which case 1st species has more heat of hydrogenation than 2nd
(a)
(c)
,
(b)
(d)
,
,
O
119
Reaction Mechanism (General Organic Chemistry)
Answer Key 1. (a), (b), (c)
2. (b), (c)
3. (b), (c)
4. (a), (b)
5. (b), (c)
6. (b), (c)
7. (a), (b), (d)
8. (c), (d)
9. (a), (b)
10. (b), (c)
11. (b), (d)
12. (a), (b), (d)
13. (a), (b), (d)
14. (a), (b)
15. (b), (c)
16. (a), (b)
17. (a), (d)
18. (a), (b), (c)
19. (a), (c), (d)
20. (a), (c)
LEVEL - II Multiple Choice Question 1. Which will have largest Ea?
(a)
(c)
+ HCl —→
Cl
(b)
+ HCl —→
Cl 2. Arrange the following in increasing order of stability.
+ HCl —→
Cl
(d) All have same Ea?
(+)
(+)
Ph 3 C( )
(1)
(+)
(3)
(2)
(a) 1 > 2 > 3 > 4
(b) 2 > 4 > 3 > 1
(4)
(c) 1 > 3 > 2 > 4
(d) 1 > 2 > 4 > 3
3. In which case 1st carbocation is not stable than 2nd.
(+)
(a)
O
(c)
(b)
(+)
& (+)
O
(+)
&
(+)
(+)
&
(+)
(d)
& (+)
4. Hexa chloro cyclohexane can have number of isomers. If one of its isomer given below is treated with three moles of alcoholic KOH,another compound [X] is formed . Select the correct statement about [X].
Cl Cl
Cl
3 moles of alc KOH
→ [X]
Cl
Cl Cl
(a) It is aromatic (c) It is tri chloro benzene
(b) Its degree of unsaturation is four (d) Reaction is not possible
120
Problems in Organic Chemistry
H (+)
→ P, P is :D
5.
(a)
(c)
(d)
OH OH This compound on nitration with nitrating mixture (HNO + H SO ) gives meta derivative because. 3 2 4
B
6.
(b)
(a) This compound takes H(+) ion from nitrating mixture and converts itself in to Ph — B O and since B O is – R group thus, meta derivatives are obtained. (b) B is electron deficient & abstracts electron from benzene ring to complete its octet thus, benzene ring acquires +ve charge at o & p positions & consequently meta derivatives are formed. (c) Nitric acid is an oxidizing agent. It oxidizes PhB(OH)2 in to PhB are obtained. (d) both (a) & (c)
O. Since B
O is –R group thus, meta derivatives
7. Rate of nitration will be maximum in:
(a)
(c)
(b)
(d)
Me
8. Which resonating structure is least stable? (+)
O
O
(a)
(b)
O (–)
(c)
O
O
(d) (b) & (c) are equally unstable.
O
(–)
9. Acetate, nitrate, Carbonate and ClO4(–) ions are stabilized by resonance. Due to resonance bond order & bond length of species affect. Select the correct statement. (a) Bond order of C — O bond in carbonate ion is equal to that of N — O bond in nitrate. (b) Bond order of Cl — O bond in ClO4(–) is equal to that of N — O bond in nitrate ion. (c) Bond order of C — O bond in acetate ion is equal to that of Cl — O bond in ClO4(–) . (d) Bond order of C — O bond in acetate ion is equal to that of N — O of nitrate ion. 10.
(+)
(1)
(+)
(2)
CD3 CD3 CD3
CD3
(+)
(+)
CD3 (3)
Arrange these carbocations in decreasing order of stabilities (a) 4 > 2 > 3 > 1 (b) 1 > 4 > 3 > 2
SiH3 SiH3 SiH3
(4) (c) 1 > 2 > 4 > 3
(d) 4 > 3 > 1 > 2
121
Reaction Mechanism (General Organic Chemistry)
11. Identify the case where -ve charge is least delocalized
(–)
(–)
(–)
(–)
CH2
CH2
CH2
CH2
CF3
CH3
(a)
(b)
(c)
(d)
CN CHO
NO2 • CH2
(–)
(+)
(+)
CH2
CH2
CH2
12.
CH3
(1) Correct order of stability (a) 1 > 2 > 3 > 4
(2)
(3)
(b) 2 > 1 > 4 > 3
(4)
(c) 4 > 3 > 1 > 2
(d) 4 > 3 = 2 = 1
13. Which is not correct about 1 & 2 :-
b
D
a
D
D
D
(1)
(a) Both are antiaromatic (c) Both are same compounds
(2)
(b) Bond lengths a ≠ b (d) Resonance decreases their stabilities
14. Which of the following has non planar geometry? (–)
(a)
(b)
(c)
(–)
(d) All are nonplanar
15. Which is not aromatic? (–)
(a)
(b)
(c)
(d) All are aromatic
(–)
16. Which is not correct about cyclo butadiene? (a) It reacts with sodium rapidly.
(b) It does not undergoes dimerisation below 4°C
(+) (+)
(c) Resonance energy of
(d) All are correct statements.
is greater than cyclo butadiene
17. You have four compounds. CH3COCH2COOC2H5
O CCl3
O
CCl3 CCl3 O
(1) (2) (3) (4) These compounds on treatment with base produce base. Identify the compound/s which can’t form planar carbanion. (a) 2, 3 & 4 (b) Only 4 (c) 1, 2, 3 & 4 (d) 2 & 3
122
Problems in Organic Chemistry
CH3
CH3
|
H( + ) Heat
|
18. CH3 — C — CH 2 — C — CH 3 → (A) Major + (B) Minor |
|
CH3
CH3
Which is correct about A & B? (a) (CH3)3CCH = C
CH3 CH3 | and (CH 3 ) 3 CCH 2 — CC H2
CH3
Minor
CH3
Major
(b) (CH3) 3CCH = C
CH3 Major
CH3
Minor
CH3
|
and (CH 3 ) 3 CCH2 — C CH 2
(c) (CH3 ) 3 CCH = C
CH3
50%
CH3 |
and (CH 3 ) 3 CCH 2 — C CH 2 50%
CH3
(d) Only one product (CH3 ) 3 CH = C
will form
CH3 19. How many products (including stereoisomers) will be formed in the following reaction?
D
H H SO
2 4→ Product 170°C
OH
(a) 1
(b) 2
(c) 3
20. How many geometrical isomers will be formed in the previous question? (a) 0 (b) 1 (c) 2 21. Which is not correct about SN2?
(a) Rate of SN2 is directly proportional to the dielectric constant of medium.
(b) Rate of SN2 is directly proportional to the nature of leaving group
(c) Rate of SN2 is inversely proportional to the steric hindrance present in substrate
(d)
Rate
Rate [base]
[substrate]
22. Which reaction is most likely to occur?
(a) RX + NaOH (in C6H6) ——→ ROH
(b) RX (in C6H6) + NaOH (aq) ——→ ROH
(c) RX + NaOH (in crown ether) ——→ ROH
(d) Both (b) & (c)
(d) 4 (d) 3
123
Reaction Mechanism (General Organic Chemistry)
23. In the given figure graphs are plotted for SN1 & SN2 in polar as well as in non polar mediums. Graph 1 is plotted in polar medium while graph 2 is plotted in non polar medium. Select the correct statement.
SN1
SN2
G1
1
G1
G2
1
G2
2
(a) (b) (c) (d)
For SN1 ∆G1 For SN2 ∆G1 For SN2 ∆G2 For SN1 ∆G2
2
> ∆ G2 because substrate is more solvated than the transition state > ∆ G2 because transition state is more solvated than the substrate > ∆ G1 because substrate is more solvated than the transition state > ∆ G1 because substrate is more solvated than the transition state 15% aq solution of ethyl alcohol
24. 2–Chloro butane → product
In this reaction 70% racemisation takes place. % of inverted product would be:(a) 30 (b) 70 (c) 35
(d) 65
Answer Key 1. (a)
2. (d)
3. (d)
4. (d)
5. (c)
6. (b)
7. (d)
8. (b)
9. (d)
10. (a)
11. (a)
12. (b)
13. (c)
14. (d)
15. (d)
16. (d)
17. (b)
18. (b)
19. (b)
20. (a)
21. (a)
22. (c)
23. (c)
24. (d)
SOLUTIONS Level - I
(AROMATICITY) Ac
Ac 1. (d)
(–)
( )
H +
aromatic and stable
2. (a) (i) is non planar because of repulsion between ‘H’ atoms as shown below H H repulsion
124
Problems in Organic Chemistry
(+)
—
3. (b)
(–)
(Both the rings follow Huckel rule & hence it is aromatic)
sp3hyd 4. (c) Non planar due to sp3 hybridization 5. (c) B is non aromatic (non planar) & A is antiaromatic 6. (b) Due to resonance (iii) becomes antiaromatic
Ph
Ph
Ph
Ph —
antiaromatic (–)
CH3
(–)
O CH3
O
7. (d) (A) is not planar (B) has 8πe– (C) has 10πe– (+) (–)
(–)
C (+)
NMe2 NMe 2
(B) 8. (b) Order of stabilities of ions produced from I, II & III is III > I > II. 9. (d) O is more electronegative so it does not give its lone pair of electrons easily for resonance hence it is least aromatic. However electronegativity of S is lesser than O & N hence resonance energy of thiophene will be more & thus, aromaticity of II will be large.(exceptional case) 10. (b) (B) follow Huckel rule and planar (C) also follows Huckel rule but not planar because S is bigger in size & does not fit in the hole created by rings. (A) is not planar due to H—H repulsion as shown below -
O H
H O Ph
Ph
Ph
Ph
(+) (–)
K
(+)
Ph
+ 2K —— H2 +
11. (d) Ph
Ph
Ph
(–)
K (+) (–) K
+ 2K —— H2 +
(+) (–) K
125
Reaction Mechanism (General Organic Chemistry)
12. (d) In these compounds tautomerism takes place.
O HN
OH
NH
N
O
O O
HO
N OH
13. (b) The products formed in 1,2 & 3 reactions are non aromatic, antiaromatic & aromatic respectively. 14. (b) does not obey Huckel rule. + 15. (a) Because after removal of H it converts in to an aromatic species
H
H
(–)
( )
H
+ aromatic
16. (c) Because it is antiaromatic (less stable) (–)
O
O
(+)
antiaromatic 17. (d) (A) is non aromatic because one of its ‘C’ is sp3 hybridised 18. (b) 19. (d) 20. (c) 21. (b) 22. (c)
O
OH SeO
PCC
2
23. (b)
aromatic
24. (a) Same as question no. 23
OH sp3
O LiAlH
4
25. (b)
1 26. (a)
2 N
3
N 4
(non planar & non aromatic)
H
H H CH3O2C
5 OCO
Rings 1, 2, 3 & 6 are aromatic. 27. (c) 6πe–, two from lone pair of e– & four from two double bonds 28. (b) Ring 1, 2, 3 & 6 have planar geometry 29. (d) Total e– present in 1, 2, 3 & 6 rings
126
Problems in Organic Chemistry
30. (c) Pentagonal ring contains 6πe– where as hexagonal ring also contains 6πe–
NaH ———→ –H
(+)
(–) Na
2
31. (c)
aromatic
32. (a) 33. (d) (–)
O
34. (a)
CH3
O
(–)
Aromatic CH3
Cl
Cl (–)
CCl2
C
(+)
Aromatic
35. (d) It undergoes tautomerism immediately. 36. (a)
(+)
(–)
(–)
(–)
(+)
antiaromatic
aromatic
(+)
nonaromatic
order of stability:- Aromatic > non aromatic > antiaromatic hence x > z > y 37. (a) N is less electronegative in comparison to O thus, for A resonance energy is high hence it is highly aromatic 38. (a)
••N
H
(+)
••N—H
H—N
••N—H (A)
A is planar & follow Huckel rule. 39. (a) See q.n. 36 40. (d) The lone pair of electron present on N undergoes resonance with double bonds & makes pyrrole planar (+)
41. (b)
——— H
(–) +
aromatic and thus, stable
42. (a) (ii) is aromatic while (iv) is non aromatic 43. (c) (i) forms less stable antiaromatic species as an intermediate while (ii) forms stable non aromatic species as intermediate.
Ph
Ph
Ph
Ph (–)
(–)
COO–t–Bu antiaromatic & less stable
COO–t–Bu nonaromatic
127
Reaction Mechanism (General Organic Chemistry)
RESONANCE (–)
O
O
O
O
(–)
(–)
(–)
O
O
O
O
O
O
(–) O
O
O
O
O
O
1. (d) (–)
(–)
(–)
2. (c) Cyclobutadiene has rectangular shape and hence does not favour resonance
(More stable)
(Less stable because double bonds are long)
3. (d) factual question 4. (c) Two tertiary butyl groups present at ortho position creates problem in resonance by disturbing the co -planarity of compound. (Steric inhibition to resonance) 5. (d) Carbon adjacent to CO groups is sp3 hybridized 6. (b) 7. (b) CH2= CH—N=SH
(+)
(+)CH
2
— CH
N — SH
8. (b) electron density will be least in I because nitro groups are electron withdrawing groups and it will be maximum in II because O(–) is +R group 9. (c) electron density will be least in (I) because F is a tautomeric group (–I & + R effects) while other groups are +R groups and order of +R effect is:- NH2 > OH > OMe 10. (a) I is more stable an it is non polar. II is more stable than III because opposite charges are close to each other. 11. (c) (III) does not show resonance hence C — N bond has maximum length. (II) & (I) shows resonance hence C — N bond acquires partial double bond character thus, bond length decreases. Out of (I) & (II) C—N bond length will be less in I because CHO (—R) group favours resonance in (I). 12 (a) 13. (a) 14. (a) Because C — O bond is a single bond and free from resonance O OH Tautomerism
15. (b)
Aromatic compound has more resonance energy. 16. (c) Because in (I) & (II) resonance takes places O ||
(–)
Oδ –
O(–) |
||
H — C— O ←→ H —C = O H —C ....... Oδ – 17. (c) In 3rd case sigma resonance takes place hence it has largest R.E. Out of 1st , 2nd & 4th last species possesses more R.E as it is aromatic in nature. (+)
(+)
sigma resonance CH CH 18. (c) First species is antiaromatic so E1 should be least. Second species is conjugated diene thus, stable and hence E2 is greater than E3.
128
Problems in Organic Chemistry
19. 20. 21. 22.
(d) Bridge head can not bear double bond. (b) (b) (c) In 1st case both the resonating structures contain-ve charge on electronegative O atom thus, it is highly stable. (III) is more stable than (II) due to more resonance. (–)
O
O ←——→
(–)
O
O
23. (c) 24. (c) 25. (c) 1st has 3 resonating structures 2nd has two resonating structure & in each resonating structure negative charge is present on electronegative oxygen atom hence 1st & 2nd are more stable than 3rd & 4th . In between 3 & 4, former is more stable due to more resonance. 26. (b) 27. (d) When lone pair of electron present on 2nd ‘N ‘undergoes resonance +ve charge comes on ‘N’ which is stabilized by + I effect of methyl group thus, lone pair of electron present on 2nd N will be more delocalized than 1st ‘N’. Lone pair present on ‘O’ will be less delocalized in comparison to that of both the nitrogens because ‘O’ is more electronegative than ‘N’. Chances of resonance will be least in ‘S’ because of less favourable 2p (carbon)–3p (sulphur) overlapping 28. (c) (II) violates octet rule as ‘Cl’ contains 10 electrons in its outermost shell. Passage-I (29 to 32)
Group C = NH is —R group & decreases electron density inside the benzene ring. NH & O are +R group & increase the electron density inside the benzene ring. Passage-II (33 to 36) 33. (a) in fluoro benzene resonance occurs effectively as 2p of F can easily overlap with 2p of ‘C’ hence due to more double bond character bond length decreases. Consequently dipole moment is least. (–) O |
O ||
( )
34. (c) CH3 — C— O — Me CH3 — C O — Me It is less favourable because electronegative ‘O’ atom acquires positive charge. 35. (b) due to less favourable resonance in phenoxide ion in comparison to other ions 36. (c) (I) is non polar hence highly stable. Out of (II) & (III) later is stable because opposite charges are closer to each other.
REACTION INTERMEDIATES AND ATTACKING REAGENT:1. 2. 3. 4. 5. 6.
(c) (a) (d) (a) (c) (a)
Besides (a) rest all are stabilized by resonance. [Stability ∝ 1 / reactivity] In BI3, B is more electron deficient because in it chances of back donation of e– is least
+R effect of NH2 stabilizes +ve charge present on CH2 group At meta position, resonance does not occur only I effect is considered. OCH3, NH2 & NO2 all have –I effect at meta position 7. (a) First spfecies is stabilised by resonance while second species is stabilised by seven hyperconjugative structures. 8. (c) H
C
H sp hybridisation
9. (a)
triplet carbene CH2 10. (d)
H3C C H
C
CH3 H
H3C H
CH2 C
C
CH3 it can undergo two types of rotations H
129
Reaction Mechanism (General Organic Chemistry)
H3C H
CH2 C
C
rotation
CH3 H
H3C C H
CH2 C
rotation
CH3 H
CH2
H3C H
C
CH3
C
H3C H
H CH2
H H 3C
C
C
H H3C
CH3 H
CH2 C
C
CH3 H
C
C
CH3 H
CH2
11. (c) In triplet carbene ‘C’ is sp hybridized. 12. (b) F3C
(+)
C
CF3
CF3 Due to – I effect of CF3 group positive charge density on ‘C’ increases consequently stability decreases. ( )
( )
13. (b) CH 2 CH — CH O CH3 CH 2 CH — C H — OCH3 It has 3 resonating structures 14. (d) Electron with drawing effects make anions stable (–)
(–)
CH2
CH2
( )
CH 2— CH
CH — OCH3
(–)
CH2
NO2
OH
NH2 (–R effect)
(–I effect)
(–I effect)
15. (b) 3rd is aromatic hence highly stable . In 1st case resonance is possible as carbon can donate its negative charge to empty d orbital of chlorine. Such kind of resonance is not possible in 2nd because d orbital is absent in ‘F’. (+)
16. (b) 17. 18. 19. 20.
aromatic & stable
(c) Due to more resonance +ve charge spreads over different ‘C’ atoms. (d) Except ‘3’ rest all shows resonance thus, delocalization of –ve charge takes place. (b) number of resonating structures ∝ stability (c) nitrene is not found in singlet & triplet states ••
(+)
hν cycohexene 21. (a) R — N — N ≡ N → R — N →
N—R
••
22. (a) number of resonating structures ∝ stability of species 23. (d) 2nd is aromatic thus, more stable than 3rd . Out of 1st & 4th former is more stable as one of its resonating structures acquires negative charge on oxygen. 24. (c) + R effect of methoxy group helps in the stabilization of positive charge. (+) CH
2
CH2
CH2
CH2 (+)
(+) (+)
OCH3
OCH3
OCH3 CH2
(+) OCH3
OCH3
130
Problems in Organic Chemistry
25. (a) This ion is least stable because -ve charge is not stabilized by resonance. 26. (b) Given carbocation is already stable because of resonance CH3
CH3 ••
••
••
(+)
C2H5 —C—C==O ←——→ C2H5—C—C≡≡ O (+)
H H 27. (d) 28. (d) See question number 11 in the topic aromaticity. 29. (a) ‘N’ can not form five bonds as it does not contain d– orbital. So tetra ethyl ammonium can not receive electron pair from nucleophile. 30. (b) 31. (a) Here resonance stabilized carbonium ion is formed. F F
C—OH F
(+)
32. (b)
C
H( ) –H 2O
(+)
(+)
F—C—F
F=C—F
F
F
is highly stable carbonium ion due to sigma resonance
3
33. 34. 35. 36. 37. 38.
(a) (d) (b) (a) (c) (c)
Carbonium ion generated by (a) is destabilized by strong (–I) effect of CF3. Stability of carbocation ∝ number of resonating structure See Q 15 & 18 In (c) & (d) resonance is not possible because of SIR effect. Substitution at ortho position cerates hindrance in resonance by disturbing the planarity of species. All are independent of resonance hence stability of free radical will depend on + I effect. + I effect ∝ Stability of free radical
Passage-III (39 to 42) (–) (–)
2Na →2e + 2Na +
→
Aromatic (A)
(X) (–)
2Na →2e + 2Na
•
+
→ (–)
(B)
(X)
(–)
(–)
•
(B)
aromatic
+
disproportionation
→ nonaromatic
(X)
(Z) or (A)
43. (d) Although carbanion has pyramidal shape yet it does not show optical isomerism because in carbanion rate of flipping of negative charge is very high. H 3C CH3 H H C C F F 44. (a) Because phenyl group is –I group. 45. (b) the carbocation in which formation of ring or opening of ring take place are referred as classical carbocations. 46. (c) Due to resonance C — Br bond acquires partial double bond character which is difficult to break. CH2=CH—Br
(–)
CH2—CH=Br (+)
131
Reaction Mechanism (General Organic Chemistry) Reduction
47. (c) (CH3) 2 C + (CH3) 2 C
CH3—CH = C2H + CH3—CH 2CH3
Oxidation
48. (c) + R effect of —O(-) is greater than that of OMe group. At meta position OMe shows –I effect thus 2nd is less stable than 1st. 49. (d) bridge carbon atom can not bear positive charge. 50. (b) same as Q.26 51. 52.
•• Carbene (C H 2 ) can show sp2 (singlet) as well as sp (triplet) hybridization. Hence in singlet carbene one ‘p’ Orbital is available while in triplet carbene two –p orbitals are available. (See Q.No.8) Compound B is benzyne & it has sp2 – sp2 pi bond.
H H H
sp 2—sp2 pi bond H
NUCLEOPHILIC SUBSTITUTION REACTIONS:1. (b) 2. (c)
–Br
Et
(+ )
(–)
|
EtOH
Ph 3CBr → Ph 3C........Br → Ph 3 C → Ph 3C ...... OH TS− 2
TS–1
(+ )
–H ( + )
(+ )
Ph 3C — OEt ← Ph 3C — OEt ...... H ← Ph 3 — O EtH TS −3
3. (c) In SN1 intimate ion pair forms which also favours the formation of inverted product. 4. (a) For SN1 reaction, stability of carbonium ion is considered. Carbonium ion formed by I is highly stable. (+) (+)
C
>
>
3
Ph Ph
(+)
CH
>
(+)
Me2CH less stable
Highly stable
5. (b) For racemisation substrate must be chiral. In (a) racemization does not occur because rearrangement in carbocation gives rise to optically inactive product. C2 H5 |
C2 H 5
SN1
C2 H5
(+)
CH 3 — CH — CH 2 Br →
CH3—C—CH 2 H
|
→ CH3 — C — CH3 ( +)
C2 H5 |
CH 3 — C — CH 3
OH
(–)
|
OH optically inactive
6. (c) Factual question st 7. (b) In 1 due to resonance C—Cl bond acquires double bond character thus , chances of nucleophilic substitution reaction will be least in it. Rate of SN2 will be maximum in 3rd because it is primary alkyl halide. Out of 2nd & 4th rate of SN2 will be more in 2nd because the ‘C’ which is attached with ‘Cl’ is less sterically hindered. 8. (d) Rate of SN1 ∝ stability of carbocation. Out of 1st & 4th rate will be more in 1st because substrate is sterically hindered & can release leaving group very easily to remove steric hindrance.
132
Problems in Organic Chemistry
CH3
9. (c) CH3—C—O—CH3 + H ——→ CH3 CH3
CH3
(–)
I ←——
CH3—C—I CH3
H
CH3
(+)
1
SN —CH3OH
CH3—C —O—CH3 (+) CH3 (+)
CH3—C CH3
10. (c) In (a) SN2 reaction will occur because base is strong & substrate is 1° alkyl halide similarly in (b) SN1 occurs hence in (a) & (b) will show inversion & racemisation respectively.
Pr
CH3
Pr C2 H5OH(weak base)
Et
OC2H5
+
→
Br (3° alkyl halide)
Pr
CH3
Et
Et OC2H5
CH3
11. (c) In SN1 carbonium ion is formed which undergoes resonance hence two products can form
CD2= CH —CH2Br
SN1 (–) –Br EtO–
CH2= CH—CD2OEt
(+)
CD2= CH—CH2
EtO–
CD2 = CH—CH2OEt
(+)
CD2 —CH = CH 2
12. (c) This reaction will occur by SN2 pathway because substrate is a primary alkyl halide thus for this polar aprotic medium is required. Dimethyl formamide (DMF) i.e. HCONMe2 is a polar aprotic medium .Another example of polar aprotic medium is DMSO. 13. (d) Substrate is 1° alkyl halide, base is strong & medium is polar aprotic, these three conditions increase the chances of SN2
O COOH 14. (d) D
(–)
O
O
C Br
OH(–) –H2O
D
C2H5
Br
O
Products
Et
Et D
15. (d) Carbonium ions formed by the release of Br (1) & Br (2) are stabilized by resonance.Carbonium formed by the release of Br (3) is less stable because it does not undergoes resonance. 16. (a) Rate of SN1 ∝ stability of carbocation 17. (b) It involves three transition states. 18. (c) It is the case of neighbouring group participation. Here epoxide formation occurs.
O(–)
OH NaOH –H2O
Br
O Br
In such case halo group can be replaced to OH group by using aq. Na2CO3
OH
OH aq Na2CO 3
Br
OH
133
Reaction Mechanism (General Organic Chemistry)
CH3
CH3
19. (b) CH3
CH2Br
1
SN (–) –Br
CH3
(+)
CH2
CH3
( +)
CH3 — C — CH 2 CH3
Methylshift
|
CH3
CH3 OEt |
C2H5OH
CH3 — C —Et |
20. (c) This is SN1 reaction. [ Rate of SN1 ∝ stability of carbocation ]
CH3
21. (c) It is SN2, thus, requires less polar solvent to avoid caging of nucleophile 22. (c) Since DMSO favours SN2 hence inversion takes place.
23. (a) Precipitation occurs via SN1 pathway. Since (a) will undergo SN1 easily hence it will easily form precipitates with AgNO3 24. (b) Since iodide is better leaving group as well as better attacking reagent hence path II is best. 25. (d) Solvolysis is a SN1 process hence solvolysis of 1st is not possible as bridge carbon can not carry positive charge. Out of 4th & 3rd later undergoes solvolysis more rapidly as iodide is a better leaving group than chloride 26. (c) SN2 mechanism occurs hence inversion in configuration takes place(Finkelstein reaction). 27. (c) In protic solvent nucleophilicity is proportional to the size of ion 28. (d) Leaving ability of any group is proportional to its stability
29. (a) Hydrolysis occurs via SN1 pathway & (I) will form highly stable intermediate(aromatic carbocation) 30. (a) 31. (d) It involves 4 carbocations & five transition states OH (+) (–)
(–)
–Br —CH2Br ———
(–)
(+)
—CH2 ———
(+)
———
———
OH
———
32. (a) in polar aprotic medium nucleophilicity is proportional to basic character. 33. (c) It is an example of neighbouring group participation.
CH—CH—S 3 2
CH—Br 2 CD2
(+)
CH3C2H—S
HOH
CH2 CD2
–Br
(–)
(+)
(+)
CH3CH 2—S —CH2 CH3C2 H—S CH2 CD2 CD2
HOH
CH3CH2SCH2CD2OH
CH3CH2SCD2CH2OH
34. (a) Carbocation so produced can not show rearrangement as it is stabilised by resonance (+)
Ph
35. 36. 37. 38. 39.
(d) (a) (c) (a) (d)
(Highly stable)
Since nucleophilic substitution reaction has ionic mechanism thus, polar medium favours nucleophilic substitution reaction. Same as question 32 & 27 2 Because it occurs via SN pathway Rate = R [alkyl halide] [OH–] First compound forms aromatic carbocation. NaCN is an ionic compound hence cyanide ion attacks from its ‘C’ site so cyanides are formed while AgCN is a covalent compound (Ag — CN) so ‘C’ site is not free to attack. CN attacks from its ‘N’ site consequently iso cyanides are formed. Similarly from NaONO, R — ONO is formed & from Ag — ONO, R — NO2 is formed.
134
Problems in Organic Chemistry
40. (d) Same as question no. 33 41. (b) Crown ether captures Na(+) easily & releases OH(–) & hence SN2 occurs easily O O (+) Na O O
(Na+ ion trapped inside crown ether)
42. (d) R—I dissociates easily to produce carbocation as ‘I’ is a good leaving group H
43. (b) H3C
•• CH3OH ••
H3C
O
CH3 HO H3C—O(+)
H
(+)
H
CH3 HO MeO
H
(B)
H CH3O
H3C
H(+) O
(+)
(–) O
O
•• CH3OH ••
H
CH3
(–)
MeO
H
CH3
(A)
HO MeO
Hence (A) forms via SN2 pathway while (B) via SN1 pathway 44. (a) In case of 20 alkyl halide reaction may occur through SN1 as well as SN2 pathway hence Overall rate = K 1 [ alkyl halide] [Nu] + K 2 [alkyl halide] 2
1
SN
SN
45. (c) In SN1 carbonium ion is formed which undergoes resonance hence two products form in SN1. 46. (c) Presence of nitro group makes benzene ring electron deficient hence nucleophile (H2O) can attack rapidly on benzene nucleus even it can be hydrolysed by warm water. 47. (b) Since bromine is present at bridge carbon atom hence hydrolysis will occur by SN2 pathway. Rate of SN2 ∝ 1 / (steric hindrance at reactive centre) Since A is less sterically hindered thus, rate of hydrolysis will be maximum in it 48. (b) Nucleophilicity is proportional to basic character. 49. (c) It is highly stabilized by resonance thus, it is a good leaving group. Et
Et O
50. (a) 51. (d)
(+)
+ H ———→
O(+)
———→
52. (b) It is an example of SN2 & thus inversion in configuration occurs.
FREE RADICAL SUBSTITUTION REACTION 1. (d)
It is more stable due to resonance & + I effect. Cl2 hν
CH2Cl
+
2. (c) 3. (a) Number of primary hydrogens are more 4. (d) All reagents can perform allylic substitution.
Cl +
+ Cl
Cl
Et (+) OH
Et (–)
I
———→
I OH
135
Reaction Mechanism (General Organic Chemistry)
Cl
Cl Cl2 h
5. (c)
H3C H3C
CH—CH
+ CH3
CH3
Cl2 hν
CH3
CH3
|
|
CH3 — CCl — CH(CH3) 2 + (CH3)2 CH — CH — CH 2 Cl
6. (d) Ease of abstraction of hydrogen depends upon the stability of f ree radical formed after the abstraction of hydrogen 7. (d) CH 3 — CH2 — CH
NBS CH3—CH—CH = CH2 = CH2 → Br
CH3—CHBrCH = CH2 CH3CH
8. (c)
h
•
NOCl N O Cl• ,
Cl
NO
NO
CH3—CH = CH—CH2
= CHCHBr
Br
HCl + N—OH
Tautomerism
9. (b) factual question
10. (c) All hydrogens are identical.
11. (c) molecular wt. of alkane = 43 x 2 = 86 i.e. C6H14 It would be 3- methyl pentane
Cl
Cl
Cl2
Cl +
+
+ Cl
12. (b) Abstraction of hydrogen by halogen radical is rate determining step.
Br
Br
13. (c)
Br CH2 Br 14. (c) This reaction occurs by electrophilic addition and follows markownikoff’s rule. Br 15. (a)
and
Br 2Cl hν
aq K 2CO3
2 → Ph — CCl Me 16. (c) Ph — Et → 2
HO
O
OH
Ph—C—Me
–H2 O
unstable
O
Ph—C—Me
Ph—CCH2Br
NBS
17. (c) All hydrogens are identical. 18. (d) TEL ( tetra ethyl lead) is a free radical generator 19. (c) Iodination of alkane is rervsible reaction because HI forms in this reaction acts as reducing agent & hence makes the reaction reversible. CH3 I + HI CH 4 + I2
136
Problems in Organic Chemistry
Hence iodination of alkane is carried out in presence of HIO3 or HNO3 (oxidizing agents) HIO
3 → CH I + HI CH 4 + I4 3
5HI + HIO3 → 3I2 + 3H 2 O 20. (c) See question no. 19 21. (c) In chlorination rate of abstraction of hydrogen is: 1° : 2° : 3° = 1 : 3.8 : 5 Cl2 (CH3 )2 — CHCH(CH3 ) 2 →(CH3 )2 hν
— CClCH(CH3 )2 + CH3 — C — CH(CH3 ) 2 |
A45.45%
CH 2 Cl
1°H / 3°H = (12 × 1) / (2 × 5) = 12 / 10 = 6 / 5 % of A = [5 / (6 + 5)] x 100 = 45.45 % 22. (c) The hydrocarbon is cyclohexane which on reaction with chlorine forms chloro cyclohexane which on elimination forms cyclo hexane Cl Cl2 NBS alcKOH h
Br 23. (b) It is 3° free radical & highly stable due to inductive effect. 24. (d) Strength of C — D bond is more in comparison to C — H bond hence Br can break C — H bond from 3° ‘C’ and form 3° free radical which is stable due to more +I effect. 25. (d) Chlorine free radical is highly reactive than bromine free radical Reactivity ∝ 1 / Selectivity 26. (d) Since C—H or C—D bond is not breaking thus rate of reaction will be unaffected.
AROMATIC ELECTROPHILIC SUBSTITUTION REACTION 1. (b) See mechanism in your text book. Ph 2. (b)
OH
(+)
(+)
H —H2O
HO O
Ph
Ph
C6H6
HO
HO (+)
O
O
OH
O Ph
3. (c)
Ph
H( )
OH
Tautomerisation
Ph
Ph
H (–)
O
O
(–)
O Ph
→
(+)
O
H
O Ph
O
→
→
H
O
Ph
4. (c) See mechanism in 3rd question 5. (d) Because both COCH3 & NO2 have same agreement at the location where Br is attached. 6. (d) Aniline, because it reacts with catalyst & benzene ring gets deactivated ••
H
( )
(–)
PhN H 2 AlCl3 Ph NH 2 Al Cl3 (Lewis base) (Lewis acid)
7. (c) Br of Br2 water can replace —R groups.
O
Ph
137
Reaction Mechanism (General Organic Chemistry)
crowded 8.
1
3
(c)
no alpha H atom no alpha– H– atom 2 4
Ring activation due to 2nd carbon is more thus; nitration should take place at 4th position 9. (c) Both (a) & (c) can form but later can form with greater ease due to lack of steric hindrance 10. (c) ZnCl2 will remove chlorine atom near to nitrogen because of the formation of more stable carbocation 11. (a)
5
(–)
3
4
2
N
(+)
(+)
N
N
(–)
H
H
H
(1)
(2)
(3)
Out of 2nd & 3rd, later is more favorable because opposite charges are closer to each other hence E+ attacks at 2nd & 5th position.
CH2 12. (b)
CH2
Cl
Pd/ –2H 2
AlCl
3
CH2 CH2
13. (b) –CCl3 is meta directing group due to reverse hyperconjugation. 14. (d) Factual question E (+) E
(+)
15. (a)
(2)
(1)
Out (1) & (2), former is more favourable than later 16. (c) It is ring activating groups and increases e– density at o & p position so incoming Nu- can not attack at o & p position and ultimately Nu– attacks at meta position.
OH H ( ) –H 2O
17. (d)
(+) C H
6 6
(+) ( )
H CH2OH
(+)
CH2
ring opening
C H
6 6
Cl (+)
AlCl
3
C H
6 6
18. (b) Factual question (see energy profile diagram of sulphonation of aromatic compounds in your text book) 19. (c) Because –I effect of iodine is least so ortho position is less deactivated (i.e. less positively charged). 20. (d) Because rate determining step of ArSE does not involve breaking of C — H or C — D bond. 21. (c)
138
Problems in Organic Chemistry
22. (a) In ‘I’ group shows electron withdrawing effect while in II & III case ‘CH2’ is present which can show hyperconjugation with benzene ring. 23. (d) Sec- butyl group is o/p directing group hence reaction will occur at para position because sec- butyl group is sterically hindered. In BrCl, Br+ serves as electrophile hence (d) is obtained. 24. (b)
H
( )
ring opening
C H ArSE
6 6
(+)
(+)
25. (d) CH3 — CH
FeCl3 CH — CH2 — CH 2 Cl C6 H 6
H ( )
(+)
Pd/
26. (c) In sulphonation SO3 a neutral electrophile attacks on benzene ring & reaction is reversible. Hence to avoid reversibility we use oleum in place of H2SO4 27. (c) Ring (C) is attached with two +R group hence it has maximum electron density. (B) has less electron density than (A) because – R effect of NO2 group is decreasing electron density inside the benzene nucleus. 28. (c) First Nitration occurs at para position & then friedel craft reaction occurs. (+)
(CH2) 4— CH2Cl
(CH2) 4 —CH 2Cl
(CH2)4 C2H AlCl
NO ( )
2
+O
NO2
(+)
O2N
O
O
29. (d)
NO2
O2N
3
O
O
AlCl H 2O
3
H ( )
(+)
COOH
CO O
Zn Hg HCl
Pd
O
30. (c) See question no. 6 31. (b) Because the benzene ring which is directly attached with ‘S’ is activated thus, nitration should take place on this ring. 32. (d) ( )
(–)
AlCl3 C6 H 6 H shift 33. (b) CH3 — CH — CH 2 Cl CH3 — CH — CH 2( ) CH3 — C — CH3 |
|
|
CH3
CH3
CH3
Br
(+)
CH3—C—CH 3 + CH3
CH3Cl/AlCl 3
Fe/Br2
CH3
34. (a) Leaving E+ is more stable. 35. (c) Bromine water can remove –R groups like NO2 from benzene ring
CH3
139
Reaction Mechanism (General Organic Chemistry)
OH
OH Br2water
Br
Br
Br NO2 + (c) Here H ion attacks on benzene nucleus & displace Me3C+ ion which on elimination gives alkene i.e. IPSO attack followed by beta elimination takes place (a) Positive charge is stabilized by + R effect of NH2 group (c) It has four resonating structure (c) – R effect of NO2 stabilizes negative charge. (b) +ve charge can be stabilized by lone pair of electrons present on nitrogen. (c) Due to more +R effect benzene ring is highly activated. (a) OMe is + R group, CHO is – R group however Cl has two effects + R & – I (b) p - G2C6H4 — G3 will have higher electron density because both G2 & G3 are electron donating groups (a) CMe3 can not show hyperconjugation hence in (IV) electron density inside the benzene ring is least thus, rate of ArSE will be small in it. 45. (b) Because 2p of F can easily overlap with 2p of ‘C’ of benzene ring hence due to more resonance electron density is maximum in fluoro benzene. 36. 37. 38. 39. 40. 41. 42. 43. 44.
(+)
(+)
46. (b)
–H
(+)
N2
H Cu
–CuH –H
47. (d) Strength of lewis acid depends upon the electron deficiency on central atom.SnCl4 is poor lewis acid as Sn has complete octet. In BF3 back donation of electrons from F to boron decreases electron deficiency of B thus, it is also a weak lewis acid 48. (d) H C C2H5
49. (c)
CH3
50. (c) It is an example of IPSO attack.
CMe3
CMe3 (+) NO 2
(+)
Me3C
+ Stable Me C Me3C CMe NO2 3 3 Carbonium ion 51. (c) Following resonating structure of pyrrole is favourable because distance between –ve & +ve is least hence ArSE occurs at 2nd & 5th position. 4 5
(+) N1
3 (–) 2
H
OH 52. (d)
(+) H
(+)
(+)
C6H6 (+)
140
Problems in Organic Chemistry
53. (a) (–) 54. (a) —O has strongest + R effect. (–)
OMe CH3
(+)
AlCl—OMe 3
CH3
AlCl3
55. (a) 1
Deactivated ring
2
Thus reaction occurs on 2nd ring 56. (b) Reaction of Ph—H with DCl is a kind of friedel craft reaction as a result of which dutereated benzene is produced which on nitration produces (b) 57. (c) alkylation should be performed al last because CH3 is o/p directing group, while in product all groups are present meta to one another. 58. (c) 1, 2, 3 & 5 are aromatic compounds & hence can undergo ArSE. 59. (c) Because nitro benzene is least sensitive towards ArSE as nitro group deactivates the benzene nucleus PCl5
60. (d)
(+)
alc. KOH
OH
Rh heat
H
Cl heat with H3PO4
61. (c) 62. (c) Br of Bromine water can replace –R group (like SO3H) 63. (b) in 3rd reaction product is sterically hindered that is carbon can not bear four phenyl groups on the other hand in 4th case generation of carbonium ion is difficult.
SO3H Sulphonation
64. (b)
SO3H
SO3H
Nitration
F.C.R
Et
NO2
NO2
OH
OH CH3
CH2C—CH 3
CH2COCH 3
(+)
(+)
65. (a)
H
–H2O
66. (c) Ring opening will occur in it. (+)
(+)
H
H (+)
(+) CH
CH2OH
2
OH O 67. (c)
OH
O H
(+)
(+)
PhOH
H
PhOH
(+) (+)
OH
HO
OH
141
Reaction Mechanism (General Organic Chemistry)
O
OH
OH ( )
||
|
H Toluene 68. (c) CH3CH 2CCH3 CH3CH 2 — C — CH3 CH3 CH2 C— ( )
Me
Me
H 69. (a)
AlCl3
O
(+) (+) O
C6 H 6
OAlCl3
H2O
OH
OAlCl3
AlCl3 70. (a) Reaction will occur on that benzene ring which is activated thus, bromium ion attacks on the ring which is attached with Me group. Since para position is sterically hindered hence ortho derivative will produce as a major product. 71. (d) CH2 CH2
72. (c)
CH= CH2
CH2
(+)
H
CH3
CH2 CH
(+)
CH
CH3
CH3 (+)
CH2 CH2
C CH2
CH3
CH3 (+)
CH2
CH2
C
CH2CH3
CH3
(+)
—H
73. (a) See mechanism in your text book.
74. (d)
75. (a) Due to presence of two NO2 groups benzene nucleus becomes e– deficient and can favour the attack of nucleophile. NO2 OH NO2 OH NO2 OH (–)
NHCH3
O2N
O2N
NHCH3
H
O2N
(+)
NO2
NO2
OH
(+)
O2N
N—CH3
H ∆
N—CH3
O2N
CH2 76. (d) CH2= CH—CH2Cl
CH2N2
77. (b) Factual question 78. (b) [X] is ortho nitro phenol.
CH2—CH—CH 2 Cl
AlCl3
(+)
C6 H 6
N—CH 3 (+)
H
142
Problems in Organic Chemistry
H (+)
O
79. (a)
O
(+)
CH2C2HOH
benzene
H
Oxirane
(CH2)2
(+)
H benzene
Oxirane
80. (b) Carbon can not bear four phenyl groups due to steric hindrance thus, Ph4C can not form. 81. (a) HI formed in this reaction is a good reducing agent which makes the reaction reversible hence during the iodination of benzene HI should be oxidized in to iodine by some oxidizing agents like HNO3 & iodic acid
O
+R group Hence ArSE will occur according to +R group
O
–R group
82. (d)
ELECTROPHILIC & FREE RADICAL ADDITION (+)
CH2
1. (d)
H
(+)
(+) (+)
OH
OH2
(+)
(+)
+ H
H2O
It involves four intermediats and thus five transition states will be achieved.
Br 2. (a) CH3—CH = CH2
Br 2
+
CH3—CH—CH 2
(–)
Cl from NaCl
Br(–)
CH3—CH—CH2Br CH3—CH—CH2Br Br
Cl
3. (d) KMnO4 gives syn hydroxylation
H3C H
C =C
CH3 H
KMnO 4 273 K
4. (c)
OH CH3 H
CH3
OH OR H
Meso isomer
H
OH H
OH CH3
Anti addition
→ Racemic mixture 5. (d) Cis alkene Syn addition
→ Racemic mixture Trans alkene
Cis alkene → Meso isomer
Syn addition
Anti addition
Trans alkene → Meso isomer All addition reactions of alkenes with halogens are the example of anti addition but fluorination is syn addition hence (d) is correct. H2(Steric repulsion) H H2(no Steric repulsion) H 6. (a)
Hexagonal ring will not show ring opening reaction because it does not possess strain.
143
Reaction Mechanism (General Organic Chemistry)
AlC4(–) Cl( ) 7. (c) AlCl3 Cl2 (+)
Cl
Cl (–)
(+)
AlCl4
AlCl3 + CH 3—CH—CH2Cl CH3—CH—CH 2 CH3—CH = CH2+ Cl 8. (c) Since carbocation forms as an intermediate in electrophilic addition reaction hence, rate of electrophilic addition reaction will depend upon the stability of carbocation. ( +)
E( +)
Ph 2 C = CH 2 → Ph 2 C — CH 2 E
................... (A) (A) Is highly stable due to resonance of +ve charge with benzene rings E ( )
(CH3 )2 C CH 2 Me2 C( ) — CH 2 E (B) is also stable due to hyperconjugation
................... (B)
( )
E ( )
Ph 2 C CHCF3 Ph 2C — CHCF3 |
E (C) is more stable than (B) but less stable than (A) because of – I effect of CF3 group hence rate of E.A.R. will be more in (1) & least in (2) i.e. 1 > 3 > 2 9. (b) Stability of alkene ∝ 1 / Reactivity ∝ 1 / Heat of hydrogenation
CH3 — CH = CH — CH3 is less the all four operation. CH2
10. (d)
H
CH3
H
CH3
+ CH2
(Molecular addition)
11. (d) All reagents will produce carbene
•• CH CH
CH
•• CH
2
2
(+)
—
12. (c)
2
H
(+)
3
Ring opening (+)
SEt
H
S
(+)
(+)
H Et
EtSH
+
13. (a) HCl does not exhibits Antimarkownikoff addition. (+)
O
OH
H
14. (b)
CH2OH
CH2OH
(+)
H 2O (+)
–H
OH (+) 2
15. (c) Cl2 + H2O gives HOCl (–)
OH
Cl
OH 16. (a)
Cl
(–)
OH –H2O
Na C O
O O Cl
OH OH
CH2OH (+)
CH3 OH
144
Problems in Organic Chemistry
CH3
CH3
|
|
CH3 CH3 ( )
CH3 CH3
|
|
|
|
( )
( )
CH3
|
|
H 2O H Re arrangement Ph —*C — CHMe2 17. (c) Ph — C H — C CH 2 Ph — C — C — CH3 Ph — C —C HCH3 (optical active)
H
|
OH (optical active)A
A contains only one chiral carbon atom. On dehydration it produces Ph(Me)C = CMe2 which can not show optical as well as geometrical isomerism CH3
CH3
|
|
Ph — CH — C
CH3 H2 CH 2 Ph Ni/
CH3
|
|
— CH — CH — CH3 only one chiral carbon atom *
18. (b) CN & Ac groups are - R groups & decrease the electron density in C == C moiety. Consequently C == C moiety favours the attack of nueleophile. Ac
C=CH—CH 3 +Br
Ac
(–)
(–)
H
C—CH—CH 3
(+)
Ac
(–)
CH—CHBrCH3 NC NC NC 19. (d) NO2 group is electron withdrawing group &Brdecrease electron density of C = C moiety hence nueleophilic addition reaction will occur on it. Consequently antimarkownikoff addition takes place.
OD (+)
(+)
20. (b)
H
D2O
Ring opening
(+)
(+)
–D
21. (a) Cl2 water gives HOCl (+) (+)
Cl
Cl
Rearrangement
(+)
Cl
(–)
OH
Cl
(+)
22. (a) 23. (c) Ph2CH—N = N—CHPh2
24. (d)
H
h –N2
2Ph2CH
2Ph2 CHBr
OD
(+)
(+)
Br 2
D2O
(A)
H D
(+)
(+)
D
OH
HO 2
D (B) 25. (c) Product of this reaction wll be OH * PH
*
*
* OH
*
* C = chiral carbon
OH
26. (c) Same as question no.1 27. (d) CH2= CH—CH=CH2
H
(+)
(+)
CH2=CH—CH—CH3 (+)
CH2—CH =CH—CH 3
Br
(–)
Br
(–)
CH2= CH—CHBrCH3 CH2BrCH= CH—CH3
OH
145
Reaction Mechanism (General Organic Chemistry) (+)
(+)
H
CH2= CH—CH=CH2
(–)
CH3–CH—CH =CH 2
Br
(+)
CH3CHBrCH = CH2
(–)
Br
CH3CH =CHCH 2Br 28. (c) When E(+) attacks on ‘1’ , a highly stable carbocation is formed. Thus, 1st double bond will be highly reactive. Out of 2nd & 3rd, later is more reactive because it is cis while 2nd is more stable & less reactive as it is trans isomer more stable. CH3—CH =CH—CH 2
H
H Me
Me
Me
(+)
29. (d)
(–)
H
Br (+)
Ph
Ph
Ph Br
30. (a) CH 2
( )
( )
( )
H 2O H CH 2 — CH 2 H CH3CH 2 OH CH 2 Slow Fast
Since slowest step contains CH2 = CH2 & H+ hence rate law is Rate = K [CH2 = CH2] [H+] 31. (c) I+ serves as electrophile as it is less electronegative than Cl 32. (a) Since electrophilicity of Br+ is greater than I+ because of less electronegativity of I than Br thus, BrCl is more reactive than IBr. Further interhalogens are less stable than halogens hence order of reactivity is: - BrCl > Br2 > IBr > I2 33. (a) In (a) C == C moiety is not attached with electron with drawing group. CH3 |
CH3 |
( +)
H 34. (d) CH3 — CH = CCH 2 CH 2 CH3 → CH3 — CH2 — CCH 2 CH 2 CH3 ( +)
[can show geometrical isomerism]
CH3 reduction
|
CH 3 — CH 2— CHCH2CH2CH3 [opticaly active] *
(+)
(+)
35. (b) O
O
H
O
OH
O
OH (+) (–)
O
OH
Tautomerism
O
Br
OH
Br
Br CH3 Br2in CCl 4
36. (a)
H Br
Br H CH3
The product of this reaction has configuration [2R, 3S]
Elimination Reactions:1. (d) At bridge ‘C’ atom double bond does not form until parent ring contains 8 ‘C’ atoms (Bredit’s rule) 2. (b) In hydrohalogenation anti elimination takes place CH3 H
H
alc KOH
—————— H
Cl H
CH3
146
Problems in Organic Chemistry (+)
CH3
(–)
3. (c) H3C—N— CH 3 OH
CH3OH + (CH 3 )3N
CH3
(+)
(+)
(+)
(+)
4. (c)
H
–H
(+)
(+) (+)
5. (b)
(+)
H –H2O
OH
Rearrangement
–H
(+)
6. (b) 7. (a) For SN1 substrate should be 3° & base should be weak.
Me I Ph
8. (c) Also see q.15
H H
Me
H
H
Me
(–)
OH
Me CH3
OH |
CH3
( )
()
|
CH3
()
|
|
H –H 9. (c) CH3 — CH — CH — CH 2 CH3 CH3 — CH CH — CH 2 CH3 CH 2 CH CHCH 2 CH3 –H 2O
CH3
CH3 (+)
CH3CH 2 CCH2CH 3
CH3CHCHCH 2CH 3
(+)
(+)
(+)
–H
–H
CH3
CH2 CH3CH2 —C—CH2CH 3
CH3CH = C—Et
10. (b) See question no. 52 in electrophilic substitution reaction. 11. (a) Anti elimination will take place.
OH
(+)
H —H2O
(+)
———— Cl
Cl
H2O
————
12. (b)
(+)
Cl
Cl
OH2 (+) (+)
–H
———— Cl OH
13. (c) Because it will produce least stable carbonium ion.
–HCl
O
147
Reaction Mechanism (General Organic Chemistry) H/Ni 2
N
N
H
H
14. (b) H2 O + N(CH3 )3 +
CH3I excess
AgOH (+)
N
H3C
(+)
∆ (+)
(–)
N OH H3C
Br H
(–)
H
D
I
H
D
CH3I AgOH
∆ N H3C
CH3 CH3
CH3
I D
15. (c)
H3C
CH3
(–)
N OH CH3
(–)
H
D
H
D
Br
H
D
D
H
OR
I
D
H
I
16. (c) Base is sterically hindered thus elimination product will produce more than substitution product. Passage (17 to19) Acid catalysed dehydration of alcohols involves formation of carbocation as intermediate Rate of dehydration depends upon 2 factors 1. Stability of carbocation formed in rate determining step 2. Acidic nature of H attached to the carbon adjacent to positively charged carbon atom 3rd will provide stable carbocation (resonance stabilized) thus, it will dehydrate more easily OH (+)
(+)
H –H2O
NMe2 NMe2 (+)NMe2 stable due to resonance 4th will provide highly unstable carbocation as positively charged carbon atom is attached with electron withdrawing NO2 group.Out of 1st & 2nd dehyration process is rapid in later as it contains more acidic hydrogen. OH (+) (+) H More acidic due to electron H withdrawing effect of NO2 group NO2 NO2
Out of 1st & 4th rate of dehydration will be more in 1st because formation of carbocation occurs in slowest step & 1st provides more stable carbocation. 20. (c) In (B) saytzeff rule is obeyed while in (C) hoffman rule is obeyed. In A anti elimination occurs as follows:-
OH (+)
H
(+)
H
More acidic due to electron withdrawing effect of NO2 group
NO2 NO2 21. (c) Since OH is a poor leaving group in basic medium thus, base will abstract ‘H’ to produce conjugate base(E1CB) 22. (d) Tertiary halide on reaction with strong base favours elimination reaction & since temperature is also large thus 100% elimination product will be obtained. 23. (b) In 1-2 elimination reaction leaving groups should be anti periplanar to each other hence rotation about single bond occurs in the following manner. Ph Ph Ph H Me Ph Ph H Br Br free rotation OR Ph Ph H about single bond H Me Me Me H H Ph
148
Problems in Organic Chemistry
MORE THAN ONE MAY CORRECT 1. a, b, c 2. b, c
In (a) Cl can receive -ve charge in its vacant d orbital. In (b) + ve charge can not be delocalized because bridge head ‘C’ can not bear double bond. (+)
–H
(+)
(+)
(not possible)
3. b, c 4. a, b
In carbonate ion bond order is 4/3 while bond order is 3/2 in acetate ion.
5. b, c 6. b, c
7. a, b, d
b & d will show aromaticity after tautomerism
8. c, d 9. a, b
because alkyl and alkenyl ions are stronger bases than NH2(–)
10. b, c
X & Y are formed nia SN1 and elimination reactions respectively.
11. b, d
II & III are antiaromatic & aromatic species respectively while I & IV are non aromatic species
12. a, b, d
In these three cases benzene ring is deactivated.
13. a, b, d
In polar proteic medium attacking base is caged solvent molecules by the help of hydrogen bonding and becomes weak.
14. a, b 15. b, c, d Cl + 2NaNH2 — 2NaCl + 2NH3 +
16. a, b
+
Cl
17. a, d
18. a, b, c 19. a, c, d
20. a, c LEVEL-II 1. (a) Trans alkene is more stable so it has less energy thus, more energy will be required to achieve transition state (+)
2. (d) 1st is stable as it can show sigma resonance
(+)
3rd is least stable as it can not show resonance it is stabilized by three hyperconjugative structures.
Antiaromatic
3. (d) (+)
4nπ electrons
4. (d) This reaction is not possible because anti dehydrochlorination is not possible
149
Reaction Mechanism (General Organic Chemistry)
H 5. (c)
H(+)
(+)
(+)
H3C
–CH3
(+)
CH3
H3C (+)
H (+)
OH
OH
B
6. (b)
B
OH
OH
6- valence electrons
8- valence electrons
7. (d) When two benzene rings are connected with each other by the help of covalent bond then resonance occurs as follows. (+)
Deactivated ring (–)
Activated ring
This kind of resonance is possible only when both the rings are present in the same plane. This resonance is maximum in (d) because CH2 group holds both the benzene rings in the same plane which is not possible in (a) & (b) (+)
O 8. (b) Lone pair present on oxygen repels negative charge.
(–)
O 9. (d) Bond order = No. of covalent bonds involved / No. of resonating structure 10. (a) + I effect of SiH3 is greater than CD3 which is further greater than CH3 11. (a) – R effect of CN will not work at meta position. 12. (b) Carbanion is stable than carbocation & radical as negatively charged carbon atom possesses 8e’s in its outermost shell. Out of free radical & carbocation former is stable than later as radical needs only one electron to complete its octet while carbocation needs 2 electrons to complete its octet 13. (c) Cyclo butadiene has rectangle shape hence bond lengths a & b are not equal & both are different compound 14. (d) All have 4nπ electrons so they all are antiaromatic & stable. Hence they loose their planarity & becomes non aromatic and stable 15. (d) π- electrons present at periphery are considered. 16. (d) Since cyclobutadiene has rectangular shape hence chances of resonance are low thus, it has less resonance energy .It is antiaromatic in nature thus, it dimerises easily at above 4°C as follows:2- moles
This is Diels -Alder reaction.
17. (b)
CCl3 CCl3
CCl3 CCl3 O
(+)
–H
O (–)
150
Problems in Organic Chemistry
Since negative charge is present at bridge ‘C’ thus, it can not be delocalized hence this carbanion will be nonplanar. 18. (b) Both the alkenes are almost equally stable CH3 CH |
A (CH3)3CCH =C
3
B (CH 3 ) 3 CCH 2 — C CH 2
and
5 hyperconjugative structure
CH3 6- hyperconjugative structure
But removal of 2nd hydrogen is more difficult in comparison to due to more steric hindrance around it thus, B is major & A is minor CH3 H2 |
|
CH3 |
CH3 — C — CH — C — CH2— H1 |
|
OH
CH3
H
D
H
D
H
D
19. (c) optically inactive
optically active
20. (a) See chapter - 3, q.n. - 6, level - II 21. (a) In polar medium (large dielectric constant) salvation of base takes place thus, nucleophilicity deceases consequently rate of SN2 decreases. 22. (c) In (a) NaOH does not dissociate in non polar medium so reaction is not possible. In (b) RX is present in a non polar medium while NaOH is dissolved in water (polar medium) so both solutions do not mix & reaction does not occur. In (c) NaOH dissociates easily because sodium ion is caged in crown ether.
O
O (+)
(–)
Na
RX + OH
ROH
O O 23. (c) In case of SN2 :RX + Nu (
δ
)
more polar & more
solvated In case of SN1:-
RX
less polar & more
δ
→ Nu .........R.......... X
more polar & more solvated
δ
δ
→ Nu .........R.......... X
more polar & more solvated
solvated 24. (d) In racemic mixture 35% d form & 35% l form will be present thus, % inverted product = 35 + 30 = 65
5
Acid & Base
Main Features H O(+) + HA + H2O 3
(acid)
A(–) (Conjugate base)
Acid Strength
∝
Stability of conjugate base
Basic Strength
∝
Stability of the species formed after protonation
Factors affecting acid or basic strength
>
Acid Strength
∝
[- I effect]
∝
1 / [+ I effect]
>
Basic Strength
∝
[ + I effect ]
>
Polarity of medium
∝
acid / basic Strength
>
% s character
∝
electronegativity
∝
acid Strength
∝
1 / basic Strength
∝
1 / [ - I effect ]
>
Aromaticity & Hydrogen bonding also affect the acidic or basic strength. Here some compound are arranged in order of their decreasing pKa values:-
Alkane > Terminal alkene > benzene > alkyl amines > NH3 > RCN > terminal alkyne > ester > ketone > alcohol (3° > 2° > 1°) > H2O > CH3NO2 > PhOH > PhSH > p - nitro phenol > RCOOH > PhCOOH > HCOOH > RSO3H > HNO3 > H2SO4 > HCl > HBr > HI
LEVEL - I
Multiple Choice Questions 1. Which among the following is least basic: (a) CH2 CH — NH2 (b) CH3 — C — NH 2 ||
2. The compound having least pKa is:
(a)
(c) CH3 — NH2
(d) Cl — CH2 — NH2
O
(b)
(c)
(d)
152
Problems in Organic Chemistry
OH
COOH NO2
COH 2
OH
3.
NO2
NO2 (1)
(4)
(3)
(2)
Correct order of acid strength is:- (a) 2 > 4 > 3 > 1 (b) 4 > 2 > 1 > 3
NH2 4.
(c) 4 > 3 > 2 > 1
NH2
NH2
CH3
(d) 4 > 2 > 3 > 1
COOH NO2
CH3
(1)
(2)
(4)
(3)
Correct order of basic strength is: (a) 3 > 1 > 2 > 4 (b) 3 > 2 > 1 > 4 (c) 1 > 2 > 3 > 4 (d) 4 > 3 > 1 > 2 5. Arrange the following compounds in decreasing order of basic strength:NH2 NH2 CH3 CH3CONHCOCH3 CH3NH2 (1)
(a) 4 > 2 > 3 > 1
(2)
(3)
(b) 4 > 1 > 3 > 2
O
(c) 2 > 4 > 1 > 3
O
6.
N
N
(4)
(d) 4 > 2 > 1 > 3
Cl
CH3
N
N
H
H
H
H
(1)
(2)
(3)
(4)
Correct order of basic strength is: (a) 2 > 1 > 3 > 4 (b) 2 > 3 > 1 > 4 (c) 2 > 1 > 4 > 3 7. Arrange the following substituted benzoic acids in decreasing order of acid strength:CO2H CO2H CO2H OMe
(d) 3 > 2 > 4 > 1
CO2H
OMe
OMe (1)
(a) 1 > 2 > 3 > 4
(2)
(b) 1> 2 > 4 > 3
(3)
(c) 1 > 4 > 2 > 3
8. Carbonic acid, H2O, Phenol, (1) (2) (3) Correct order of acid strength is: (a) 1 > 4 > 3 > 2 (b) 2 > 1 > 3 > 4 (c) 2 > 4 > 1 > 3
(4)
(d) 2 > 1 > 4 > 3
FCO2H (4)
(d) 4 > 1 > 3 > 2
2
O 9. H–N 3
NHCl N 1
Correct order of basic character of these nitrogens is:(a) 1 > 2 > 3 (b) 2 > 1 > 3
(c) 3 > 2 > 1
(d) 2 > 3 > 1
153
Acid & Base (3)
HOOC 10.
(1)
OH (2) OH
1, 2 & 3 are H+ donating sites. Correct order of ease of removal of H+ is:- (a) 3 > 2 > 1 (b) 3 > 1 > 2 (c) 1 > 3 > 2
O (1)
11.
(d) 3 > 1 > 2
(2)
N H
NO2
N
(3)
H
N
Highly basic nitrogen is:(a) 3
(b) 2
(c) 1
(d) 1 = 2 > 3
12.
N
N
N
N
H (1)
H (2)
H (3)
H (2)
Correct order of basic strength is:(a) 2 > 1 > 4 > 3 (b) 2 > 1 > 3 > 4
(c) 1 > 2 > 4 > 3
(d) 1 > 2 > 3 > 4
(c) 1 > 2 > 3
(d) 1 > 3 > 2
(1)
NMe 13. (3)
(2)
Nac
Bu—N
Correct order of basic strength of these nitrogens is:(a) 2 > 1 > 3 (b) 2 > 3 > 1
14. Identify the least acidic alcohol
OH H
NO2
H
H
H
H
H
NO2
H (a) 2
H
H
H
H NO2
H
(3)
(1)
OH
OH
(2)
(b) 1
(c) 3
(d)
all are equally acidic
15. The compound having least pKb or large pKa is:NH2
NH2
(a)
H2N
16. From the following reactions NH3 CH CLi CH CH LiNH 2
|
H2N—C CH 2 C = NH 2 (b) H2N—CH—NH 2 (c)
CH CH LiOH CH CLi H 2 O
|
NH2
(d) Ph—N—NH 2 |
Ph
154
Problems in Organic Chemistry
Predict which of the following orders regarding base strength is correct?
(a) CH C OH NH 2
(c) OH NH 2 CH
(–)
(–)
(–)
(–)
(–)
(–)
(–)
(–)
(b) CH C NH 2 OH (–)
(–)
C
(d) OH CH
(–)
(–)
C
NH2
17. Consider the following reactions
RONa H 2 O ROH NaOH
RONa NaNH 2 RONa NH3
Predict, which of the following order regarding base strength is correct
(a) OH NH2 RO
(–)
(–)
(–)
(–)
(–)
(–)
(–)
(–)
(–)
OH RO NH 2 (b) NH 2 OH RO (c)
(–)
(–)
18. Correct order of acid strength is:-
H CH=C H H C
C
H (3) CH (2)
H(1)
(a) 1 > 2 > 3
(b) 2 > 3 > 1
(c) 2 > 1 > 3
(d) 1 > 3 > 2
19. Correct order of ease of removal of H+ is:(+) (1) OH
C—H
(+)
NO2H
(a) 3 > 1 > 2 > 4
COOH
(3)
HO (b) 2 > 4 > 3 > 1
(c) 3 > 2 > 1 > 4
(d) 2 > 3 > 4 > 1
20. Which will have least pKa?
(a)
SH (+) 2
(b)
COOH
(c)
OH
(d) H2SO4
(d) Phenol
(+)
21. Which will have maximum pKa? (a)
(b)
SH
(c)
OH
22. Strongest base among the following is: (a) BuLi (b) LiNH2 23.
COOH (c) CH3ONa
Which reaction is not possible? (a) 2CH3COONa + H2SO4 ——→ Na2SO4 + 2CH3COOH (b) HCOOH + CH3OK ——→ HCOOK + CH3OH (c) CH ≡ CNa + CH3OH ——→ CH ≡ CH + CH3ONa (d) LiNH2 + CH2 = CH2 ——→ CH2 = CHLi + NH3
24. Which is strongest acid among the following? (a) HCOOH (c) FCH2COOH
(b) 2, 4, 6 trinitro phenol (d) CH3COOH
(–)
(d) OH RO NH2
(d) HCOONa
155
Acid & Base
25. FCH2COOH Cl2CHCOOH CH3COOH HCOOH ClCH2COOH (1) (2) (3) (4) (5) Correct order of acid strength is: (a) 1 > 5 > 2 > 4 > 3 (b) 5 > 1 > 2 > 4 > 3 (c) 2 > 1 > 5 > 4 > 3 (d) 5 > 2 > 1 > 4 > 3 26. Which statement is not correct? (a) Oxalic acid is stronger acid than formic acid (b) Formic acid is stronger acid than succinic acid (c) Fumaric acid is stronger acid than maleic acid (d) Fluoro ethane is stronger acid than Chloro ethane 27. Which of the following ‘O — H’ bond has least bond energy? (4)
COO—H (1) CH—O—H 2 Cl H—O—CH2 C (3)
(a) 1
CH3 O—H
(b) 2
(c) 3
(d) 4
28. Correct order of case of removal of marked ‘H’ will be:1
3
H H
H COOH
CH2
4
2H
(a) 1 > 2 > 3 > 4 (b) 1 > 4 > 2 > 3 (c) 4 > 1 > 2 > 3 + 3 29. Out of 1, 2, 3 & 4 which can with draw H from sp hybridized carbon of propene
(d) 4 > 1 > 3 > 2
(–) (2)
COO (–) (3)
C
(–)
O
(b) 2
(–) (1)
C
(a) 1
(c) 3
30. 31.
Strongest base among the following is:(a) Me3N (b) Me2NH (c) Me4NOH Identify the correct statement? (a) p – nitro phenol is less acidic than ortho nitro phenol (b) Picric acid is less acidic than formic acid (c) pKa of maleic and fumaric acid is same (d) Dichloro acetic acid is stronger acid than fluoro acetic acid
OH OH
(d) 4 (d) NaOH
OH
OH Cl
32.
Cl (1)
Correct order of p is:(a) 4 > 3 > 2 > 1
Cl (3)
(2)
(4)
Ka
(b) 4 > 2 > 1 > 3
(c) 4 > 1 > 2 > 3
(d) 4 > 3 > 1 > 2
156
Problems in Organic Chemistry
OH
OH OH
OH OMe
33.
OMe (1)
OMe (3)
(2)
Highly acidic compound is:(a) 1
(b) 2
(4)
(c) 3
(d) 4
34. Which is not a monoacidic base? (a) Urea (b) Methyl amine
(c) Ethylene diamine
(d) NH2CH2CO NH2
3 5. C NH
m — CH3 — C 6H 4C — NH2
p — NO2 — C 6H 4C — NH2
|
||
NH2
NH (3)
(2)
(1)
CH3 CONH2
||
NH
Correct order of basic strength is:(a) 3 > 1 > 2 > 4 (b) 4 >3 > 1 > 2
(c) 3 > 1 > 4 > 2
(4)
(d) 4 > 3> 2 > 1
36. Which reaction is not possible?
(a)
+
+
N
N
(+)
H
N
H
H
(+)
H ()
N
(–)
(–)
(b) ( NH3 )HSO 4 Me2 NH MeNH 2 HSO4 NH 2
(c) C NH H
()
()
C N H2
|
|
NH2
NH2 (–)
(d)
+ OH
H2 O +
N
N
(+)
H 37. Consider the following amines:-
NH2
Me
NMe2 Me
NMe2
(2)
(3)
Me NH2 Me
(1)
Correct order of basic strength is:(a) 4 > 2 > 3 > 1 (b) 2 > 4 > 3 > 1
(c) 3 > 4 > 2 > 1
38. Weakest base among the following is: (a) n-BuLi (b) AcCH2(–) (c) FCH2(–)
(4)
(d) 3 > 2 > 4 > 1 (d) i - Pr carbanion
39. Consider the following compounds:N Et3N FNH2 Me2NH (1) (2) (3) (4)
Correct order of basic strength is:(a) 1 > 4 > 2 > 3 (b) 4 > 1 > 3 > 2
(c) 4 > 2 > 1 > 3
(d) 2 > 1 > 4 > 3
157
Acid & Base (–)
(–)
40. Which statement is not correct about OH & SH :(–)
(–)
(a) SH is better leaving group than OH (–)
(–)
(b) OH is stronger base than SH
(–)
(–)
(c) Conjugate acid of OH is stronger acid than conjugate acid of SH
(d) Nucleophilicity of SH is large in comparison to OH in polar protic solvents
(–)
(–)
41. The decreasing order of relative basic strength of following: Me3CO(–) MeO(–) Me2N(–) BuLi (1) (2) (3) (4) (a) 1 > 2 > 3 > 4 (b) 4 > 1 > 2 > 3 (c) 4 > 3 > 1 > 2 (d) 3 > 1 > 2 > 4 42. Consider the following compounds:O
O
O
Ac3CH Me3COH
(1) (2) (3) (4)
Correct order of acid strength will be:(a) 3 > 4 > 2 > 1 (b) 3 > 4 > 1 > 2
(c) 3 > 1 > 2 > 4
(d) 4 > 1 > 3 > 2
43. Which among the following is highly basic: (a) N
NH NH (c) O
(b)
(d)
N
Me
H
NH
NH O
44.
N
N H
Me (3)
(2)
(1)
Me
Correct sequence of basic strength is:(a) 1 > 4 > 3 > 2 (b) 1 > 4 > 2 > 3
(4)
(c) 4 > 1 > 2 > 3
(d) 4 > 3 > 1 > 2
45. Which among the following is least basic:-
NO2
(a)
N
O Cl ( ) (b) C5 H5 NMe (c) N
Cl
Cl
(d)
N Cl
46. Identify the wrong statement (a) N O (c)
is basic but not
O O
O N (b) NH
O is less acidic than
O
is less basic then
CF3 OH is less acidic the CH3COOH F3 C (d) CF3
N
158
Problems in Organic Chemistry
47. Correct sequence of acid strength of the following four compounds will be:-
Me
CO2H
CO2H
CO2H
CO2H
Me Me
Me
NO2
NO2
SO3 H
NO2
(1)
(2)
(3)
(4)
(a) 3 > 1 > 2 > 4
(b) 3 > 1 > 4 > 2
(c) 1 > 3 > 2 > 4
(d) 3 > 2 > 1 > 4
48.
Which statement is correct regarding the acid strength of the following organic and inorganic compounds? (a) Sulphonic acids are more acidic than carboxylic acids (b) BCl3 is more acidic than BF3 (c) Acetone is more acidic than acetaldehyde (d) All are correct
49.
Pick out the wrong statement. (a) (CH3)3SiCH2COOH is more acidic than Me3CCH2COOH (b) Fumaric acid is more acidic than maleic acid. (c) Chloroform is more acidic than Fluoroform. (d) In CH3CONH2, Hydrogen attached with N is more acidic than hydrogen attached with C
50. Which among the following is highly basic?
(a) O
CH3
(b)
N H
(c) (d) N N N H CH3 Ac
51. Which among the following is least acidic? (a) FCO2H (b) FSO3H (c) FOH (d) FCH2OH 52. If pKb of CH3CN, CH3CHO and CH3OCH3 are x, y & z respectively then (a) x > y > z (b) y > z > x (c) z > x > y
(d) z > y > x
53. Which is not expected to be a good carbon acid?
O
O (a) CH3COCH2COOC2H5 (b)
O
O (c)
(d) NO2 — CH(COOCH3)2
54. Which has maximum pKa?
CO2H
CO2H
CO2H (a) N 55.
CO2H
(b)
N
(c)
(d)
N
Select the wrong statement. (a) Acetamide is less basic than aniline (b) Pyridine -2-carboxylic acid is more acidic than Pyridine -3-carboxylic acid (c) BCl3 is stronger acidic than AlCl3. (d) GaCl3 is stronger acid than ZnCl2
56. Which among the following is strongest base? (a) Cyclohexa – 2, 4 – dinone (c) Acetone
(b) Propene (d) 4 – Oxa cyclohexa 2, 4 dienone
NO2
159
Acid & Base
Cl H 57.
The reagent which can abstract H is:(a) HCOOK
(b) CH3OK
(c) FONa
(d) CH3Li
58. Salicylic acid is stronger acid than p - hydroxyl benzoic acid because of (a) Resonance effect (b) Steric hindrance (c) Intra molecular – H– Bonding (d) Hydration energy. 59. Match the following Acid CF3 FC (A) 3 OH CF3
pKa (p) 5.40
(B) 2, 4, 6 - trinitro phenol (C) AcOH
(q) 0.7 (r) 6.60
(D)
(s) 4.76
(b) A → r, B → q, C → s, D → p (d) A → p, B → q, C → s, D → r
COOH O (a) A → q, B → q, C → p, D → s (c) A → r, B → q, C → s, D → p
60. Match the following Acid O HO (A) Squaric acid HO O (B) RSO3H (C) HClO4 (D) H3BO3 (a) A → p, B → r, C → q, D → s (c) A → s, B → r, C → q, D → p
Approx pKa (p) 1 (q) –1 (r) 0.0 (s) 9.27 (b) A → s, B → q, C → r, D → p (d) A → p, B → q, C → r, D → r
61.
Assertion: - para chloro phenol is more acidic than para-fluoro phonol. Reason: - (–I) effect of F is greater than Cl. (a) Assertion is True, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true
62.
Assertion: - salicylic acid is stronger acid than p-hydroxy benzoic acid Reason: - salicylate ion is stabilized by H-bonding. (a) Assertion is True, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true
63.
Assertion: - 2nd dissociation constant for sulphuric acid is smaller than 1st dissociation constant. Reason: - resonace stabilization of sulphate ion is lesser than that of bi sulphate ion. (a) Assertion is True, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true
160
Problems in Organic Chemistry
64.
Assertion: - para chloro benzoic acid is more acidic than benzoic acid. Reason: - +R effect of Cl group is not prominent thus due to –I effect of Cl para chloro benzoic acid is more acidic (a) Assertion is True, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true
65.
Assertion: - Hydroxide ion can withdraw hydrogen from benzene . Reason: - basic strength of alkenyl carbanion is greater than hydroxide ion (a) Assertion is True, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true
Answers Key 1. (b)
2. (c)
3. (c)
4. (a)
5. (b)
6. (a)
7. (d)
8. (a)
9. (a)
10. (a)
11. (a)
12. (a)
13. (a)
14. (b)
15. (a)
16. (d)
17. (c)
18. (c)
19. (a)
20. (d)
21. (b)
22. (a)
23. (d)
24. (b)
25. (c)
26. (d)
27. (d)
28. (d)
29. (c)
30. (c)
31. (d)
32. (a)
33. (b)
34. (c)
35. (a)
36. (a)
37. (b)
38. (b)
39. (a)
40. (c)
41. (c)
42. (c)
43. (c)
44. (d)
45. (b)
46. (c)
47. (d)
48. (d)
49. (a)
50. (c)
51. (d)
52. (a)
53. (b)
54. (a)
55. (b)
56. (d)
57. (d)
58. (c)
59. (d)
60. (a)
61. (b)
62. (a)
63. (a)
64. (a)
65. (d)
Multiple Choice Questions (More Than One May Correct) 1. In which case 1st compound is more acidic than 2nd. (a) CF3COOH, HCOOH
(b) PhOH, CH CH
(d) F
(c) Me3CCOOH, CCl3COOH
COOH Cl
OH
2. In which case 1st compound is more acidic than 2nd
COOH
COOH
(a) Cl
(c)
OH F
OH
OH O2N
OH
(b)
COOH (d)
NO2 3. Which sequence of basic strength is not correct? (a) PhO(–) > CH3O(–) > NH2(–) > HCOO(–) (c) (CH3)3CO(–) > CH3COO(–) > HCOO(–) > HSO4(–)
NH3
NH2
(+)
H3C
CH3 (b) CH3O(–) > NH2(–) > CH C(–) > PhO(–) (d) CH3(–) > CH2 = CH(–) > NH2(–) > OH(–)
COOH
161
Acid & Base
4. Which statement is wrong about the following compound? 4
OH 1 2
COOH
OH 3
HO
NO2
(a) 4th is least acidic hydrogen (b) Most acidic hydrogen is 1st (c) 2nd hydrogen is more acidic than 3rd hydrogen (d) 4th hydrogen remains unreactive when 1 mol of this compound reacts with 3 moles of NaOH.
5. In which case 1st compound is more basic than 2nd:-
NH2
(a)
NH2
(b)
NMe2
NMe2
CH3
NO2 (c) NH = C — NH2 , |
H 2 N — CHNH 2 |
NH2
(d) (CH3)3N, Et2NH
NH 2
CH3
O
6.
N
N
H
H
(I)
(II)
N (III)
Which is/are correct statement/s? (a) III is more basic than I & II. (c) I is more basic than III
7.
In which of the following reactions, forward reaction is favoured:CH3C ≡ CLi + BuH (a) CH3C ≡ CH + BuLi HCO 2 Na + EtOH (b) HCOOH + CH3CH 2 ONa (c) PhSO3 Na + CH ≡ CH PhSO3 + CH ≡ CNa
H 2SO 4 + OH (–) (d) H 2S O 4 + H 2 O
(b) I & II are aromatic compounds (d) II is more basic than III
(–)
8. Which among the following will produce antiaromatic species after the removed of H(+)
Ph Ph (b) (c) (d) (+) Ph (–) Ph
(a)
9. In which case 1st is stronger acid than 2nd.
O2N (a)
COOH
COOH
OH COOH
NO2
(b)
OCH3
OCH3
NO2
COOH
COOH
COOH
(c)
COOH CH3
CH3
CH3
(d)
Cl
162
Problems in Organic Chemistry
Answer Key 1. (a), (b), (d)
2. (a), (d)
3. (a), (b)
4. (c)
6. (a), (b)
7. (a), (b)
8. (a), (c)
9. (a), (b), (c), (d)
5. (b), (c)
LEVEL - II
Multiple Choice Questions 1. Arrange the following in decreasing order of acid strength. O
O O
O
O
O
O
F
Cl
(1)
(2)
O
(3)
O
O
(4)
(5)
(a) 2 > 1 > 3 > 4 > 5 (b) 1 > 2 > 3 > 4 > 5 (c) 1 = 2 > 3 > 5 > 4 2. Which is true regarding the basic strength of the two amines given below?
N
N
(1)
(2)
(d) 2 > 1 > 3 > 5 > 4
(a) Both are equally basic. (b) 2nd is less basic than 1st because in 1st case amine inversion is not possible. (c) 1st is more basic than 2nd because in 2nd case N is sterically hindered. (d) 2nd is more basic than 1st because in 1st case amine inversion takes place.
3. Your have three species PhNH3(+) (1), CH3NH3(+) (2) & CH3COOH (3) Correct order of their pKa will be : (a) 2 > 3 > 1 (b) 2 > 1 > 3 (c) 3 > 2 > 1 (d) 3 > 1 > 2 4. Strongest carbon acid among the following is:F
S
(a) S
NH (b) N
(c)
O
(d)
F
O
H
H1 5.
2H
Most acidic hydrogen
3H
H4 (a) 1 (b) 2 (c) 3 6. Correct sequence of bond energies of C — H bonds in the following compound will be:O H2 H1
H4
(a) C — H4 > C — H2 > C — H3 > C — H1 (c) C — H4 > C — H1 > C — H3 > C — H2
(d) 4
H3 (b) C — H2 > C — H3 > C — H4 > C — H1 (d) C — H2 > C — H1 > C — H3 > C — H4
163
Acid & Base
7. Which statement is correct regarding the acid strength of following three compounds? (1) HO O O—H(2) HO
(a) (b) (c) (d)
O (3) HO
O
H (4) O
H(3)is more acidic than H(4) H(1) is less acidic thanH(2) H(4) is more acidic than H(2) H(3) is less acidic than H(1) but more acidic than H(2)
8. In which case Ist is more acidic than 2nd OH OH (a)
OH
(b) &
&
Cl
F
OH
CMe3
OH
(c)
OH
Et
(d) & MeO OMe
&
Cl
F
9. In which case Ist compound is more basic than IInd compound.
NH2
(a)
&
N
(b) CH3CONHCOCH3 & CH3CONHCOPh
H N N
(c)
NH=CH
&
NH2 NH3
(d) H2O & NH3
10. Which nitrogen will pick up H(+) rapidly? H
NH2
N (a)
(b)
(c)
(d) MeCONHCOCH2ME
N H 11. Which will not form zwitter ion? (a) p – NH2C6H4SO3H (b) p – NH2C6H4 — COOH (c) p – CH2NH2C6H4SO3H (d) NH2CH2COOH Hint: - The ion which contains positive as well as negative charge is called zwitter ion for e.g.
Answer Key 1. (a)
2. (b)
3. (b)
4. (a)
5. (b)
11. (b)
6. (b)
7. (c)
8. (b)
9. (a)
10. (a)
164
Problems in Organic Chemistry
SOLUTION Level – I 1. (b) Resonance makes the NH2 group less basic however resonance also occurs in (a) but resonance in (b) is more favourable because electro negative ‘O’ atom acquires negative charge. Hence due to more extent of resonance (b) is least basic. (+ )
CH3 — C—N H 2←→ CH3 —C = NH 2 ••
||
|
O
O(–)
(+)
2. (c)
H
aromatic & stable
+ (–)
3. (c) Phenols are less acidic than carboxylic acids hence 1 & 2 are less acidic than 3 & 4. Out of 1 & 2, later is more acidic because NO2 is electron with drawing group & helps in the dissociation of OH bond. Similarly out of 3 & 4th later is more acid due to more –I effect as well as – R effect of NO2 group at ortho position. 4. (a) The methyl group raises the electron density more at ortho & para position than at meta. Thus, the carbon atom para to the methyl group has a high electron density, & the lone pair on the nitrogen atom therefore is prevented, to some extent, from entering in to resonance with the ring. A methyl group raises the electron density at meta position to a very small extent. Hence resonance with the ring of the NH2 group at this position is prevented less than for the p- position. Consequently, a methyl group in the ring increases the basicity of the aniline, more so from the p- position than from meta NH2 NH2 CH3NH2 NH2 5. (b) NH3 Cl ( +I effect) +R & –I effect ortho effect (1)
(2)
(4)
(3)
4 > 1 > 3 > 2 3rd is placed after 4 & 1 because Cl in ring deactivating group. 6. (a) 3 & 4th are less basic than 1 & 2nd because lone pair of electron present on N is delocalized. Out of 3 & 4th former is more basic them later because electron –I effect of Cl decreases electron density on ‘N’. Out of 1 & 2nd is more basic due to + I effect of methyl gruoup. 7. (d) 2nd & 1st are almost equally acidic because –I effect in 2nd will be equal to that of +R & ortho effect in 1st. 8. (a) acid strength ∝ stability of conjugate base 9. (a) –R effect of CO group makes 3rd nitrogen least basic. Out of 1st & 2nd later is less acidic because (–I) effect of Cl group decreases electron density on N atom. 10. (a) Carboxylic acid is more acidic than phenols. Out of 1st and 2nd later is less acidic due to presence of intra molecular –H bonding 11. (a) 1st is least basic because lone pair of electron present on ‘N’ undergoes resonance with CO group. Out of 3rd and 2nd former is more basic as its lone pair undergoes resonance only with one double bond 12. (a) 3rd is least basic because lone pair of electron present on N is delocalized. Out of 1, 2 & 4th, 4th is least basic because N is sp2 hybridized thus, it is more electronegative & can not donate its lone pair of electrons easily. 2nd is more basic than 1st because of more +I effect. 13. (a) 3rd is least basic as lone pair of electron present on N is not localized and takes part in resonance with CH3CO group(Ac). 14. (b) 1st is least acidic due to the presence of intramolecular H bonding 15. (a) Because after protonation intermediate so formed is stabilized by resonance |
(+ )
••
NH2
NH2
NH2 (+ )
|
||
H 2 N — C = NH → H 2 N —C = N H 2 ←→ H 2 N—C — N H 2 H
••
••
( +)
••
165
Acid & Base (–)
(–)
→ NH3 + C CH ≡ CH + N H2 ←
(–)
≡ CH
(More basic)
16. (d) (–)
(–)
NH 2 > C ≡ CH (–)
(–)
CH ≡ C
→ OH + CH ≡ CH CH ≡ C + H 2 O ← Therefore overall sequences will be:-
(less basic) (–)
> OH
(More basic)(less basic)
(–)
(–)
(–)
NH 2 > CH ≡ C > O H
17. (c) Same as previous. 18. (c) % S character electronegativity. Triple bonded carbon atom is more electronegative than double & single bonded carbon atom due to 50% S character so it can withdraw electrons very easily from C—H bond. Consequently 2nd is most acidic 19. (a) Stability of conjugate base acidic character. 20. (d) Mineral acids (H2SO4, HCl, HNO3 ) are stronger acids than other. 21. (b) Alcohols are less acidic than thiols because ‘S’ is bigger in size thus ‘S—H’ bond has less bond energy than ‘O—H’ bond 22. (a) Alkanes are poor acid but their anions are strong base. 23. (d) CH2 = CH(–) is stronger base than 24. (b) Because electron with drawing effect of 3-NO2 groups helps is dissociation of ‘O—H’ bond. 25. (c) Negative I effect enhances acidic character (–)
(–)
26. (d) CH 2 Cl is stabilized by resonance while CH 2 F is not because Cl can receive electrons of –ve charge in its empty d orbital. 27. (d) Acidic character ∝ 1 / Bond energy 28. (d) Carboxylic acid is more acidic than alkane & alkene.1st is more acidic than 2nd & 3rd because it forms an aromatic anion. 29. (c) Because alkyl carbanions are more basic than alkenyl carbanion 30. (c) Me4NOH is more ionic in comparison to NaOH because of less polarization as the size of Me4N+ is so large in comparison to Na+ 31. (d) Stability of conjugate base ∝ acidic character 32. (a) Out of 2 & 3 former is more acidic as Cl can not show resonance at meta position 33. (b) At meta position OMe group shows electron withdrawing effect. 34. (c) Because it contains two NH2 groups (basic groups) although urea & NH2CH2CO NH2 also contain two NH2 groups yet lone pair of electron present on one NH2 group undergoes resonace with CO group. 35. (a) 4th is least basic because lone pair of electrons present on ‘N’ take part in resonance with CO group. Out of 1, 2 & 3, 3rd is more basic because +I effect of CH3 increases electron density on ‘N’ 2nd in less basic than 1st because of the presence of electron withdrawing NO2 group. Thus sequence of basic strength is 3 > 1 > 2 > 4 H N
N 36. (a) Because
is more basic than
37. (b) 2nd is more basic then 3rd because groups present at ortho position create hindrance in resonance (steric inhibition to resonance) thus, lone pair of electrons present on ‘N’ does not undergo resonance with benzene ring easily & available for protonation. 38. (b) –ve charge undergoes resonance which is more favourable as electronegative element ‘O’ acquires negative charge. O O(–) ||
(–)
|
CH3 — C— CH 2 → CH3 —C = CH 2
39. (a) 1st is more basic than 2nd & 4th because in 1st case amine inversion takes place. 40. (c) H2S in stronger acid than H2O
166
Problems in Organic Chemistry
41. (c) CH3OH is stronger acid than Me3COH Me2 NH & BuH hence methoxide ion is least basic. BuH is least acidic hence its conjugated base i.e will be strongest base. Since Me3COH is stronger acid than Me3NH hence (–)
(–)
(CH3 )3 CO will be stronger base than (CH3 ) 2 N 42. (c) Ac3C(–) is highly stabilized by resonance. O
O (–)
CH3 — C — C — C — CH3 O == C — CH3 43. (c) Because lone pair of electron present on ‘N’ are localized & (c) is 2° amine which is more basic than 3° amine (a) 44. (d) 2nd is least basic because of delocalization of lone pair of electrons present on N with CO group. 4th & 3rd are more basic than 1st as 1st is 3° amine 45. (b) ‘N’ does not contain lone pair of electron. O (–) (–) O O O 46. (c) is stabilized by extended resonance while is not
47. (d) Sulphonic acid is more acidic than carboxylic acid. 2nd is more acidic than 1st due to ortho effect. 4th is least acidic because at meta position methyl group shows + I effect. 48. (d) BF3 is less acidic than BCl3 due to back bonding. Acetone is more acidic than acetaldehyde because base
(–)
CH3CHO → C H 2 CHO ←→ CH 2
= CH — O(–)
less substituted alkene & less stable
(–)
CH3COCH3 → CH 2 COCH3 ←→ CH 2 = C — CH3 base
more substituted
|
alkene & more stable
O(–) 49. (a) + I effect of (CH3)3 Si- is greater than Me3C– 50. (c) Lone pair of electrons present on nitrogen is localized. 51. (d) Anion formed after the removal of H+ ion is not stabilized by resonance (alkoxide ion PhCH2O(–)). 52. (a) N of CH3CN is sp hybridized hence more electronegative while oxygen of MeCHO & MeOMe are sp2 & sp3 hybridized respectively. So oxygen of MeOMe is least electronegative. Electronegativity ∝ 1 / basic character 53. (b) Conjugate base is not stabilized by resonance as bridge carbon atom can not bear double bond.
54. (a) Due to the presence of intra molecular H—bonding
N
HO
55. (b) Same as previous question. (+)OH
O
OH
OH
OH
(+)
(+)
56. (d)
CO
H
O
O
(+)
O
O
O (+)
highly stabilized by resonance 57. (d) Same as q.n.22 58. (c) Salicylate ion is stabilized by H– bonding as:(–)
COOH OH
CO—O (+)
–H
O—H intra mol—H bonding
167
Acid & Base
59. (d) Acid strength ∝ Stability of conjugate base 60. (a) Acid strength ∝ Stability of conjugate base 61. (b) See question 49 62. (a) See question 58 63. (a) Resonance stabilization of sulphate ion is lesser than that of bi sulphate ion 64. (a) + R effect of Cl group is not prominent as 3p of Cl does not overlap effectively with 2p- of carbon thus due to – I effect of Cl para chloro benzoic acid is more acidic. 65. (d) hydroxide ion is poor base than alkenyl anion More Than One May Correct 1.
(a), (b), (d)
2. (a), (d)
3. (a), (b)
4.
(c) Second hydrogen is less acidic than 3rd hydrogen because 2nd hydrogen forms intramolecular hydrogen bond with NO2 group
5. (b), (c) In (b) 1st is more basic than 2nd amine because in 1st amine SIR (Sterric inhibition to resonance) takes place in (c) 1st is more basic than 2nd amine because in 1st amine after protonation resonance takes place (+ )
H( + )
NH = C — NH 2 → NH 2 |
NH2
6.
(a), (b)
(+ )
= C| — NH 2 ←→ NH 2 — C| = NH 2 NH2
7.
NH2
(a), (b)
8. (a), (c)
(–)
Ph (+) Ph
Ph Antiaromatic (+) Ph
(+)
H
+ (–)
(+)
H
Aromatic
+ (–) (+)
Ph Ph (–)
9.
H
Antiaromatic
+ (–)
Ph Aromatic Ph (–)
(+)
H
+
(a), (b), (c), (d)
LEVEL - II 1. (a)
2.
(b) 2nd is less basic than 1st because 2nd amine undergoes inversion
i—Pr i—Pr i—Pr
i—Pr N
N
i—Pr i—Pr
3. (b) Check the stability of the anion formed after the removal of H+
S
4. (a)
(–)
S
sulphur can receive electrons of negative charge in its d orbital
168
Problems in Organic Chemistry
5. (b)
6. 7. 8. 9.
(–) (–)
aromatic ring
(b) Acid strength ∝ 1 / Bond energy of C-H bond (c) Acid strength ∝ Stability of conjugate base (b) CMe3 does not has hyper conjugation effect (a) Pyrrole is less basic because its lone pair undergoes resonance & makes the ring aromatic. Thus its lone pair of electron is least available for protonation.
H
H
N 10. (a)
(+)
(+)
H
N
H
antiaromatic & less stable
nonaromatic & stable
11. (b) Lone pair of amino group is delocalized and COOH attached with benzene ring is least acidic due to -R effect
6
Hydrocarbon (Alkane, Alkene & Alkyne) Main Features REACTION CHART FOR ALKANE Preparation
Properties NaOH CaO, ∆
AlCl3 HCl
RX
Na/Ether
Pyrolysis
RX
Zn/Ether
R—COONa or RCOOH
RCOR or RCHO
Al2O3 or Mo2O3 600°
Chain isomers (at least 4 C atoms)
Lower alkane & alkene
Aromatic Compound
Zn/Hg +HCl NH2 NH2 NaOH, ∆
RCOONa
RMgX
Current
H2O
(+)
RX
RCOR or RCHO or ROH
Zn + H
Red P HI
A L K A N E
X2 hvν
RX
O2 ∆
CO2+ H2O
conc. HNO3 400°C
R—NO2
conc. H2SO4 ∆
RSO3H
170
Problems in Organic Chemistry
REACTION CHART FOR ALKENE Preparation
(+)
RCHXCH2X
Zn dust heat
RCH2CH2X
alc. KOH heat
R—CH2CH2OH
H /heat
Properties
H2O/H X2 HX
(+)
R—C CH R—CH—COOK
A L K E N E S
Partial Reduction
R—CH—COOK
Electric Current
RCH2CHX2
Zn dust heat
HBr Peroxide cold KMnO4 CH2N2 hot KMnO4 O3/H2O Zn, ∆ O3/H2O, ∆
Hoffman elimination MeCH2OCOCH3
Pyrolysis
Antimarkownikoff addition
A. M. A
Markownikoff addition
M.A.
High P & T Catalyst B2H6 H2O2/OH(–) H2/Ni 200-300°C S2Cl2
REACTION CHART FOR ALKYNES Preparation CHC 3(2-moles)
Ag. heat
CH2Br—CH2Br
alc. KOH or NaNH2
CH3CHBr2
alc. KOH or NaNH2
CHBr=CHBr Cal. or Mag. Carbide
CH
CNa
CH3—C
CH
CH3X (i) CH3Mgl. (ii) CH3X
Markownikoff addition Antimarkownikoff addition
RCH(OH)CH2OH [Syn addition] CH2 R—CH — CH2 Carboxylic acid aldehyde & ketone Carboxylic acid Polymers RCH2CH2OH (A.M. A.) alkane CH2ClCH2SCH2CH2Cl Mustard gas Alcohol (M.A.)
per acid hydrolysis
RCH(OH)CH2OH [Anti addition]
CH2 = CHCH2Br allylic substitution
Properties H2/Ni
alkane alkene
(+)
A L K Y N E S
Hg+2/H H2O HOCl
Fe/heat Trimerisation Fe/heat Dimerisation Cold KMnO4 Hot KMnO4 Zn Ozonolysis X2
M. A A. M. A
RCHXCH2X RCHXCH3 (M.A.) RCH2CH2Br (A.M.A.)
Hg(OAc)2 NaBH4 Br2/500°C or NBS
H2(1mole)
Zn dust heat H2O
Alcohol (M.A.)
AsCl3 CO + H2O Ni(CO)4
Aldehyde or ketone (M. A.) CI2CHCHO (M. A.) Benzene [For acetylene] [For acetylene] CHO—CHO or vicinal di ketones Carboxylic acid Carboxylic acid RCX2—CHX2 CHCl=CHAsCl2 Lewsite gas R—CH = CHCO2H
NH4OH + CuCl2
R—C
CCu
Only for terminal alkynes Tollen’s reagent R—C CAg NaBH4 Only for terminal alkynes B2H6/H2O2 aldehyde or ketone (A.M.A.) NaOH 2HX RCX CH (M.A.) 2
Hg(OAc)2 NaBH4
3
aldehyde or ketone (M.A.)
171
Hydrocarbon (Alkane, Alkene & Alkyne)
LEVEL - I Multiple Choice Questions (i) NBS / hν (ii) alkaline KMnO 4 &heat
→
1.
X
(i) Current (ii) aq KOH 623K,200atm
→
Y, Y is:-
(a) CH3CHO
(c) HOOC —CH = CH — COOH
2.
Ph C COOH NaOH Hg 2+ ¾¾¾¾ ®[X] ¾¾¾¾ ® || Current H3O(+) CH3 C COOH
(b) CH2 = CH — CH2 — OH OH (d) HOOC HOOC
Final product of this transformation is:O OH Ph Ph (a) (b) (c) (d) Ph Ph O
OH CH3
CH3 (i) NaNH
MnO(-)
H / Pd / BaSO
2 ® ¾¾¾¾¾¾ 2 4 ® ¾¾¾¾ 4 ® 3. CH3CH 2 C º CH ¾¾¾¾¾ (ii) EtBr ice cold
Final product of this reaction (a) is optically active (c) is optically inactive
(b) contain three chiral ‘C’ atoms (d) is a racemic mixture
4. Which will have least rate of dehydrobromination?
Br Br
(a)
(b)
(c) Br
Br
(d)
CH3
Cl Na/Et2O
5.
(A) (A) & (B) respectively are:-
Br (i) Mg/Et2O (ii) H2O
Cl
Cl
(a)
,
Br Br
(B)
(b)
,
Br
Br
(c)
(d) ,
, Cl Zn
6. (A) → Me2O
Zn
(B) → ∆
But - 2 - ene (A) will be:-
CH 2 — CH — CH — CH3 (b) CH3 — CH2Cl (a) |
Cl
|
Cl
Cl
(c) CH3CHCl2
(d) CH3CCl3
172
Problems in Organic Chemistry
Me 7.
Me C=C
H
OsO / H O
(A),
OH
OH
4 2 →
H
(a) Me H
H
(A) would be:-
OH
Me
H
OH (c)
(b) H Me
Me
Me
H
Me
H
(d) All of these
OH
OH
8. Which product is unexpected in the following reaction? (i) O
3 → Product, CH3 — C = C — CH3 (ii) H O, ∆ 2
||
||
OH |
(c) CH3 — C—C = CH 2 (d) MeCHO
(a) CH3 — C— C— CH3 (b) AcOH ||
O
O
O
9. Pick out the reaction which is not stereospecific (–)
KMnO4 / OH (a) Cis – 2 – butene → Product
3 → Product (c) Trans – stilbene (+ )
Hg(OAc) NaBH 4
2 → Product (b) Trans – 2 – stilbene
(i) PhCO H
Br
2 → (d) Cis – stilbene Product in CCl
(ii) H 2O / H
4
CH N in presence of N
2 2 2 → Product 10. Cis – stilbene
Ph Ph H H H Ph — C C C—C (a) (b) (c) C—C H H Ph Ph Ph H CH2 CH CH2 2
1 mol H2
→
11.
(a)
12.
dilute H 2SO4
→
(d) Both (a) & (b)
(A), (A) would be:–
(b)
(c)
(d)
[X] (Major)
OH
H2SO4 OH
(a)
(b)
HSO4
(c)
(d)
173
Hydrocarbon (Alkane, Alkene & Alkyne) hν
Peroxide
13. CH 4 + CI2 →(A) → (B), (B) would be:– Cyclo octa-1, 5-diene 1
: 3
(a)
(b)
CCl3
(c)
(d) CCl3
CCl3
CCl3
14. CH3 — CH = CH2 + HI (Solid) ——→ [X] will be:– (a) CH3CH2CH2I (b) 2 – Iodo propane (c) Both (a) & (b) (d) No product will form
15. How many structural isomeric alkenes are possible for the fomula C3H4Br2? (a) 3 (b) 4 (c) 5 (d) 2 16. Buta – 1, 3 – diene can be converted in to But – 2 – ene by:– (a) LiAlH4 / Ether (b) Na / Liquid NH3 (c) Na / C2H5OH (d) B2H6 / AgNO3 17. Which alkene will produce CH3COCOCH2CHO & OHC — CH2 — COCH2CH2CHO on ozonolysis followed by hydrolyses? (a)
(b)
(c)
(d)
18. Which among the following will have largest heat of hydrogenation?
(a) Trans –2–butene
(b) Isobutene
(c)
(d)
19. Which among the following will undergo dehydrochlorination rapidly:– (a) 2 – chloro pentane (b) 3 – chloro but – 2 – ene (c) 1, 5 – dichloro pentane (d) 1 – chloro – 4 – Phenyl pentane
20.
21.
These compounds are dehalogenated by Zn /∆ . The correct order of ease of these compounds for dehalogenation is:– (a) 1 > 2 > 3 (b) 1~ 2 > 3 (c) 3 > 2 > 1 (d) 3 > 2 ~1 Which will give four products when subjected to mono chlorination. (only structural isomers) (a) n –pentane (b) Iso – butane (c) Methyl cyclo pentane (d) 1, 2 – Dimethyl cyclo propane
22. Identify [Y] in the following reaction. CH3CH 2 MgCl +
[Y]
→ 44800ml of alkane at S.T.P + other products
(unknow)
(a) Ethylene glycol (c) HO3S — CH2 — CH2 — SO3H
23. Cl2C = C = O ••
(a) C Cl2 ,
hν → A
(b) Succinic acid (d) All of these B,
A & B are:–
CHCl2 (b) C2Cl2, Cl
••
(c) C Cl2 , Cl
CCl2
Cl ••
(d) C Cl2 , Cl CCl2
174
Problems in Organic Chemistry
24. Arrange the following hydrocarbons in decreasing order of their acid strength Propyne, Ethyne, But – 2 – yne (1) (2) (3) (a) 2 > 1 > 3 > 4 (b) 2 > 1 > 4 > 3 (c) 1 > 2 > 4 > 3
Ethene (4) (d) 1 > 2 > 3 > 4
25. 1 mol of hydrocarbon ‘A’ gives 4 moles CO2 on combustion. It can decolourise Br2 water. A gives B when treated with Pd/C. B can also decolourise Br2 water. A can react with Tollens reagent but B can not. A & B are:– (a)
(b)
,
(c)
,
,
(d)
,
KOH in C H OH
2 5 26. ΦCH 2 Cl →[X] , [X] may be:–
(2 − Moles)
(a) ΦCH2 — CH2 Φ (c) ΦCH = CH Φ Φ — CH — CH 2 — Φ (b) |
(d) 2 moles of PhCH2OH
Cl 27. Which product will not form in the following reaction? O3 ,H 2O CH3 — C — CH 2 — CH = CH 2 → Product ∆
||
CH 2
(a) HCOOH
(b) CH3COCH2COOH
(c) Me2CO
(d) All are possible
CHO O3 ∆ NaOH Pd 28. A + maleic acid → → → →3 | (A) will be:– ∆ CaO / ∆ H 2O + Zn CHO
CH 2 (a) || (b) (c) CH 2
(d) Ethene
29. Which product is not possible in the reaction given below? H( + ) D 2O
CH2 = CH — CH = CH2 → Product (a) CH3 — CH — CH
CH 2
(b) CH3 — CH = CH — CH2OD
OD (c) CH 2 D — CH — CH = CH 2
(d) Both b & c
|
OD 30. Which among the following will liberate two moles of CO2 gas on oxidative ozonolysis followed by heating? (a)
(b) (c) (d)
Mg(1mol) Ether
31. BrCHDC º C - CHD - Br ¾¾¾¾¾ ® Product
Product of this reaction (a) can release alkane with ROH (c) can show optical isomerism
32. Butan – 1, 4 – di – ol ••
(b) can show geometrical isomerism (d) can show geometrical as well as optical isomerism A ∆
H PO ∆
→
3 4→
•
MnO (–) ∆
4 →
(a) CH3 C H 2 (b)
NaOH
→ Butane. (A) is:– CaO, ∆ (c) C2H2
(d) C2H4
175
Hydrocarbon (Alkane, Alkene & Alkyne)
33. Which reaction is / are not possible among the following? + CH2=CH2
∆
+ CH2 =CH2
...................(1)
∆
...................(2)
O
Heat
+ iso butene
...................(3)
O +
∆
...................(3)
O
(a) 1 & 3
(b) 1, 2 & 4 (i) NaNH
(c) only 1
(d) 1 & 4
0°C
(i) NaNH
2 → X 2 → Y → Z , Z is:– 34. Acetylene (–) (ii) (CH ) Br (ii) Ni –B 2 4
2
CHOHCH2OH
(a) (CH2)4
CHOHCH2OH
MnO4
2
(b) (CH2)4
C=O (c) Br(CH 2 )4 CH — CH 2 OH (d) C O
=
|
OH
OH
35. Correct order of boiling points of n – hexane, n– pentyne, iso – pentyne, iso pentene, iso pentane is:–
(a) n – hexane > n – pentyne > iso pentyne > iso pentene > iso pentane
(b) n – pentyne > iso pentyne > iso pentene > n – hexane > iso pentane
(c) n – hexane > iso pentyne > n – pentyne > iso pentene > iso pentane
(d) n – pentyne > iso pentyne > n – hexane > iso pentene > iso pentane CN
36.
∆
+ CH3 CH3
[X],
X is:–
CH3 CN CN
(a) Me
CH3
Me
Me
(b)
Me
CH3
CN Me
(c) Me
CH3
(d) Me Me
CH3
CN
37. Which of the following alkene is highly statble?
(a)
(b)
(c)
OH
(d) (a) & (b)
176
Problems in Organic Chemistry
38. Consider the following reactions:–
........................ (1)
...................... (2)
...................... (3)
The reaction /s in which alkene will not form is /are:– (a) 2 (c) 1, 2, 3, 4
...................... (4)
(b) 1 & 2 (d) Alkene will produce in all cases
39. 2 moles of acetylene are passed through red hot iron tube followed by heating with acetylene gives X. The product obtained by the ozonolysis of X will be:– CHO
OHC (a) OHC
(b)
CHO |
CHO
CHO
(2 moles)
(c) OHC — CH2 — CH2 — CHO
COH
(d)
|
C OH (3moles) NBS 40. CH CH = CHD → 3
(a) CH2Br CH = CHD
product of this reaction would be:– (b) CH3 CH = CHBr
(c) CH2 = CH CHDBr
(d) both a & c
H CH3 CH3ONa(in DMSO) → X, heat Br
41.
X is:–
H D
H H
H
CH3 (b)
(a)
CH3 (c)
CH3 (d) CH3
H H
D
D
D
H
H
CH3 Br I(–) → X, H in Me2CO
H 42. Br
X is:–
CH3 CH3
CH3
H (a) I
I H
H (b) H
I I
CH3
(c)
C H
CH3
CH3
C
CH3 H (d) C H H
CH3 C H
177
Hydrocarbon (Alkane, Alkene & Alkyne)
COONa Current
¾¾¾¾ ® [X]
43.
dim erisation ¾¾¾¾¾® [Y]
COOK COONa Current ® [Z] COOK ¾¾¾¾
O Aromatic products are (a) All
(b) Y & Z
(c) X & Y
(d) Z
(i) Na / liq NH (ii) A / ∆ (iii) Pd / ∆
3 → ortho-Xylene, A is:– 44. A
(a)
(b) CH2 = CH2
(c) CH3 — CH = CH — CH3 (d) CH2 = CH — CH = CH2
45. The reaction in which anti addition occurs is:–
Ni (a) RCH = CHR + H2 → RCH2CH2R
(c) RCH = CHR → RCHOH CHOHR
Cold KMnO4
(b) RCH = CHR + F2 (in CCl4) → RCHF — CHFR PBA
(d) RCH = CHR → RCHOH CHOHR (+ ) H3O
CH3 H NAOEt → product Br
46. H H3C
H
The product of this reaction is:– CH3
CH3
H
(a)
(b) CH3 H
CH3
H (d)
(c) CH3
H
CH3
47. Which is ordered correctly for heat of hydrogenation:–
(a)
>
(c)
>
(b)
>
(d) >
48. By the ozonlysis of cis – jasmone propanal & 3, 4, 7 – trioxo heptanal are produced, the structure of cis jasmone will be:– O
(a) O
(b)
(d)
O
(c)
178
Problems in Organic Chemistry
O
49.
→
This conversion can be performed by:– (a) PBA (b) Br2 / water & NaOH
(c) air / Ag / ∆
(d) All of these
50. An alkene A on ozonolysis gives dicarbaldehyde. When A is subjected to hydrolysis followed by acid catalysed dehydration compound B is produced which on ozonolysis gives diketone. A will be:–
(a)
(b)
(c)
(d)
CH = CH2
51. In which alcohol acid catalysed dehydration is not possible via E1 mechanism:– (a) (CF3 )2 — COH
|
(b) CF2MeOH (c) Φ2CMeOH (d) CH3 — CH — OH |
CH3
CH3 52. In which case 1st compound has higher melting point than 2nd. Cl Cl Cl (a)
&
(b) CH3COONa & NaCl
Cl
Cl
(c)
&
(d) both (a) & (b)
Cl Cl
Cl 53. The compound with highest value of heat of combustion is:–
(a)
(b)
(c)
(d)
(+ )
H3O 54. ΦCH (OH) C ≡ CH → CH =CHCHO Which intermediate will not form during the course of above reaction? +
(a) ΦC H C ≡ CH (b) ΦCH= C=
(+ )
(+ )
C H (c) ΦCH(OH) C
= CH 2
(d) All are correct
55. Identify the unexpected product in the following reaction:– Me
Br in NaCl
2 →
Product
Br
Me
Cl Me (b)
(a)
Me (c)
Br (d) All products are possible
H H
H
Br Br
Br
Cl
2 → A, Product A is:– in CCl
56.
4
H
Cl
H
Cl
Cl
Cl H
(a)
Cl (b)
H
Cl
(c)
(d) None
179
Hydrocarbon (Alkane, Alkene & Alkyne)
COOR
H3C
∆
COOR
The end product of this reaction is:– Me
heat strongly
→ →
+
57.
Me
CO2R
(a)
CO2R
(b)
CO2R
CO2R Me CO2R
(c)
Me
CO2R
(d)
CO2R
58.
N
& CH 2
= CH 2
CO2R
CH Cl(excess) AgOH/∆
(i) C H / ∆ (ii) Pd / ∆
3 2 4 → → E
CH3
The end product ‘E’ of the reaction is:–
(a)
(b)
(c)
(d)
59. Which isomer of pentane will produce three mono chloro derivatives (including stereoisomers)
(a)
(b)
60.
→ in MeOH/∆
MeO(–)
Br
(c)
(d) None
[X]
H
Br MeO(–)
→ in MeOH/∆
[Y]
H
(X) & (Y) are (a) Same alkyne (c) Alkyne & cumulated diene respectively
(b) Cumulated diene & alkyne respectively (d) Same cumulated diene
61. Wurtz reaction will be observed in:– OH
OH
(a) CH ≡ C — CH 2 Cl (b)
(c) CH3CHCl2
(d) Cl
Br
62. A Hydrocarbon ‘A’ (C6H10) on reduction with limited quantity of H2 gives another hydrocarbon B. B on mono chlorination gives two isomers. A is :– CH3
(a)
CH3 (b)
CH3
(c)
(d) All of these
180
Problems in Organic Chemistry
63. Rank the following in decreasing order of rate of E2. Et Cl
Cl
Cl
(1)
Et
Et
Et
(2)
(a) 1 > 2 > 3
(b) 2 > 3 > 1
(3)
(c) 3 > 1 > 2
(d) 2 > 1 > 3
(c) C > B > A
(d) A > B > C
NH2 Li / liq NH
H 2 / Ni 3→ (B) → CH OH
64.
3
(C)
(A)
Correct order of basic strength of A, B & C is:– (a) B > C > A (b) B > A > C
65. Styrene (Phenyl ethylene) + Chloro methoxy methane ——→ Product P. P would be:– (a) Ph—CH(OCH3)CH2CH2Cl (b) PhCH(CH2Cl)CH2OCH3 (c) Ph–CH(CH2OCH3)CH2Cl (d) Ph–CHCl(CH2)2OCH3 Hg(OAc)2, THF NaBH4
A Me H H
OH (–)
66.
THF/BH3
X
H2O2, OH
C Me H
Me
H
H
OH
CH3COOH
B
Me H
Out of A, B & C which product is wrong? (a) A (b) B
(d) All are wrong products
(CH2)3Br
67.
(c) C
CH2Br
Sequence of reagents required for this conversion is:– (a) Anh.AlCl 3, Pd/D (c) Mg, ether, H3O+, Pd, heat
(b) Na / ether, Pd / Heat (d) both (a) & (b)
68. Identify the product of following rearrangement CH3 H SO
O
2 4
CH3
(a) O
(b) OH
(c) OH
(d)
181
Hydrocarbon (Alkane, Alkene & Alkyne)
Passage– I LiAlH
dil H SO
H( + ) ,Heat
4 → B 2 4 → C → D A
Compound D on ozonolysis gives heptan – 2, 6 – dione Answer the following questions from 69 to72 69. Compound D would be:– (a)
(b) (c) (d)
70. Compound C would be:–
OH
(a)
(b)
OH (c)
OH (d) OH
71. Compound B would be:–
(a)
(b)
(c)
(d)
(b)
(c) (d) Both (a) & (c) Cl Cl
72. Compound A would be:– (a)
Cl
Passage–II An alkyl halide (X) C4H8Br2 gives (A) when treated with alcoholic KOH. A gives B when fused with KOH at 197°C .When (A) & (B) are treated with NaNH2 (C) & (D) are formed respectively. (D) On treatment with HCl gives (E). (E) On reaction with (C) can produce (D). Compound (A) & (B) have same degree of unsaturation (A) can react with Tollen’s reagent but (B ) can not. However both (A) & (B) forms same compound (F) on hydration. Answer the questions from 73 to 79 73. Compound X is (a) 2,3 – di bromo butane
(b) 1,1 di bromo butane
(c) 1, 4 di bromo butane
(d) All of these
74. Compound A is:– (a) Butyne
(b) But–2–yne
(c) Buta–1, 3–diene
(d) But–2–ene
75. Compound B is (a) Butyne
(b) But–2–yne
(c) Buta–1, 3–diene
(d) But–2–ene
76. Compound C is:– (a) Cis–but–2–ene
(b) Trans–but–2–ene
(c) Butylide ion
(d) None of these
77. Compound D is:– (a) Cis–but–2–ene
(b) Trans–but–2–ene
(c) Butylide ion
(d) Butene
78. Compound E is:– (a) 2 – chloro butane
(b) 2, 2 di chloro butane
(c) Both (a) & (b)
(d) 1 – chloro butane
79. Compound F is:– (a) Butanal
(b) Butanol
(c) Butanone
(d) Butan–2–ol
182
Problems in Organic Chemistry
Passage – III Consider the following three reactions. Reaction - 1 Reaction - 2 Reaction - 3
H O( + )
3 → A
B H
2 6 → B (–)
H 2O2 , OH
oxymercuration → C
Answer the questions from 80 to 84 80. The reaction which occurs via rearrangement in carbocation. (a) 1 and 3 (b) 1 only (c) 3 only 81. Pick out the correct statement about A, B & C? (a) A and B are same and different from C (c) C and B are same and different from A
(d) 2 only
(b) A and C are same and different from B (d) All are different
82. C is heated with sulphuric acid and product so obtained is shaken with dilute phosphoric acid to get another product D (a) A and D are same (b) B and D are same (c) C and D are same (d) A B C D all are different 83.
Select the correct statement. (a) A B & C will produce same product on acid catalysed dehydration (b) B & C will produce same product however A will produce different product on acid Catalysed dehydration (c) A & C will produce same product however B will produce different product on acid catalysed dehydration (d) B & A will produce same product however C will produce different product on acid catalysed dehydration
84. Which is regeoselective? (a) Acid catalysed hydration of alkene (c) Hydroboration oxidation of alkene
(b) oxymercuration of alkene (d) All of these
Passage – IV Consider the following process NaNH
A
2→ A B → C If A is a chloro compound can decolorize boromine water & on vapourisation 1.49 g of A gives 448 ml vapour at S.T.P. then
Answer the question from 85 to 90 85. Molecular wt. of A is:– (a) 37.25
(b) 74.5
(c) 98.5
(d) 76
86. Compound A is:– (a) 2 – chloro propene
(b) 1– chloro propene
(c) 3–chloro propyne
(d) 1– chloro propyne
87. Compound B is:– (a) Cl CH2C ≡ CNa
(b) Cl CH2CH = CH2
(c) Propyne
(d) CH2 = C = CH2
88. Compound C is:– (a) 1.4 di chloro hex – 2 – ene (c) 6–chloro hexa–1, 4–di–yne
(b) 6–chloro hexa–1, 4–di–2–yne (d) 1– chloro hex – 2–yne
183
Hydrocarbon (Alkane, Alkene & Alkyne)
89. A will not show:– (a) Reaction with Tollen’s reagent (c) Ozonolysis 90.
(b) Wurtz reaction (d) Hydration reaction
Which is correct about A & C? (a) A can react with ammonical cuprous chloride but C can not (b) C can react with ammonical cuprous chloride but A can not (c) A can react with Tollen’s reagent while C can not (d) Both A & C can react with Tollen’s reagent as well as ammonical cuprous chloride
Passage – V (A)
Fe/∆
2 moles
Answer the questions from 91 to 93 91. Compound (A) is:– (a) Propyne
A Heat O3/Zn/ H2O
(B)
(C)
A Heat
(D)
2 moles of glyoxal
(b) Ethyne
(c) Ethene
(d) Propene
92. Compound (C) is:– (a)
(b)
(c)
(d)
93. Compound (D) is:–
(a)
(c)
(b) (d)
94. Match the following Characterstic of Compound 1. Optically active alkane with least molecular wt 2. Alkene with least mol.wt which can not exhibit geometrical isomerism & becomes optically active after hydrogenation 3. Alkene with least mol.wt & optically active (a) 1 – B, 2 – A, 3 – C (b) 1 – A, 2 – B, 3 – C (c) 1 – C, 2 – B, 3 – A (d) 1 – C, 2 – A, 3 – B
Molecular wt A 98 B 100 C 84
95. Which is / are not correctly matched?
A
HOOC B
C (a) A
gives Diels alder with ethane COOH
gives meso tartaric acid with cold KMnO4
gives trans but 2 – ene on reaction with Pd / C / H2 (b) B & C (c) A & C
(d) All are mismatched
96. Consider the following compounds:–
(1)
Correct order of their heats of combustion will be (a) 1 > 2 > 3 (b) 2 > 1 > 3
(2)
(c) 3 > 2 > 1
(3)
(d) 3 > 1 > 2
184
Problems in Organic Chemistry
OH CH2 C CH2
97. CH3
(+ )
H NH2 →(X) product
CH2
Heat
CHO
X(major product) would be:– CH2 C
(a) CH3
CH
CH CH3 NH2 (b)
CH2
C CH CH2
NH2
CHO
CHO
NH CH2 C CH3
CH2 C CH2
(c) CH3
NH2
CH
(d) CH3
CH2
CHO
CHO CH3MgBr 98. CH3 — CH —CH — CH3 → (A) |
OH
∆
|
Br
2 CH3 — CH = CH — CH3 → (C)
Select the correct option. (a) A = B = C ≠ D
Cl water aq K 2CO3
(b) A = B = D ≠ C
CH3 — CH
Br2 water → (B) CH — CH3 aq KOH
PAA CH3 — CH = CH — CH3 → (D)
(c) B = D ≠ A ≠ C
99. Identify the product which is least likely to form:– Br in NaCl
2 → Product
Br
(a)
Br
Br
(b) Br
(c)
Cl
Br
(d) All are possible
(d) A ≠ B ≠ C ≠ D
185
Hydrocarbon (Alkane, Alkene & Alkyne)
100. Identify the alkene which will not provide the following alcohol upon oxymercuration – demercuration. OH
(a)
101. CH 2
CH 2
(b)
(c)
(d) None of these
H SO
2 4
Major product of this reaction would be:-
(a)
(b) (c) (d) AlCl —HCl 25°C
3 102. Assertion: – CH3 — CH2 — CH3 → No product:–
Reason: – For product formation temperature should be 300°C. (a) Assertion is true, Reason is true: Reason is a correct explanation for assertion (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true 103. Assertion: – In pyrolysis of alkane lower alkenes and alkanes are formed Reason: – Disproportionation of free radial takes place in this reaction (a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is False (d) Assertion is false, Reason is true 104. Assertion: – Methyl isonitrile can not be reduced to Dimethyl amine by Sabatier senderson reduction Reason: – Sabatier senderson reduction.is a kind of syn – molecular addition while above reduction requires nascent H (a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true Na Ether
105. Assertion :– (C6H5)3CCl + chlorobenzene (C6H5)4C → (C6H5)4C The product (C6H5)4C does not form in this reaction Reason: – In chloro benzene resonance occurs and C – Cl bond acquires double bond character which is difficult to break. (a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion. (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true 106. Assertion: – Methane has largest Calorific value among all hydrocarbon nevertheless it is not used in L.P.G. Reason: – When it comes out from the cylinder it does not convert it self in to gas and remains in liquid state (a) Assertion is true, Reason is true: Reason is a correct explanation for assertion (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true
186
Problems in Organic Chemistry
107. Assertion: – Hyrdrogenation of compound (given below) by Pd / H2 (1mol) gives an optically inactive compound. Me
H H
H Reason: – It is a kind of anti addition (a) Assertion is true, Reason is true: Reason is a correct explanation for assertion (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true
108. Assertion: – n – Butane on mono chlorination gives three products which on distillation gives 2 products, Reason: – On distillation association takes place (a) Assertion is true, Reason is true: Reason is a correct explanation for assertion (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true 109. Assertion: – Addition of bromine on fumaric acid yield enantiomers. Reason: – Bromination of alkene is an example of electrophilic addition. (a) Assertion is True, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true OH
(+ )
H /∆ → A (Product)
110. Assertion : –
(A) does not give addition with Cl2 / CCl4 (in dark)
Reason: – A contains phenyl group which does not show addition with Cl2 in dark. (a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true
Answer Key 1. (a)
2. (d)
3. (b)
4. (b)
5. (d)
6. (c)
7. (d)
8. (d)
9. (d)
10. (d)
11. (b)
12. (c)
13. (d)
14. (d)
15. (a)
16. (b)
17. (a)
18. (c)
19. (b)
20. (c)
21. (c)
22. (d)
23. (d)
24. (b)
25. (b)
26. (c)
27. (b)
28. (c)
29. (c)
30. (a)
31. (b)
32. (d)
33. (c)
34. (d)
35. (b)
36. (d)
37. (c)
38. (d)
39. (a)
40. (d)
41. (a)
42. (d)
43. (a)
44. (d)
45. (d)
46. (b)
47. (d)
48. (c)
49. (d)
50. (c)
51. (a)
52. (a)
53. (c)
54. (c)
55. (c)
56. (a)
57. (d)
58. (b)
59. (c)
60. (a)
61. (c)
62. (b)
63. (c)
64. (c)
65. (d)
66. (a)
67. (b)
68. (c)
69. (a)
70. (c)
71. (a)
72. (d)
73. (b)
74. (a)
75. (b)
76. (c)
77. (b)
78. (a)
79. (c)
80. (b)
81. (d)
82. (a)
83. (a)
84. (c)
85. (b)
86. (c)
87. (a)
88. (c)
89. (b)
90. (d)
91. (b)
92. (d)
93. (a)
94. (a)
95. (c)
96. (a)
97. (a)
98. (b)
99. (b)
100. (c)
101. (a)
102. (c)
103. (a)
104. (a)
105. (b)
106. (a)
107. (c)
108. (c)
109. (d)
110. (a)
187
Hydrocarbon (Alkane, Alkene & Alkyne)
Multiple Choice Questions (More Than One May Connect) 1. Which among the following compound can produce two moles of ethane when 1 mol of the compound reacts with excess of EtMgBr? OH (a) CH3CH = CH2 (b) (c) (d) Buta -1,3-diene OH 2. Maleic acid on neutralization by NaOH followed by electrolysis gives ‘X’, ‘X’ on reaction with: (a) Cl2/H2O can produce dichloro ethanal (b) MnO4(–) / 0°C can produce glyoxal (c) H2O/H(+) can produce ethanol (d) O3 followed by hydrolysis can produce acetic acid 3.
Which statement is true about addition reactions? (a) Unsymmetrical alkenes on reaction with HCl give markownikoff product in presence of sun light (b) C2H5SH shows KHARASCH effect in presence of active peroxide (c) When addition reaction occurs via the formation of free radicals, antimarkownikoff product is obtained as major product (d) C C bonds on addition reaction with I2 can not produce vicinal di iodide
4. Which among the following reagent is not suitable for the following transformation? H |
CH3 — C ≡ C — CH3 → CH3 — C = C— CH3 |
H
(a) Na/liquid NH3
(b) P - 2 Catalyst
(c) Pd / C
(d) H2 / Ni
OH CH3
R
5.
H
R may not be:-
→ H
CH3 OH
(a) HCOOOH / H3O(+)
(b) OsO4 / H2O2
(c) Cold MnO4(–) (d) CH3COOH / H3O(+)
→A + B 6. Buta1, 3 diene + HBr (1 mol)
A & B are primary and secondary allyl bromides. Select the true statement/s:(a) A is thermodynamically controlled product (b) B is thermodynamically controlled product (c) A is kinetically controlled product (d) B is kinetically controlled product
7. In which case 1st compound has higher value of heat of hydrogenation than 2nd compound. (a)
(c)
,
(b)
, CH3CH = CH—CH3
(d) None of these
8. In which of the following case optically active compound is produced. CH3
CH3
|
→ Product (a) C2 H5 —C H —= CH CH 2 + HBr CH3
|
(c) CH3 — CH — CH =
|
2 → Product (b) CH3 — CH — CH = CH 2
Hg(OAc) NaBH 4
CH3 Cl2 → Product CH 2 in CCl4
|
dil H 2SO4 (d) C2 H5 —CH — CH = CH 2 → Product
188
Problems in Organic Chemistry
9. Which of the following can show Diels - Alder reaction? (a) O
(b)
(c)
(d)
10. In which case meso isomer is produced. in CCl4 + Cl2 → Product
(a)
H |
KMnO4 → Product 0°C
(c)
KMnO
4 → Product 0°C
(b)
Cl
2 → (d) Ph — C = C— Ph Product in CCl 4
|
H
Answer Key 1. (b), (c)
2. (a), (b)
3. (a), (b), (c), (d)
4. (b), (c), (d)
5. (b), (c)
6. (a), (d)
7. (d)
8. (b), (c)
9. (a), (b), (c), (d)
10. (c), (d)
LEVEL - II Multiple Choice Questions 1. Predict the product of the following reaction. dil H SO
2 4→ Product
OH
(a)
OH
(b) (c) (d) HO OH
2. Which intermediate will not involve in the following reaction. H( + )
2mec = CH 2 → |
Me
Ph
Me
Me
PH
CH2
Ph
Me2 C (+ ) (+ ) (+ ) | (a) (b) (c) (d) Me2 C — C —CH 2 Me2CPh MeC — CH 2 — CMe + |
|
Ph
Me
|
Ph
|
C–Me Ph
Ph
3. (1)
(2)
(3)
Arrange the following compounds in decreasing order of their calorific values. (a) 1 > 2 > 3 > 4 (b) 4 > 3 > 2 > 1 (c) 4 > 2 > 1 > 3
(4)
(d) 4 > 2 > 3 > 1
189
Hydrocarbon (Alkane, Alkene & Alkyne)
4.
Oxygen is free radical scavenger or eater because. (a) It is paramagnetic in nature (b) Bond dissociation energy of O = O bond is small (c) Free radicals form stable paramagnetic compounds with oxygen (d) It decreases velocity of free radicals
5. Which is not correct about wurtz reaction? (a) Rate of reaction is maximum in 1° alkylhalide & least in 3° alkyl halide because in 3° alkyl halide chances of disproportionation is more (b) Dry ether is used so that Na does not react with it (c) By the addition of O2 reaction does not stop but slows down (d) All are correct 6. Select the reaction with wrong product Br
(a) Br
(c)
Br
Na dry Ether
→
Br Na THF
7. ICHDC2CHDI →
(a) (b) (c) (d)
Br
(b)
Br
Na dry Ether
→ (CH2)n
Na dry Ether
→
(d) All are correct
X, [X] is:-
An alkyne and can react with tollen’s reagent Optically active alkyne and can not react with tollen’s reagent Not optically active but can show geometrical isomerism
8. Select the reaction with wrong produced COOK Current
(a)
→ COOK
Current (b) (CH3)3CCOONa → (CH3 )3 CC(CH3 )3 100%
Current
(c) KOOC
COOK →
KOOC
(d)
Current
→ COOK
Current 9. CH3 — C H — CH 2 — COONa → |
[X],
[X] would be:-
Cl
(a) CH3 — CHClCH2CH2CHClCH3 (b) CH3 — C H — CH 2 Cl |
CH3 —CH—CH 2Cl Cl |
CH3 — C = CH 2 (c)
(d) CH3 — CH — CH3 |
Cl
190
Problems in Organic Chemistry
10. How many structural alkenes on hydrogenation can give following alkane?
(a) 11
(b) 19
(c) 10
(d) 12
11. Which will not decolourise bromine water? (a) (b)
(c)
(d) CF2 = CF2
Matrix Match Questions Each question contains statement given in two columns which have to be matched. Statement (A, B, C , D) in column I have to be matched with statements (p, q, r, s) in column II. The answers to these questions have to be appropriately bubbled as illustrated in the following examples. If the correct match are A – p, A – s, B – r, B – r, B – q, C – q, D – S, then the correctly bubbled 4 x 4 matrix should be as follows. A
p
q
r
s
B
p
q
r
s
C
p
q
r
s
D
p
q
r
s
12. Compounds (A) Mg2C3 (B) CaC2 (C) CH2CO (D) CH2 = CH— CH = NH 13. Column - I
Hybridization on Carbon (p) sp (q) sp2 (r) sp3 (s) sp3d Column - II
HOCl
→ Product (A) CH2 = CClCHO H O( + )
3 → Product (B) CH2 = CH — SiH3
Br hν
2 → Product (C) CH2 = CH — CH2 — CH3
CCl Br
3 → (D) CH2 = CH — CH3 Product Peroxide
14. List - I (Oxidation)
(p) optically active (q) Nucleophilic addition reaction (r) Electrophilic addition (s) Markownikoff addition List - II (Product)
O ,H O
3 2 (A) CH3COCH2CH = CHCH3 → heat
(p) CH3CHO
hot KMnO4 (B) (CH3)2C = CHCH3 →
(q) CH3COOH
O ,Zn,H O heat
3 2 → (C) (CH3)2C= CHCH3
O ,Zn
3 → (D) (CH3)2C = CHCH2COCH3 H O, heat 2
(r) CH3COCH3 (s) CH3COCH2CHO
191
Hydrocarbon (Alkane, Alkene & Alkyne)
15. Column - I Column - II (A)
(p) Na / liq NH3 / Paraftin, heat
(B)
(C)
(D)
(q) Solid KOH (r) Na / liqNH3 / CH3OH (s) Na / liq NH3
Answer Key 1. (a)
2. (b)
3. (c)
4. (a)
5. (d)
6. (d)
7. (d)
8. (b)
9. (b)
11. (d) Answers Matrix Match 1 2. (A) - p, r (B) - p (C) - p, q (D) - q 14. (A) - q, r (B) - q, r (C) - r, p (D) - s, r
13. (A) - q, (B) - p, r, s (C) - p, (D) - p 15. (A) - p, q, (B) - r, (C) - s, (D) - r
SOLUTION Br KMnO4
NBS
1. (a)
Br (–)
Kolbey’s COO(–) CH2=CH—Br electrolysis COO
Tautomerism
CH2CHO
CH2=CH—OH
KOH 623 K, 200 atm
2. (a) [X] = Ph - C º C - CH3 and last product is Ph - CO - CH 2 CH3 Et-Br
® CH3CH 2C º C - Et 3. (c) CH3CH 2C º CH + NaNH 2 ® CH3CH 2C º CNa ¾¾¾¾
(meso isomer) 4. (b) On dehydrobromination it will produce cyclobutadiene which is less stable because it is antiaromatic. Cl Na / Et O
2 →
5. (d) Br
Wurtz reaction
RBr is more reactive than RCl towards Grignard’s reagent formation
10. (a)
192
Problems in Organic Chemistry
Cl
Cl
Cl
Br
Mg → Et 2O
H 2O → –MgBrOH MgBr Wurtz reaction
Wurtz reaction
Cl
Cl
| | Zn Zn / ∆ 6. (c) 2CH3CHCl2 → CH3 — CH — CH — CH3 → CH3CH = CH — CH3 Ether
7. (d) OsO4 gives syn hydroxylation. cis alkene on syn addition produces meso isomer; options (a) (b) & (c) all are meso isomers hence all are correct. O O
8. (d) CH3—C C—CH3
O3 H2O
H2O2
CH3—C—C—CH3
Tautomerism
2CH3COOH O O CH3—C—C=CH2
9. (d) Halogenations of alkene is anti addition. cis alkene on anti addition forms mixture of enantiomers i.e. racemization takes place hence (d) is not stereo specific. Note: - See question no. 5. topic electrophilic and free radical addition in chapter - 4 10. (d) In the atmosphere of N2 triplet carbene is formed. (see question no. 10 topic reaction intermediate in chapter -04 ) 11. (b) Less substituted alkene is less stable and hence undergoes reduction easily. Benzene ring does not undergo reduction because after reduction it loses its Aromaticity (+) (+)
(+)
ring opening
H
12. (c)
13. (d)
H2O
OH
• RCOOOCOR → 2RCO O •
•
•
•
RCO O → R + CO 2 R + CHCl3 → RH + C Cl3 CCl3
CCl3 RH –R•
•
+ C Cl3
•
•
CCl3 or
CCl3
14. (d) HI solid does not undergo addition reaction with alkene. Br Br
15. (b) Br
Br
Br
Br
Br
Br
(+)
16. (b) CH2 = CH—CH= CH2 CH3—CH =CH—CH3
Na
(–) Na
CH2 = CH —CH—CH2
NH3 –NaNH2
(+) (–)Na
CH2—CH=CH—CH3
NH3 –NaNH2 Na
•
•
CH2=CH—CH—CH3
CH2—CH=CH—CH3
193
Hydrocarbon (Alkane, Alkene & Alkyne)
17. (a) O
ozonolysis O → CH3COCOCH2CHO + OHCCH2COCH2CH2CHO O
O O O
18. (c) Heat of hydrogenation ∝ Number of double bonds 19. (b) Because on dehydrohalogenation conjugated alkene will form which is stable due to resonance. Cl 20. (c) 3rd will undergo dehalogenation rapidly because after dehalogenation resonance stabilized alkene i.e. CH2 = CH — CH = CH — CH = CH2 forms. 1st will undergo dehalogenation slowly because it forms cyclo propane which is less stable due to ring strain. 2nd also gives alkene on dehalogenation but resonance stabilization of CH2 = CH — CH = CH2 is less in comparison to the alkene given by 3rd compound. Thus, rate of dehalogation is 3 > 2 > 1 1 2
21. (c)
3 4
22. (d) 1mol of gas at S.T.P. occupies 22400 ml volume. Moles of gas liberated = 44800 / 22400 = 2 Hence Y should have two active hydrogen’s glycol, succinic acid & ethylene disulphonic acid, all contain two active hydrogen’s Hence (d) is correct. hν
23. (d) CCl2 CO →
••
CO + C Cl2
Di chloro carbene
Dichloro carbene undergoes addition reaction with alkene. + CCl2
Cl
Cl 24. (b) Acid strength ∝ 1 / stability of conjugated base
25. (b)
To Reallens gen t
(A)
(B) Pd/C Partial reduction
CAg
(A) decolourises Br2 water which indicates presence of multiple bond. (A) reacts with Pd / C which indicates presence of triple bond & since (A) reacts with Tollen’s reagent hence triple bond must be present at the terminal. alc 26. (c) Ph — CH 2 Cl → KOH
(–)
Φ — CH — Cl
Stabilised by resonance
Cl (–)
ΦCHCl + Φ —CH2—Cl
(–)
Φ—CH2—CH—Φ
Cl
alc KOH Elemination
dicarboxylation Heat
→ CH3COCH3 + CO2 27. (b) CH3COCH 2COOH 28. (c)
COOH
CH—COOH Diels Alder
+ CH—COOH
COOH CHO 3
CHO
O3/H2O/Zn
NaOH CaO/∆
Pd/∆ –2H2
Φ—CH=CH—Φ
194
Problems in Organic Chemistry
29. (c) See question no. 27 topic - Electrophilic & free radical addition, Chaptor -04. 30. (a) 31. (d) This product is cumulated diene and can show geometrical isomerism H( + )
32. (d) CH 2 — (CH 2 )2 — CH 2 → CH 2 ∆ |
|
OH
= CH — CH = CH 2
OH (–)
COOH COOH
MnO4
D. Alder Reaction
+
33. (c) Trans alkenes do not undergo Diels Alder reaction.
Decarboxylation –2CO2
34. (d) CH
(–)(+)
C Na CH
NaNH2
CH
(–)(+) Br(CH ) Br CNa –NaBr2 4
CH
C(CH2)4Br
NaNH2
CH
OH 0°C MnO4
C—(CH2)4—Br
C Ni2–B
(CH2)4
(CH2)4 CH OH Y 35. (b) boiling point ∝ van-der-waal’s forces of attrations ∝ surface area
– NaBr
C X
36. (d)
37. (c) Bredt’s rule: - A bridged bicyclo compound can not have a double bond at bridged carbon untill ring contains at least 8 carbon atoms. 38. (d) In 1st reaction base is sterically hindered thus alkene will produce via elimination reaction In 2nd case alkene will produce due to iodination - deiodination. In 3rd & 4th cases dehydrobromination & debromination take place respectively. 39. (a) CH = CH CH = CH CH = CH
dimerisation
CH = CH
dimerisation
OHC
CHO
OHC
CHO
40. (d) CH3CH=CHD
NBS
•
CH2CH=CHD
O3/Zn/H2O •
CH2=CHCHD HBr
HBr
BrCH2CH=CHD
Br
41. (a)
H
H
H E2 anti elimination
Me
Me
H H D
CH2=CHCHDBr
195
Hydrocarbon (Alkane, Alkene & Alkyne)
42. (d) Trans alkene will produce via anti elimination reaction. 43. (d) NaNH
2 44. (d) CH2 = CH — CH = CH2(A) → CH3 — CH = CH — CH3 (But-2-ene) For this conversion See question no. 16. This but-2-ene undergoes Diels alder with A to produce o-xylene i .e. o-methyl toluene o - xylene Pd/∆ –2H2
+
o-xylene
(A)
OH
O (+)
45. (d) R—CH= CH—R
PBA
R—CH—CH—R
H3O
R—CH—CH—R
OH 46. (b) Same as question no. 41 47. (d) Cyclo butane has more strain in its ring hence it is more reactive than cyclo hexane. (a) is incorrect because cis form will have greater heat of hydrogenation than trans form. (b) is incorrect because former alkene is more stable than later because of resonance. (c) is incorrect because benzene is more stable due to Aromaticity. 48. (c) 49. (d) O3 H2O/Zn/∆
50. (c)
CHO CHO (+)
(+)
(+) H2O
H
OH
(+)
H /∆
(+)
O Ozonolysis
O
51. (a) Because (a) gives less stable carbocation (F3C)2
(+)
C—CH3 (unstable due to strong -I effect of CF3)
52. (a) Melting point ∝ closed packing Out of cis and trans forms later has higher mp than former because molecules of trans isomer are closely packed. ⇒ Trans - but-2-ene
⇒ Cis-but-2-ene (poor packing)
Thus, option (b) & (d) are wrong. The para substituted benzene has linear geometry & such molecules are closely packed in their solids state. Cl Cl Cl
⇒ Closely packed
Cl Cl
Cl
Cl Cl
Cl
⇒ Poor packing Cl
Hence (a) is correct
Cl
Cl
196
Problems in Organic Chemistry
53. (c) As it has high molecular weight 54. (c) PhCHOHC
(+)
(+)
H –H2O
C
PhCHC
PhCH=CH CHO
(+)
PhCH=C=CH
CH
H2O
PhCH=C=CH OH
55. (c) Bromination of alkene is a kind of anti addition. So bromine atom should acquire axial position (c) option is incorrect because in it Br atoms acquires axial & equitoral positions i.e. syn addition Ha 1
Note:-
H e
2
H e
Ha
At 1, 2 position (a, a) → Trans, (e, e) → Trans
(a, e) → Cis, (e, a) → Cis
56. (a) Although in chlorination of alkene anti addition takes place yet in the given reaction syn addition occurs due to steric hindrance. (–)
Cl
H
Cl
(+)
Cl
Cl H
Cl
2→
steric hindrance →
57. (d) If product of Diels -Alder reaction is heated ‘C’ chain forming bridge in bicyclo compound comes out from compound in the form of alkene.
CO2 R
COOR
Me
Me
(+)
H
+
Me
CO2R + CH2 = CH 2 CO2 R
COOR CO2 R
58. (b) Hoffman elimination followed by Diels - Alder reaction See question no. 14, topic Elimination reaction, chapter - 04. 59. (d) 60. (b) 61. (c) Wurtz reaction is carried out by Na which is very sensitive towards active ‘H’. Except (c) rest all compounds contain active hydrogen (acidic ‘H’) so (a), (b) & (d) will remove hydrogen on reaction with Na metal.
ONa OH +
Na
1/2 H2 +
Br
Br (–) ( + )
CH ≡ C — CH 2 Cl + Na → ClCH 2 — C ≡ C Na + 1/ 2H 2
OH + Cl
ONa + 1/2 H2
Na Cl
197
Hydrocarbon (Alkane, Alkene & Alkyne)
62. (b)
CH3
CH3
CH3
H2/Ni
CH3
Cl
Cl2/hv
+
CH3
CH3
(A)
CH3
CH3
CH3
CH3 CH3
CH2Cl
63. (d)
NH2 64. (c)
NH2
NH2 H2/Ni
Birch reduction
(A) (B) (C) (c) is highly basic because lone pair of electron present on ‘N’ are localized and does not take part in resonance. (A) & (C) both have delocalized lone pair of electron but (A) is less basic than (B) because of more delocation of electron due to the presence of three double bonds. 65. (d) CH3OCH2Cl can dissociate by two ways
(–) ( + )
First way:- CH OCH Cl → CH3 O + C H 2 Cl 3 2 (+)
••
CH—Cl 2
••
(+)
••
CH2 = Cl
(+ )
(–)
Second way:- CH — OCH Cl ←→ Cl + CH O CH 3 2 3 2 •• ( + )
(+ )
CH3 — O CH2 ←→ CH3 — O ••
= CH 2
(B)
Resonance stabilization of (B) is greater than (A) because 3p of Cl will not overlap with 2p of carbon effectively (+ )
(+ )
Cl(–)
PhCH = CH 2 + CH 2 OMe → Ph C HCH 2 CH 2 OMe → PhChClCH 2 CH 2 OMe (Styrene)
66. (a) (A) is wrong because in oxymercuration, Markownikoff rule is followed. 67. (b) Wurtz reaction followed by dehydrogenation. 68. (d)
Passage - I (69 to 72) (+) (+)
LAH A
Cl
H
(+)
B H2 O
LAH
O
(+)
O (+)
H ∆
ozonolysis
A Cl
D
OH C
198
Problems in Organic Chemistry
Passage - II (73 to 73.9)
Since A is terminal alkyne hence it can react with Tollen’s reagent.
Passage - III (80 to 84) Reaction 1st is hydration reaction. Here carbocation is formed as an intermediate which may show rearrangement as follows:(+)
(+)
HO 2
OH (+) Reaction 2nd is hydroboration. Here antimarkownikoff addition of water takes place. hydroboration
OH rd Reaction 3 is oxymercuration where markownikoff addition (without rearrangement) takes place. oxymercuration
OH
Passage - IV (85 to 90) 48 ml vapours has wt. = 1.49 g 22400ml vapours will have wt. = 1.49 x 22400 / 448 = 74.5 g Thus, molecular wt of A = 74.5g A os a chloro compound hence it should have Cl atom (at. wt. = 35.5 g) Therefore remaining carbon chain will have wt. = 74.5 - 75.5 = 39 g Thus, compound A should be 3- chloro propyne. Cl NaNH (+) Cl A 2 Na (–) A B C Cl
199
Hydrocarbon (Alkane, Alkene & Alkyne)
A can not show Wurtz reaction because it contains acidic hydrogen which can liberate hydrogen gas with sodium metal. A & C both can exhibit substitution reaction with Tollen’s reagent as well as with ammonical cuprous chloride as both are terminal alkynes.
Passage - V (91 to 93)
C2 H 2(2 moles) A
∆ Fe
B
A
B
∆
∆
C
ozonolysis
D
2 moles of CHO–CHO CH 2CH3 |
94. (a) Optically active alkane with least mol. wt. (100)
CH3 — C H — CH 2 CH 2 CH3 CH 2 ||
Least molecular wt. (98) can not show geometrical Isomerism but becomes optically active after reduction
CH3CH 2 — C — CH 2 CH 2 CH3 CH3 |
CH 2 = CH 2 — C H — CH 2 CH3
alkene with least mol.wt(84). & optically active 95. (c) In compound A both double bonds are anti to each other hence it can’t show Diels- Alder reaction H H Pd/C/H 2
syn addition
(C) 96. (a) Heat of combustion ∝ 1 / stability of compound 97. (a) When acid reacts with OH group, A is formed which undergoes hydride shift to produce more stable carbocation B (+)
(+)
CH2CCH 2 CH3
CH2CHCH NH2
CH2
CH3
A
B
CHO Thus due to the formation of B, product (a) is obtained.
O
98. (b)
CH3 MgBr
CH3MgBr
A Br
O
Br OH Br2 water
aq KOH
B O
Br OH
OH K2CO3
Cl2 water
Cl PAA
CHO
(–)
OH
D O
NH2
CH2
C OH
200
Problems in Organic Chemistry
Br (+)
99. (b)
Br2 (–)
(–)
Br
Cl
Br
Cl Br
Br
100. (c) Oxymercuration
OH
101. (b) 102. (c) Propane can not exhibit isomerisation reaction because for chain isomerism molecule should have at least 4-carbon atoms. isomerisation
isomerisation no isomer will form 103. (a) Disproportionation of free radial produces alkane & alkene
LIAlH
4 → CH NHCH 104. (a) CH3 NC 3 3 This reduction is not possible by H2 / Ni because in this reduction unsymmetrical addition of hydrogen occurs on triple bond for which nascent hydrogen is required. 105. (b) The product Ph4C is sterically hindered thus, difficult to form. 106. (a) When CH4 comes out from the cylinder it does not convert it self in to gas and remains in liquid state. 107. (c) It is a kind of syn addition & after reduction moslecule becomes achiral because of absence of chiral carbon atom.
108. (c)
H
H
Me
Me Pd/H 2
H
H H
H mono chlorination
H
H
+ A
B
Cl
Cl optically active
distillation
B
d & l isomers
COOH
H COOH 109. (d) HOOC
Br2 anti addition
Br
COOH
HOOC
H
H
Br
H
Br
OR
Br COOH 110. (a) A is ethyl benzene & benzene ring does not undergo addition reaction easily. MORE THAN ONE MAY CORRECT 1. 2. 4. 5.
(b, c) hydrogen of [= C - H ] is not acidic in nature (a, b) X is acetylene 3. (a, b, c, d) (b, c, d) only sodamide provides anti addition (b, c, d) only performic acid [HCOOH] provides anti hydroxylation
201
Hydrocarbon (Alkane, Alkene & Alkyne)
6.
(a, d)
+
+ HBr
Br
7. 8. 9.
[A]
Br [B]
(d) Heat of hydrogenation ∝ 1 / Stability (b, c) In (a) 3 - bromo - 3 - methyl pentane is formed while in (d) 3 - methyl pentan - 3 - ol is formed & both are optically inactive as they do not contain chiral carbon atoms. (a, b, c, d 10. c, d
LEVEL - II
(+) (+)
(+)
1. (a)
2. (b)
(+)
(+)
H
MeC =CH 2
HO
H2 O
H
MeC — CH3
Ph (A)
Ph Me
Me
Me—C—CH2
Me—C—CH2
+
C—Me
C—Me
(+)
(+)
Ph
Ph
Me Me
Ph Me
(+)
–H
3. (c) Calorific value ∝ heat of combustion ∝ Molwt ∝ Van-der-waal forces of attractions 4. (a) Free radicals and O2 both are paramagnetic in nature because of presence of unpaired electron thus they combine to form diamagnetic species. 5. (d) See wurtz reaction. 6. (d) In wurtz reaction if more than 6 C atoms are present & halogens are present at terminals then polymer is formed other wise cylo alkanes are formed. Br
7. (d) I—CHD—C
Br
Na dry Ether
C—CHD—I
Na
(CH2)n •
CHD—C
H
H C=C = C D
Current 8. (b) (CH3)3CCOONa →
•
C—CHD
D
disproportionation
+
202
Problems in Organic Chemistry
9. (b) If α –carbon possesses halogen then rearrangement in free radical takes place.
CH3—CH—CH2COONa
Current
•
CH3—CHCH2 COO Cl
Cl dimerisation
•
CH3CHCH2Cl
CH3—CH—CH 2
10. (a) Double bond can be placed at numbered positions
Cl
9 2 1
3 4
5
10 11
6
7 8
11. (d) Due to strong -I effect of fluorine CF2 = CF2 becomes electron deficient thus, can not undergo electrophilic addition reaction with bromine water.
7
Alkyl Halide & Grignard’s Reagents Main Features ALKYL HALIDE & GRIGNARD’S REAGENT Preparation
Properties LAH LAH
X2
alkene or alkyne
Moist Ag2O
HX Red P + X2
ROH
PCl5/PCl3/SOCl2 Anh ZnCl2 + HCl
ROH (2°, 3° alcohols) RX + NaI (for alkyl iodides) RX + AgF (for alkyl fluoride) RCOOAg + Br2 (for alkyl bromides) alkene ( For allylic halides) alkane
alkane [For 1°& 2°halides] Ether
inCCl4
alkene or alkyne
ROH
Ag2O
alkene [For 3° halides]
Acetone
heat NBS
X2 hv
A L K Y L H A L I D E
AgCN NaCN Na / Pb alc KOH aq KOH Mg / Ether
Alcohol RNC RCN [For 1°& 2°halides] Et4Pb Alkene Alcohol RMgX
Na / Ether
alkane
Zn / Ether
alkane
(+)
Zn / H
AgNO2
alkane RNO2
NaNO2
RONO [For 1°& 2°halides]
RONa
ROR [For 1°& 2°halides]
204
Problems in Organic Chemistry
Multiple Choice Questions 1.
Consider the following reaction:– → Br (–) + CH3OH (1) CH3Br (0.1M) + OH (–)(0.2M) → Ph3COMe + Br (–) (2) Ph3CBr (0.1M) + CH3OH (0.2M) → CH3CH2OH + Br (–) (3) CH3CH2Br (0.1M) + OH (–)(0.4M) (–) → Ph3COH + Br (–) (4) Ph3 CBr (0.1M) + OH (0.3M)
If R1, R2, R3 and R4 are the rates of the reactions 1, 2, 3 & 4 respectively then:– (a) R2 = R1 (b) R3 = R4 (c) R2 = R4 (i) Mg/Ether
Ag/∆ → (A) → 2. CH3Cl (+ ) (ii) HCHO/H3O (iii) NaOCl
(B) 2 mole
(d) R1 ≠ R2 ≠ R3 ≠ R4
(B)
Fe/∆ → (E) → (C) ∆
The end product E is:–
(a)
(b)
(c)
(d)
3. Here four reactions are given below for the preparation of alkyl bromide. HBr in CCl4
HBr in CCl4
1. CH2 = CH2 → EtBr
3. CH3CH = CHCH3 → MeCHBrEt
2. CH3 — CH = CH2 → i — PrBr HBr
HBr in CCl
→ Me2CBrCH3 4. (CH3)2C = CH2 in CCl 4
If these reactions have rates R1, R2, R3 & R4 respectively then correct order of rates of reaction will be:– (a) R1 < R3 < R2 < R4 (b) R1 < R2 < R3 = R4 (c) R1 < R2 < R4 < R3 (d) R1 < R4 = R3 < R2
4. In the following reaction the most probable product is :Cl H
CH3 H
CH3
NaOH
→ in DMSO
n-Pr CH3
OH H CH3 (b) CH3 H
H (a) CH3
CH3
H CH3 (d) n-Pr H
OH
H (c) H CH3
n-Pr
n-Pr
OH
n-Pr
CH3 H
OH
5. Which reaction is the example of entertainment technique? RX
(a) (X = F,Br,I)
CH Cl Ether
3 → RMgX + Mg
CH F
3 → RMgX RX + Mg Ether (X = Cl,Br,I) (b)
CH I
CH3Br 3 RX + Mg → RMgX RX + Mg → RMgX Ether Ether (X F,Cl,Br) (c) (d) = (X = Cl,F,I)
6. Correct order of boiling points of the following halides would be:– CH3Cl, CH2Cl2, CHCl3, CCl4 (1) (2) (3) (4) (a) 1 > 2 > 3 > 4 (b) 3 > 2 > 1 > 4 (c) 4 > 3 > 2 > 1 (d) 2 > 3 > 1 > 4
205
Alkyl Halide & Grignard’s Reagents
OH lucas reagent
→ Product is:–
7.
CH3
Cl Cl
(a)
(b) (c)
(d)
CH3
CH3
8. Which is thermodynamically controlled product for the following reaction?
O + CH3MgCl
H3O
(+)
Product is :(Major)
OH
O
CH3 (a)
(b)
(c) CH3
(d) All are possible
9. Which among the following will have highest boiling point:– (a) CH3CH2F (b) CH3CHF2 (c) CH3CF3 10. Correct sequence of dipole moment is:– (a) CH3F > CH3Cl > CH3Br > CH3I (c) CH3F > CH3I > CH3Br > CH3Cl 11. CH3COOAg
(b) CH3I > CH3Br > CH3Cl > CH3F (d) CH3Cl > CH3F > CH3Br > CH3I
∆
Br2 → AgBr + CO2 + CH3Br
This reaction is called Borodine Hunsdecker reaction. If in this reaction Br2 is replaced by I2 then the product will be:– (a) CH3COI (b) CH3I (c) CH3COOCH3 (d) No reaction
12. CH3F
(–)
OH
(i) NaI (ii) OH
+
(d) CF3CF3
(–)
CH3OH ...................(A) CH3OH ...................(B)
If R1 & R2 are the rates of reaction A & B respectively then (a) R1 = R2 (b) R1 > R2
(c) R1 < R2
(d) R1 R2
13. Product and mechanism of the following reaction would be :H Br Finkelstein reaction
→
t-Bu
(a)
H H
H
H
H , SN1 (b)
I , SN1
I
(c)
H, SN2
I
(d) t-Bu
I
t-Bu
t-Bu
H H
H t-Bu
, SN2
206
Problems in Organic Chemistry
14. Which among the following will not show haloform test:– (a) CH3COCH2COOC2H5 (b) (CH3)3COH
(c) 2 – Butanol
15. List the following alkyl halide in decreasing order of SN2 reactivity:– Me Me
(d) Both (a) & (b)
Me Cl
Cl
(1)
Pr (3)
(2)
Cl
Pr (4)
I
D (5)
Cl
(a) 3 > 2 > 4 > 5 > 1
(b) 2 >3 > 5 > 1 > 4
(c) 5 > 3 > 2 > 1 > 4
(d) 5 > 3 > 2 > 4 > 1
16. Which among the following will not produce ethane when treated with ethyl magnesium halide:– (a) CH3SH (b) Propyne (c) C2H5CH = CH2 (d) C2H5OH 17. Identify the product of following reaction. Br
+ AgNO3 + HO2
Br
Product
OH O
(a)
(b)
OH
(c)
Br
NO3
O (d) OH
Ac
18.
NaOCl
A + B electric current C (ppt)
The compound (C) is:–
(a)
& (b) (c)
(d) both a(major) & c(minor)
19. tert – butyl chloride on treatment with lithium aluminium hydride will provide:– (a) Iso butane (b) Iso butene (c) 2, 2, 3, 3 – Tetramethyl butane (d) no reaction takes place CH3 |
Hoffman
→ (B) 20. CH3 — CH —CH — CH3 E limination |
Cl
CO2 H3O(+) C2H5OH
(D)
Correct order of rate of SN1 for A, C & D will be:– (a) A > C > D (b) C > D > A
21. RMgX (A)
(B)
(C)
HBr/hv
(A)
HBr
(A)
(c) A = C > D
(d) C > A > D
(C) + other product
(E) + other products
Which statement is correct;– (a) (C) is an ester (c) (C) & (E) are same compounds
(b) (E) & (B) are same compounds (d) (E) is ether
207
Alkyl Halide & Grignard’s Reagents
22. In Finkelstein reaction when Acetone is replaced by water then
(a) Reaction occurs in forward direction via SN1 pathway
(b) Reaction occurs in forward direction via SN2 pathway
(c) Reaction occurs in backward direction because sodium halide formed in right hand side is soluble in water & can not ppt out
(d) Reaction occurs in backward direction because sodium halide formed in right hand side is insoluble in water
(i) NBS (A)/ ∆ 23. (A) →(B) →
(B) will be
(ii) alc KOH
(a)
(b)
(c)
(d)
HBr(I − mole)
24.
→ A Product A will be :
Br Br
Br
(a)
(b)
(c)
(d) both a & c
Br
25.
Mg
PhOH
→[X] →[Y] THF The product [Y] of this reaction is O
(a)
(b)
(c)
OH
(d)
26. Which of the following reaction is least likely to occur? Moist Ag 2O
(a) EtBr → EtOH
→ EtNO2 (c) EtBr
Sod.Nitrite
(b) EtBr → EtNO2
Silver nitrite
Pb/ Na
(d) EtBr → Et4Pb
(i) B H /OH(–)
2 6 → A, Product A is :– Me (ii)SOCl
27.
2
H (a) H
Me Cl
(b)
Me
Me (c) H
(d)
Me Cl
H
Cl
Cl
Cl H
208
Problems in Organic Chemistry
28. What will be the product of given reaction?
CH3 H
OH HBr
Br
→ Product
H CH3
CH3
CH3
H (a) H
Br Br
(b)
H
OH
H
Br
CH3
CH3
CH3
(c)
H (d) Br H
H Br
CH3
OH
Br
H CH3
CH3
29. 0.037 gm of alcohol was added to CH3MgBr and the gas evolved has volume equal to 11.2 cm3 at S.T.P. This alcohol gives haloform test ,thus, alcohol would be:–
(a) CH3CH2OH (b) CH3 — CH — CH3 |
(c) iso butyl alcohol
(d) n — Propyl alcohol
OH 30. Which reaction results in the formation of a pair of enantiomers? Br
H H O
2 →
(a)
CH OH
3 →
(b) H
(c)
Br
NaOH
→ in DMSO
Br NaCN
→ in DMF
(d) H
Br
CH3 CH3
→ H2O + RBr 31. ROH + HBr
In which case first alcohol is more reactive towards HBr than second one:OH
(a)
and
OH
CH2OH and
(b)
OH OH
O
(c) and Me 2 C – CH 2 CH3 and CH3 – CH – Et (d) |
OH
|
OH
OH
32. Consider the following compounds NH2CHBrCH3 AcNHCHBrCH3 NH2CHBrNH2 NH2CHBrOH (1) (2) (3) (4)
If K1, K2, K3 & K4 are the rate constant for the solvolysis of 1, 2, 3 & 4 respectively then (a) K1 > K2 > K4 > K3 (b) K4 > K3 > K1 > K2 (c) K3 > K4 > K1 > K2
33. CHCl3 can be converted in to methyl chloride by using:– (a) Zn / H2O (b) Zn / HCl (aq)
(c) Zn / HCl (alc)
(d) K3 > K4 > K2 > K1
(d) H2 / Ni
209
Alkyl Halide & Grignard’s Reagents
34. Match the following:–
Reaction
End Product current
(A) CHCl3 + KOH(Excess) → X →
(1) Mesitylene
(i) alcKOH(ii) NaNH (iii) Butadiene/∆
2 (B) Ethylene di chloride →
(2) Hydrogen gas
Na/liq NH3 (C) Bromo benzene →
(3) Cyclohexene
2→ (D) 1,1–di chloro propane
EtOH
(i)alc KOH & NaNH (ii) Fe/heat
(4) 1–Bromo cyclo hexa – 1, 4 – diene
→ 2; B → 4 ;C → 1; D → 3 (a) A
→ 2; B → 3; C → 1; D →1 (b) A
→ 2 ; B → ;3 C → 4; D → 1 (c) A
→ 1 ;B → 2 ; C → 1; D → 4 (d) A
35. Consider the following E1/SN1 reaction Br
CH3
H
H O + C H OH
2 2 5 → Products
∆
CH3
Which product is not possible? CH3
OH
(a) H
OH
CH3 (b) H
(c) H
CH3
CH3
(d) H
CH3
CH3
36. Identify the reaction in which benzene does not form as an end product:–
(i)Ag/ ∆ (a) Acetone →(A) ↓ → (–)
(c) 1, 3 – Butadiene + ethene →
OI
37.
(a)
(i)H O
2 (b) CaC2 → (ii)Fe/ ∆
(ii)Fe/ ∆
(i) ∆ (ii)Pd/ ∆
(d) In all cases benzene is formed.
Moist Ag O
CH2NH2
NOCl 2 →(P), → Product P is :– (+ ) H
/∆
(b) CH—O 2 2
(c)
(d)
210
Problems in Organic Chemistry
COCH3 (–)
OI → ppt + (A), (A) will be :–
38.
COCH2I (–)
(–)
COO
(–)
COCH3
COO
COO (a)
(b)
(c) (d)
ICH2OC
(–)
COO H NaC AgC N
39. CH3—CH—CH2Cl
CH3
COCH2 OI
COCH2 OI
(A)
(A) & (B) would be : (Major)
(B) CH3
CH3
(a) CH3 —C—CN & CH3 —C—NC (b) CH3 —CH—CH2CN & CH3—CH—CH2NC
CH3
CH3
CH3
CH3
CH3
CH3
CH3
(c) CH3 —C—CN & CH3—CH—CH2NC (d) CH3—CH—CH2CN & CH3—C—NC CH3 CH3 CH3
40. When Ethylene dichloride is treated with alc KOH followed by NaNH2 (A) is formed. (A) on treatment with tertiary butyl chloride forms (B) which on reduction with H2 / Ni gives (C), (C) will be:– (a)
(b)
(c)
(d)
H H
H Br
41.
H (a) Ph
CH3ONa → (A) + (B) , (A) & (B) are :– in CH3OH
H C =C
H
Major Minor
D &
Ph
Ph H Ph (b) & C =C H H H D
H
H C =C
C =C
H
Ph Ph H H Ph H H Ph C =C C =C C =C (c) C = C & (d) & H H D D H H H H 42. An alkyl Magnesium bromide on reaction with ethyl alcohol forms a hydrocarbon ‘X’ which occupies 0.2872 lit per gm at S.T.P. Grignard reagent will be :– (a) C2H5MgBr (b) PhMgBr (c) CH2 = CHCH2MgBr (d) (CH3)2CHMgBr 43. Which graph is incorrect?
→ Product For Ph3CBr + OH(–)
(a) Rate
Base
211
Alkyl Halide & Grignard’s Reagents
(–)
Rate (b)
For (CH3)2CHBr + O H (in acetone) → Product
Base
→ Product For PhCHDBr + EtOH
(c) ∆G
Reaction progress
→ Product For Me2(Et) COTs + NH3
(d) ∆G
Reaction progress 44. Correct order of rates of dehydrohalogenation of following halides is:–
Cl
(I)
(II)
CH3 — CH2 — CH2Cl
(III) CH3 — CH — CH 2 — CH3 (IV) CH2 = CH — CH2 — CH2Cl |
Cl
(a) III > II > IV > I
(b) I > IV > III > II
(c) III > IV > II > I
(d) I > III > IV > II
45. At 25°C chloro benzene can be converted in to acetophenone by treating it with:– (a) KCN & CH3MgBr / H3O(+) (b) CH3COCl / Na / Ether (+) (c) Mg / Ether & CH3CHO / H3O , mild oxidation (d) All of these
Hint–2° alcohols on mild oxidation produce ketone
46. 1 – Bromo – 4 – chloro butane is treated with Mg (1 mol) / Ether to give (A).Which among the following represents A ? MgCl
MgBr
(a) Cyclo butane
(b) Cl
(c) Br
(d)
47. 2 – Bromo – 3 – chloro butane on Finkelstein reaction will provide. (a) 2 – chloro – 3 iodo – butane (b) 1, 2 – Di methyl ethylene (c) Di methyl acetylene (d) 2, 3 – Di iodo butane 48. ∆H1 , ∆H 2 , ∆H3 & ∆H 4 are the enthalpies of dissociation of C–X bond of the following four halides:–
Cl
→
I
→
Cl (+)
∆H1 Br
(+)
∆H3
→
(+)
→
(+)
∆H2 ∆H4
212
Problems in Organic Chemistry
Which order is correct? (a) ∆H1 > ∆H3 > ∆H2> ∆H4 (c) ∆H2 > ∆H1 > ∆H3> ∆H4 Q.49
(b) ∆H1 > ∆H2 > ∆H3> ∆H4 (d) ∆H2 > ∆H3 > ∆H4> ∆H1
→ 1-chloro-1,2-dimethyl cyclo hexane (A) can be :A + HCl
OH (a)
(b)
(c)
(d) All of these
Passage – I Compound (A) is optically active and gives (B) on treatment with (KOH + H2O) compound (B) is also optically active. (B) On heating with phosphoric acid gives (C) which does not show geometrical isomerism. When (C) is subjected to ozonolysis (O3 / Zn,H2O, heat) (D) & (E) are obtained. (D) & (E) on treatment with OI(–) give precipitates. These precipitates are removed by filtration and filtrates of (D) and (E) are then treated with soda lime, (D) gives (F) while (E) gives methane .Compound (F) can be obtained by heating cyclo hexane with Al2O3 | Cr2O3. Now answer the questions from 50 to 53. 50. (F) would be:–
(a)
(b)
(c)
(d)
51. (A) and (B) are:– Φ 2 CHCCl(CH3 )2 , Φ C(Me) — CHMe2 (b) (a) ΦC2 H5 — CClCHMe2 , | OH
(c) Φ — CH — CClMe2 , Φ — C(OH)CHMe2 (d) Φ3C — CClMe2 , Φ 2 C — CHMe2 |
|
|
Me
OH
Me
52. (C) would be: – (a) Φ 2 C = CMe2 ,
Ph (b) Ph (CH3) C = CMe2 (c) C = C Ph
53. (D) and (E) are:– (a) PhCOPh, CH3COCH3
(b) Ph2CO, CH3COPh
Ph
Ph Me
(c) PhCOCH3, CH3CHO
Passage – II Consider the following halides:– CH3 — CHCl — CHCl — CH3 CH 2 — CH 2 —CH 2 — CH 2 |
|
CH3 — CH2 — CH2CHCl2
Cl Cl (1) (2) These halides can undergo substitution and elimination reaction now
(3)
(d)
Me C =C
Et
Ph
(d) PhCOCH3, CH3COCH3
213
Alkyl Halide & Grignard’s Reagents
Answer the question from 54 to 59 54. On treatment with Zn/Ether cyclo alkane will be formed by:–
(a) 1 & 2
(b) 3
(c) 2
(d) 1
55. (3) can be converted in to (1) by:–
(a) Zn / Ether, Zn /∆, O3 / Zn, H2O, PCl5
(b) aq KOH, LiAlH4 / Ether , H3PO4 /∆ ,Cl2 in CCl4
(c) alc KOH(Excess), heat , Pd / C , Cl2 / in CCl4
(d) All of these
56. Which two alkyl halides on treatment with KOH (alc) will give same product:–
(a) 1 & 2
(b) 2 & 3
(c) 3 & 1
(d) None
57. Which alkyl halide will form higher alkene on treatment with Zn / THF / heat :–
(a) 1
(b) 2
(c) 3
(d) 1 & 2
58. The alkyl halide which on treatment with alc KOH followed by CH3MgBr can liberate CH4
(a) 1
(b) 2
(c) 3
(d) None
59. Halide which can give two different bromides when treated with alcoholic KOH followed by HBr (1 mol) is / are –
(a) 1
(b) 3
(c) 1 & 2
(d) 2
Passage – III
ICl
B
Finkelstein reaction
(i) Mg in ether (ii) H2 O
A
C
H3O (+)
D
P, Cl2, heat
HCl
G
E
E
Answer the questions from 60 to 63 60. Compound B would be:–
Cl
Cl
I
(a)
(b)
I (c) I
I
Cl
Cl (d)
Cl
61. Compound C would be:–
I (a)
Cl
(b)
I
(c)
(d)
I
I
62. Compound E would be:–
OH (a)
I
(b)
(c)
OH
(d)
63. Which is wrong statement about the given passage?
(a) E & F are identical
(b) F & G are position isomers
(c) E & G are identical
(d) A & C are identical
Cl
214
Problems in Organic Chemistry
Passage–IV NBS
Propyne
1 mol HBr
(A)
(F)
PhMgBr
(B)
Mg Ether
(C)
Succinic Anhydride (Diels Alder)
(D)
CO2
(E)
H3O
(+)
Answer the Questions from 64 to 67 64. Compound (B) is:–
Br
Br Br Br
Br
Br (a)
Br
(b)
Br (c)
65. Compound (C) is:– Br
Br (a)
(d)
Br
Br
(b) Ph
Ph Ph (d) Ph
(c)
66. Compound (E) is:–
COOH
HOOC
Ph CO2H
(a) COOH (b) (c)
(d)
67. Compound (F) is:–
O
OH O
OH (a)
COOH
(b) HOOC
OH
OH CH— 2
CH— 2
O
HOOC
O
OH
OH (c) PhCH2
OH
(d) PhCH2
OH
COOH Current
68. (A) + Brine →
(a) CH2 = CH2
Acetone/KOH
(B) → Hypnotic medicine, (A) will be:– (b) CH3COONa
(c) C2H5OH
(d) CH3CH2COONa
69. Identify the reaction in which alkyl chloride is not form as product:– (i) AgOH (ii) X 2 / ∆
.......... (1)
FeCl
......... (3)
CH3COOH →
3→ Φ — H + Cl2
(a) 1
(b) 4
SOCl
.......... (2)
P + Cl / ∆
.......... (4)
2→ CH3CHOHCH3 2 → Φ —CH 2 OH
(c) 1 & 3
(d) 1, 2 & 3
215
Alkyl Halide & Grignard’s Reagents
70. The reaction in which carbene does not form as an intermediate:– (a)
(Me3CO)3 Al + CHCl3 →
CCl2
(b) Reimer Tieman reaction
(c) Carbyl amine reaction → Chloretone (d) CHCl3 + CH3COCH3 + KOH (alcoholic)
71. Assertion – Tert butyl chloride on rection with AgCN produces tert Bu –CN
Reason – this reaction occurs via SN1 pathway
(a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is False (d) Assertion is false, Reason is true 72. Assertion – Tert butyl chloride on rection with NaCN produces tert Bu –CN
Reason – Cyanide ion is a strong base & favours elimination reaction
(a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is False (d) Assertion is false, Reason is true 73. Assertion – Alkyl halides on reaction with moist Ag2O forms alcohols & not ethers
Reason – Ag2O undergoes hydrolysis to form silver hydroxide
(a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is False (d) Assertion is false, Reason is true 74. Assertion – R– butan–2–ol on reaction with SOCl2 becomes R– 2–chloro butane
Reason – This reaction occurs via SN1 pathway
(a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is False (d) Assertion is false, Reason is true 75. Assertion: – Boiling point of CH3F is lesser than CH3I
Reason: – Boiling point of alkyl halides is proportional to polarisability and I has good polarizing power in comparision to that of F.
(a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is False (d) Assertion is false, Reason is true 76. Assertion: – Dipole moment of CH3Cl is greater than CH3–F
Reason: – F has less electron affinity than Cl
(a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is False (d) Assertion is false, Reason is true
216
Problems in Organic Chemistry Mg
77. Assertion: – PhCH2Br → PhCH2MgBr Ether This reaction is not possible. Reason: – PhCH2Br Undergoes nucleophilic substitution reaction with PhCH2MgBr to produce PhCH2CH2Ph (a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true 78. Assertion: – In proteic solvent order of nucleophilicity of halide ions is I(–) > Br(–) > Cl(–) > F(–) Reason: – In polar proteic medium ion–dipole attraction takes places. F(–) is strong base hence due to more ion dipole attraction it is arrested by solvent molecules. (a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is False (d) Assertion is false, Reason is true 79. Assertion: – The relation nucleophilicity proportional to basicity holds in protic solvent but not valid in aprotic solvent like DMSO Reason: – If the solvent is protic it forms H – bond with nucleophile but aprotic solvent like DMSO con not form H – bond with nucleophile. (a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is False (d) Assertion is false, Reason is true 80. Assertion: – In SN1 reaction 100% racemisation occurs. Reason: – Here initmate ion pair is formed (a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is False (d) Assertion is false, Reason is true 81. Assertion: – DMF is aprotic solvent Reason: – C –H bond is not polar (a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is False (d) Assertion is false, Reason is true
Answer Key 1. (c)
2. (b)
3. (a)
4. (b)
5. (d)
6. (c)
7. (b)
8. (b)
9. (a)
10. (d)
11. (c)
12. (c)
13. (d)
14. (d)
15. (d)
16. (c)
17. (c)
18. (d)
19. (b)
20. (d)
21. (c)
22. (c)
23. (d)
24. (b)
25. (b)
26. (b)
27. (c)
28. (c)
29. (c)
30. (b)
31. (c)
32. (c)
33. (b)
34. (c)
35. (d)
36. (d)
37. (d)
38. (b)
39. (d)
40. (b)
41. (b)
42. (b)
43. (d)
44. (b)
45. (c)
46. (a)
47. (b)
48. (a)
49. (d)
50. (c)
51. (c)
52. (b)
53. (d)
54. (c)
55. (c)
56. (a)
57. (c)
58. (c)
59. (c)
60. (b)
61. (d)
62. (d)
63. (b)
64. (b)
65. (d)
66. (c)
67. (a)
68. (c)
69. (c)
70. (d)
71. (d)
72. (d)
73. (a)
74. (c)
75. (a)
76. (c)
77. (a)
78. (a)
79. (d)
80. (d)
81. (a)
217
Alkyl Halide & Grignard’s Reagents
Multiple Choice Questions More Than One May Correct 1. Which of the following reactions will produce an aromatic product?
O
(a)
Fe heat
Pr opyne →
(b)
+ HBr (1 mol)
O Cl (c)
alc KOH
VO 773K,10 − 20atm
(d)
2 5 n − hep tan e →
O 2. The method which can not be used for the preparation of propane is:– (a) Wurtz reaction (b) Frank – land reaction (c) Finkelstein reaction
(d) Corey – House reaction
3. The substrate which can produce a di iodide on finkelstein reaction is:–
Cl
Cl
Cl (b)
(a)
Cl (d) I
(c)
I I
I 4. Which of the following can exhibit Wurtz Reaction? OH
O
(a)
(b)
(c) Cl — CH2 — NH2
Cl
Cl
(d) CH3 CH — C |
CH = CH 2
5. Which statement is true about the products A & B present in the following reaction scheme? I(–) Acetone
→
Br
(A)
CH = CH—CH Cl ZnCl2
2 2 →(B)
I
(a) A can decolourise Br2 water (c) (B) on hydrolysis produces optically active product.
(b) B decolourise Br2 water (d) (A) is a di iodide
••
6. CHCl3 + (X) → C Cl2
(X) would be:– (a) alc NaOH
(b) Me3CO(–)
(c) Pyrrole
(d) All of these
→ Sweet smelling liquid 7. (X) + CaOCl2 + H2O ‘X’ would be:–
O O
(a)
(b)
Ac
(c)
(d) CH2ClCOCH3
218
Problems in Organic Chemistry
8. Which of the following reagent is not required during the following transformation?
Cl I
COOH (a) NaOH / CaO, ∆
(b) NaI / acetone
(c) H2 / Ni / heat
(d) KOH + H2O
9. Permanent dipole moment is shown by:– O (a)
(b) Br
Cl
(c) Cl
Cl
Cl Cl (d)
10. Pick out the correct equations
→ C2 H 6 (a) CH3OCH3 + CH3Br + Zn
CH3CH — CH3 + C2 H5OH + KOH → propene (b) |
Br
AlCl + HCl
3 (c) CH3CH 2 CH 2 CH 2 Br + Zn + H ( + ) →(A) →
→ (CH3)2COHCCl3 (d) CHCl3 + KOH + CH3COCH3 + C2H5OH
11. X + NaOH + I2 → Yellow ppt + sodium adipate,
(a) octan – 2, 7 – dione
X would be:
(b) 7 – oxo heptanoic acid
(c) HOOC
(d) O
` O
O
Answer Key 1. (a), (b, (c), (d)
2. (a), (b), (c)
3. (b), (c)
4. (a), (d)
6. (a), (b)
7. (a), (b), (c), (d)
8. (d)
9. (a), (b), (d)
5. (b), (c) 10. (a), (b), (c), (d)
11. (a)
LEVEL – II (+ )
(+ )
→ R ( + ) + H 2 O .....................(1) 1. R — OH 2 R N 2 → N 2 + R (+ )
If these reactions have rates R1 & R2 then:– (a) R1 = R2 (b) R1 > R2
(c) R2 > R1
...................(2)
(d) R1 R2
2. Select the compound which can give substitution reaction with aq NaOH at room temperature (a) PhCl (b) MeI (c) Me3CI (d) Me3CBr 3.
The rate of nucleophilic substitution of an alkyl halide RX in 75% EtOH & 25% acetone is given below:– Rate = 1.67 × 10–6 [RX] [Base] + 0.05 × 10–6 [RX] Calculate fraction of SN2 when [Base] = 1M (a) 90% (b) 3% (c) 10% (d) 97%
219
Alkyl Halide & Grignard’s Reagents
4. The reaction in which 4 products can form (including stereo isomers) is:– ICl →
(a)
HCl
(b)
CD3 → (c)
5. Match the following Reaction Product Br
+ HBr
1.
1. (a) 1 → r, ; 2 → p; 3 → q
(d)
DCl
→
% yield
Br
2.
HCl
→
Br 3. (b) 1 → q; 2 → p; 3 → r
(p) 2 (q) 52 (r) 46 (c) 1 → p; 2 → q; 3 → r
(d) 1 → r; 2 → q; 3 → p
Comprehension Consider the following reactions
Cl
(a)
Me
Cl al KOH → A
al KOH
(b)
Cl
Me Cl
Me
Me alc KOH (excess)
(c)
→ B
Me
→ C
alc KOH (excess)
→ D
(d)
Me
Cl
Cl
Cl Me (e)
Me
alc KOH (excess)
→ E
Cl Answer the following questions:– 6. Which product is least stable? 7. Which product has least value of heat of hydrogenation? 8.
LiNH2 CH3OH
[X]
Out of A, B, C, D & E which is similar to that of [X]
9. Arrange A, B, C, D & E in increasing order of their stabilities
Matrix Match Questions Each question contains statement given in two columns which have to be matched. Statement (A, B, C , D) in column I have to be matched with statements (p, q, r, s) in column II. The answer to these questions has to be appropriately bubbled as illustrated in the following examples. If the correct match are A – p, A – s, B – r, B – r, B – q, C – q, D – S, then the correctly bubbled 4 x 4 matrix should be as follows.
220
Problems in Organic Chemistry
10.
Reaction
Ph (A) Et
A
p
q
r
s
B
p
q
r
s
C
p
q
r
s
D
p
q
r
s
Mechanism / Stereochemistry Product
C—Br + NH3
(p) SN1
Me (B)
Br
2 in CCl4 Product
(C) CH3CH2Br + KOH (aq) Product (D) 11.
Cl
2 → Product in CCl 4
List – I (Reaction)
(q) SN2 (r) Racemisation (s) Meso product List – II (Production)
OH (A)
Br HBr
→
(p)
Br (B)
HBr → hν
(C)
HBr → peroxide
(D)
→
(q)
Br
12.
NBS hν
(r)
(s)
Br
Reaction Reagent
(A) (CH3 )3 CBr → iso butene
(p) LiAlH4
(B) i — pr — I → Pr— H
(q) KOH + C2H5OH
(C) i — Pr— Br → i — Pr— I
(r) I(–) / Acetone
(D) iso pentene → Iso pentane
(s) Wilkinson Catalyst
13. Alkane
Number of monochloro derivatives produced (only structural isomers)
(A)
(p) 1
(B)
(q) 2
221
Alkyl Halide & Grignard’s Reagents
(C)
(r) 3
(D)
(s) 6
14. Alkane (A)
Number of mono chloro derivatives Produced (p) 2
(B)
(q) 3
(C)
(r) 4
(D)
(s) 5
Answer Key 1. (c)
2. (b), (c), (d)
6. Product A 7. Products a & b
3. (d)
4. (a)
5. (d)
8. None
9. A < B < E < C = D
Answers matrix match 10. (A) – p, r (B) – r (C) – q (D) – s 12. (A) – p, q, (B) – p (C) – r (D) – s 14. (A) – p, (B) – q, (C) – p, (D) – q
11. (A) – p, q, s (B) – p (C) – q (D) – r 13. (A) – r, (B) – s, (C) – r, (D) – q
SOLUTION 1. (c) Reaction 2nd & 4th are going on by SN1 pathway which is independent of concentration of base hence both will have same rate because concentration of alkyl halide is same. 1st and 3rd are going on by SN2 depends upon concentration of base hence R1 will be different from R3 because concentration of base is different. (+)
2. (b) CH3Cl
Mg. Ether
CH3MgCl
HCHO, H 3 O CH
CH3CH2OH CH
Fe, heat
∆
(E)
Haloform
CH
CHCl3
Ag. heat
CH(B) (2 mol)
(C)
3. (a) Rate of reaction ∝ Stability of carbocation 4. (b) In DMSO SN2 reaction takes place 5. (d) CH3I is very sensitive towards Grignard’s reagent formation. Hence to increase the rate of formation of Grignard’s reagent small amount of CH3I is added. This is called entertainment technique. 6. (c) As no. of halogens increases boiling point decreases (Except fluorides)
222
Problems in Organic Chemistry
7. (b) Anh ZnCl2 + HCl is called Lucas reagent
OH (+) (+)
Cl
O 8. (b)
Cl
(–)
(+)
H –H2O
OMgCl CH3
OH H3O
+ CH3MgCl
CH3
(+)
Kinetically controlled product O
OMgCl
OH
CH3MgCl
H3O
O
(+)
Tautomerism
CH3
CH3
CH3 Thermodynamically controlled product
9. (a) See question no. 6 10. (d) Dipole moment = charge on dipole (q)
× distance between opposite charges (d)
Br & I have low electro negativities hence value of q is too small thus; dipole moment is less. Out Cl & F, C — Cl bond has large bond length is comparision to C — F bond however q for F is greater than that of Cl but product q × d for CH3Cl is found to be greater than that of CH3F 11. (c) If Br2 is replaced from I2 in Hunsdecker reaction ester is formed. This reaction is called Smith simonini reaction. 12. (c) Because nuclophilicity of I is greater than that of F. I is a good attacking reagent as will as good leaving group. 13. (d) 14. (d) See mechanism of haloform reaction in your text book. 15. (d) 1st is 3° alkyl halide so rate of SN2 will be least in it. Rest all are 2° alkyl halide but 3rd contain ‘I’ which is a good leaving group hence rate of SN2 is maximum it 3rd case. Out of 2nd & 4th, later is sterically hindered halide because near to Cl two methyl groups are present. So formation of pentavalent transition state is difficult in it in comparison to 2nd. similarly Cl of 5th is less stericallly hindered than that of 4th hence rate of SN2 will be 3 > 2 > 5 > 4 > 1 16. (c) It does not contain acidic hydrogen.
Br
5
17. (c)
(+)
Br Br
Ag –AgBr
6
4 3
Br
1 1
2 (+)
6
(+)
Br
3
5
2
OH H2O
4
18. (d) Haloform gives CHCl3 (A) & cyclo hexyl carboxylate ion (B) which undergoes Kolbey’s electrolysis as follows:–
• COO
current
(–)
COO
Disproportionation
Cyclohexyl free radical →
Dimerisation Cyclohexy Free radical →
–CO2
• Cyclohexyl free radical
+
223
Alkyl Halide & Grignard’s Reagents
19. (b) 3° alkyl halides undergoes elimination reaction by LiAlH4 LiAlH
4 → (CH ) C = CH (CH3)3CCl 3 2 2
20. (d) CH3CH(CH3)CHClCH3 .……….….………(A) (CH3)2CHCH2CH2Br …………………..… (D) (CH3)2CBrCH2CH3 …….………….….… (C) Since rate of SN1 depends upon the stability of carbocation hence (C) will undergo SN1 rapidly because it will provide highly stable carbocation while D will undergo SN1 with slower rate because it gives primary carbonium ion as intermediate which is less stable. 21. (c) (B) = RCOOH RCOOH + RMgBr → RH + Other Product (B)
(C)
(A)
RMgBr + C2 H5OH → RH + Other Product (E) 22. (c) In this reaction NaCl or NaBr or NaF is formed along with alkyl iodide RX
in water
+ NaI → NaX + RI
[X = Cl,Br,F] In water NaCl , NaBr & NaF are soluble thus can not precipitate out. While in acetone these are insoluble due to their more ionic character thus; precipitate out. in acetone
RX + NaI Soluble in acetone
NaX
+ RI
insoluble in acetone
Br 23. (d)
NBS
alc KOH
+
D-Alder
(A)
24. (b) Product formed is stabilized by resonance. THF
25. (b) PhBr + Mg → PhMgBr PhMgBr + PhOH
+ PhOMgBr
(–) • •
26. (b) O N O nitrite is an ambident nucleophile because it can attack from its O as well as N site. NaNO2 is an ionic compound so O site of nitrite is free to attack on halide hence ethyl nitrite forms as:– (formNaNO )
2 EtBr + (–) ONO → Br + EtONO Ag – ONO is a covalent compound so nitrite can not attack from its oxygen site because covalent bond is present between silver and oxygen. Hence nitrite attacks from its nitrogen site & nitro alkane is formed ••
EtBr + Ag—ONO
Et—NO2+ AgBr
224
Problems in Organic Chemistry
Me 27. (c)
Me
Me
B2H6
H
Me
H2O2/Base
H
H BH2
BH2—H
H OH Me SOCl2
H H Cl
CH3 28. (c)
H Br
CH3
CH3 OH H
(+)
H
H
OH2
CH3
(+)
(–)
Br
(+)
Br H
•• • • Br ••
CH3
H Br
Br H
H CH3
CH3
CH3 29. (c) Volume of gas = 11.2 cm3 Moles of gas = 11.2 / 22400 = 0.0005 0.0005 moles of gas are obtained from = 0.037 gm alcohol 1 mole of gas will be obtained from = 0.037 / 0.0005 = 74 Hence molecular wt of alcohol = 74 gm Therefore alcohol will be (CH3)2CHCH2OH 30. (b) 31. (c) 32. (c) Solvolysis is SN1 reaction. Observe the stability of carbocations formed by 1, 2, 3, & 4 33. (b) Zn / HCl reduces chloroform partially. Zn / HCl (aq) → 4(H) → 2(H) Zn / HCl (alc) Zn/HCl(aq)
→ CH3Cl + 2HCl CHCl3 + 4(H) Zn/HCl(aq)
CHCl3 + 2(H) → CH2Cl2 + HCl Zn/H O or H /Ni
2 2 → 3HCl + CH4 CHCl3 34. (c) 35. (d) During elimination reaction more stable or more substituted alkene is produced. 36. (d) (c) is Diels–Alder reaction
(a) CH3COCH3
C6H6
Haloform
Fe, heat
CHCl3
CH ≡ CH
Ag heat
Hydrolysis (b) CaC2 → ACETYLENE → 6 6
225
Alkyl Halide & Grignard’s Reagents
37. (d) NOCl is called Tilden reagent and can convert amine in to halide.
NOCl
Ag2O moist
CH2NH2
CH2 —Cl heat H
(+)
CH2OH Ring opening
(+)
(+)
(+)
CH2
H –H2O
38. (b) Haloform reaction (–) •• 39. (d) C N
is an ambident nucleophile
NaCN is an ionic compound so CN can attack from ‘C’ site. In NaCN, CN behave as strong base hence reaction occurs via SN2 pathway therefore CH3CH(CH3)CH2CN is formed. In Ag–CN, C is joined with Ag by covalent bond so CN attacks from lone pair present on N atom. Now CN becomes weak base and reaction occurs via SN1 pathway 1
CH3 — C H — CH 2 — Cl ?
(+ ) (+ ) Rearrangement CH3 — CH — CH 2 ————————→ CH3 — C — CH 3
SN
|
|
CH3
|
CH3
CH3
CH3
•• AgCN
|
CH3 — C — NC |
+ Ag ( + )
CH3 alcKOH
40. (b) CH 2Cl — CH 2Cl → Acetylene (–) ( + )
CH ≡ CH + NaNH 2 → CH ≡ C N a + NH3 (–)
CH3
| E limination CH C + (CH3 )3 C – Cl → CH3 — C Reaction
≡
CH3 |
= CH 2 → CH3 —C H — CH3 H 2 / Ni
41. (b) Bond energy of C — D bond is greater than that of C — H bond so former is more difficult to break in comparison to later. The alkene in which D is present will be major product. 42. (b) 0.2872 lit of gas has weight = 1 gm 22.4 lit of gas will have weight = 22.4 / 0.2872,
=
78
Hence molecular weight of gas RH = 78 Mol. Wt. of R = 77 Hence R will be C6H5 (12 × 6 + 5 = 77) Hence Grignard reagent is C6H5MgBr RH + other product → RMgBr + C2H5OH Mol.wt = 78 43. (d) this is SN1 reaction hence1st step should be RDS and endothermic but in the proposed graph 1st step is exothermic. 44. (b) rate of dehydrohalogenation ∝ stability of product formed
226
Problems in Organic Chemistry
Ph 45. (c) PhCl + Mg (Ether)
PhMgCl
CH3CHO
CH3—CH—OMgCl Ph
PhCOCH3
mildoxidation
H3O
CH3—CH—OH
(+)
Cl Mg (Ether)
+
46. (a) Br
(+)
Cl
MgBr(–)
(A)
Acetone 47. (b) CH3 — CHBr — CHClCH3 + 2NaI → CH3CHICHICH3 → CH3CH = CHCH3 + I2 Unstable
48. (a) Enthalpy of dissociation ∝ 1 / ease of formation of carbocation
OH
(+)
(+)
H –H2O
49. (d)
Cl
(+)
Methyl shift
Passage – I 50 to 53 Al2O3, heat
(F) Hence (D) should be PhCOCH3 (acetophenone) haloform
sodalime
PhCOCH3 → PhCOO(–) → PhH (F)
(D)
haloform
sodalime
CH3COCH3 → CH3COO(–) → CH 4
(E)
Hence (C) would be:–
CH3 Ph
CH3 C=C
Ozonolysis
PhCOCH3 + CH 3 COCH3 (E)
CH3 CH3
CH3
|
|
Ph — C — CH(CH3 ) 2 → Ph — C = C(CH3 ) 2 H3PO4
|
(C)
OH CH3 |
Ph — *CH — CMe2 |
(A)
Cl
1
SN
CH3 |
*
→ Ph — *C — CHMe2 , Here C = Chiral carbon atom KOH |
OH
(B)
227
Alkyl Halide & Grignard’s Reagents
Passage - II (54 to 59) Zn, heat 54. (c) ClCH 2CH 2CH 2Cl →
55. (c) CH3—CH2CH 2CHCl 2
alc KOH
KOH (alc)
heat
Cl Pd/C H2
Cl2
(1) Cl alc KOH
56. (a) CH3CHCl—CHClCH 3
heat
CH2= CH—CH=CH2
alc KOH
CH2CH 2CH2—CH 2 Cl Cl
57. (c) It will form oct–4–ene.(Frankland reaction followed by dechlorination) 58. (c) (3) on treatment with alcoholic KOH forms terminal alkyne (See question no. 55) CH3CH2C ≡ CH + CH3MgBr ——→ CH4 + CH3CH2C ≡ CMgBr 59. (c) From question no. 56 it is clear that by alcoholic KOH halides form butadiene. This butadiene on reaction with HBr (1 mol) can form two products as:–
CH2=CH—CH=CH3
H
(+)
(+)
CH3—CH—CH=CH2 Br
(–)
(+)
CH3—CH=CH—CH2 Br
CH—CH—CH=CH 2 3
(–)
CH3CH=CHCH2Br
Br
Passage – III (60 to 63)
I
Cl
A
(+)
H3O
finkelstein
ICl
I
B
I
C
Mg in ether
Cl
Cl H2O
E
Cl MgI
G
P, Cl2, heat
OH D
228
Problems in Organic Chemistry
Passage – IV (64 to 67) A is 3– bromo propyne which on reaction with 1 mol of HBr gives (B) BrCH2—CBr = CH2 Now allylic bromine of B undergoes substitution reaction with PhMgBr to produce (C) Which is Ph CH2—CBr = CH2. Compound (D) is grignard’s reagent Ph CH2—CMgBr = CH2 Which on reaction with carbon dioxide produce Ph CH2—C(COOH) = CH2 68. (c) By the electrolysis of aq NaCl (brine), Cl2 librates & NaOH is Produced. C2H5OH + Cl2 ——→ CH3CHO + 2HCl CH3CHO + 3Cl2 ——→ 3HCl + CCl3CHO CCl3CHO + NaOH ——→ CHCl3 + HCOO Na CHCl3 + acetone ——→ Chloretone
(Hypnotic medicine)
69. (c) 1→ Borodine– Hunsdecker reaction. In it alkyl bromines are formed. 3→ Here aryl halides is formed. 70. (d) Here tri chloro carbanion is formed which attacks on carbonyl group of acetone . 71. (d) In this BuNC will form(See q. n 39) 72. (d) Substrate is sterically hindered & base is strong so cyanide ion will give elimination product with 3°–BuCl thus; instead of alkyl cyanide alkene is formed 73. (a) Ag 2 O + H 2 O → 2AgOH
→ AgX + ROH AgOH + RX
74. (c) Here intimate ion pair is formed thus, retention occurs. 75. (a) Boiling point of alkyl halides is proportional to polarisability and I has good polarizing power in comparison to that of F. 76. (c) Bond length of C—Cl bond is larger than bond length of C—F bond. 77. (a) PhCH2Br undergoes nucleophilic substitution reaction with PhCH2MgBr to produce PhCH2CH2Ph 78. (a) In polar proteic medium ion–dipole attraction takes places. F(–) is strong base hence due to more ion dipole attraction it is arrested by solvent molecules. 79. (d) In protic solvent nuceophile is caged by solvent molecules by H—bonding thus, in protic solvent Nuceophilicity ∝ 1 / Basicity 80. (d) Here initmate ion pair is formed. Hence racemisation occurs along with retention. 81. (a) Di methyl formamide (DMF) does not contain acidic hydrogen (C–H bond is not polar) HCON(CH3)2
.........D.M.F
More than one may correct:–
O
OH H
1. a, b, c, d
OH
(+)
Br (+)
O
OH
O
(–)
Br O
OH
Br O
OH (Aromatic)
229
Alkyl Halide & Grignard’s Reagents
CH3 Fe Propyne heat (Aromatic)
H3C
CH3
Cl
O
alc KOH
O
OH
(Aromatic)
2.
a, b, c
3.
b, c
In (a) but–2–ene is formed while in (d) benzene is formed
4.
a, d
(b) & (c) contain active hydrogen
5.
b, c
Compound A is benzene while B is PhCH2CH = CH2
6.
a, b
Pyrrole is not a strong base
7.
a, b, c, d CHCl3 is sweet smelling liquid and the proposed reaction is haloform reaction. Since all a, b, c & d contains CH3CO linkage hence all will undergo haloform reaction.
Cl 8. d)
NaI Acetone
I
H/Ni 2 ∆
NaOH/CaI ∆
COOH
COOH
9. a, b, d (c) is a symmetrical trans alkene hence it has zero dipole moment. 10. a, b, c, d 11. a LEVEL –II 1. (c) N2 is inert gas thus good leaving group. 2. (b) Due to resonance C — Cl bond acquires double bond character & hence Ph—Cl does not undergo substitution. 1.67 × 10 –6 [RX] × 1 × 100 = 97% 1.67 × 10 −6 [RX] × 1 + 0.05 × 10 –6 [RX]
3. (d) % SN 2
H H
4. (a)
I Cl
I Cl
(D.D)
H H
H Cl
(L.L)
+
H
(L.D)
(D.L) (+)
(+)
Br Br
(–)
46%
5. (d) Br
52%
Br
(–)
(+)
(+)
resonance (More Stable)
I H
I H
2%
H Cl
230
Problems in Organic Chemistry
Comprehension
H H H
Cl
Cl alc KOH → (Anti elimination)
H Me
(A)
Me
Me
alc KOH
→ H (Anti elimination)
H
(B)
Me Me
Similarly C is
Me D is
Me E is
Me Me
Compound X is formed by Birch reduction as:– 6. (a) 7. (a) & (b) 8. None
LiNH CH 2OH
2 →
9. A < B < E < C = D
Me
Alcohol, Ether, Epoxide Grignard’s Reagents
8
Main Features Reaction Chart for Alcohol Preparation Alkene Aldehyde or ketone or ester or acid halide or carboxylic acid
H2O/H
(+)
CrO3 or PCC
LAH
KMnO4
Sugar Starch
Fermentation
Alkene Alkene Ester
aldehyde or ketone (mild oxidation)
or PDC
Fermentation
RNH2
Properties
NaNO2+HCl O3/H2O NaBH4
ROH/H
(+)
RCOCl / OH
RCOOR ( Esterification) (–)
(RCO)2O / OH
RCOOR (Acylation) (–)
NH3/Al2O3
B2H 6
RCOOR ( Acylation) RNH2
(+)
H2O2 / OH(–) H 2O / H
A L C O H O L S
RCOOH ( strong oxidation)
H 170°C
RCH = CH2
(+) (+)
H
O CH2–CH2 + RMgX
H3O
RCHO or RCOR + RMgX
H3O
(+)
140°C
RCH2CH2OCH2CH2R
( not for 3°alc) PCl5 /PCl3 /SOCl2
(+)
Red P + HI
RCl alkane
232
Problems in Organic Chemistry
Reaction Chart for Ether & Epoxide Preparation
Properties
(+)
H / 140°C
ROH
HI (cold & dil)
Ag2O
RX
HI (hot & conc)
E T H E R S
RONa
RX (1° or 2°)
RI + ROH H R—O—R
H2SO4 H2O / H
2RI (+)
(+)
2ROH
PCl5 / heat SOCl2 or PCl heat
2RCl RCl
Reaction Chart for Epoxide Preparation CH2=CH2 + Ag + 1/2O2
H3O
heat
E P O X I D E S
PAA
CH2=CH2 CH2—CH2 Cl OH
Properties
aq KOH
(+)
CH2OH—CH2OH
HCl NH3 / H
CH2Cl—CH2OH (+)
CH2NH2—CH2OH
RMgX (+) H3O PCl5 heat
RCH2CH2OH
CH2Cl—CH2Cl
LEVEL - I Multiple Choice Questions CH2NH2
1.
(i)Tildent reagent
→ + (ii)aqKOH (iii) H ,heat
A, Compound (A) will be:-
CH2OH (a)
(b)
(c)
2. OH group of ROH is a good leaving group in: (a) Basic medium (b) Acidic medium OH
Br
3.
OH
(d)
CH3
(c) Neutral medium
(d) Presence of sun light
(c) CH4
(d) Both b & c
CH MgBr heat
3 → X
In this reaction X would be:- Br
O (a)
(b)
233
Alcohol, Ether, Epoxide Grignard’s Reagents
4. How many intermediate carbocations are involved during the following transformation? H( + ) Heat
OH → Alkene (major product)
(a) 1
(b) 2
(c) 3
(d) 4
5. In which case epoxide will form as product:-
OH
OH
H( + )
OH
(a)
→
OH
(c)
NaOH + H O
2 →
(b)
H( + )
→
(d) All of these
6. Most Stable form of 4-nitro cyclo hexanol is:-
NO2
(a) H
NO2
H
(b)
OH
OH
NO2
OH NO2
(c) H
H
(d) OH
H H
H PO
3 4→ Products?
7. Which product is not expected for
OH (a)
OH
(b)
O
COCH3
(c)
(d) (b) & (c)
COCH3
(AcO) Pb
4 → (B) 8. (A) → + ve haloform test (A) may be: (a) (CH3CH2CHOH)2 CH2 (b) CH3CHOHCHOHCH3 (c) CH3— CHOHCOCHOHCH3 (d) (b) & (c)
H H( + )
OH → ?
9.
OH
H Product of this reaction will be:(a) Cyclo pentan carbaldehyde (c) Cylco hexanone
(b) 2 - Hydroxy Cyclohexanone (d) All of these
234
Problems in Organic Chemistry
10. Correct order of acid catalyzed dehydration to produce alkene in the following compounds will be:-
OH (1)
OH | CH 2 = CH — CH = CH — CH — CH3 (2) CH 2
Φ3COH (3) (a) 1 > 2 > 3 > 4 (b) 3 > 1 > 2 > 4
CH — CH — CH3 OH (4)
(c) 3 > 2 > 1 > 4
(d) 1 > 2 > 4 > 3
11. Consider the following reduction. NaBH 4 O →
D
Product of this reaction will be:(a) Optically active (c) Product forms with partial racemisation
ether,H( + )
Product
(b) diastereomer (d) meso
12. Consider the following reduction. NaBH 4 O → Product
D
Product of this reaction will be:(a) Optically in active diastereomeric pair (c) Product forms with partial racemisation
ether,H( + )
(b) Optically active diastereomeric pair (d) Product forms with 100% racemisation
13. In which reaction product will not exhibit haloform test.
lead tetra acetate (a) CH3COCH2CHOHCHOHCH2COCH3 → Product
4 (b) CH3CH2OH → Product
H(+) /KMnO
CH 2CHO |
(c) CH3 C = CCH3 |
CH 2CHO
O3 Product → H 2O/Zn, heat
(d) Products of all reactions will not show haloform test.
OH
O
14.
This oxidation can be performed by:(a) Aluminum tert- butoxide (b) Hot KMnO4
(c) Jones reagent / 35°C
(d) (a) & (c)
OH |
15. Φ 2 — C — CH3 + Ac Cl → X + Y, X & Y are : CH3
CH3 | (a) Φ 2 C — Ac + HCl (b) Φ 2 COAc & HCl |
CH3 |
CH3 |
Φ 2 C — COOH + CH3Cl (c) Φ 2 C — Cl & AcOH (d)
235
Alcohol, Ether, Epoxide Grignard’s Reagents
H SOCl
2 → CH3CH—Pr OH in dioxane | Cl
Pr
CH3 16. Configuration of product will be: (a) R (b) S
(c) Racemic mixture
(d) Unpredictable
17. Consider the following reagents: Al2O3 / 350°C H2SO4 / 1400C (CH3)2SO4 / NaOH CH2N2 (1) (2) (3) (4) Methyl alcohol can be converted in to dimethyl ether by using (a) 1, 2, 3 & 4 (b) 1, 2 & 3 (c) 1 & 2 (d) 1, 2 & 4
CH3 18.
SOCl2
Configuration of ‘A’ & ‘B’ will be:-
OH—— P + Br2
H
(a) R, S
(b) S, R
Φ
(c) S, S
(d) R, R
Φ
|
|
19. CH3 — CH — CHOHCH3 → CH3 — C — CH 2 — CH3 , Here A will be:(A)
|
Cl
(a) Sarret reagent
(b) Lucas reagent
(c) SOCl2
(d) All of these
Passage I An organic compound A (C7H13OCl) contains one four membered ring whose two opposite corners are substituted by two group, one of which is Cl. This Compound reacts with lucas reagent immediately. This compound gives B & C on treatment with aq Na2CO3 & aq KOH respectively. Compound (B) can react with Sarret reagent. (A) On reaction with dilute sulphuric acid gives D(minor) which can react with lucas reagent with in 5 minutes. When (D) is subjected to heat in presence of H(+) (E) is formed which can decolourise Br2 water. (E) On treatment with KMnO4, heat gives (F). (F) Gives haloform test & on treatment with excess of CH3MgBr followed by hydrolysis (F) gives (G) which can not react with Sarret reagent. Answer the questions from 20 to 26. 20. Compound (A) is:-
CH—Et
OH
OH (a)
(b)
Cl
Cl
OH
(c)
OH
(d)
Cl
Cl
21. Compound (B) is:-
OH
CH —Et OH
(a)
OH
(b)
OH
(c)
O
OH
(d)
OH
236
Problems in Organic Chemistry
22. Compound (C) is:-
O
(a)
O (b)
(c)
OH
OH
OH
OH
OH
(a) Cl
Cl (b)
OH
(d)
OH
23. Compound (D) is:-
OH
Cl
(c)
Cl (d)
24. Compound (E) is:-
(a) Cl
Cl (b)
(c) Cl
(d) Cl
25. Compound (F) is:-
COCH3
COCH3 COCH3
CHO
(a) Cl
COCH3
(b) Cl
CHO
Cl (c)
(d)
None
26. Compound (G) is:COH(CH3)2
Cl (a)
COHMe2 COHMe2
CHOCH3
(b) Cl
COCH3 COHMe2
CHOHMe CHOHMe
(c) Cl
(d) Cl
Passage II An organic compound (A) on hydrolysis gives (B) along with ethyl alcohol & on treatment with LiAlH4 (A) gives (C) & ethyl alcohol. Compound (B) can be converted in to (C) by the help of LiAlH4. Compound (C) gives two moles of CH4 while (B) gives one mole of CH4 when treated with excess of CH3MgBr.Compound (C) does not give lucas test under ordinary condition. Compound (A) on treatment with excess of CH3MgBr/H3O+ gives (D) which gives haloform test but when (D) is treated with PCC followed by CH3MgBr(excess) / H3O+ compound (E) is formed which does not give haloform test.
237
Alcohol, Ether, Epoxide Grignard’s Reagents
Answer the questions from 27 to 31. 27. Compound (A) is: (a) HO
COOC2H5
(b) EtO2C
CO2Et OH
(c) OHC
COOEt
(d)
(b)
COOH
(d)
CH—CO2Et
28. Compound (B) is:-
OH (a)
COOH
(c) HOOC
OHC COOH CH2OH
HOH2C
29. Compound (C) is:CH2OH
(a) HO2C
COOH
CH2OH
(c) HOH2C
(b) HOOC
(d) HOCH 2
OH
30. Compound (D) is:-
COOMgBr (a) OH
OH (c)
(b) HO
CHOHCH3
(d)
CHOHCH3
CHO Ac
31. Compound (E) is:
(a) Me2HOC
(c) HOH2C
HO COHMe2 (b) COHMe2
COHMe2
(d) None
Passage III An organic compound (A) can decolourise Br2 water but does not react with Pd / CaCO3 / H2. (A) gives (B) on treatment with hot KMnO4 followed by reaction with NaBH4 / Ether. (A) Gives (C) on treatment with cold KMnO4. (B) Gives blue colouration in victor maeyer test. (B) gives (E) when heated with H3PO4 followed by treatment with hot KMnO4. When alkaline solution of (E) is electrolysed ethane is formed. (C) On reaction with H3PO4 gives (D) which does not give victor maeyer & lucas test. Answer the question from 32 to 36. 32. Compound (A) will be: (a)
(b)
(c) (CH3)2C = C(CH3)2
(d) EtC ≡ CEt
238
Problems in Organic Chemistry
33. Compound (B) will be:-
OH
CH3
OH
(a)
(b)
O
(c)
CHOH (d) CH3
34. Compound (C) will be:-
OH
OH OH (a)
(b)
OH
OH
OH
(c)
(d) Me2COHCOHMe2
35. Compound (E) is: (a) Pentanoic acid
COOH
(c) HOOC
(b) Propane 1,3- di carboxylic acid
(d) HOOC — CH = CH — COOH
36. Compound (D) will be:-
O (a)
(b)
O
(c)
(d)
O
Passage - IV An organic compound (A) gives 3 moles of CH4 when one mole of it reacts with excess of CH3MgBr. Compound (A) on dehydration gives (B). (B) gives +ve Br2 water test but -ve haloform test. (B) on reaction with CH2N2 gives (C) which on reaction with LiAlH4 followed by H3O+ gives cyclo propyl methanol. Compound (B) on reaction with cold KMnO4 gives (D). (D) Can be converted in to (A) by treatment with LiAlH4/Ether/H+. Answer the question from 37 to 40. 37. Compound (A) is:-
OH
OH
OH
(a)
(b) OH
OH
HO
(c)
OH
OH
OH (d)
CH2OH
HO
38. Compound (B) is:
(a)
OH
(b) CH2 = CH—CHO
O (d)
(c)
OH
OH
239
Alcohol, Ether, Epoxide Grignard’s Reagents
39. Compound (C) is:-
OH
(a)
CHO (b)
(c)
OH
OH
40. Compound (D) is:-
O
O
(a)
(b)
HO
HO
OH
(c)
HO
OH
(d)
O
(d) CH2OHCHOHCHO
CHO
41. Ether (A) when reacts with cold HI gives alcohol & iodide. Alcohol does not give haloform test but iodide can give haloform test (A) Will be:
(a) AcCH2– O
(c) Ph—O—CH2
O—CH2
(b)
COCH2I
(d) All of these
Ac 42. Which is not an example of SN2 reaction:-
Br
Br
(a)
O
OH OEt
Cl
HI NH2 (b) CH3OC2 H5 → CH3I + EtOH
NH3 → Methanol
AgNO3 → EtONa
(c)
(d) PhONa + CH3Br
PhOCH3 + NaBr
43. Correct order of boiling point is: (a) n — BuOH > i — BuOH > 2° — BuOH > 3°— BuOH (b) n — BuOH > 2° — BuOH > i — BuOH > 3° — BuOH (c) n — BuOH > 2° — BuOH > 3° — BuOH > i — BuOH (d) n — BuOH > i — BuOH > 3° — BuOH > 2° — BuOH 44. 2 — Cyclohexyl butan — 2 — ol can’t be prepared by:-
MgBr
Ac
Ether
Ether
→ (b) → + EtMgBr + Ethyl methyl ketone (+ ) (+ )
(a)
H3O
CHOHEt (c)
H3O
MeMgBr
Ag, heat → (d) → H O( + ) 3
OH 45. One mole of Ac
I
Mg THF
→ X
excess
COCl
2→ (+ )
H3O
NO2 COCHI2
OH
is treated with NaOH / I2 and then acidified, the product obtained after the removal of precipitates is treated with 3 moles of NaOH and another salt (A) is formed (A) will be:-
240
Problems in Organic Chemistry
NO2
ONa
(a) NaOOC
COCHI
(b) NaOOC
COONa ONa
ONa
ONa
NO2
OH
ONa
NO2
(c) NaOOC
COONa
NO2
(d) HOOC
COONa
OH
ONa
H CrO
2 4 46. R2CHOH → R2CO In this oxidation colour of the solution changes from (a) Orange to Red (c) Orange to Blue - Green
(b) Red to Blue (d) Orange to white
47. Which is not correctly matched:DMSO (a) RCH 2 OH → RCHO (Swern oxidation)
PCC
(b) R 2 CHOH → R 2 CO(Sarret oxidation) in CH Cl 2
CrO Pyridine
2
H 2CrO4 3 RCH 2 OH → RCHO (Colling oxidation) (d) (c) RCH 2 OH → RCHO (Robin oxidation) in acetone
48.
D
C OOH |
Br KOH DMSO → (A) → (B) DMSO H
CH 2
hot C H2 = CH — CH 2CH 2OH → KMnO 4
H
|
CH 2 —COOH
OH
(a) (A) & (B) both are
(b) (A) is
D H
& (B) is
H
(c) (A) is
OH
H
D
& (B) is
H
O
(d) (A) is
D OH
OH & (B) is
H
H
CH = CHCHO
(a) LiAlH4 / Ether / H+
(A)
PhCH2CH2CH2OH,
(b) H2 / Ni
D H H
49. Consider the following reagents: (i) LiAlH4 / Ether / H+ (ii) CH2N2 / H3O+ (iii) NaOH / CaO, mono chlorination, aq KOH (iv) B2H6 / THF For the conversion of cyclo propane carboxylic acid to cyclo propyl methanol we can use. (a) (i), (iv) (b) (ii), (i), (iii), (iv) (c) only (i) 50.
O
H
H D
D
(d) (i) & (iv)
Here (A) is:(c) NaBH4 / Ether
(d) (a) & (b)
51. Which reaction is least likely to occur ? (–)
Me2 CHCO (a) CH3 O + CH3COOH → CH3OH + CH3COO(–) (b) (–)
(–) + Me COH → Me3CO(–) + Me2 CHOH 3
(−)
(–)
(c) NH 2(–) + EtOH → Et O + NH3 (d) CH 2 = CH + BuOH → Bu O + CH 2
= CH 2
241
Alcohol, Ether, Epoxide Grignard’s Reagents
52. An alcohol (A) on heating with Cu / 300°C gives (B). (B) on oxidation gives (C). (C) on reaction with CH3MgBr / H3O+ gives (A). (A) will be:OH OH (a) 53.
CH3 CH3
OH (b) (c) OH (d) O CH3MgBr Cu → (A) → C—CH2 (+ ) 300°C (B) Here (B) is:H3O
O
(a) (CH3)3CCHO (b) CH3 — C = CHCH3 |
||
CH3 — C CH 2 CH 2 CH3 (c) (CH3)2CHCOCH3 (d)
CH3 C2H5SH, MeOH CH3MgBr/H3O(+)
54.
(C) (B)
MeOH, H(+)
(A) O Out of (A), (B) & (C) the compound which will not give victor maeyer test is:-
(a) (A) & (C)
(b) (B) & (A)
(c) (C)
(d) (B) & (C)
55. Which among the following reaction is not possible: (a) CH 2
O3 = CHCH 2CH 2OH → HCHO + CHO Me S |
2
(CH 2 )2 OH (b) CH 2
DMSO = CH — CH 2CH 2OH → CH 2 = CHCH 2 — CHO
(c) CH 2
MnO2 = CH — CH 2CH 2OH → CH 2 = CHCH 2 CHO
C OOH (d) CH 2
|
hot C H2 = CH — CH 2CH 2OH → KMnO 4
|
CH 2 —COOH
56. What will be the order of reactivity of following compounds with Grignard’s reagent?
CH3CHO
(I) (II) (a) I > II > III (b) III > II > I (c) II > I > III O
(III) (d) II > III > I
CH O(–)
TsCl 3 OH → (B) →(C)
57. (A)
O
Me
Pyridine
Identify the correct statement (a) Configurations of (A) & (C) are same (b) When (A) converts in to (B) configuration does not change (c) From (A) to (B) configuration changes but when (B) converts in to (C) configuration remains same (d) All have same configuration
242
Problems in Organic Chemistry
58. Consider the following two reactions having R1 & R2 rates of reactions respectively. ¾¾¾¾ ® H3O(+)
(1)
O RMgX ¾¾¾¾ ® (2) R H3O(+) OH
RMgX
(a) R1 > R2
(b) R2 > R1
OH
(c) R1 = 2R2
(d) R1 = R2
OH Phosphoric acid
Br → [X], product
59. CH3
R
[X] would be:-
CH3 O
O (a) CH3
Br
(b)
Br
O
O (c) CH3
(d) CH3
Br
Br 60. When cyclo hexane is added in ethanol then which of the following will happen (a) Heat is evolved & boiling point does not change (b) Heat is absorbed and boiling point increases (c) Heat is evolved and boiling point decreases (d) Heat is absorbed and boiling point decreases 2CH MgBr
3 61. (A) → (B) If B is acetone then A would be:-
(a) CH3COCl
(b) CH3COOC2H5
O
62. CH3
CH3 (1)
O
(2)
(d) All of these
O H
CH3
(c) CO(OC2H5)2
H
H (3)
Correct order of bond angle is:(a) 1 > 2 > 3 (b) 2 > 3 > 1
(c) 3 > 2 > 1
63. The common cyclic ether containing least number of carbons is: (a) Oxolane (b) Oxetane (c) Oxane
(d) 2 > 1 > 3 (d) Dioxane
243
Alcohol, Ether, Epoxide Grignard’s Reagents
O H( + ) H 2O
→
64.
Product, Product of this reaction will be:-
CH2OH
(a)
CH3
OH
OH CH2OH
(b)
(c)
CH3 OH
(A),
H SO
2 4→
CH3
OH
OH
65.
(d) OH
(A) is:-
o
HO
(a)
(b) OH
(c)
OH
(d) OH
66. Identify the correct esterification reaction, Here O18 = O* H( + )
* *
* *
1 (a) RC O OH + R1OH RC O OR + H 2 O
H * * 1 (c) RCOOH + R1 O H RCOOR + H 2 O
(+ )
OH
67.
HI excess
Zn Ether
→ →
(+ )
H * * 1 (b) RCO 2 H + R1 O H RCO O R + H 2 O
(d) all are correct
P, Product P is:-
OH
(a)
68.
(b) (+)
O
+ H3O
CH3OH +
(c)
(d)
CH2OH CHO
OMe
In this reaction which intermediate will not form during the mechanism.
(a)
O
H—OMe (+)
CH2OH
(b)
(c)
O—H
(+)
HO
OCH3
OCH3
(d) None of these
244
Problems in Organic Chemistry
COOCH3 Ph
Ph
H( + )
69.
→
A,
Product (A) would be:-
Ph
Ph
(a)
Ph
Ph
COOH Ph
CO
(b) Ph
Ph
Ph Ph
Ph O
OCH3
Ph
Ph
(c)
Ph
(d)
Ph
Ph
70. Which ether will not react with dil HI at room temperature ?
(a) (CH3)3COC (CH3)3
(b)
Φ2O CH2OH (c)
(d) All of these
O C H
Anh
6 6 → →(A) + HCHO + HCl (B), (A) & (B) respectively are:ZnCl dil H SO
71.
2
CHO
2
4
CHO
CH2OH CH2
(a)
&
2
Ph CH2OH
CH2OH
(c)
(b) CHO
&
&
(d)
CHO &
Cl Ph 72. The product obtained by the acid catalysed hydrolysis of cyclo propyl oxirane is:-
Ph
OH CH2OH
(a) HOCH2
OH (b)
CHOH
(c) both (a) & (b)
(d)
CH2OH
73.
OH
H( + ) Heat
→[P] major
CH3
[P] would be:-
(a) Toluene
(b)
(c) CH3
CH3
(d) Reaction is not possible
245
Alcohol, Ether, Epoxide Grignard’s Reagents
74. Match the following
Compounds
Dipole moment
(A) Et2O
(i) 1.18 D
(B) n - Pr - Cl
(ii) 2.10 D
(C) n - Pr - CHO
(iii) 2.72 D
(D) n - BuOH
(iv) 1.63 D
Correct option is:-
(a) A→ (i) ; B→ (ii) ; C → (iii) ; (D) → (vi)
(b) A→ (iv) ; B→ (iii) ; C → (ii) ; (D) → (i)
(c) A → (iii) ; B → (ii) ; C→ (iv) ; (D) → (i)
(d) A → (iii) ; B → (ii) ; C → (i) ; (D) → (vi)
75. Assertion- Benzyl ethyl ether on reaction with dilute & cold HI produces benzyl iodide
Reason- This reaction occurs via SN2 pathway
(a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true R O
2 → RMgX 76. Assertion: - RX + Mg
OR2
R
Here RMgX is present in the form of
Mg X
OR2
Reason: - Due to-I effect of halogen Mg is electron deficient thus, receive electrons from ether.
(a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true 77. Assertion- acylation of alcohol by acid halide is carried out in basic medium Reason- base neutralizes HCl produced in the reaction and avoid the formation of alkyl halide (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true 78. Assertion- OH group is better leaving group than I in acidic medium Reason- In acidic medium OH of alcohol picks up proton and comes out in the form of water (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true 79. Assertion: - Rates of acid calatysed dehydration of CH3CH2OH & CD3CH2OH are almost equal.
Reason: - Breaking of C—H or C—D bond is not rate determining steps.
(a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true
246
Problems in Organic Chemistry
80. Assertion: - Solubility of butyl alcohols follows following order. CH3
CH3 |
|
CH3 — C — OH > CH3CH 2 — CH — OH > CH3 — CH 2CH 2CH 2OH |
CH3
Reason: - As branching increases surface area decreases and solubility increases. (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true 81. Assertion: - Melting point of cyclo hexanol is greater than that of hexan – 1 – ol Reason: - Cyclo hexanol has less molecular wt. than hexan – 1 – ol (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true 82. Assertion: - Dipole moment of cyclo hexanol is lesser than phenol Reason: - In phenol resonance takes place & C - O bond acquires partial double bond character. (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true OH
83. Assertion: -
OH
does not undergo pinacole-pinacolone rearrangement.
Reason: - Due to resonance one C - OH bond becomes C = O (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true
Answer Key 1. (a)
2. (b)
3. (d)
4. (b)
5. (d)
6. (c)
7. (c)
8. (d)
9. (a)
10. (d)
11. (d)
12. (b)
13. (d)
14. (a)
15. (c)
16. (b)
17. (b)
18. (a)
19. (b)
20. (c)
21. (b)
22. (a)
23. (d)
24. (a)
25. (c)
26. (b)
27. (c)
28. (b)
29. (c)
30. (c)
31. (a)
32. (b)
33. (b)
34. (c)
35. (c)
36. (a)
37. (b)
38. (b)
39. (a)
40. (d)
41. (d)
42. (c)
43. (a)
44. (d)
45. (b)
46. (c)
47. (d)
48. (c)
49. (d)
50. (d)
51. (b)
52. (c)
53. (b)
54. (d)
55. (c)
56. (d)
57. (b)
58. (b)
59. (d)
60. (d)
61. (c)
62. (a)
63. (b)
64. (a)
65. (c)
66. (b)
67. (b)
68. (a)
69. (c)
70. (c)
71. (b)
72. (b)
73. (a)
74. (a)
75. (c)
76. (c)
77. (a)
78. (a)
79. (a)
80. (a)
81. (b)
82. (d)
83. (a)
247
Alcohol, Ether, Epoxide Grignard’s Reagents
Multiple Choice Questions (More Than One May Correct) 1. Which of the following will not release H2 on reaction with Na metal? C
(a) Me2O
(b) n - BuOH
CH=CH2
CH
(c)
(d)
O
OH
2. Which of the following will give yellow precipitate with NaOI? (a)
OH
(b)
O
(c)
(d)
O
OH
3. Which of the following transformation is not possible by NaBH4? (a)
O
OH
(b)
COOEt
CH2OH + EtOH OH
O
(c) CHO
CH2OH
O
(d)
O EtOH
O
4. Which will undergo oxidation by IO4(-)? OH
(a)
OH
OH
CH 2OH
OH
(b)
|
C = O (c) | OH OH CH 2OH
(d)
OH
5. In which of the following case alkane is producing.
→ (a) Glycerine + EtMgBr
(c) CH3 — CH — CH3 →
Red P + HI
|
110°C
→ (b) Glycerine + oxalic acid (i) SOCl (ii) Na / Me2O
2 →
(d) OH
OH 6. Which is not the preparation of epoxide:Cl
Cl / Water
2 → aq Na 2CO3 (b) →
(a) HO
(c)
PAA
→
7. Which among the following will not be oxidized by CrO3? (a) 2-Methyl propan 2- ol (c) 3-Phenyl propan-2-ol
(d) C H 2 ||
CH 2
heat
+ Ag + 1/ 2O 2 →
(b) PhOH (d) CH3CH2OCH2CH3
248
Problems in Organic Chemistry
8. In which case products formed are not according to reaction? → PhCH 2CH 2CH 2OH (b) PhCH = CHCHO + LAH
∆ (a) Gycerine + conc H 2SO 4 → CH3CHO
D
(c) PhCOOCH3 + NaBH 4 → PhCH 2OH + CH3OH
O
(d)
LAH
→ (+ )
OH
D
9. Which of the following is not the preparation of alcohol? (a) Sucrose + H3O( + ) → (c)
(b)
O / H O / NaBH
3 2 4 →
Cl
aq KOH
OH
→
(d) Sucrose + yeast →
10. Which of the following will not show Lucas test?
(a) Phenol
(c)
(b)
OH
(d)
OH CF3
CF3
CF3
OH
Answer Key 1. (a), (d)
2. (c), (d)
3. (b), (d)
4. (a), (b)
5. (a), (c), (d)
6. (a), (b)
7. (a), (b), (d)
8. (a), (c), (d)
9. (a), (b)
10. (a), (b), (d)
LEVEL - II CH2OH
1. CHOH + CH3MgBr → 44.8 lit CH4 + [X], [X] would be:CH2SH (1 mol) CH2OMgBr
CH2OH
CH2OMgBr
CH2OMgBr CHOMgBr (c) CH2OMgBr (d) CH2SH CH2SMgBr
(a) CHOH CHOMgBr (b) CH2SMgBr
CH2SMgBr
2. Which among the following will consume more than two moles of Grignard’s reagent? O
(a) O O
OH
O
O
(b) (c) OH O COOH
OEt
(d) Br
3. Dehydration of alcohol is performed by concentrated H ion. It involves 3 transition states (TS)1 (TS)2 & (TS)3. If free energy (+)
changes for these transition states are ∆G*1, ∆G*2 & ∆G*3 respectively them:-
∆G*1 > ∆G*2 > ∆G*3 (b) ∆G*2 > ∆G*3 > ∆G*1 (c) ∆G*3 > ∆G*2 > ∆G*1 (a)
* * * (d) ∆G 2 > ∆G 1 > ∆G 3
249
Alcohol, Ether, Epoxide Grignard’s Reagents
Et
4.
CH MgBr
3 ¾¾¾¾¾ ® Product H O( +) 3
O Products in this reaction will be
(a) Emnantiomers
(b) Stereoisomers
(c) Diastereomers
(d) Geometrical isomers
O CH MgBr
5.
3 ¾¾¾¾¾ ® [X] ( +)
H3O
Select correct statement regarding [X] (a) It contains two chiral ‘C’ atom (b) It can show geometrical isomerism (c) On acid catalysed dehydration [X] givens [Y] which can not show geometrical isomerism (d) [X] contains plane of symmetry
6. Consider the following reactions CH3 |
1. CH3CH 2CH 2 — C — CH 2OH |
(+ )
H → Major Product ∆
CH3 CH3 |
CH3 — C — CH 2 — (CH 2 )2 — CH3 2. |
H( + ) ∆
→ Major Product
OH H( + ) ∆
3. i — Bu — CH 2 — CH — CH3 → Major Product |
OH CH3 |
H( + )
4. CH3 — CH —CH — CH 2 Et → Major Product |
OH
Reaction with same major product is / are:(a) 1, 2, 3 & 4 (b) 2, 3 & 4
(c) 1 & 3
(d) 2 & 4
7. Consider the following reactions. CH2OH
1.
OH H( + ) ∆
→ Product
H( + ) ∆
OH →
2. HO
Product
Cl
3.
OH H( + ) → Product ∆ OH
Cl
Cl
Zn dust heat
→ Product
4. Cl
Cl Cl
250
Problems in Organic Chemistry
Cl Cl
Cl Zn dust heat
→ Product
5. Cl
Cl Cl
Reaction in which benzene is formed as a product is / are:(a) 1, 2, 3, 4 & 5 (b) 1, 3 & 5
(c) 1 & 5
(d) 1, 2, 3 & 4
8. Consider the following acid catalysed transformation. OH
H(+) ∆
OH
→ O
Following intermediates are expected in this transformation
OH
(+)
1.
2. OH
(+)
3. OH
4.
(+)
(+)
OH OH
5.
(+)
The intermediate which is not produced during this transformations is / are :(a) 2, 3 & 5 (b) 3, 4 & 5 (c) 2, 3, 4 & 5
(d) Only 4th
9. Alcohols are prepared in following reactions.
D( + ) H 2O
D( + ) D 2O
(+ )
H( + ) H 2O
H 1. CH3CH = CH2 → 2. CH3CH = CH2 → 3. CH3CH = CH2 → 4. CH3CH = CH2 → D 2O
Select the correct statement regarding these reactions. (a) Except 4th reaction two products are formed in rest of the reactions. (b) In 1st & 2nd reactions two products are formed (c) In 2nd reaction two products are formed (d) In 1st and 3rd reactions two products are formed CH3 NaNO2 + HCl → [ X] CH2NH2
10. i–Pr i–Pr
We have following statements about the proposed reaction. 1. [X] is optically active 3. [X] shows haloform test
2. [X] gives red colour in victor maeyer test H( + ) Heat
ozonolysis
4. [X] →[Y] →
O +
CHO
251
Alcohol, Ether, Epoxide Grignard’s Reagents
Out of these four statements correct statements are:(a) 1 & 4 (b) 1, 2, 3 & 4
(c) 1,3 & 4
(d) 2,3
11. Nucleophilic addition of Grignard’s reagent is not possible in O O || || CH C CH C - CH3 (c) CH COCOCH 3 2 (b) 3 3
OH
(a)
O C–OH
(d)
Matrix Match Questions Each question contains statement given in two columns which have to be matched. Statement (A, B, C , D) in column I have to be matched with statements (p, q, r, s) in column II. The answer to these questions have to be appropriately bubbled as illustrated in the following examples. If the correct match are A – p, A – s, B – r, B – r, B – q, C – q, D – S, then the correctly bubbled 4 × 4 matrix should be as follows. A
p
q
r
s
B
p
q
r
s
C
p
q
r
s
D
p
q
r
s
12. Select the reagents form list -II by which you can differentiate the pairs of compounds given in List I List - I List - II (A)
& OH
(B)
OH
& OCH3
&
HO—CH—CH3
13.
(B)
LiAlH
O
4→ (+ )
H
CH MgBr
O
(s) Anhydrous ZnCl2 + HCl
CH2OH
List - I (Reaction)
(A)
(r) KOI
OH
& OH
(q) Br2 water
CH2OH
(C)
(D)
(p) Na
3 → (+ )
H3O
List - II (Product) (p)
OH
(q) OH
252
Problems in Organic Chemistry
(C)
3 → (+ )
OH
(r)
CH MgBr
O
OH
H3O
O
LiAlH
4→ H( + )
(D)
14.
(s)
List - I (Reaction)
OH
List - II (Missing Reagent) CH3
CHO ?
(A)
(p) Zn / Hg + HCl
?
→ Ph(CH2)3OH (B) PhCH = CH — CHO
→ PhCH = CH—CH2OH (C) PhCH = CH—CHO
(q) H2 / Ni
?
?
→ CH3CH2—CH2CH2OH (D) CH3CH = CH—CHO 15.
(r) NaBH4 / Ether / H(+) (s) LiAlH4/Ether/H(+)
List - I (Compond) OH
(A)
List - II (Test)
(B)
OH
(p) Evolve H2 with Na metal
(q) Yellow ppt with KOI
CH2OH
(C)
(r) +ve victor Maeyer test
(D) OH (s) +ve lucas test at room temperature 16. Organic compounds W, X, Y and Z are formed in following chemical reactions. CH3 |
NaNO + H SO
2 2 4 CH3 — C — CH 2 NH 2 → W
|
AgNO
3→ X Ph 2CIC(OH)(CH3 )2
CH3 AgNO3 →Y Ph 2CIC(OH)(CH3 )2 → X W heat Ag
cold and dilute HI
Me3COCHMe2 → alkyl Iodide + Z
Select the reactions from list II which is exhibited by W, X, Y and Z List - I (Compounds) List - II (Reactions) (A) W (p) Can give yellow ppt with KOBr (B) X (q) infinitely soluble in water (C) Y (r) Decolourise Br2 water (D) Z (s) Evolve H2 with sodium metal.
Answer Key 1. (b)
2. (a), (b), (c)
3. (b)
4. (b), (c), (d)
6. (a)
7. (d)
8. (b)
9. (d)
11. (a), (b), (d)
5. (b), (c), (d) 10. (a)
253
Alcohol, Ether, Epoxide Grignard’s Reagents
Answers matrix match 12. (A) - s, (B) - p, (C) - r, (D) - q,s 14. (A) - p, (B) - q, s, (C) - r, (D) - q 16. (A) - s, (B) - p, (C) - r, (D) - p, q, s
13. (A) - p, (B) - q, (C) - s, (D) - s 15. (A) - p, (B) - p, r, s (C) - p, r (D) - p, q, r, s
SOLUTION NOCl
1. (a)
aq KOH
CH2NH2
CH2—Cl
CH2OH
(+)
(+) Ring opening
heat (+) –H
(+)
CH2
H –H2O
2. (b) Because in acidic medium it removes H2O & generates carbocation (+ )
(+ ) → R OH 2 → H 2O + R ( + ) (Carbocation) ROH + H
OH
OMgBr
3. (d)
CH3MgBr
Br
O Br
CH4 +
O Heat ring opening (+) (+)
H
4. (d)
OH
1
(+)
hydride shift
2 (+)
–H
H
OH
5. (d)
H( + ) –H 2O
→
OH
(+)
O
→
–H( + )
O
Cl
Cl NaOH
→
→ ••
••
OH
(+)
OH + H
(+)
H
(+)
O
H (+)
–H
O
6. (c) Due to intra molecular H– bonding. ON
••
→
O(–)
OH
(+)
••
OH
••
O2H H Intramolecular–H–bonding
254
Problems in Organic Chemistry
7. (c) It is an example of pinacole - pinacolone rearrangement (+)
H
H
(+)
OH
OH
(+)
(+)
OH
OH
(+)
(+)
OH
OH
(+)
–H
(+)
–H
O
COCH3 (a)
(b) (AcO) Pb
4 → 2CH CHO → + ve haloform test 8. (d) CH3CHOHCHOHCH3 3
(AcO) Pb
4 → + ve haloform test CH3CHOHCOCHOHCH3 → 2CH3CHO + CO 2
H
H
(+)
OH
9. (a)
H
(+ )
→
OH
OH
H H Here H - migration does not occur because always anti migration takes place hence contraction in ring occurs. H
OH (+)
–H( + )
→
CHO
(+)
→
OH H
10. (d) Acid catalyzed dehydration occurs by E1 pathway hence for arrangement stability of carbonium is considered. 3rd can not produce alkene on acid catalyzed dehydration. 11. (d) Product is optically inactive due to the presence of plane of symmetry. D
H
D
OH
H
H
OR H
OH OH
H
D
12. (b)
D H
OR
OH
13. (d) In a & c active methylene compounds are produced while in b acetic acid is produced . Both active methylene compounds & acetic acid are inactive towards haloform test. 14. (a) At - tert - butoxide can oxidize 2° alcohol in to ketone without affecting double bond jones reagent at 0° C behaves as mild oxidizing agent but above 25° C it can oxidize multiple bond.
255
Alcohol, Ether, Epoxide Grignard’s Reagents
OCOCH3 |
15. (c) Φ 2C(OH)CH3 + CH3COCl → Φ CCH3 + HCl The ester formed in this reaction is easily attacked by H+ of HCl (+ ) H– O COCH3
OCOCH3 |
(+ )
|
Φ 2 C — CH3 + H ( + ) → Φ 2 — C — CH3 → CH3COOH + Φ 2 C CH3
(resonance stabilized)
(+ )
Φ 2C CH3 + Cl(–) → Φ 2 C CH3 |
Cl
16. (b) Alcohols can be converted in to halides by SOCl2, PCl5, PCl3 & P + X2. In case of SOCl2 configuration of the reactant does not change after reaction (Retention)
Hence in the given question configuration of reactant & product will be same 4
H 2
1
Pr
Cl ‘S’
3CH3
17. (b) CH2N2 can not methylate less acidic hydrogen of aliphatic alcohols 18. (a) Alcohols can be converted in to halides by SOCl2, PCl5, PCl3 & P + X2. In case of SOCl2 configuration of the reactant does not change after reaction (Retention) while in case of P + Br2 inversion in configuration takes place (see q.n. 16) 19. (b) Anhydrous ZnCl2 + HCl is called Lucas reagent H
(+)
(+)
(+)
H CH3CH(Ph)CHOHCH3 –H O 2
CH3C—CH—CH3
rearrangement
Ph
CH3—C—CH2—CH3
Cl
Ph (–)
CH3—C—CH2—CH3
Cl
Ph
Passage - I (20 to 26) PCC in CH2Cl2 is called sarret reagent. It can oxidize 1° and 2° alcohols in to aldehydes and ketones respectively. (D) reacts with Lucas reagent with is 5 minutes. It means it is 2° alcohol (D) on dehydration gives (E) which can decolourise Br2 water. Hence E is alkene. F gives haloform test hence F must have ‘CH3CO’ linkage. G does not react with sarret reagent (PCC in CH2Cl2) hence it is a 3° alcohol. (A) reacts with Lucas reagent immediately hence it is 3° alcohol.
256
Problems in Organic Chemistry
Passage - II (27 to 31) (C) does not gives Lucas test under ordinary conditions it indicates that it is primary alcohol. (D) gives haloform test hence (D) must have CH3CO linkage. (D) should have either 1° or 2° alcoholic group because it is oxidized by PCC. E does not give haloform test hence it should be 3° alcohol & (D) should be 2° alcohol. (A) on hydrolysis gives alcohol hence it may be an ester. Since (B) given are mole & (C) given two moles of CH4 on treatment with CH3MgBr hence (B) contains 2OH groups COOEt
OHC
Hydrolysis
COOH + C2H5OH
OHC (B)
(A) LiAH4 LiAlH4
CH2OH + C2H5OH
HOCH2 (C) OH
(A)
CH3MgBr (Excess) H2O(+)
CH3
CH3CH
C—CH3 (D)
O CH3—C
OH
OH
CH3 C—CH3
+ve haloform test
OH
OH CH3MgBr (Excess) H3O(+)
CH3—C CH3
–ve haloform test
C—CH3 (E)
CH3
257
Alcohol, Ether, Epoxide Grignard’s Reagents
Passage - III (32 to 36) A decolourises Br2 water but does not react with Pd/CaCO3 hence it may be an alkene. B gives blue colour in victor maeyer test thus B must be a 2° alcohol. Alkaline solution of (E) on eletrolysis gives ethane thus, (C) should be a carboxylic acid (D) does not give victor maeyer test and lucas test hence (D) is not an alcohol. O OH • Hot KMnO4
(+)
NaBH4
H
(A)
heat
(B)
cold KMnO4
OH
,
CH2 = CH2 + 2CO2 OH
OH H
1 4
1 4
Hot KMnO4
(E)COOH
(+)
2
3
(+)
COOH
OH(–), Electrolysis
5
5
(C)
(+) OH 2
1
(D)
3
Passage - IV (37 to 40) A contains 3 - active hydrogen’s because it produces 3 moles of CH4 when reacts with Grignard’s reagent (B) Gives -ve haloform test but positive Br2 water test it indicates that (B) contains C - C multiple bond. B is obtained by the dehydration of A hence (B) must be an alkene. CH2OH CHOH
CH2 H2SO4
CH2OH (A)
CH
CH2N2
CHO (B)
CH2OH Cold KMnO4
41. (d)
CH2—CH—CHO OH
LiAlH4
CH2OH
(C)
(B)
CHO
–H
cold HI
CH2
(+)
LiAlH4
OH (D)
AcCH2—O
(+)
H2O
(+)
H
CH2—CH—CH2 OH
OH OH (A)
AcCH2I + OH +ve haloform test
O—CH2
cold HI
COCH2I
ICH2
+ phenol COCH2I
+ve haloform test
O
258
Problems in Organic Chemistry
O—CH2
ICH2
cold HI
+ phenol
Ac
Ac
+ve haloform test Cl
42. (c)
OEt
(+) Ag
EtONa
(+)
–AgCl, SN1
Hydride shift
43. (a) Boiling point ∝ surface area OH I
44. (d)
MgI
Mg
COCl2
H3O(+)
THF
excess
45. (b) OH adjacent to NO2 is less acidic due to intramolecular H–bonding. OH
OH
NO2
Ac
COCHI2
haloform and acidification
OH
OH
HO2C
COONa
COOH OH
NO2
NaOOC
NO2
3 moles of base
ONa
46. (c) Orange (H2CrO4) to Blue - Green 47. (d) H2CrO4 in acetone in called jones reagent, thus, it is jones oxidation 48. (c) In presence of DMSO, SN2 reaction occurs Br
H KOH,DMSO
D H
DMSO
D
D
O
OH
H H H 49. (d) 50. (d) However LiAlH4 does not reduce C = C bond but LiAlH4 reduces C = C bond when it is present in conjugation with phenyl group. CH = CH—CHO
LiAH4
CH2—CH2—CH2OH
H2/ Ni
51. (b) Because Me3CO(-) is stronger base than Me2CHO(-). So Me2CHO(-) can not remove H(+) from Me3COH (+)
52. (c)
OH (A)
Cu, 300°C
OH (B)
Oxidation
OH (C)
CH2MgBrH3O
(A)
259
Alcohol, Ether, Epoxide Grignard’s Reagents
53. (b) CH3
O C—CH2
CH3
CH3MgBr SN2
OMgBr C—CH2CH3
CH3
OH
CH3
(+)
H3O
C—CH2CH3
CH3
CH3 Cu,300°C
CH3—C = CH—CH3 CH3 (+)
H , CH3OH SN
(A)
1
OCH3
CH3MgBr, H3O
54. (d)
SN
OH
(+)
(B)
2
O
OH
C2H5SH, MeOH SN
(C)
2
H5C2S
OH
Since B & C are tertiary alcohols hence they will not exhibit victor maeyer test 55. (c) Because MnO2 oxidises allylic alcohol. 56. (d) O
O OH
O OTs
Me (–)
CH3O
TsCl, Pyridine –HCl
57. (b)
(–) + OTs
2
SN
(A) Me
(B) Me
(C) OCH3
From (A) to (B) configuration does not change. From (B) to (C) inversion in configuration occurs because reaction occurs via SN2 pathway. 58. (b) This reaction is an example of SN2 reaction. In first substrate steric hindrance is more, thus, rate of reaction will be less 59. (d) Migrating aptitude of phenyl group is greater than p–Br–C6H4– OH OH OH H
Br
CH3
(+)
(+)
Br
CH3
OH (d)
–H
(+)
(+)
CH3
Br
H |
C 2 H5 |
C 2 H5 |
60. (d) C2 H5 — O.......H — O.......H — O..... ( Before addition of cyclo hexane) C2H5OH
C2H5OH (After addition of cyclo hexane)
It is an example of non ideal solution with positive deviation. In ethanol hydrogen bonds are present when cyclo hexane is mixed in it H–bonds get broken down for this heat is absorbed.
260
Problems in Organic Chemistry
Boiling point ∝ H - Bonding Hence due to breaking of H - bonds, heat is absorbed & b.p. decreases O
O
||
||
CH3MgBr
CH3MgBr 61. (c) H5C2 O — C OC2 H5 → CH3 — CO 2 H5 → Acetone(B)
62. (a) In each case oxygen contains two lone pairs of electrons thus, due to lp–lp repulsion contraction in bond angle takes place . This contraction in angle will be maximum in water & minimum in di methyl ether because size of methyl group is larger than hydrogen lone pair lone pair repulsion ••
••
O Group Group O
63. (b)
O
O
O Oxolane
Oxetane
O Dioxane
Oxane
(+)
64. (a)
O–H
O
CH2OH
CH2OH
(+)
H
H2O
(+)
–H(+)
OH (+)
(+)
H
65. (c)
OH
(+)
O
OH –H
(+)
OH (+)
O
OH OH OH (+) 1* 1* ROH ROH 66. (b) R — C— O — H + H ( + ) R—C—OH R—C—OH2 R—C—OH 1 O(+) O—R 1 * * R H OH O ||
* 1
(+)
* 1
–H R—C—OR R—C—OR (+)
67. (b)
OH OH
HI (Excess)
I
(+)
–H
HI
I
–I2
I (Frank land reaction)
Zn, Ether
261
Alcohol, Ether, Epoxide Grignard’s Reagents (+)
68. (a)
H
O
(+)
O
H
H2O
MeO
OMe
OMe
–H2O
OH
(+)
OH MeO
OH2
OH unstable
CH2OH
–MeOH
CHO H
(+)
COO
COOCH3 Ph
Ph
Ph
(+)
Ph
Ph
Ph ArSE
Ph
Ph
Ph
Ph (+)
••
O—Ph ••
70. (c)
O
–CH3OH
H
69. (c)
(+)
CO
CH3
O—Ph
•• (–)
Due to double bond character in C - O bond this ether does not react with HI 71. (b) HCHO + HCl → CH 2OHCl (+ )
→ ZnCl3(–) + CH 2OH CH 2OHCl + ZnCl2 (+)
(+)
CH2OH
+ CH2OH
CH2
(+)
H
C6H 6
A
Ph2CH2 B
H (+)
O
O
72. (b)
OH OH
•• H2O
••
(+)
H
CH—CH2
CH—CH2
CH—CH2 CH3
73. (a)
(+)
OH CH3
(+)
H
–H
(+)
(+)
CH3
CH3
74. (a) 75. (c) Reaction occurs via SN1 pathway because PhCH2+ is formed. 76. (c) Solvation of grignard’s reagent takes place by ether 77. (a) Base neutralizes HCl which is produced during acylation and avoid the formation of alkyl halide → Me3COCOCH3 Me3COH + CH3COCl (+ )
–CH COOH
(+ )
Cl(–)
3 Me3COH + H ( + ) (from HCl) → Me3C — OCOCH3 →(CH3 )3 C → Me3CCl
|
H Hence to avoid the formation of alkyl halide base should be added to neutralized acid (HCl) (+)
78. (a) In acidic medium OH (alcoholic group) converts itself into O H 2 & release water. 79. (a) Breaking of C – H or C – D bond is not rate determining steps. In slowest step breaking of C– O bond takes place
262
Problems in Organic Chemistry
80. (a) As branching increases, surface area of hydrocarbon part (hydrophobic part)decreases so solubility increases 81. (b) Molecules of cyclohexanol are more closely packed in comparison to that of hexan–1–ol (+)
••
O—H ••
82. (d)
O—H
•• (–)
Due to resonance bond length of C– O bond decreases thus dipole moment of phenol is found to be lower than that of cyclo hexanol. 83. (a) Due to resonance it acquires C = O bond hence, no need to show pinacole-pinacolone rearrangement ••
(+)
OH
••
OH
OH
OH (–)
More than one may correct:1. a, d 2. c, d (a) can not show haloform test as it is 30 alcohol while is (b) CH3CO linkage is absent. 3. b, d 4. a, b (c) & (d) are not vicinal diols so these alcohols can not show oxidation with IO4(-) 5. a, c, d 6. a, b 7. a, b, d CrO3 can oxidize 1° and 2° alcohols 8. a, c, d 9. a, b 10. a, b, d Except (c) rest all give unstable carbocation LEVEL -II 1. CH2OH
More acidic hydrogen
CHOH
less acidic (2°alc)
CH2OH
More acidic hydrogen
Since 44-8 litre CH4 is liberated thus 2 active hydrogens are removed by grignard’s reagent. Hence (b) option is correct. 2. (a) a, b, c 3. (b) (TS)2
(TS)3
(TS)1 P.E
(+)
(+)
ROH2 R
Thus ∆G*2 > ∆G*3 > ∆G1*
alkene
263
Alcohol, Ether, Epoxide Grignard’s Reagents
4. (b) b, c, d OH
O
CH3
CH3
OH
+ Me Mg Br
Et
5.
(+) → H3O
Et
Et
(b, c, d) O
OH
R Rmgx
→ H O
[X]
2
(+)
H OH → D
R
[X] can show geometrical isomerism as : -
Because [X] contains plane of symmetry hence it can not show optical isomerism 6. (a) In 1st reaction propyl shift occurs. CH3
CH3
(+)
(+)
H
CH3CH2CH2 —C—CH2OH
CH3CH2CH2–C–CH2 CH3
CH3 CH3
CH3
(+)
C=CH—CH2CH3
(+)
–H
(+)
CH3
C—CH2CH2CH3
CH3
(+)
CH2OH
CH2
(+)
7. (d)
(+)
H –H2O
–H (+)
(1)
OH
OH (+)
OH
OH
H –H2O
H
(+)
(+)
OH
CH2OH (+)
OH OH OH
–H
(+)
OH –H
(+)
OH
OH
(1) (+) (+) In 4th Case anti elimination is not possible hence here benzene does not form?
(3)
264
Problems in Organic Chemistry (+)
(+)
H –H2O
8. (b)
••
OH
OH
OH
••
+
(+)
–H
+
O
O
H (+)
*
D 9. (d) (1) CH3 —= CH CH 2 → CH3 — CH — CH 2 D H 2O
|
OH
(opticallyactive)(d & 1 isomers) (+)
H (2) CH3 —= CH CH 2 → D 2O
CH3 —CH — CH3 |
OD
Optically inactive (Only one product) D( + ) D 2O
*
(3) CH= → CH 2 —CH — CH 2 D 3CH CH 2 |
OD
(Opticallyactive) (d & 1 isomers) H( + ) H 2O
(4) CH= → CH3 — CH — CH3 3CH CH 2 |
OH
(Optically inactive)
CH3
CH3
10. (a) i-pr
CH2NH2
NaNO2 + HCl
CH3 (+)
i-pr
i-pr
CH2
(+)
i-pr
i-pr
CH3 i-pr
CH2 i–Pr
optically active OH (+)
CH2I
11. (b)
CH2 OH
Ag
(+)
OH OH
O
H2O
CH2 i–Pr
9
Aldehyde & Ketone
Main Features Reaction Chart for Aldehyde & Ketone Preparation
Properties
266
Problems in Organic Chemistry
LEVEL - I Multiple Choice Questions H+
NaBH
4 → (B) → no reaction, Here (A) is :1. (A) D Ether
|
↓
Blue colour in victor Maeyer test
O
(a)
CHO (b)
(c) (d)
CO2C2H5
O 2. Choose the answer that has the following compounds located correctly in the separation scheme. Propanone - A , Methanol - B, AcOH – C. NaHSO
Pass through
3 → ppt + solution A + B + C → ppt + Solution Ca(OH)2
(1)
vapours ppt (1) ppt (2) (a) A B (b) B C (c) C B (d) C A
(3)
solution (3) C A A B
LiAH / Ether
3.
COPh
4 →(X), (X) wiil are:(+)
H
OH
CH—Ph
(a)
CHOHPh
(b)
O
Ph
(c)
CHOHPh
O
(d)
OH
4. Cyclo petntanone can be converted in to 5 – Hydroxy pentanoic acid by: (a) hot KMnO4 (b) SeO2, H3O+ (c) NH2OH/H3O+ 5. OHC 6.
COCH3
OHC
(d) CH3CO3H & H3O+
CH2CH3
It can be performed by:(a) glycol / H+, Zn / Hg + HCl, H3O+ (b) N2H4 / OH– followed by H+ – (c) OH / ∆ , H2 / Ni (d) (a) & (c) An organic compound reacts with Tollen’s reagent but undergoes Baeyer villager oxidation. The compound is:O (a) C2H – CH2 – COCH3 (b)
(c) CH3 – CHO
(d) CH3COCH3
267
Aldehyde & Ketone
7. Consider the following processes
Process - I
B2H6 / H2O2 / OH–, Al (OEt)3
Process - II
Pd - C / H2, Hot KMnO4, CH3OH / H+
Process - III
HgSO4 / H+, CH3COOH / O3
Propyne can be converted in to methyl ethanoate by using the process:-
(a) I, II, III
(b) II, III
(c) I, III
(d) II
8. Identify the reaction which is not the preparation of ketone Gilaman reagent
(a) CH3COCl → Product Hot KMnO 4 Product →
(b)
Al(OEt)3 CH3MgBr(excess) (c) CH3CHO → → (+) H3O
(d) End product of all reactions is ketone
9. The reaction in which hydride shift does not occur is:H O( + )
(a)
OH
3 →
H( + ) D
(c)
→
(–)
OH (b) ΦCHO → ΦCH 2 IG + ΦCO 2(–)
(d) None
OH Ca(OH)
10. CH3CHO + HCHO 2 → Products of this reaction will be:D (1
:
4)
(a) (CH2OH)3CCOO– + (CH2OH)3CCH2OH (c) CH3COO– + CH3OH
(b) (CH2OH)4C + HCOO– (d) (CH2OH)3CCHO
11. Which reaction is not correct? (–)
OH (a) CHO — CHO → HOCH 2 — COO(–) (–)
OH (b) PhCOCHO → PhCHOCOO(–) n
NaOH Sol (c) 2Me2 CHCHO → Me2 CHCO 2(–) + Me2 CHCH 2 OH
(d) All are correct
200°C
12.
(1) & (2) are respectively:(a) CH2 = C = O & Br2 in CCl4 (c) Na|NH3 & CHBr3 + KOH
(b) H2|Pd|C & CHBr3 + KOH (d) CH2N2 & HBr (2 moles)
268
Problems in Organic Chemistry (+)
(i) ΝΗ3/Η
(i) CH3MgBr/H3O(+) (i) Φ3P = CH2
13. Cyclopentanone
(A) (B), which statement is correct :-
(+)
(ii) H3O (+)
LiAIH4/ Ether/H
(a) (b) (c) (d)
(C)
(A) is ketone however (B) & (C) are alcohols (B) &(C) gives blue colour however (A) does not produce any colour in victor maeyer test (B) & (C) are same compounds and give blue colour in victor Maeyer test however (A) is primary amine (A), (B) & (C) all are identical
14. Arrange the following four compounds in order of their rate of addition with NaCN / H(+) CHO CHO
CHO NO2
OMe
1
CHO MeO
3
2
Correct order of rate of addition with HCN is:(a) 2 > 1 > 3 > 4 (b) 1 > 2 > 3 > 4
4
(c) 2 > 3 > 1 > 4
(d) 3 > 4 > 1 > 2
15. Cyclo pentene is treated with cold KMnO4 followed by lead tetra acetate to give (A) when (A) is heated with Ba(OH)2 the product obtained will be:(–)
(a)
COO
OH
(b)
CH2OH
COO
(–) (c)
(d) CHO
16. HCHO can be separated from the mixture of HCHO, CH3COCH3 & CH3CHO by treating the mixture with; (a) NaHSO3 (b) 2, 4 DNP (c) Semi carbazide (d) [Ag (NH3)2] (+) 17. Pyroligneous acid does not contain (a) HCHO (c) CH3CH2OH 18.
(b) CH3CHO (d) All of these are absent in pyroligneous acid
Which will not show reducing property with tollens & fehling solution? I - Hemi acetal of acetaldehyde II - acetal of formaldehyde III - Hemi ketal of acetone IV - formic acid (a) III & II (b) II, III & IV (c) I, II, III & IV
19. Which base catalysed reaction is least likely to occur? O
(a)
EtONa D
→
+
O
(b)
+
EtONa D
→
O
(c) CH3COCH2COOC2H5 +
(d) (a) & (b)
EtONa D
→
CO2Et |
Ac – C = CMe2
(d) I & II
269
Aldehyde & Ketone
20. Which will not exhibit cannizaro reaction? CHO CHO
(a) Ph – COCHO
(c)
(b)
(d) All of these can show cannizaro reaction
21. Which among the following is called Tollen’s reaction?
OH (–)
→ Me2 C — CH 2 OH + HCOONa (a) (CH3 )2 CHCHO + HCHO 1
:
|
2
CH2OH
CH3CHO + [Ag(NH3 )2 ]( + ) OH (–) → Other + Ag ↓ + CH3COO(–) (b) Product OH
OH (–) H 2O
→
(c) O
COOH
O
(d) CH ≡ CH + Ag 2 O → H 2 O + AgC ≡ CAg
Passage - I An organic compound (A) [C8H8O] gives (B) & (C) on reaction with NH2OH / HCl. (B) & (C) give (D) & ( E) on reaction with sulphuric acid with formula (C8H9NO). When (D) is boiled with alcoholic KOH an oily compound (F) separates out. (F) Rapidly reacts with acetyl chloride to reproduce (D). (E) on boiling with KOH followed by acidification gives a white solid (G). If benzoic acid is produced by the oxidation of (A) then: Answer the questions from 22 to 27. 22. (B) & (C) is the mixture of:
(a) Syn and anti PhCH2CH = NOH
(c) Syn and anti
C3H7
(b) Syn and anti Ph(CH3)C = NOH
NOH
(d) both (a) & (b)
23. Compound (D) is: (a) CH3CONHPh
(b) PhCONHCH3
(c) PhCH2NHCHO
(d) PhCH2NHCH3
24. Compound (F) is an : (a) aliphatic amine
(b) aromatic amine
(c) aliphatic aldehyde
(d) aliphatic ketone
C4H9
25. Compound (A) when subjected to heat with aq KOH another compound (H) is formed. Which is correct about the newly formed compound (H) : (a) (H) can show addition with sodium bi sulphite (b) (H) can decolourise bromine water (c) (H) can not exhibit haloform test (d) All of these 26. Compound (G) is: (a) Benzoic acid
(b) Acetic acid
(c) Phenyl acetic acid
(d) None of these
270
Problems in Organic Chemistry
27. Compound (H) on ozonolysis can produce: (a) Benzophenone (b) Benzoic anhydride
(c) Acetic anhydride
(d) none of these
Passage - II For Aldol / Ketol condensation aldehyde and Ketone must contain a – H – atom. This reaction is base catalysed. Base removes acidic a – H – atom and forms carbanion. This carbanion produces aldol / Ketol when attacks on other aldehyde or ketone on the other hand cannizaro reaction is exhibited by those aldehydes which do not contain a – H – atom. Answer the question from 28 to 33. 28. The compound containing most acidic a – H – atom is:
(a) AcOH
(b) CH3NO2
(c) CH3SO3H
(d) CH3CHO
29. Which will not show aldol as well as connizaro reaction? O
CH3
O
(a)
(b) PhCOCHO
H
(c)
(d) All of these
Me H
30. The compound which can undergo cross aldol with HCHO is:-
NO2
(a) PhCOCHO
(b)
O
(c) CH3NO2
(d) None of these
31. Intra molecular aldol / ketol is not possible in:
(a) H5C2OOCCOCOCOOC2H5
(b) PhCOCHO
(c) CH3COCHO
(d) All of these can not show intra molecular aldol
32. (A)
OH ( − ) D
LAH
→ (P) → (Q)
If Q does not decolourise bromine water then A would be:O (c) CH3COCHO (c)
(a) PhCH2CHO
(d) CH3NO2
OH ( − )
→ Product, The product of this reaction would be:33. CH3CHDCHO D |
OH |
(a) CH3 —C — CH — CHDCH3 |
CHO CH3 (c) OHC
H |
(b) CH3 — C — CHOHCHDCHO |
CHO CHO
C = CHCHDCHO
|
(d) CH3 —CH — CHOD CHD CHO
271
Aldehyde & Ketone
Passage - III An organic compound (A) is highly volatile. (A) On reaction with HI gives two products (B) & (C). (C) gives red colour in victor Maeyer test but when (C) is heated with H+ followed by dilute sulphuric acid, it gives (D) which immediately reacts with Anhydrous ZnCl2 + HCl. (B) on treatment with C2H5OH + KOH followed by O3 / Me2S gives (E). (E) can gives haloform test. (E) When heated with NaOH gives (F). (F) Also gives haloform test & can decolourise Br2 water. Answer the question form 34 to 39. 34. Compound (A) is:O
(a)
(b)
O
O–CH3
(c)
(d)
O—CH2—Ph
35. Compound (C) is:OH
OH
(a)
(c) PhCH2OH (d) OH
(b)
36. Compound (D) is: (a)
OH
(b)
OH
OH (d) Ph — C — CH3
(c)
|
OH
37. Compound (E) is: (a)
O
CHO
(b)
COCH3 (d) CHO CHO
CHO
(c)
CHO
38. Compound (F) is:O CHO
(a)
Ac (b)
CHO
(c)
(d)
39. Compound (B) is:-
(a)
CH3 I
I
(b)
(c)
I
I
(d)
Passage - IV
CH2OH
HO
(A)
CH2OH
MeO
OH
(V)
OH
CH–CHOH
MeO
(B)
(D)
CH–CHCO2Et
MeO
Me
(C)
Me
Answer the questions from 40 to 43 40. A would be:
(a) CH3OH / H+
(–)
O H/ MeCl (b) CH3OH / Al2O3 (c)
(d) All of these
272
Problems in Organic Chemistry
41. B would be: - O
||
(a) CH3 – CH2COOH (c) PCl5, Mg / Ether
(b) CH3 — C CH 2 OH (d) MnO2
42. Compound (C) is: (a) H+
(b) CH3CH2COOEt/OH–
(d) CH3 — C— CO 2 Et / OH −
(c) H3O+
||
43. Compound (D) is: - (a) NaBH4 / Et2O / H+ (b) LiAlH4 / Et2O / H+
O
(c) DlBAL – H at – 78° C
(d) (b) & (c)
44. An organic compound (A) can react with tollen’s to give (B) which on heating gives (C). Compound (C) does not give haloform test. (A) Would be:CHO (a) CH3COCH2CHO (b) (c) Both (a) & (b) (d) None CH2 CHO O (i) Glycol (1mole)/H (+)
→ (A) Product, (A) will be; – (ii) NaBD 4 / H 2O
45.
(iii) H3O( + )
O
O
OD
(a)
OH
O D
(b) (c) (d) OD
O
OH D
O
46. Identify the compound for which rate of hydration is least O O (b)
(a)
(d) MeCHFCOCHFMe
(i) dil H 2SO 4 → (A)
47.
(ii) Al2O3 / D
O
(c) MeCOCOCOMe
O
Which statement is not correct about (A) (a) It will react with Benedict’s solution and 2, 4 – D.N.P. (c) It will react with Br2 water but not with [Ag(NH3)2](+)
(b) It will decolourise Br2 water. (d) It will not show iodoform test
48. Two unknowns, X & Y both having molecular formula C4H8O, give the following results with four chemical tests. Br2 Water Na metal Chromic acid Lucas reagent X decolourise Bubbles Orange to Green No reaction Y no reaction no reaction no reaction no reaction X & Y are:OH CHO (b) CHO (a) & & OH OH
(c)
&
CH3CH2COCH3 (d)
OH
&
CH3CH2COCH3
273
Aldehyde & Ketone
49. An organic compound (A) on hydrolysis by aq KOH gives another compound B. B gives cannizaro reaction with NaOH. What is (A) if (A) can exhibit haloform test? (a) PhCCl2CHO (b) CHCl2CHO (c) CH3CCl2CHO (d) CH3COCHCl2 50. Acetaldehyde can be converted in to an optically active compound (a) By treating it with LiAlH4/Ether/H+ (b) By treating it with (i) KCN (ii) H3O+ (c) By heating it with aq NaOH (d) By treating it with OH(–) / D followed by H2 / Ni 51. By the help of formaldehyde we can not prepare (a) Urotropin (b) Bakelite
(c) Farmose
(d) Phorone
52. In which of the following cannizaro reaction products are not proposed correctly.
OH (–)
→ HCOO– + PhCOO– + CH3OH + PhCH2OH (a) HCHO + PhCHO (–)
CHO
CH2OH
CO2
(–)
(–)
OH
(b) PhCHO +
PhCH2OH +
OMe
(c)
(–)
CHO
OH
→
|
CHO
+ PhCO2 + OMe
OMe
COO( −) |
CH2OH
(−)
OH (d) PhCOCHO → PhCHOCOO(–) H O18
2 53. CH3COCH3 → P (+)
trace of H
Product P of this reaction would be:(a) No chemical reaction takes place but acetone dissolves in water due to H – Bonding (b) CH3CO18CH3 OH |
(c) CH3 —C— CH3 |
OH
18
(d) Reaction occurs & the end product P is CH3COCH3
54. The order of acidity of the compounds A–C is:O O O NO2 CO2Et
A
B
C
(a) A > C > B
(b) C > B > A
(c) B > C > A
(d) B > A > C
55. Out of A, B & C the product which will not show Cannizaro reaction with NaOH is:
2 →(A) (a) CH ≡ CH
SeO
aq KOH
(b) CHCl2
O H 2O
3 →(C) (c) Ethene
(d) None of these
(B)
274
Problems in Organic Chemistry
OH (i) N 2O 4 / CHCl3 CH—Φ →(P) n (ii) Zn/Hg Conc HCl, D
56. OH
Product P of this reaction will be:OH CH—Φ
(a)
(c)
CH2—Φ
(b) CH2—Φ
(d)
Cl
OD CH2OH
57.
CHD
This conversion can be performed by:(a) MnO2, LiAlH4 / D2O (c) MnO2, NaBD4 / H2O
(b) N2O4 / CHCl3, LiAlH4 / D2O (d) N2O4 / CHCl3, NaBD4 / D2O
OH
58.
(1) Mg/ether
+ HI (excess)
[Y]
[X]
PCC
[Z]
(2) HCHO
OH
(3) H3O(+)
(a) [Z] can show cannizaro reaction (c) [Z] can show self aldol condensation = (i) CH CHCO Et / D (ii) O3 , CH3 –S–CH3 (iii) OH (–)
59.
2 2 →
(b) [X] is a di iodide (d) [Z] contains two CHO group.
[X]
Product [X] of this reaction would be:OH
CO2Et CH2OH
(a)
(b)
(–)
CO 2 Et
COO
(c)
O
(d) None
60. The reaction in which disproportionation takes place is:
Na
(a) (CH3)3CBr → Ether (c) Ketol condensation
OH (–)
→ (b) CH3CHO + PhCHO (d) (a) & (b)
275
Aldehyde & Ketone
61. In a cannizaro reaction the intermediate which is the best hydride ion donor is:H
H
|
|
p — NO 2 — C6 H 4 — C— O(–) (a)
(–) (b) p — OCH3 — C6 H 4 — C — O
|
|
O(–)
O(–) OH
(–)
O CH
(–)
O—C—H
(–)
O
(c) NH2
(d) NO2
O2N NH2
62.
NO2
Which is wrong about cannizaro reaction:(a) Dianion can form as intermediate (b) It follows 3rd as well as 4th order kinetics. (c) It requires concn base and hydride ion transfer occurs in it. (d) Attack of base (OH–) is the rate determining step.
63. Select the correct order of reactivity of following compound towards RMgX (a) CH3COCH3 > CH3CH2CHO (b) CH3CONH2 > CH3COCl
(c) CH3CO
SO3H > CH3CO
(d) CH3COOCH3 > CH3CON(CH3)2 + CH3COCH = CH 2 → (P)
64.
NaOH D
O
OCH3
Here P is:OH
CH = CH2
C
(a)
CH3
(b) CHCOCH=CH2
O
(c)
(d) O
O
65. Which among the following will not show cross aldol with formaldehyde? (a) Glyoxal (b) nitro methane (c) cyclopentadiene
(d) 3–oxo butanal
66. Which will have least pKa? (a) Di chloro acetaldehyde
(d) Acetaldehyde
(b) Tri fluro acetaldehyde
(c) Fluro acetaldehyde
67. Match the following Compound Properties (A) Methyl cyclo hexanol (1) disproportionation by caustic soda (B) Glyoxal (2) Haloform as well as tollen’s test (C) 2- oxo propanal (3) gives alkene with Ag/ 673K (D) Nitro methane (4) negative bendict’s test but gives aldol with HCHO (a) A-3 B-4 C-2 D-1 (b) A-3 B-1 C-2 D-4 (c) A-4 B-3 C-2 D-1 (d) A-4 B-3 C-1 D-2
276
Problems in Organic Chemistry
68. Match the following Compound Tests (A) Di chloro butan-2-one (1) –ve haloform & fehling tests but +ve 2,4 DNP test (B) Formic acid (2) +ve haloform but -ve fehling test (C) Di ethyl ketone (3) gives Hg on reaction with corrosive sublimate (a) A–3 B–1 C–2 (b) A–3 B–2 C–1 (c) A–1 B–3 C–2 (d) A–2 B–3 C–1 69.
Product of this reaction would be:-
(a)
(b)
(c)
(d)
OH heated with silver metal
70.
(i) NH OH
2 →Y X (ii) H3PO 4
Compound Y is:(a) An imine (c) cyclic secondary amine containing one C=C bond
(b) cyclic amide (d) alkane
Passage - V Answer the questions 71 & 72 from the following reaction scheme. (i) C H / D
2 4 (A) ← (+) (ii)H /D 71. Compound (A) is.
(a)
NaOH D
→(B)
O O
(b)
Et (c)
Et
Et (d)
OH OH
OH OH 72. Compound (B) is.
(a)
Et
Et
(b)
(c)
(d)
COOD CF COOH 73. CH3CO CH CH3 ∆ [X] [Y] Identify wrong statement. (a) Both [X] & [Y] are optically active
(b) [Y] is an ester
(d) [Y ] is optically inactive ester
3
(c) [X] is an optically active ketone
O 74.
This conversion can be performed by:(a) (i) Ph3PCH2, (ii) LiNH2 with CH3OH (c) (i) HCHO / OH(–), (ii) Red P + HI
(b) (i) LAH (ii) 1 Mol H2 / Ni (d) (i) LAH, (ii) H+ / heat , (iii) NaNH2 with CH3OH
277
Aldehyde & Ketone
75. The compound which will not show geometrical isomerism after acidification: (a) Ph(CH3)CO (b) Ph(CH3)CNH (c) CH3CHO
(d) CH3CHS
76. The compound which can show geometrical isomerism after acidification is :O
CH3 Et CH—C—(i–Pr) (d) O (b) (c) S CH3
(a)
NH
77. Which is correct for following reactions?
O K
K
1→ ←
K2
O
(b) K3 > K1
(a) K1 > K3
3→ ←
OH CN
K4
(c) K3 = K1
(d) K4 = K2
(i) LiAlH / Ether / H ( + )
4 78. PhCH = CH—CHO → (A) (ii)PCC in CH Cl (iii) OH 2
(–)
OH
2
(B)
Here (A) and (B) are:- (a) Same, PhCH = CH — CH2OH + PhCH = CHCO2 anion COH OH
(b) Same, Ph — CH 2 — CH — CH — CH 2 CH 2 Ph
|
|
CHO
|
(c) Different, A is PhCH 2 —CHCH — (CH 2 )2 Φ & B is ΦCH = CHCH2OH + ΦCH = CHCO2 anion |
OH
(d) Different, B is ΦCH 2 — CH — CH — (CH 2 )2 Φ & A is ΦCH = CHCH2OH + ΦCH = CHCO2– |
|
CHO OH OH (–)
→ Product, Product of this reaction is:79. CH3 — CH = CH — CHO D
(a) CH3CH2 = CHCOO–
&
CH3CH = CHCH2OH
CH3 — CH 2 — C = CH — CH = CH — CH3 (b) |
CHO
(c) CH3 — CH = CH — CH = CH — CH = CH — CHO (d) Reaction is not possible
80. Assertion - Ketones have slightly higher bp than the isomeric aldehydes Reason - Presence of two + I group in ketone makes C = O bond polar (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true 81. Assertion-Neo butyl alcohol can show Haloform test Reason- Halogens are weak oxidizing agents & can not oxidize tert- alcoholic group (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true
278
Problems in Organic Chemistry
82. Assertion- Tri chloro ethanal can not exhibit cannizaro reaction Reason -It undergoes haloform test (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true 83. Assertion- Tri chloro ethanal can show hydration reaction & can form gem di ol Reason -strong -I effect of chlorine atoms makes carbon of carbonyl group more electropositive (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true 84. Assertion- formic acid can show reducing properties Reason - It is stronger carboxylic acid than other non substituted aliphatic carboxylic acid (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true
O / H O / Zn Boil
85. Assertion:-
O
O
3 2 →
O
O
+ H
(A)
(B)
H
Along with (A) & (B) CH3COCHO also forms Reason: - A & B on reaction with Zn / H2O undergo rearrangement to given CH3COCHO (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true 86. Assertion: - Cyclo propanone undergo addition with HCN more easily in comparision to that of acetone Reason: - Cyclo propanone contains strained ring. (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true
O 87. Assertion:- CH3—C—CH 2 —CH3
OH CH3 —C = CH—CH3 (A)
OH CH2= C—CH2—CH3 (B) Out of A & B, (B) is less favourable tautomeric form than (A). Reason: - A is more substituted alkene & hence more stable (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true
279
Aldehyde & Ketone
OH
O
O
(+)
H heat
→
88. Assertion:-
+ OH (A)
OH
(B)
Out of A & B, B does not form. Reason: - H+ attacks on C = O and not on OH group because in C = O oxygen acquires –ve charge due to resonance. (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true O
89. Assertion: -
∆
[X], [X] can not release CO2 on reaction with NaH CO3
COOH Reason: - [X] does not contain COOH group (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true
90. Assertion: - Cl2C(OH)2 is unstable and produce phosgine along with H2O but CCl3CH(OH)2 does not loose water. Reason: - In CCl3CH(OH)2 intera-molecular H - bonding presents. (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true LiAlH
4→ 91. Assertion: - CH2 = CH — CHO
LiAlH 4 PhCH = CH — CHO →
(A)
(B)
(A) gives Br2 water test but (B) does not. Reason: - (B) does not contains CHO group. (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true 92. Assertion: - Zn/Hg/HCl converts Acetone to propane but Mg/Hg/H2O can not convert acetone to propane. Reason: - Mg/Hg/H2O converts acetone to pinacole (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true 93. Assertion: - R2C(OH)(OMe) easily looses methyl alcohol along with ketone but R2C(OMe)does not loose di methyl ether along with R2CO Reason: - Di methyl ether is more volatile than MeOH (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true
280
Problems in Organic Chemistry
Answer key 1. (c)
2. (d)
3. (c)
4. (d)
5. (a)
6. (a)
7. (b)
8. (c)
9. (c)
10. (b)
11. (d)
12. (b)
13. (b)
14. (c)
15. (d)
16. (a)
17. (d)
18. (a)
19. (a)
20. (d)
21. (a)
22. (b)
23. (a)
24. (b)
25. (d)
26. (a)
27. (b)
28. (b)
29. (a)
30. (c)
31. (d)
32. (a)
33. (a)
34. (b)
35. (a)
36. (b)
37. (d)
38. (b)
39. (a)
40. (c)
41. (d)
42. (b)
43. (b)
44. (b)
45. (c)
46. (b)
47. (c)
48. (d)
49. (b)
50. (b)
51. (d)
52. (a)
53. (c)
54. (c)
55. (c)
56. (d)
57. (d)
58. (a)
59. (a)
60. (a)
61. (c)
62. (d)
63. (c)
64. (c)
65. (a)
66. (a)
67. (b)
68. (d)
69. (b)
70. (b)
71. (b)
72. (b)
73. (d)
74. (a)
75. (b)
76. (b)
77. (a)
78. (c)
79. (c)
80. (a)
81. (d)
82. (a)
83. (b)
84. (b)
85. (c)
86. (b)
87. (a)
88. (a)
89. (a)
90. (a)
91. (b)
92. (a)
93. (b)
Multiple Choice Questions (More Than One May Correct) 1. Which among the following can show Tollen’s test? OH
(a) CH3CHO
(b) PhCHO
|
CH3—CH (c) CH3 — CH — F (d)
2. Which among the following can’t show Fehling test? (a) CH3 —CH — OCH3
(b) CH3COCH3
|
(c) PhCHO
(d)
3. CH3CH = N NH2 is called acetaldehyde hydrazone. Which is not true about it? (a) It can produce methane on reaction with conc. H2SO4. (b) In can not show geometrical isomerism (c) This compound can show geometrical isomerism after protonation. (d) On heating with NaOH it will produce ethane 4. In the context of following road map which statement is correct?
O CF3CO3H
(X)
LiAlH4 D2O
(B)
Ph3PCH 2
(C)
(a) Reaction (X) to (A) is Baeyer villiger oxidation
(b) Reaction (X) (C) is tollen’s reaction
OH D while (C) is
O (d) while (B) is O 5. Which of the following compound is not required in the preparation of urotropin? (a) HCHO (b) CH3CHO (c) NH3
(c) (B) is
O O (d) CH3NH2
6. In which of the following case plastic is not formed. H( + )
(a) n(HCHO + PhOH) →
(c) CH3COCH3 →
dry HCl gas
high P & T
→ (b) n(CH2 = CHCl) very dilute alkali
→ (d) n HCHO
OH
CHO
OH
(A)
OCH 3
281
Aldehyde & Ketone
7. Which among the following reagent are required for the transformation of ethene in to CH4? (a) O3 / H2O / Zn / ∆ (b) Zn / Hg + HCl (c) H2 / Ni (d) NaOH / CaO / ∆ Cr O 2 −
(+)
(–)
PCC OH H CH 2 OH → (A) → (D) → (E) | CH2OH
2 7 8. (B) ←
Select the correct statement:(a) (D) is CH2OH – COO(–) (c) (A) is glyoxal
(b) (B) & (E) are same compounds (d) (E) on oxidation gives (B)
9. LiAlH4 can reduce:O
O
(a)
O
||
O ||
(b) CH3 — C— CH 2 — C— CH3 O
||
(c) CH2 = CH — NCH3
(d) CH3 — C— OCH3
Answer Key 1. (a), (b), (c), (d)
2. (b), (c)
3. (a), (b)
4. (a), (d)
5. (b), (d)
6. (c), (d)
7. (a), (b)
8. (a), (c), (d)
9. (a), (b), (d)
LEVEL - II Comprehensions Comprehension - I
1. Which hydrogen is most acidic? (a) 1 (b) 2
(c) 3
2. Arrange these hydrogens in decreasing order of pKa values 3. Write the mechanism of the proposed reaction and identify the product formed in it.
(d) 4
282
Problems in Organic Chemistry
Comprehension - II Consider the following aldehydes CH3 — CH = CH CHO
CH2 = CH — CHO
(A)
(B)
4. Identify the correct reactions Base
→ C (b) B annizaro reaction
Base
→ Aldol reaction (d) B
→ C (a) A annizaro reaction
→ Aldol reaction (c) A
→ Cannizaro reaction (e) A + B
Base Base
Base
Base
→ Aldol reaction (f) A + B
5. Write a mechanism for the following transformation. Ba(OH) D
2 →
A + PhCH = CHCHO
Ph(CH = CH)3CHO (87%)
6.
3 → C + D B + CH3MgBr Select the correct statements (a) Both C & D can show haloform test (c) Out of C & D, one can show Tollen’s test
H O( + )
(b) Both C & D can show Victor maeyer test (d) Both C & D can react with Br2 water
7. Provide the missing reagents & products.
CH2OH
HO
C
MeO
CH2OH
A
PCC
CHO
HO
D ∆ CH2(COOEt)2 E
B (–)
(–)
COO
O & (–)
O
CH2OD
8.
NH ( − )
O (limited) Zn, H 2O,
3 2 →(A) →(B)
What is B? Write mechanism for the transformation (A → B)
9. You have following aldehydes
NO2 CCl3CHO Ph3CCHO
F
(1) (2) Select the compounds which can show cannizaro
CHO
CHO
F
NO2 (3)
(4)
283
Aldehyde & Ketone
Matrix Match Questions Each question contains statement given in two columns which have to be matched. Statement (A, B, C , D) in column I have to be matched with statements (p, q, r, s) in column II. The answers to these questions have to be appropriately bubbled as illustrated in the following examples. If the correct match are A – p, A – s, B – r, B – q, C – q, D – S, then the correctly bubbled 4 × 4 matrix should be as follows. A
p
q
r
s
B
p
q
r
s
C
p
q
r
s
D
p
q
r
s
10. List - I (Reaction)
List - II (Missing reactant & reagent)
(A) CH3CHO + ? → penta-2,4-dienal (p) NH 2 NH 2 / OH (–) / D O COOH
(B)
COOH
?
Cl
(q) Zn / Hg + HCl Cl
OH (C)
O
OH Ph
Ph
? Ph
Ph
(D) CH3CHO → But − 2 − en − 1 − al 11. List - I (Reaction) 2, 4- dimethyl hepane (A) Product
(–) (r) OH / D
(s) CH 2
= CH — CHO / OH(–)
List - II (Type of Reaction & Product) (p) Decarboxylation
Al2 O3–CrO 3
∆
(B)
(C)
(q) Mesitylene
dry HCl gas (D) Acetone → Product D
12. List - I (Reaction)
(r) Phorone
(s) Aromatization
List - II (Reagent)
Ethene → Oxirane (A)
(p) Ag, air, heat
Acetone → Methylacetate (B)
(q) Ph3PCH2
Ethanal → Ethyl acetate (C)
(r) Me3CO(–)K(+)
Acetone → isobutene (D)
(s) PAA
284
Problems in Organic Chemistry
13. Select the reagents from list II by which you can differentiate the pairs of compounds given in list I List - I List - II CHO (A)
[Ag(NH3 )2 ]( + ) (p)
& O
(B)
CHO &
CHO
Ac (C)
&
CHO &
14. Pair of Compounds OH (A)
(r) Cu +2 / OH (–)
O (D)
(q) KOI
OH &
(s) 2, 4 – DNP Reagent by which distinction can be made
(p) Na
(B) OH and Acetone O OH (C) and
(q) (r) NaBrO
OH
(D)
and
(s) PhNHNH2
Answer Key 1. (d)
2.
2>3>1>4
4. (b), (c) & (f)
6. (c) & (d)
7. A is PCC (or other mildoxidizing agent), B is OD(–)
D = MeO
CHO C = CH2N2
E = MeO
CH = CH(COOEt)2
8. Compound B is
OH
9. None
Answers matrix match 10. (A) – s, (B) – q, (C) – p, (D) – r 12. (A) – p, s, (B) –s, (C) – r, (D) – q 14. (A) – r, (B) – p, s, (C) – p, r, s, (D) –p, q, r
11. (A) – q, s, (B) – q, (C) – p, q, (D) – r 13. (A) – r, (B) – p, r (C) – p, q, r, (D) – s, p
285
Aldehyde & Ketone
SOLUTIONS LEVEL-I
OH
O H,(+) heat
NaBH
4→
1. (c)
No Reaction
(B)
(A)
Blue colour in victor Maeyer test
Alcohol (B) can not undergo dehydration because bridge ‘C’ can not bear double bond. 2. (d) In ppt(1) acetic acid separates as cal.acetate 2AcOH + Ca(OH)2 ——→ 2H2O + Ca(OAc)2 After separation solution contains MeOH & propanone. Propanone separates as propanone sodium bi sulphite (ppt-2) SO3Na O ||
|
NaHSO
3→ CH3 — C— CH3 CH3 — C — CH3 ppt–2
|
OH Thus, solution contains methyl alcohol. 3. (c) Reduction followed by pinacole pinacolone rearrangement takes place.
Ph
H
COPh
CHPh
H
OH
CHPh (+)
OH OH
OH
(+)
→
LiAlH / Ether
4 → (+)
ring opening
(+)
OH Ph
(+)
–H
OH
O
O
O 4. (d)
5. (a) OHC
CH3CO3H
COOH OH
H3O( + )
→
→
COCH3
O
OH (+) OH, H
CH
COCH3
O
OHC
Et
H3O
(+)
Zn–Hg + HCl
O CH
Et
O Tollen’s Reagent
6. (a) HC ≡ C — CH2COCH3 → AgC ≡ C — CH2COCH3 Since it has ketone group hence it will undergo Baeyer villiger oxidation. 7. (b) In II-process propyne undergoes partial reduction to give propene which on oxidation by KMnO4 Gives acetic acid which undergoes esterification with methyl alcohol to give methyl acetate.In III- process propyne gives acetone on hydration which undergoes Baeyer villager oxidation to produce methyl acetate. 8. (c) CH3COCl + (CH3)2CuLi (Gilman reagent) ——→ CH3COCH3 + CH3Cu + LiCl O hot, KMnO 4
O Al(OEt)
CH MgBr(excess)
3 3 CH3CHO → CH3COOC2 H5 →(CH3 )COH Tishenko reaction H3O( + )
286
Problems in Organic Chemistry
9. (c) Here methyl shift takes place. (+)
H
∆ –H(+)
(+)
methyl shift (+)
OH (–)
O(–)
(–)
|
H 2O OH HCHO 10. (b) CH3CHO → CH 2 CHO → H — CH — CH 2 CHO → HOCH 2 — CH 2 CHO –H 2O
CH2O(–)
CH2OH |
|
H O
(–)
HCHO
2 HOH C — CHCHO ← H 2 C — C H — CHO ← HOH 2 C — CHCHO 2
(–) OH, HCHO
(–)
OH –H2O
CH2OH |
HOH 2 C — C — CHO (P) |
CH2OH This is Cross aldol condensation. Now (P) does not contain a - H atom so further addition of HCHO will lead to cross cannizaro reaction OH (–)
(HOH 2 C)3 CCHO + HCHO →(HOCH 2 )4 C + HCOO(–) 11. (c) is exceptional case 12. (b)
NH NH3
NH2 CH3MgBr H3O
CH3
(+)
(A)
O 13. (b)
Ph3P
=CH 2 LiAlH4
OH
H3O (+)
(B)
OH (C)
Since (B) & (C) are 2° alcohols hence they will produce blue colour in victor maeyer test (A) is primary amine which is inactive towards victor Maeyer test. 14. (c) Addition of HCN on carbonyl compound is nucelophilic addition reaction Rate of nucleophilic addition ∝ electropositive character of ‘C’ of carbonyl group 15. (d)
OH cold KMnO 4
(AcO) 4 Pb
CHO
Aldol
CHO CHO OH 16. (a) HCHO on reaction with NaHSO3 produces white precipitates of formaldehyde sodium bi sulphite 17. (d) By the destructive distillation of wood pyroligneous acid is formed along with wood tar & waste gases. This acid contains acetone, methyl alcohol & acetic acid.
287
Aldehyde & Ketone
18. (a) Hemi acetal of acetaldehyde is unstable & dissociates in to alcohol & acetaldehyde. Thus, it shows reducing properties. O || HCOOH also shows reducing properties because it contains CHO as well as COOH group H — C— OH (–)
→ EtOH + 19. (a) + EtO(–) Formation of cyclo propenyl anion is not possible as it is antiaromatic & unstable. 20. (d) Since all compounds do not contain ‘a’ H atom thus, all can show cannizaro reaction 21. (a) The reaction in which aldol and cannizaro reactions both occur simultaneously is called Tollen’s reaction.
Passage - I (22 TO 27)
Passage - II (28 To 33) 28. (b) Became after the removal of α – H atom carbanion is formed which is stabilized by resonance. (–)
(–)
H 2 O + CH 2 NO 2 CH3 NO 2 + O H (–)
O(–)
O ||
CH 2 — N — O ←→ CH 2 ( + )
|
=(N+ ) — O
29. (a) Because it is phenol 30. (c) See question no. 28. O(–) OH O || (–) | | H 2O 31. (d) H — C— H + CH 2 NO 2 → H —CH — CH 2 NO2 → CH 2 — CH 2 NO 2 (–) –OH
(–)
OH ,heat 32. (a) PhCH 2 CHO → Ph — C Aldol
|
LiAlH 4 → Ph — C H — CH 2 CH 2 Ph = CHCH 2 Ph |
CHO (P)
CHO
Compound (P) can be reduced to by LiAlH4 here it should be noted that double bond in conjugation of phenyl group can be reduced by LAN. Since (P) does not contain C = C bond hence it can not decolourise Br2 water 33. (a) It is aldol condensation base will break C -H bond & not C -D bond as we know that former has less bond energy than later.
288
Problems in Organic Chemistry (–)
(–)
H 2 O + CH3 CDCHO CH3CHDCHO + O H O(–)
O ||
(–)
|
H O
2 → CH CDCHOHCHDCHO CH3CDCHO + CH3CHD — C— C → CH3 — CHD —C HCD — CHO 3
|
|
CH3
CHO
Passage - III (34 TO 39) A is highly volatile & can react with HI to give two products hence a may be an ether. (C) gives red colour in Victor maeyer test hence (C) is 1° alcohol. (D) reacts immediately with Lucas reagent hence D is 3° alcohol. E & F can show haloform test hence both should have CH3CO linkage.
Passage - IV (40 TO 43) (–)
OH
CH2OH
MeO
OH –H2O
(–)
MeCl
O
CH2OH
CHOHCHCO2Et
CH3CH2COOEt, OH Aldol
MeO
CH2OH
MeO
CHO
MnO2
Me
LiAlH4
Product CHO
44. (b) CH2
CHO Tollen's reagent
heat –CO2
CH2
CHO (A)
CHO (B)
CH3COOH
–ve haloform test
289
Aldehyde & Ketone
Steric hindrance O
O
OH
OH D
D (+)
(+)
45. (c)
NaBD4 H2O
glycol, H
O
O
O
H3O
O
O
O
46. (b) (c) & (d) forms stable hydrates as they are stabilized by intramolecular H- bonding . Thus rate of hydration will be maximum in these two cases. Although hydrates of both (b) & (c) are not stabilized by H bonding yet hydrates of (a) will be more stable because of decrease in angle strain. Hence rate of hydration will be least in (b) H3O
47. (c)
OH
HO
O
H
(+)
OH
OH OH
OH
–H2O
OH
CHO
OH
OH
(+)
H, heat
CHO (A) Contains double bond hence it will decolourise Br2 water. Due to the presence of CHO it will show all reducing properties.with Tollen’s & Bendict’s Solution Due to the presence of CHO group it will undergo condensation with 2, 4 - DNP & semicarbazide. 48. (d) CH2 = CH – CH2CH2OH will decolourise Br2 water because it has C = C bond. It will react with Na because it has OH group. OH group can be oxidized by chromic acid. Since it has 1° alcoholic groups hence it will not react with Lucas reagent CH3CH2COCH3 can not decolourise Br2 water & can not react with Na because of the absence of C = C bond & OH group chromic acid can not oxidise keto group. (–) (–) OH 49. (b) CHCl2CHO 2KOH HCCHO –H O CHOCHO OH OOC–CH2OH –2KCl 2 (Cannizaro) OH OH (–)
OH (–)
→ CCl3CHO → CHCl3 + HCOO(–) CHCl2 CHO + Cl2
O( − ) |
KCN
50. (b) CH3CHO → CH3 CHCN
OH |
H3O( + )
→ CH3 — CH — COOH (A)
Compound A is lactic acid and optically active due to the presence of chiral carbon atom 51. (d) Phorone is prepared by acetone, by treating it with dry HCl gas (dehydrating agent) 52. (a) Reactivity of base is more towards aliphatic aldehyde than aromatic aldehyde. (–)
O
O (–)
H—CH—OH
H—C—H + OH
O( −)
(–)
O
O
|
Ph—C—H + H —CH—OH → HCOOH + Ph — C — H → HCOO(–) + PhCH 2 OH |
H (+)
OH
||
53. (c) CH3 — C— CH3 + H ( + ) → CH3—C—CH 3
18
H2•O•
••
O
OH CH—C—CH 3 3 (+)
18 OH2
OH (+)
–H
CH3—C—CH 3 18 OH (P)
Product P will not loose H2O because C — O & O — H bonds have high bond energies 54. (c) 18
55. (c) CH 2
18
O3 → HCOOH → –ve cannizaro reaction = CH 2 H 2O
2 → CHO — CHO CH ≡ CH → +ve cannizaro reaction
SeO
290
Problems in Organic Chemistry
56. (d) N2O4/CHCl3 can oxidise 2° benzylic alcohol in to ketone.
OH
CHOHPh
N2 O4/CHCl 3
PhCO
OH
Cl
O N 2O 4CHCl3
57. (d) PhCH 2 OH → Ph
O(–)
(–)
||
Zn/Hg+HCl
CH2Ph
D from → Ph — C— H NaBD 4
OD
|
D 2O
|
→ Ph —CHD —C — H |
D
58. (a)
59. (a)
OHC CH2
Diels-Alder
+
ozonolysis
CHCO2Et OHC
CO2Et CO2Et CH2OH
CO2Et (–)
OH Cannizaro
COO(–)
60. (a) In wurtz reaction free radical generation takes place thus, in reaction (a) free radical will undergo disproportionation 61. (c) +R effect of NH2 group helps in the dissociation of C — H bond 62. (d) Migration of hydride ion is rate determining step. Rate = (aldehyde)2 (base) & some times kinetic studies show that Rate = (aldehyde)2 (base)2 63. (c) 64. (c) It is an example of cross aldol condensation. (–)
O
(–)
OH –H2O
O H2O
CH2 = CH—C—CH 3
(–)
O CH O
O
3
3
(–)
O CH O
O O
(–)
(–) H2O
(–)
OH, heat
2
OH –H2O
O CH OH
O CH O 3
291
Aldehyde & Ketone
65. (a) Glyoxal & formaldehyde both are independent from alpha hydrogen atom thus, these two will show cross aldol condensation (−)
base
→ C Cl2 CHO (highly stabilised by resonance as well as –I effect of Cl) 66. (a) CHCl2 CHO 67. (b) A is tertiary alcohol hence undergoes dehydration by Ag / 573 K. B does not contain alpha H atom thus, it will show disproportionation (cannizaro). C has MeCO linkage thus, it will exhibit haloform. D does not contain CHO group so it will not react with bendict solution but can show cross aldol as follows:(−)
OH (–)
CH3 NO 2 → CH 2 NO 2 + H 2 O stabilized by resonance O(–)
O H—C—H
OH
|
(–)
|
H 2O
OH (–) Heat
→ H — C — H → H — C — H → CH 2 CH2NO2 |
|
CH2 NO2
= CH 2 NO2
CH2 NO2
68. (d) Formic acid can show reducing properties as it contains CHO as well as COOH group HCOOH + 2HgCl2 ——→ 2HCl + CO2 + Hg2Cl2 Hg2Cl2 + HCOOH ——→ 2Hg ↓ + 2HCl + CO2
69. (b).
OH 70. (b)
O
(+)
NOH
(+)
NH2OH
Ag/573K
H backman rearrangement
H O
N
N
OH
HO 2
N
(+)
71. (b) & 72. (b) Solution of these two question is given below O O
C2H4 ∆
tautomerism
OH OH
H
(–)
∆
O O intramolecular aldol
+
(A) O CH= CH2 (B)
OH OH
N—OH2
292
Problems in Organic Chemistry
73. (d)
O wattig reaction
74. (a)
Ph 75. (b)
NH
H
Birch reaction
Ph
(+)
CH3
H N (+)
CH3
can not show geometircal isomerism
H
76. (b) Same as previous. 77. (a) Attack of cyanide ion is easy on cyclo propanone as carbon CO group is more electropositive. 78. (c) (B) forms via cannizaro reaction LiAlH
PCC
4 → PhCH CH CH OH = PhCH CH — CHO → Ph(CH 2 )2 CHO 2 2 2
OH
PhCH 2 CH — CH(CH 2 )2 Ph | (A) | CHO OH (–)
(–)
(–)
OH 79. (c) = CH3CH CH — CHO → CH 2 CH = CH — CHO –H 2O
O
(–)
O
(–)
CH3—CH = CH—CH—CH2CH=CH—CHO
CH3CH = CH—C—H + CH2CH = CH—CHO
OH
(–)
H2O
OH, heat CH CH=CH—CH=CH—CH=CH—CHO –H O CH3CH = CH—CHCH2CH= CH—CHO 2 3 80. (a) 81. (d) Neo butyl alcohol is 3° alcohol & can not be oxidized in to ketone by mild oxidizing agents like halogens thus, can not exhibit haloform test. 82. (a) See mechanism of haloform reaction from your text book 83. (b) Hydrate of tri chloro ethanal is stabilized by intra molecular H - bonding Cl H–O
Cl—C ——
C—H
H–O Cl 84. (b) See Question Number 68 O 85. (c)
O
O3/H2O/Zn Boil
O
H (B)
H
+ (A) O
O3/H2O/Zn Boil
O
O +
CH3COCHO
H H 86. (b) Because of less steric hindrance around CO group. 87. (a) A is more stable due to more hyperconjugation. 88. (a) H+ attacks on C = O and not on OH group because in C = O oxygen acquires –ve charge due to resonance. 89. (a) 90. (a) See Question Number 83
293
Aldehyde & Ketone
91. (b) Double bond present in conjugation with Ph group can be reduced by LAH thus, B is PhCH2CH2CH2OH & can not show bromine water test. 92. (a) See acyloin condensation in your text book 93. (b) OH is attracted by OMe by H bonding so bond angle reduces thus, to remove strain R2C(OH)(OMe) looses MeOH More than one may correct:1. a, b, c, d F CH3—CH HF + CH3CHO (+ve tollen’s test) OH (unstable)
OCH3
CH3CH
CH3OH + CH3CHO
OH
(+ve tollen’s test)
(unstable) 2. b, c
5.
3. a, b
4. a, d
b, d 6HCHO + 4NH3 → (CH 2 )6 N 4 + 6H 2 O (urotropin)
6.
c, d In (c) phorone is formed while in (d) farmose is formed (mixture of saccharides)
7.
O3 / H 2O / Zn Zn / Hg + HCl a, b C2 H 4 → HCHO → CH 4
8.
D
a,c,d COOH—COOH
Cr2O7
2–
CH2OH—CH 2OH
PCC
CHO—CHO
CH2OH—COOH 9.
NaOH
(–)
CH2OH—CO 2
acidification
a, b, d LAH can not reduce C—C multiple bond.
LEVEL -II 1. (d) Acid strength ? stability of Conjugate base 2. 2 > 3 > 1 > 4 pKa α 1 / acid Strength
3.
O
O
(–)
O
(–)
PhCCH2CH3 + CH2 = C—C—CH 2
PCC
CH3 PhC=CHCOC 4. (b) (c) & (f)
Et
= CH 2
CH3 (–)
OH ∆
(–)
Ph—C—CH2CO—C = CH2 OH
CH2CH3
Ph—C—CH2CO—C = CH2
(–)
CH2C3H
OH 5. = CH3CH CHCHO → CH = 2 — CH CHCHO –H 2O
O (–)
PhCH = CH—C—H + CH2CH
CH3
= CHCHO
CH3
H2O
294
Problems in Organic Chemistry (–)
O
OH
PhCH = CH—CH—CH2CH
H2O
=CHCHO
PhCH=CH—CH—CH2CH =CH—CHO (–)
PhCH=CH—CH =CH—CH = CH—CHO
OH ∆
O CH2= CH—C—H+CH3MgBr
6. (c) & (d)
OMgBr
OMgBr
CH3—CH 2 —CH=C—H
CH2= CH—C—H CH3
(+)
(+)
H3O
H3O
OH CH3CH2—CH = CH—OH
CH2= CH—CH—CH3 Can show haloform & vector Maeyer test
CH3CH2CH2CHO can show tollens test
7. A = PCC (or other mild oxidizing agent) B = OD(–) D = MeO
CHO
E = MeO
CH=CH(COOEt)2
8. Double bond is more sensitive towards ozonolysis
O C—H
(–)
O
NH 3
OH
(B)
(–) st nd th 9. None of these can show cannizaro reaction. 1 gives haloform, in 2 & 4 CHO group is sterically hindered. In 3rd case migration of hydride ion is difficult because of presence of electron withdrawing groups.
10
Carboxylic Acid & Its Derivatives Main Features REACTION CHART FOR CARBOXYLIC ACID ACID DERIVATIVES
Preparation Properties
RCH2CN
RCH2CONH2 RCH2COCl (RCH2CO)2O
H3O H3O
(+)
(+)
H3O
Aldehyde
ROH/H
(+)
hot KMnO4
KMnO4
RCH2CONH2
PCl3/PCl5/SOCl2
H2O
Alkane / alkyne RCH2CH2OH
NH3 heat
RCH2COOH CARBOXYLIC ACID
(+)
RCH2COOR
N3H H2SO4
RCH2NH2
P2O5 heat
(RCH2CO)2O
Na
RCH2COONa + 1/2 H2 KMnO4
FeCl3
Ketone
KMnO4 heat
NaHCO3
RCH2CX3
aq KOH H(+)
NaOH / CaO
Ester
RCH2COCl
H3O
(+)
LAH P + X2 H2O
Fe(OOCCH2R)3 (Blood red colouration) CO2+H2O+RCH2COONa RCH3(Decarboxylation) RCH2CH2OH RCHXCOOH [ HVZ ]
296
Problems in Organic Chemistry
Multiple Choice Questions
1.
O
(–)
O
(i) OH
(+)
(A), Compound (A) is:-
(ii) H
CH2CO2H OH O (a)
O
OH (c) O
(b)
OH
O
CH2COOH
(d)
O
OH O
O
2. Consider the following compound, COOH
S COOH
HOOC — CH2 — (CH2)3CONH
COOH
O
How many chiral ‘C’ atoms will be present in the compound after heating? (a) 5 (b) 6 (c) 4
(d) 3
3. Which reaction product is not expressed correctly? – EtO( )
CH 2O + 2CH 2 ( CO 2 Et ) 2 (a) ( +) → H
EtO2C
CO2Et
O
(b)
O
EtO(–)
+ CH 2 ( CO 2 Et ) 2 ( +) → H3O
O O
– EtO( )
(c) HCHO + 2CH3COCH 2CO 2 Et ( +) → H
CO2H ( +)
H3O (d) CH 2 ( CO 2 Et ) 2 → CH3COOH Heat
OH (4) O
H
4. Given the structure of ascorbic acid (vitamin C)
Which proton is most acidic?
(a) 1
HOH2CCO (3) HO (2)
(b) 2
(c) 3
O OH (1)
(d) 4
297
Carboxylic Acid & Its Derivatives NaOBr
5. CH3CONHCH3
→
Product ……………………… (1)
CH3CON (CH3)2 6. 7. 8. 9.
NaOBr Products ……………………… (2) → Products of reaction (1) & (2) are (a) CH3NH2 & CH3NH2 (b) CH3NH2 & CH3NHCH3 (c) CH3CONBrCH3 & No reaction (d) CH3CONBrCH3 & CHCl3 + (CH3)2NCOO(–) Formic acid can be distinguished from acetic acid by:(a) NaHCO3 (b) NaOBr (c) NaOH + CuSO4 (d) Na An organic compound does not show haloform test but reacts with sodium to give H2 gas. Compound gives blood red colouration with neutral FeCl3. Organic compound is:(a) (CH3)3COH (b) HCOOH (c) acetylene (d) AcOH Calomel can be converted in to mercury by:(a) Lactic acid (b) Oxalic acid (c) Pyruvic acid (d) Formic acid γ - hydroxy butanal can be converted in to γ - lactones by:(a) H+ / Cr2O72–, H3O+ (b) KMnO4/H+, H3O+ (c) Alkaline CuSO4, H+ / heat (d) All of these D + 2 – acetyl propenoic acid → product
10.
Product of this reaction will be:Ac
COOH
(a)
Ac
(c)
(b)
(d)
COOH COOH
Ac
EtONa 11. CH3COOC2H5 → Product, Product of this reaction would be:-
( +) ( – ) Na (a) CH3COCH2COOEt (b) CH3COCHCOOEt (c) CH3COCH2 — COH(OEt)2 (d) CH2(COOEt)2 12. In which of the following case only one molecule of active methylene compound is formed:EtONa
→ (a) CH3COOCH3 + EtCO2Et + H( )
(c) CH3COOMe + CH3COOEt O
O 13. (A) ← O (a) OH
HO
(d) (CH3)3CCOOC2H5
H
EtONa
→ + H( )
LiAlH / Ether
NaBH 4
(c)
EtONa
→ + H( )
EtONa
( +) →
(b) HCOOEt + CH3COOEt
O
4 → ( +)
H
(B), (A) and (B) are:CH2OH
CH2OH
, OH HO
CH2OH
(b)
, O
O
OH
O
, O
CH2OH HO
CH2OH HO OH
(d) Both are
CH2OH HO
298
Problems in Organic Chemistry
14. Ethyl acetate can be converted in to acetone by: (a) CH3MgBr (Excess), H3O+ (c) OH– / H2O, CaO/NaOH , heat
(b) EtONa / H+, H3O+ / heat (d) HCHO / OH(–) / heat
15. Rate of decarboxylation of following carboxylic acid will follow :COOH COOH COOH CH COOH 3
O
O
(1)
(2)
(3)
(a) 1 > 2 > 4 > 3
O
(4)
(b) 4 > 3 > 2 > 1
(c) 4 > 2 > 1 > 3
(d) 4 > 2 > 3 > 1
(1) SO3 16. CH MgBr ¾¾¾¾¾ ®[X] 3 ( +) (2) H3O
(1) CO (2) dil H 2SO 4
2 CH3MgBr ¾¾¾¾¾¾ ®[Y]
Select correct statement (a) [X] and [Y] are same compounds (b) pKa of [X] is lesser than that of [Y] (c) [X] can release CO2 on reaction with NaHCO3 while [Y ] can not (d) [Y] can release CO2 on reaction with NaHCO3 while [X] can not.
17. (A)
P2O5 / heat H3O
(B), (A) would be :-
(+)
(a) (CH3CO)2O (c) CH3COOH
or or
CH3CONH2 CH3CONH2
(b) (CH3CO)2O or CH3COOH (d) CH3CONH2 , or C2H5OH
COBr (i) Ag2O
18. CHO
O
(ii) H2O / Heat
[X]
(i) Pd / BaSO4
[Y]
(ii) H2CrO4 / Heat (i) H2O / Heat
[Z] (–)
(ii) N2H4 / OH
/ Heat
O
COBr
X ⇒ (a)
Y ⇒
Z ⇒
O
CH3 O
(b) X & Y are
(c) X, Y &
(d) X= Z ≠ Y
& Z all are
Z
is
299
Carboxylic Acid & Its Derivatives
CHO
19. Which statement is not correct about
? CONH2
It can be converted in to amino acid by treating it with NaOBr, Ag2O, H+ Its IUPAC name is 2 – formyl cyclohexanamide It shows bi uret test. It is amphoteric in nature
(a) (b) (c) (d) O
20.
OH
CH3COEt
(1)
PrCOOH
(2)
(3)
Correct order of boiling points would be:(a) 1 > 3 > 2 (c) 3 > 1 > 2
21.
(R) – 2 – butanol is treated with Ac2O to get another product [X], [X] will have:(a) R configuration (b) S – configuration (c) Equimolar mixture of R & S will be obtained (d) Both R & S will form will be produced in larger amount.
(b) 2 > 3 > 1 (d) 3 > 2 > 1
Passage – I Ethyl acetoacetate and malonic ester are used for the synthesis of various important organic compound like ketone & substituted carboxylic acid. It contains active methylene (CH2) group. If contains acidic hydrogen and can be replaced as:.. (–) Base CH3COCHCO2Et CH3COCH2COOC2H5 R CH3COCHCO2Et
R–X
This alkyl substituted ethyl acetoacetate on hydrolysis produces acids CH3COCHCO2Et
dil NaOH (+)
H
R
CH3COCH—COOH R
Now answer the questions from 22 to 27 22. CH3COCH 2CO 2 Et (1 mole)
(i) EtONa / (+)
(ii) H3O
(a) CH3COCH2CH2CH2COCH3
Br Br
(P),
P is:-
/∆
(b)
Ac
COOH
(c)
(d) 5 – Hydroxy pentan – 2 – one
Ac
23. CH2(COOC2H5)2 is also an active methylene compound. It can be converted in to adipic acid (butane-1,4-di carboxylic acid) by: (a) EtONa, CH2I2, NaOH / H+, heat (b) EtONa, CH2ICH2I, H3O+ / heat (c) H3O+, heat , EtONa , (CN)2 (d) H3O+ 24. Malonic ester CH2(COOEt)2 can be converted in to CH3CH = CH – COOH by treating malonic ester with (a) EtONa / CH3CHO / heat, H3O+ / heat (b) EtONa / CH3Cl, H3O+ / heat (c) EtONa / CH2Cl2, H3O+, heat, alc KOH (d) All of these
300
Problems in Organic Chemistry (+)
25. CH3COCH2COOC2H5
EtONa RX
CH3COCHCOOC2H5
H3O –C2H5OH
X
O3
H2O
R CH3COCOOH + HCOOH + CH3COCH2CH2COOH
RX would be:XCH = CH — C = CH2 (a) CH3
(c) XCH2 — C — CH = CH — CH3
(b) XCH2CH = CCH3CH = CH2 (d) XCH — C = CH — CH = CH 2 2
CH3
CH3
26. CH3COCH2COOC2H5 Can be converted in to 1, 3 di ketone by treating it with (a) EtONa / CH3COCl, H3O+ / heat (b) EtONa / CH3COCH2COCl, H3O+ / heat + (c) EtONa / CH3CH2COCl, H3O / heat (d) EtONa / CH3COCH2CH2COCl, H3O+/ heat 27. Which will form active methylene compound when treated with CH3OH / H+ O COOH
(a) CH3COCH2CH2COOH
(b)
(c) CH3CH2COOH
(d) (CF3)3CH2COOH
Passage - II An organic compound (A) containing six membered ring gives (B) when treated with KOH + C2H5OH. (B) Gives (C) when subjected to heat with KMnO4. (C) Releases CO2 with NaHCO3 & gives (D) & (F) when treated with [CaO / heat] & [CH2N2/heat] respectively.(D) gives (E) when heated with Ca(OH)2. (E) further gives (D) when treated with [N2H4/ heat] followed by ozonolysis. On the other hand (F) gives (G) when treated with EtONa/ H+. (G) Also gives (D) by [H3O+ / heat]. If compound (A) on reduction with LiAlH4 gives hydrocarbon then Answer the questions from 28 to 34. 28. Compound (A) is:Br O
(a)
(b)
O
Br Br
(c)
(d)
(b)
(d)
29. Compound (B) is:- (a)
OH COOH O
(c)
30. Compound (C) is: (a) Adipic acid (c) Succinic acid
(b) Glutaric acid (d) Oxalic acid
301
Carboxylic Acid & Its Derivatives
31. Compound (D) is:O
(a)
(b)
O
(d) MeCOMe
O
(c)
32. Compound (E) is:O O
(a)
(b) O
O
(c) 33. 34.
OH
(d)
Compound (F) is:(a) Dimethyl adipate (c) Dimethyl Succinate Compound (G) is:CO2CH3
(a)
(b) Dimethyl Glutarate (d) Dimethyl Oxalate CO2Me
(b)
O CO2Me O
(c)
O
(d) None
Passage - III 2A + B
+ H( )
→ C
(A) Can not show haloform test but can reduce schiff’s reagent.A can not show aldol condensation but its oxidized form can show reducing properties. Compound B can be obtained by partial reduction of the alkyne which can not show tollen’s test. If compound B contains 4 carbon atoms then Answer the questions from 35 to 42. 35. Compound (A) is: (a) 36. 37.
CHO
(c) PhCHO Compound (B) is:(a) CH3CH2CH = CH2 (c) Iso butene Compound (C) is:-
O (a)
O
O (c)
O
(b) CH3—CH—OMe OH (d) HCHO (b) CH3CH = CH CH3 (d) CH2 = CH—CH = CH2
(b)
(d)
O
O
O
O
302
Problems in Organic Chemistry
38. Which intermediate will not form during the formation of (C) from (A) & (B)? (a)
OH (+) O
O (c)
(+)
(b)
(d) H
CH2OH
(+)
H
(+)
O
O
39. How many optical isomers are possible for (Y) ?
( B)
+
( CN) 2
+ H O( )
3 → ( X ) →
(a) 2
( Y)
(b) 3
(c) 4
(d) 5
40. Which statement is not correct about (Z) & (W)? Hot KMnO 4 (B) → ( Z) , (A)
(a) (b) (c) (d)
( i) alkaline CuSO ( ii) H
4 → ( W) ( +)
pKa of (W) is lesser than that of (Z) Both can show reducing properties (Z) gives blood colouration with neutral FeCl3 (W) produces mercury on reaction with corrosive sublimate
41. What is (P) in the following reaction?
NaOH / CaO → ( I) ( Y)
(a)
+ HCl
3 → ( P)
AlCl
(c)
(b)
(d)
O O
O CH3
CH3
42. (Y) is subjected to heat to produce another organic compound R. Which is correct about (R)? (a) It is a carboxylic acid (b) It is optically active (c) It is an anhydride (d) It is an ester 43. Consider the following reagent: NH3 / heat
NH2OH
(1)
(2)
PCl5 / heat
Na / heat
(3)
(4)
Which among the above reagents will react with both the functional group of lactic acid? (a) (3) (b) (4) (c) (1), (2), (3) & (4)
(d) (3) & (4)
44. An organic compound on treatment with NH3, NaOBr & NaNO2 + dil H2SO4 gives cyclo pentanol.
Organic compound will be:-
(a)
(b)
COCl
(d) Both (b) & (c)
COOH
(c)
CH2COCl
303
Carboxylic Acid & Its Derivatives
45.
( A)
(1 mol)
+ H O( )
3 →
( B)
(1 mol)
Na
→ 2 moles of H gas 2 CO2Et
COOEt
(a)
(b)
CN OH OH
HO
(CH2)3Ac
(c)
HO
(CH2)4Ac
(d)
HO
EtO2C CH2CH = NH
CN
Br (1) 1 mol Mg
46.
[X]
(2) D2O I [X] would be:D
D
(a)
(b) I
D D
(c)
(d) None of these
Br
47. Consider the following processes.
Process (1) C2H5OH / H+, LiAlH4, PCC
Process (2) NaBH4, Cu / 300°C
Process (3) PCl5, LiAlH4, CrO3 / H2SO4
Process (4) SOCl2, Pd / BaSO4
The process by which acetic acid can be converted in to acetaldehyde is/are:(a) 1, 2, 3, 4 (b) 1, 4 (c) 1, 3, 4 (d) 2
48.
O
LiAlH D O
→ (X), (X) would be:O CH2OD
CH2OD
(a)
CH2OH
(b)
CD2OD
(c)
CH2OD
CH2OD CHDOD
(d)
CH2OD
304
Problems in Organic Chemistry
CONH2
49. (CH2)2 (A)
(+)
P2O5 ∆
(B)
H3O H2O2
CONH2
(–)
(C) (D)
OH (+)
H3O
NH3 ∆
(E)
(F)
50.
Which statement is not correct? (a) (F) and (A) are same (b) (C) and (E) are same and can be converted in to (A) by NH3 / Heat (c) (D) and (A) are same and can be converted in to (C) by H3O+ (d) All are correct The reaction in which ester does not form is -
(a) AcBr + BuOH
+ H( )
→ PCC
→ (b) Amyl alcohol
Product (X)
Al( OEt )
3 → Product
(c) HO — CH2 — CHOHCH2OH + HNO3 → (d) In each case ester is produced 51. Consider the following four compounds. O O O
O
CH3 — C — NH2
52.
(1) (2) Decreasing order of bond lengths of C = O bond would be:(a) 2 > 3 > 1 (c) 3 > 2 > 1 Consider the two carboxylic acids
Product :-
CH3 — C — O — C — CH3
(3) (b) 1 > 3 > 2 (d) 2 > 1 > 3
(CH3)3C COOH
(CH3)3 SiCOOH
(1)
(2)
CH3 — C — OCH3
Select the correct statement (a) (pKa)2 > (pKa)1 (c) (pKa)1 = (pKa)2
(b) (pKa)1 > (pKa)2 (d) None of these
O CN
53. HOOC
COOH
(+)
H3O ∆
[X]
ClOC O Select the correct statement (a) X will give effervescence of CO2 with NaHCO3 (b) X will produce alkane when treated with Zn / Hg + HCl (c) X will produce symmetrical diketone when heated with sodalime (d) All are correct Assertion: - Rate of ethanoic acid is greater than propanoic acid. . of decarboxylation . Reason: - CH3 is less stable CH3CH2 (a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true
54.
305
Carboxylic Acid & Its Derivatives
55. Assertion:- Vont Hoff’s factor for carboxylic acid is more in aqueous medium but least in benzene Reason: - In benzene RCOOH is present in its associated form while in water it is gets dissociated in to caboxylate and hydronium ions (a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true
NaOH 56. Assertion: - RCOOH → RH
heat
This reaction requires the presence of CaO
Reason: - CaO absorbs the released CO2 in the reaction and forms CaCO3. (a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true C2H5 heat
→ 57. Assertion:- CH3 — C — COOH
Product
COOH In this reaction two products are formed
Reason: - Due to rearrangement CH3 — C ( COOH ) converts in to the mixture of 2 | C2 H5
CH3 — C ( COOH ) 2 | C2 H5
and
HOOCCH 2 — CH — COOH | C2 H5
(a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true 58. Assertion: - rate of decarboxylation of fluoroacetic acid is greater than that of chloro acetic acid Reason: - fluoro methyl carbanion is less stable than chloro methyl carbanion. (a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true 59. Assertion: - decarboxylation by sodalime in 4-methoxy phenyl acetic acid occurs rapidly in comparison to that of 4- nitro phenyl acetic acid. Reason: - Rate determining step involves the formation of carbanion. (a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true
306
Problems in Organic Chemistry
dil H 2SO 4 soda lim e 60. Assertion: = CH 2 CH — OCOEt → ( A ) + ( B) → Et − H
A does not release H2 on reaction with Na
Reason: - A is an aldehyde. (a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true COOH soda lime
→ Product
61. Assertion:-
Br This reaction gives decalin as a product
Reason: - carbanion is formed as intermediate which undergoes SN2 reaction (a) Assertion is true, Reason is true: Reason is a correct explanation for assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true
Answer Key 1. (a)
2. (c)
3. (b)
4. (a)
5. (d)
6. (c)
7. (d)
8. (d)
9. (c)
10. (c)
11. (b)
12. (b)
13. (c)
14. (b)
15. (a)
16. (b)
17. (c)
18. (b)
19. (b)
20. (c)
21. (a)
22. (b)
23. (b)
24. (a)
25. (b)
26. (a)
27. (d)
28. (c)
29. (b)
30. (a)
31. (a)
32. (a)
33. (a)
34. (c)
35. (d)
36. (b)
37. (c)
38. (c)
39. (b)
40. (b)
41. (b)
42. (c)
43. (d)
44. (d)
45. (d)
46. (d)
47. (b)
48. (b)
49. (c)
50. (d)
51. (b)
52. (a)
53. (b)
54. (b)
55. (a)
56. (a)
57. (c)
58. (d)
59. (d)
60. (a)
61. (a)
Multiple Choice Questions More Than One May Correct 1. Which among the following can’t show Fehling test? (a) HCOOH (c) CH3CH2COOH
(b) CH3COOH (d) PhCOOH
2. Which is not the preparation a chiral molecule? CH3CH — COOH (a) | CH3
Red P + Br H 2O
2→
HOCH 2 — CH — COOH (c) | CH3
LiAlH
NBS
= → (b) CH 2 CH — CH 2 — CH3 Red P + Br H 2O
2→ 4 PhCH 2COOH ( + ) → (d)
H
307
Carboxylic Acid & Its Derivatives
3. Which is not correct about X ? COOH
OHC
(a) (b) (c) (d)
∆
COOH O
X
X can show haloform test X can give brisk effervescence with NaHCO3 X can not show Fehling test X contains one CHO one keto and one COOH group
4. In which case 1st compound is more acidic than 2nd. (a) PhCOOH, CH3COOH (c) CH3CH2CH2COOH, CH3OH
(b) CH3SO3H, HCOOH (d) CH3OH, CH3NH2
5. Which among the following is not the preparation of ester? ∆
(a) CH3COOAg + I2
PAA
CH3COCH2CH3 (c)
(b) CH3CN
(d)
H2O2 (–)
OH
OH + CH3COCl
6. In which case products formed are not according to reaction? (+)
H
CH3COOH + CH3OH (a)
H2O + CH3COOCH3 (+)
H
(b) CH3COOH + CH3OH
H2O + CH3COOCH3 (+)
H
(c) CH3COOH + CH3OH
H2O + CH3COOCH3 (+)
H
CH3COOH + CH3OH (d)
H2O + CH3COOCH3
7. Which is correct statement about A, B & C ? (–)
CH3CH2CH2OH
(A)
MnO 4
(B)
(+)
H
Red P + Br2 H2O
(C)
(a) A is more acidic than (B) (c) C can not show Wurtz reaction
(b) C is more acidic than A & B (d) B is more acidic than A & C
8. The reaction by which acetic anhydride can be prepared is:Acetyl chloride + Sod. acetate (a)
heat
P2O5
Acetic acid heat (b) +
260°C
(c) Glycerine + oxalic acid
(d) CH3COOEt + CH3OH
H
9. In which case products formed are not according to the reaction:
(a) Glycerine + fatty acid (c) Glycerine + fatty acid
+
H +
H
Oil & fat
(b) fatty alcohol + fatty acid
Wax
(d) fatty alcohol + fatty acid
10. Which among the following can not show HVZ reaction? (a) PhCOOH (c) CH3CBr2COOH
+
H +
H
Oil & fat Wax
(b) CH3CHBrCOOH (d) PhCOCOOH
Answer Key 1. (b), (c), (d)
2. (a), (c)
3. (a), (b), (c), (d)
4. (a), (b), (c), (d)
6. (a), (b), (c)
7. (b), (c)
8. (a), (b)
9. (b), (c)
5. (b), (d) 10. (a), (c), (d)
308
Problems in Organic Chemistry
LEVEL - II Comprehension 1. Write the product (organic compounds) of following dehydration (CH2)2 (a)
COOH COOH
∆
COOH (CH2)3 P1 (b) COOH
(CH2)4 (c)
COOH COOH
∆
P3
∆
P2
2. Write mechanisms of (a), (b) & (c) (Q.No. 1 ) 3. CH2Br
(+)
Mg (1 Mol) THF
CH2Br
(A)
Blood red Colouration
H3O neutral FeCl3
(B)
C
(D)
Identify A, B, C & D
Matrix Match Questions Each question contains statement given in two columns which have to be matched. Statement (A, B, C, and D) in column I have to be matched with statements (p, q, r, s) in column II. The answer to these questions has to be appropriately bubbled as illustrated in the following examples. If the correct match are A — p, A — s, B —r, B — q, C — q, D — S, then the correctly bubbled 4 x 4 matrix should be as follows. A
p
q
r
s
B
p
q
r
s
C
p
q
r
s
D
p
q
r
s
4.
List - I Comprise of reagent for given compounds (X) List - II indicates the possible products (s) COOC2H5
match List - I with List - II
O OH
Cl O
List - I
List - II (CH2)2OH
OH (A) LiAlH4 (p) OH
CH2OH
COOC2H5
(B) Luke warm water
(q)
O OH COOH
309
Carboxylic Acid & Its Derivatives
(C) NaBH4
(r)
COOEt OH CH2OH
OH COOC2H5
(D) Cu/300°C
(s)
O Cl O
5.
List - I (Reaction)
(A)
List - II (Reagent required)
COOH
(p) CH3MgBr (2 mol) / H3O(+)
NH2
O
(B)
OMe
OH
O
O
(C) MeO
(q) EtMgBr / H3O(+)
(r) NH3 / D then KOBr
(s) N3H / H2SO4
OMe
O (D)
(t) MeMgBr (2 moles)
OH
6. Column-I O CN
Column-II
(A)
(p) Product contains ketonic group
dil H 2SO 4 → ∆
O
(B)
2 4→
(q) Product contains CONH2 group
(i) KCN / H SO (ii) H 2O 2 / NaOH
O
H O( +) ∆
3 →
(C) NC
CN
(r) Product contains COOH group
CN NH2 OH
(D)
HONO ∆
→
(s) Product contains OH group
310
7.
Problems in Organic Chemistry
Reaction (A) CBr3COCH3 + NaOH → (B) CH3CONH2 + NaOH + Br2 → (C) CH3COCH3 + KOH + Br2 → (D) CBr3CHO + NaOH →
Name of reaction (p) Haloform reaction (q) Hoffman degradation (r) Aldol condensation (s) Cannizaro reaction
Answer Key O
O
1. P1
=
P2
O
O
O
=
P3
O
=
O
3. A = ethane
B = ethanol
C = KMnO4
D = (CH3COO)3Fe
Answers matrix Match 4. (A) — p (B) — q (C) — r (D) — s
5.
(A) — r, s (B) — p (C) — t (D) — q
6. (A) — p, r (B) — q, s (C) — p, r (D) — p
7.
(A) — p (B) — q (C) — p (D) — p
SOLUTIONS LEVEL - I O OH OH H H
O 1. (a)
O
OH basic hydrolysis
OH esterification
MeCHO + (–)
O OH
CH2CO2
CH2CO2H
2. (c) Because compound will lose two molecules of carbon dioxide one from left hand side and other from right hand side to produce :HOOC—CH2—(CH2)3CONH *
*
S *
3. (b) CH2(CO2Et)2
O Base
*
(–)
CH(CO2Et)2
COOH
311
Carboxylic Acid & Its Derivatives
O
O
(–)
OH H2O
(–)
+ CH(CO2Et)2 CH(CO2Et)2 O
(–)
O
(–)
CH(CO2Et)2 O
Base
C(CO2Et)2
C(CO2Et)2
CH(CO2Et)2
(–)
OH
(+)
H2O
H3O
CO2H
C(CO2Et)2
OH
CO2H
4. (a) Because when it removes -ve charge comes on oxygen which is stabilized by resonance. 5. (d) CH3CONHCH3
OH –H2O
CH3CONCH3
Br2
CH3CONBrCH3 (Hoffman Bromamide)
In CH3 — CON(CH3)2, H is absent on ‘N’ hence it can not undergo Hoffman Bromamide reaction however it undergoes haloform reaction because it contains CH3CO linkage:CH3 CON(CH3)2
NaOBr or NaOH + Br2
(–)
CHCl3 +(CH3)2NCOO
CHCl3
6. (c) Because formic acid contains CHO as well as COOH group & can show reducing properties with fehling solution (CuSO4 + NaOH). O acid group
H—C — O – H aldehyde gp.
7. (d) Since compound gives blood red colouration with neutral FeCl3 hence it should have COOH group. Option b & d may correct but HCOOH does not react with neutral FeCl3 hence (d) is correct. 8. (d) HgCl2 + HCOOH → 2Hg + 2HCl + CO2
( Mercury)
(Calomel)
CHO
COO
CuSO4 + NaOH
9. (c) OH
(–)
(+)
COOH
H
OH
OH O O
10. (c)
+ Ac
COOH
Diels-Alder heat
+ CO2 Ac
312
Problems in Organic Chemistry
11. (b) This is claisen condensation: (–) CH3COOC2H5 + EtONa ——→ EtOH + CH2COOC2H5 (–)
O
O
(–)
CH3 — C — OC2H5
CH2COOC2H5 + CH3 — C — OC2H5
CH2CO2Et (–)
CH3 — COCH2CO2Et + EtO
If we do not add H(+) at the last of the reaction then EtO(–) will remove H(+) from CH3COCH2CO2Et (–)
(–)
as:- CH3COCH2COOEt + EtO
CH3COCHCOOEt + EtOH
12. (b) Except (b) in all cases mixture of active methylene compound will form however in (d) condensation is not possible O
13. (c)
O
OH
O
NaBH4
O
O
OH
LiAlH4
HO
CH2OH
NaBH4 can not reduce ester group but LiAlH4 can reduce ester as well as ketone group. (+)
14. (b) CH3COOC2H5
CH3COCH2COOC2H5
EtONa, H Claisen Condensation
(+)
heat – CO2
CH3COCH3
H3O – C2H5OH
CH3COCH2COOH
15. (c) Carbanion is formed as an intermediate in decarboxylation. Since (a) gives more stable carbanion thus decarboxylation of 4th acid occurs rapidly & it is too slow in 3rd case 16. (b) [X] = CH3SO3H 17. (c) CH3COOH
[Y] = CH3COOH
P2O5, heat
H3O
(+)
COBr
18. (b) OHC
(CH3CO)2O
COBr
Ag2O
O
COOH
H2O
O
O
∆ – CO2 (X)
HOOC
HOOC CHO
Pd, BaSO4
OHC
H3O
+
O
OHC
O
HOOC COOH O
∆
COOH
H2CrO4
(–)
∆
N2H4,OH
OHC
O (Z)
O
313
Carboxylic Acid & Its Derivatives
19. (b) Its IUPAC name is 2 — formyl cyclo hexan carboxamide 20. (c) Boiling point of Carboxylic acid will be higher due to H — bonding HO
O Pr—C
C—Pr O—H
O
Here two H — Bonds are present in between two molecules. After 3rd, 1st will have higher boiling point than 2nd due to H — bonding. H O
O OH
O
21. (a) Because configuration changes only when bond breaks and forms in new direction. H4
H4 1
1
Ac2O –AcOH
OH
H3C 3
OAc
H3C 3
CH2CH3
CH2CH3
2
2
[R]
22. (b)
[R]
(–)
EtONa
CH3COCH2CO2Et
CH3COCHCO2Et
BrCH2CH2Br
CH3COCHCO2Et CH2CH2Br
Ac
COOH
(+)
(–)
H3O
CH3CO—CCO2Et
CH3COCCO2Et
EtONa
CH2CH2 – Br heat – CO2
Ac
23. (b) CH2(COOC2H5)2
EtONa
(–)
CH(COOC2H5)2
CH2I -CH2I (1 mol)
2-moles
COOH CH2CH
CH2CH(COOC2H5)2 COOH COOH
2
H3O(+) –4C2H5OH
CH2CH
SN
CH2–CH(COOC2H5)2 COOH CH2CH2COOH
heat
(Adipic acid) CH2CH2COOH
(–)
O
24. (a) CH2(COOEt)2
EtONa –EtOH
(–)
CH2(COOEt)2
CH3CHO
CH3—CHCH2(COOEt)2
OH (+)
H3O
heat
EtOH
CH3—CHCH2(COOEt)2 CH3CH=CH(COOEt)2 COOH heat CH3CH=CH COOH –CO CH3CH=CHCOOH (Crotonic acid) 2
314
Problems in Organic Chemistry OH H
25. (b) CH3COCH2COOC2H5
EtONa RX
(+)
H3O
CH3COCHCOOC2H5
CH2CH=CMeCH=CH2 CH3COCH(COOH)CH2CH=CMeCH=CH2 X ozonolysis
CH3COCOOH + HCOOH + CH3COCH2CH2COOH
26. (a) CH3COCH2CO2Et
EtONa RX
(–)
CH3COCHCO2Et
CH3COCl
CH3COCHCO2Et COCH3 (+)
CH3COCH2COCH3 1 2 3 (1, 3 di Ketone)
heat –CO2
CH3COCH—COOH
H3O
COCH3
27. (d) (CF3)3C — CH2 — COOCH3 active methyl group Electron with drawing groups
Passage - II (28 to 34) (C) releases CO2 with NaHCO3, Thus, (C) must be a carboxylic acid. Since (A) on reduction with LiAlH4, gives hydrocarbon hence (A) must be a 1° or 2° alkyl halide. COOH (Adipic acid) COOH (C)
KMnO4,heat
alc KOH
Br (B) CaO
CH2N2
O
Aldol, Ca(OH)2
(D)
COOCH3 COOCH3
COO Ca COO
heat –CaCO3
(F)
(F) +
ozonolysis
EtONa, H Claisen Condersation
CO2Me
O (–)
N2H4, OH
(E)
O (G)
315
Carboxylic Acid & Its Derivatives
Passage - III 35 to 42 Since A can reduce schiff’s reagent hence it should contain CHO group & since its oxidized form can show reducing properties thus, it should be formaldehyde. Since B is obtained by reduction of the alkyne which can not reduce thus, B should be 2-butene Pd / C
(B)
→
Z & W are acetic acid & formic acids respectively. I is butane & on isomerisation it gives isobutene (P)
HCHO + H
••
(+)
(+)
CH2OH
CH2OH
B
A
CH2OH •• (+)
(+)
O=CH2
H O (+)
O (C) O CN
(+)
H3O
B + (CN)2 CN
43. (d)
OH
O COOH
(Y) COOH it contains two chiral carbon atoms
ONa COOH
Na
COOH Cl COCl
44. (d)
NH3
COCl
OH
NaOBr
H2O
CONH2
NH2
(+)
NaNO2+ H2SO4
NH3
NaOBr
CH2COCl H2O
OH
CH2CONH2
CH2NH2
(+)
NaNO2 + H2SO4 (+)
CH2
316
Problems in Organic Chemistry
OH
OH (CH2)4Ac
HO
45. (d)
(CH2)4Ac
HO
4-moles of Na
acidic hydrolysis
HO2C
EtO2C CN
COOH ONa (CH2)4Ac
NaO
+ 2H2
NaO2C COONa
46. (d)
Rest all amide produce resonance stabilized amines. Hence (a) will provide least yield of carboxylate ion on hydrolysis. 47. (b) See chemical reactions in your text book CHO O(–)
(–)
H from LiAlH4
48. (b) O
(–)
O
O
H CH2OD D2O
O (–) H O CH OD
CHO OD
(–)
H
CH2OD CONH2
49. (c) (CH2)2
CONH2
CN
P2O5 heat
(CH2)2
CO2H
(+)
H3O
(CH2)2
CN
(A)
CO2H
(B)
(C) CO2H
(–)
H2O2OH
(CH2)2 CO2H (D)
COOH
(+)
H3O
NH3, heat
(CH2)2 COOH
CONH2 (CH2)2 CONH2
(E)
(F)
50. (d) In (c) tri nitro glycerine is produced which is a nitro ester. 51. (b) 52. (a) Because electron donating effect of Me3Si is greater than Me3C. O
O CN COOH
53. (b) HOOC ClOC
COOH COOH
(+)
H 3O ∆
HOOC HOOC
[X]
O
O O Heat
Zn-Hg+HCl
[X] O
D2O
317
Carboxylic Acid & Its Derivatives
54. (b) In decarboxylation reaction carbanion is formed as an intermediate O (–)
(–)
CO2 + R
R—C—O
(–)
(–) RH + OH R + H2O Stability of carbanion ∝ rate of decarboxylation. 55. (a) In benzene two molecules of RCOOH are associated with each other by the help of H Bonding. O (–)
56. (a) R—C—O(–) (–)
R
(–)
RH + OH
+ H2O
CaCO3
CO2 + CaO
COOH
57. (c) C2H5 C—COOH CH3
58. (d) 59. (d)
CO2 + R
heat
CH3 C2H5
H
COOH
COOH C2H5
+ H
CH3
See question number 54 See question number 54
H OH 60. (a) CH2=CH—OCOEt
dil H2SO4 hydrolysis
CH2=CH—OH + EtCO2H
soda lime
Et—H
tautomerism
CH3CHO COOH (–) soda lime
61. (a) Br
Br
More than one may correct:1 b, c, d 2. COOH OHC COOH
a, c ∆
OHC
O O 3. a, b, c, d 4. a, b, c, d Organic compounds can be arranged in their decreasing order of acid strength as follows:RSO3H > HCOOH > RCOOH > PhOH > H2O > ROH > RNH2 H O / OH (–)
2 2 → CH3CONH 2 b, d CH3CN OH + CH3COCl OCOCH3 + HCl
5. 6.
Cl +CH3COOH
a, b, c See mechanism of esterification in your text book Br
7.
OH
b, c
(–)
MnO4 (+)
[A]
H
HVZ
COOH [B]
Order of acid strength is C > B > A 8. a, b 9. b, c 10. a, c, d In (a), (c) & (d) alpha ‘H’ atom is absent
COOH [C]
318
Problems in Organic Chemistry
LEVEL -II O
1 & 2 (CH2)2
O
C—OH
(CH2)2
C—OH
O
(+)
C—OH2
O
(–)
COO
O
O O (CH2)3
P1
O
C—OH
(CH2)3
C—OH
O
(+)
C—OH2
O
(–)
O
COO
P2
O
(CH2)4
CH2Br 3. CH2Br
O
O
C—OH
C—OH2
(+)
(–)
C—OH
C—O
O
O
O H2O + CO2 +
(–) (+) Mg (1 mol) (THF)
CH2Mg Br CH2—Br
CH2=CH2+ MgBr2 (A)
(+)
CH2 = CH2
H3O
EtOH
KMnO4
CH3COOH
neutral FeCl3
(CH3COO)3Fe (Blood red)
P3
11
Amines
Main Features Reaction Chart for Amines Preparation RNO2 (for 1° amine) RX RCONH2 (for 1° amine) Imine/schiffs base
LAH or H2/Ni or Sn+HCl alc NH3 LAH or N2/Ni
LAH or H2/Ni
Gabriel Phthalimide (for 1° amines) RCN(for 1° amine)
LAH or H2/Ni
ROH(for 1° amine)
NH3 Al2O3
RCONH2(for 1° amine)
KOBr
RCOOH(for 1° amine)
N3H H2SO4
RNC(for 2° amine)
LAH
RMgX + ClNH2
Properties NaNO2 + HCl
Alcohol / nitroso amine [ 1°amine / 2° amine]
CH3COCl
Acylation (for 1° and 2° amines)
CS2/HgCl2
A M I N E S
CHCl3 KOH PhSO2Cl KOH CH3Br (in excess) H2O2 or H2SO5 COCl2
RNCS (Mustard oil test) (for 1° amine) RNC (for 1° amines) Hinsberg test (for 1° and 2° amine)
(CH3)4NBr (for all amines) R3N→O (for 3° amine) RNCO (for 1° amines) Hoffman elimination
320
Problems in Organic Chemistry
LEVEL - I Multiple Choice Questions C H ONa 110°C
2 5 1. Urea (A) + Diethyl malonate → (B)
Which is correct about the basic strength of A & B? (a) (pKb)A = (pKb)B (b) (pKb)A > (pKb)B
(c) (pKb)A < (pKb)B
(d) (pKb)A ; (pKb)B
2. Identify (X) in the following reaction:OH
C H OH
COCH3
2 5 + NH2NHCONH2 → (X) 75°C,8hr
OH
ONHNHCONH2
(a)
COCH3
(b)
C= NHNCONH2 CH3 OH
ONHCONHNH2
(c)
COCH3
(d)
C=NCONHNH2 CH3
3. Rank the following compounds in order of basic strength. O
O N
N
(1) CH3
O N (3) H
(2) CH3
(a) 2 > 3 > 4 > 1
N
(b) 2 > 3 > 1 > 4
(4) CH3
(c) 3 > 2 > 1 > 4
(d) 3 > 2 > 4 > 1
CH3 |
(i) LiAlH
4 → Product 4. CH3 — C(NO 2 )C(OH)Me2 (ii) NaNO + H SO 2
2
4
Product of this reaction would be:CH3
CH3 CH3
CH3 CH3
|
|
|
| | CH 2 = C — C = CH 2 (a) (b) CH3 — C — C — CH3 (c) Me3CAc (d) | CH3 — C = C — CH3 | | CH3 OH OH
5. Cyclobutyl Ethanoic acid is treated with N3H / H2SO4 and the product formed in this reaction is further treated with potassium nitrite & dilute sulphuric acid to give [X] , [X] would be: (a) Cyclobutyl Methanol (b) Cyclo pentanol (c) Cyclobutyl ethanol (d) Nitrosocyclo butyl methanamine 110°C
→ (X), (X) will not react with 6. Carbamide + N2H4 Amyl alcohol
(a) Carbinol
(b) AcH
(c) Formalin
(d) Acetophenone
321
Amines
7. Consider the following reactions. NaN / ∆
KBrO
3 EtCONH 2 → Product .............(1) CH3CH 2COCl → Product .............(2) H O
LiAlH 4 3 → Product .............(4) AcNH 2 → Product .............(3) CH3CHO (ii) LiAlH
The reaction in which ethyl amine forms as a product is/are:(a) 1, 3 & 4 (b) 1, 2 & 4 (c) 1 & 3
2
(i) NH
4
(d) 1, 2, 3 & 4
8. Caffeine is found in tea leaves. It contains four nitrogens. Correct sequence of basic strength of these nitrogens is:O
(1)
(4)
N
CH3—N
CH3
(2)
O
N (3)
N
CH3
(a) 1 > 2 > 3 > 4
(b) 2 > 1 > 3 > 4
(c) 1 > 4 > 3 > 2
(d) 1 > 3 > 4 > 2
9. 3° — butyl alcohol can be converted in to 3° butyl amine conviently by: (a) PCl5, NaNH2 (b) HCN + H2SO4 (c) AcCl / C2H5N & NH3 / Heat (d) All of these 10. The reaction which can not be performed by H2 / Ni is:
→ CH3NH2 (a) CH3NO2
→ CH3CH2NH2 (b) CH3CONH2
→ ΦCH 2 — CH 2CH 2 NH 2 (d) CH NC (c) — Φ CH = CH — CN → CH3NHCH3 3 11. (A)
NaNO2
(C)
–ve victor maeyer test
+ve mustard oil test Blue colour in vector maeyer test (D)
(B) NaNO2 HCl
(A) and (B) are respectively. CH3
CH3
NH2
|
(a) CH3 — C — NH 2 , CH3—N— CH3 |
|
CH3
Et
CH3
|
, CH3 — C — NH 2 | CH3
(b)
NH2
|
(c) CH3 CH — CH 2 NH 2 ,
(d) , CH2NH2
Et |
CH3 —CH — NH 2
12. Which product will not form when primary amine reacts with sodium nitrite & HCl ? (a) Alcohol (b) Alkylhalide (c) Alkene
(d) Chloramine
13. Which of the following is the weakest bronsted base ? N
(a) O
(b)
(c)
(d)
322
Problems in Organic Chemistry
14. Mitomycin C is an anticancer agent & used to treat stomach and colour couch. Its structure is 2
O
1
CH2OCONH2
H2N
OMe
3 N 4 NH O Which is correct about mitomycin C?
(a) It contains 5 chiral carbon atoms (c) It contains 3 chiral carbon atoms
(b) 3rd nitrogen is highly basic (d) 4th nitrogen is highly basic
15. Arrange the following compounds in decreasing order of boiling point (a) (CH3)3N < CH3NHEt < PrNH2 < PrOH (b) (CH3)3N < PrNH2 < PrOH < MeNHEt (c) PrNH2 < MeNHEt < Me3N < PrOH (d) Me3N < MeNHEt < PrOH < PrNH2 16. Strongest acid among the following is:H NH2 N SO NH (b) (c) (d) 2 2
(a)
N
17. Consider the following reactions (i) L.A.H
dil H 2SO 4 ® [B] CH 2 = CH 2 + (CN)2 ¾¾¾¾¾ ® [A] Benzonitrile ¾¾¾¾¾¾¾ (ii) MeCl / NaOH
OH
NH2
O (i) AgNO 2
+HI (excess) → → [C] (+) (ii) Fe / H
+
H ( )
H / Pd
2 [D]
Out of A, B, C & D the product (s) the product which can show Hoffman mustard oil test is/are:(a) B & D (b) only D (c) A, B & D (d) Only C OH
18. Identify (X) in the following reaction. (+)
(i) LaH (ii) H / heat (iii) NaOH
ClCOCH 2CH 2CH 2 NH 2 →(X)
H H N N
(a)
CH3 N—CH3 (b)
CH3 (c)
CH3 (d) N—H
19. Identify [Y] in the following reaction O
O
+ O
[Y]
NH2 O
O
OH
OH
NH
(a)
O
(b) NH
NH O
(c)
NH OH
(d)
O OH
323
Amines
20. The reaction in which ketone does not form. (i) KMnO
(i) CH3MgBr 4 (CH3 )2 CHNH 2 (a) CH3CN + → Product (b) + → Product (ii) H3O
(ii) H3O
(i) NaNO 2 + HCl (c) i − Propylamine → Product
(d) In all cases ketone is formed
(ii) Cu / ∆
Passage - I Reserpine is a natural product belonging to the family of alkaloids. Clinical application of it is in treatment of hypertension and nervous and mental disorders N
MeO
N H
H
H H
MeO2C
O OMe (Reserpine)
OMe
O—C—
OMe OMe
Reserpine breaks in to products when subjected to hydrolysis as follows OH OH
Reserpine + H3O + → A + 6MeOH + HOOC
OH
Answer the question from 21 to 23 21. Which is correct about the basic nature of nitrogen atoms ? (a) ‘N’ of pentagonal ring is more basic than that of hexagonal ring (b) ‘N’ of pentagonal ring is less basic them that of hexagonal ring (c) Both are equally basic (d) ‘N’ of hexagonal ring can show amine inversion 22. Chiral ‘C’ atoms present in reserpine are: (a) 4 (b) 5
(c) 6
(d) 7
23. When 2 moles of aqueous NaOH are added in to one mole of (A) then product formed would be:-
(a) HO
N
N
(–)
Na+ NaO2C
(c) HO
OH
OH
N
N H
N
(+) (–) N (b) Na O H
NaO2C
(+) (–)
(d) Na O
OH
OH
N
N
(–)
(–) (+)
NaO2C
O Na
OH
NaO 2C
OH OH
324
Problems in Organic Chemistry
Passage - II CH I
∆
Ag O H 2O
H SO
heat
3 → P → 2 5 → S 2 N − methyl peperidine Q → R →T
Answer the question from 24 to 28 24. Compound P is:(+)
(a) (+) N
N I(–) (b) (c) (–) (–) N (+) I I CH3 H3C
(+)
(d)
N I(–) H3C
CH3
25. Compound Q is:
(a)
(b) (+) N
(+)
N
(c)
(–) (d) (+) (–)
(+)
OH
N
N OH CH3 H3C
(–)
(–)
CH3
H 3C
CH3 O
CH3 O
26. Compound R is: (a)
(b)
(c)
N
N
(d) NMe2
NMe2
27. Compound S is:- (a) (+) N
O (b) (–) N O
O
(c)
(d)
N
28. Compound T is (a) Buta - 1, 3 - diene (c) Penta - 1, 3 - diene
(+) (–)
N—O
(b) Penta - 1, 4 - diene (d) (CH3)2 N+ = CH — CH2 — CH = CH2
Passage - III ∆
OH (–) , heat
H O
2 2 → A → B → C CHI(CH2)3NHMe ∆
N
Answer the question from 29 to 31 29. Compound A is:I CH
(a)
(+)
N
N
CH3 H
(CH2)3
CH=CH(CH2)2NHMe
(+)
N
(b) N
CH3
H (c)
N
(d) None of these
325
Amines
30. Compound B is:OH
I
CH
(a)
N
N
CH
(CH2)3 (b)
(+)
(+)
N
CH3
N
Me
N
(CH2)3 (c) N
H
CH3
(+)
(d)
N
CH3
N
31. Compound C is:OH
O CH2CH=CH2 (b)
(a) N
N
N OH
N
(c) N
N
(d) OH N
OH
Passage - IV A nitrogen containing compound (with four carbon atoms) can have two isomers (A) & (B). (A) & (B) On treatment with Fe + HCl give (C) & (D) respectively. Both (C) & (D) produce same compound (E) on treatment with sulphuric acid and sodium nitrite. Compound (C) on treatment with KMnO4 followed by hydrolysis gives (F). (E) And (F) don not give haloform test. (F) Gives (G) when treated with NaBH4 & thionyl chloride. Now (E) is treated with aqueous NaOH & then with (G) to produce (H). (H) Further gives (E) when mixed with dilute sulphuric acid.If H can decolourise bromine water then:Answer the question from 32 to 40 32. Compound (A) is: (a)
NO2 (c) NO2 (b)
NO2 (d)
NO2
33. Compound (B) is:
(a)
NO2 (c) NO2 (b)
NO2 (d)
NO2
34. Compound (C) is:
(a)
NH2 NH2 (b) (c)
NH2 (d)
NH2
35. Compound (D) is:
(a)
NH NH2 (b) 2 (c)
NH2 (d)
NH2
36. Compound (E) is: (a)
OH
OH OH (c) (b) (d) OH
37. Compound (F) is: (a)
CHO NO (c) (d) CHO (b) O
326
Problems in Organic Chemistry
38. Compound (G) is: (a)
CH2Cl (c) Cl (b)
CH2Cl
(d) NHCl
39. Compound (H) is: (a)
CH2—O (b)
OCH2 (d)
(c)
40. Compounds which exhibit enantiomerism (a) (F) & (G) (b) (B)
(c) (H)
(d) None
41. Consider the following amines. Et
CH3 | CH3 — C — Pr (2) (1) | CD3CHCH 2 NH 2 NH2 |
CH2NH2
(3) CHD2 – CH2 – CHDNH2 (4)
Amine/s which will exhibit optical isomerism after the treatment with NaNO2 + H2SO4 is / are:(a) 1, 2 (b) 1, 3 (c) 1, 3, 4 (d) Only 1
42. Match the following: Reaction Product (i) HONO
→ (A) PrNO2 (ii) NaOH
(B) MeNO2
→ (C) EtNO2 (ii) HONO (iii) KOCl
(D) Me2CHNO2 →
Cl + NaOH
2 →
(i) Sn + HCl
(i) HONO (ii) NaOH
(1) Blue Colour (2) Red Colour (3) Chloropicrin (4) Chloroform
A B C D A B C D (a) 5 3 4 1 (b) 2 3 4 1 (c) 2 4 3 1 (d) 2 4 1 3 43. The method which does not offer preparation of 1° amine is:
(a) Gabriel Phathalimide ∆
H
2→ (CH3 )2 CO + NH3 → (c) Pd
H
2 → (b) CH3CHO + CH3 NH 2 Pd / ∆
(d)
CH3 CH3
(i) KMnO
4→ CH—NH2 (ii) H ,Pd 2
44. The compound whose one mole will react with three moles of Ac2O is:O HN (a) O
CN NH2 (b) (c) COOH O N HO
NH
(d) None of these
OH
45. Least basic amine among the following is:N—H
(a)
N••
(b) Pr3N: (c) (d)
••
N H
327
Amines
46. A primary amine on treatment with NaNO2/HCl gives alcohol which on dehydration followed by ozonolysis gives two compounds. One of them gives +ve iodoform test but negative tollen’s test whiles other gives +ve tollen’s test but negative iodoform test. If alcohol does not produce any colour in victor maeyer test then structure of amine will be:-
(a)
(b)
NH2
(c)
NH2 (d)
NH2
NH2
47. Weakest base among the following is:H
H
N
N
N
N (b) CH3 CH (c) CH3 3
(a)
(d) H
C2H5 C2H5
48. Which of the following is the IUPAC name of C3H7NC?
(a) Propyl isocyanides
(b) Carbyl amino butane
(c) Carbylamino propane
(d) Butyro isonitrile
(c) Isocyanic acid
(d) Cyanogen
49. HNCO on tautomerism will produce
(a) Cyanuric acid
(b) Cyanic acid
50. The reaction in which amine does not produce is:(i) P + I
–AgI KOH 2 → Product CH3OH (a) CH3I + AgNCO → X → Product (b) (ii) NH 3
(i) CH NH
KOH (c) CH3 NCO → Product
3 2 Product (d) CH3COCH3 → (ii) H / Ni
51. Me4N + OH– is: (a) Amphoteric (c) Basic in nature but lesser than NaOH
(b) stronger base than NaOH (d) Neutral in nature
nBu
2
nPr
|
|
nitrous acid 52. CH3 — C — CH —C — CH3 ¾¾¾¾¾ ® major product (X) | | | Et NH2 Et
[X] would be:nBu
nPr
|
|
(a) CH3 — C — CH —C — CH3 |
Et
|
OH
nBu |
|
nBu
OH
|
|
(b) CH3 — C — CH —C — CH3 |
|
|
Et
Et nPr
Et
nPr
OH
nPr
|
|
|
CH3 — C —CH —C — CH3 (c)
(d) CH3 —C— CH —C — CH3
OH Et
Et nBu
|
|
|
Et
|
|
53. If boiling points of CH3NO2, CH3ONO and CH3COOH are x, y & z respectively then (a) z > x > y (b) y > x > z (c) z > x = y
|
Et
(d) z > y > x
328
Problems in Organic Chemistry
O (CH2)2CN
54.
H Pd.heat
2 → Product,
Product of this reaction will be:OH
(a)
OH
O (CH2)3NH2
(b)
(CH2)3NH2
(c)
(CH2)2CN
(d)
N H
D 14
55.
KBrO CONH2 → Product
C O H
Product of this reaction will have:(a) ‘R’ configuration (b) ‘S’ configuration (i) H O /OH(–)
2 2 56. (A) (ii) KBrO
(B) (C) KBrO
(i) H2O/H2SO4 (ii) N3H, Heat
(c) Both R & S configuration (d) product is optically inactive
[X]
By the help of [X] carbonyl compounds can be identified select the correct statement. (a) (A) is carbamide & B is cyanogens (b) (B) is NH2CN & A is Cyanogen
(c) (A) is Cynogen & (C) is NH2CN
(d) (B) is NH2CN but (C) is Carbamide
57. Assertion- aliphatic amines can not form RN2 + Cl– with NaNO2 + HCl while aromatic amines form diazonium salts Reason- Aliphatic diazonium salts are not stabilized by resonance (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true 58. Assertion-aniline can not be prepared by Gabriel phthalimide reaction while phenyl methyl amine can Reason- Aniline is less basic than phenyl methyl aniline (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true 59. Assertion- acetamide is amphoteric in nature Reason- Due to resonance NH2 group acquires positive charge and oxygen acquires negative charge so when acid is added it attacks on oxygen and when base attacks it removes water from NH2 (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true 60. Assertion- Although acetamide contains CH3CO linkage yet it does not form iodoform Reason- C—H bond is more polar than N—H bond (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is frue, Reason is false (d) Assertion is false, Reason is true
329
Amines
61. Assertion- 2nd acylation of primary amine is difficult Reason- After first acylation N of primary amine becomes sterically hindered (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true 62. Assertion- Aniline (aqueous solution) can precipitate FeCl3 in to ferric hydroxide but same is not true for methyl amine Reason- In aniline the lone pair of electrons take part in resonance in benzene ring while lone pair of electron of methyl amine is localized. (a) Assertion is true, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is true, Reason is false (d) Assertion is false, Reason is true
Answer Key 1. (c)
2. (b)
3. (d)
4. (c)
5. (b)
6. (a)
7. (d)
8. (b)
9. (b)
10. (d)
11. (c)
12. (d)
13. (a)
14. (b)
15. (a)
16. (d)
17. (d)
18. (c)
19. (d)
20. (d)
21. (b)
22. (c)
23. (b)
24. (a)
25. (c)
26. (c)
27. (a)
28. (b)
29. (b)
30. (c)
31. (c)
32. (b)
33. (a)
34. (b)
35. (a)
36. (b)
37. (a)
38. (b)
39. (a)
40. (d)
41. (b)
42. (b)
43. (b)
44. (a)
45. (b)
46. (a)
47. (b)
48. (c)
49. (b)
50. (a)
51. (b)
52. (d)
53. (a)
54. (a)
55. (a)
56. (c)
57. (a)
58. (b)
59. (a)
60. (c)
61. (c)
62. (d)
Multiple Choice Questions More Than One May Correct 1. Water insoluble amine is:
(a) Me2NH
(b) PhNH2
(c) Ph2NH
(d) Ph3N
2. Which statement is not correct about acetamide? (a) It is amphoteric in nature
(b) On hydrolysis it produces primary amine
(d) On heating with P2O5 it gives carbyl amine
(c) On heating with NaOH it gives NH3
3. Which among the following pair can not be differentiated by Hoffman mustard oil test?
(a) Ph2NH, Me3N
(b) MeCONH2, Me2NH
(c) MeNH2, i-Pr-NH2
(d) EtNH2, Me2NH
(c) CH3NHC2H5
(d) CH3NH CH3
4. Which is resolvable?
(+ )
(a) CH3CH2CH2CH(Et)NH2 (b) CH3 NHD |
Et 5. Which is not the preparation of secondary amine? heat
LaH
(a) CH3CHO + NH 2OH → →
LAH (c) CH3CONHCH3 →
(b) Gabriel Pthalimide reaction (d) CH3 NH 2 + CH3Br(2moles) →
330
Problems in Organic Chemistry
6. Aliphatic 1° amine:
(a) is more soluble in water in comparison to alcohol of comparable molar mass
(b) has pyramidal shape
(c) has more boiling point than RCOOH
(d) has sp3 hybridised ‘N’ atom
7. Which is correct about the following road map?
NaNO + HCl
LAH
2 →[A] →[B]
NO2
(a) A is optically inactive
(b) B is optically inactive
(c) A is optically active
(d) B is optically active
8. Which of the following reaction is least likely to occur? (a) (CH3 )2 NH + CHCl3 + KOH →(CH 2 )2 CNC NaBH 4 (b) CH3CONH 2 → CH3CH 2 NH 2
(c) CH3 NH 2 + CH3COCl → CH3CONHCH3 + HCl pyridine
(d) PhNHCOCH3 + CH3COCl →(CH3CO)2 NPh + HCl 9. Which is not in the favour of Gabriel pthalimide reaction?
(a) It is the method for the preparation of pure primary amine
(b) Aromatic amines can not be prepared by this method
(c) All aromatic amines (except aniline) can be prepared by this method
(d) 2° & 3° amines can also be prepared by this method H KOBr
CONH 2 → product
10. CH3
(a) Product of this reaction can show geometrical isomerism
(b) Product of this reaction contain plane of symmetry
(c) Product of this reaction can show both geometrical & optical isomerism
(d) Product of this reaction is a secondary amine. O
11.
NC
CN
NC
3 CN → [X]
NC
(i) H O(+) (ii) N3H +H 2SO 4
CN CH3
Which is correct about [X]? (a) It is optically active
(b) It bears 3 chiral ‘C’ atom
(b) It contains only ketone as functional group
(d) It contains one ketone & one NH2 group
331
Amines
OH
NH2
NO2
CH2NH2
12. OH (1)
(2)
(3)
(4)
(a) (1) is strongest acid while (4) is strongest base (c) (3) is strongest acid
(b) (4) is strongest base (d) (3) is weakest acid while (4) is strongest base
Answer Key 1. (b), (c), (d) 6. (a), (b), (d)
11. (a), (d)
2. (b), (d)
3. (a), (b), (c)
4. (a), (b)
7. (a), (d)
8. (a), (b), (d)
9. (c), (d)
5. (a), (b), (d) 10. (a), (b)
12. (b), (c)
LEVEL - II Comprehension Comprehension - I NH3 SOCl H ( ) CH3 MgBr *CO 2 [A] [B] 2 [C] [E]
Precipitates
NaOI
H / Pd / BaSO 4
2 [D]
1. What is ‘F’ ? 2. Can ‘G’ show Hoffman mustard oil test 3. Out of A, B,...............G, which contains isotopic carbon i.e. *C 4. Which will show Tollens’s test among B, C, D, E & F ? 5.
RCONH 2 + Br2 + KOH ¾¾ ® RNH 2 x
:
y
: z
What is the ratio x:y:z ?
Comprehension - II Consider the following reactions.
OH |
KOBr
CH3 — CHCONH 2 →(A) NH 2CONH 2 → (B)
[F]
BrO()
[G]
332
Problems in Organic Chemistry NaNO 2 + HCl → (C)
N H
OH |
(i) CrO , ∆
3 → (D) CH3CHCH 2CONH 2 (ii) NaNO + HCl, ∆
(a) A
2
(b) B
(c) C
(d) D
6. A can undergo condensation with: (a) B (b) C
(c) D
(d) None
7. B can undergo condensation with: (a) A (b) C
(c) D
(d) All of these
8. Which is yellow coloured solution ? (a) A (b) B
(c) C
(d) D
(c) Homomers
(d) Homologues
B NaOH,∆
9. A → E B
→ F D NaOH,∆
E & F are:(a) Optical isomers
(b) Geometrical isomers
Comprehension - III Glycerine
10. Which is highly aromatic? (a) C
KHSO4 ∆
(+)
(A)
Aniline
(E)
[O]
(b) D
11. Total number of chiral carbon atoms present in C is: (a) 2 (b) 3
(B)
H
(D)
–H2O
(C)
(c) E
(d) All are equally aromatic
(c) 4
(d) 1
Answer Key 1. CHI3
2. yes
3. A, B, C, D & E
4. (d)
6. (a)
7. (a) & (c)
8. (c)
9. (d)
11. (d)
5. x : y : z = 1 : 4 : 1 10. (c)
333
Amines
SOLUTIONS 1. (c) CO A
NH2
EtOOC +
CH2
EtOOC
NH2
–2EtOH
NH—CO CH2 B
CO NH—CO
In B lone pair of electron present on N are more delocalized in comparison to that of urea because each lone pair takes part in resonance with two carbonyl group, hence B is less basic than A. OH
2. (b)
O
C
•• •• •• NH2NHCONH2 1
2
3
OH
C2H5OH 75°C, 8 hr
(–)
O C—NH2NHCONH2 (+)
CH3
CH3 OH
OH
C=NHNCONH2 CH3
OH C—NHNHCONH2
∆ –H2O
CH3
Out of 1st 2nd & 3rd nitrogens, 2nd & 3rd nitrogens can not participate in condensation as lone pair of electrons present on nitrogens is deloclised & take part in resonance with CO group. 3. (d) 1st & 4th are less basic as lone pair of electron is delocalized (resonance with CO group). 3rd is more basic than 2nd because secondary amines are more basic than tertiary amines. 4. (c) It is a kind of pinacole-pinacolone rearrangement CH3
CH3
CH3—C(NO2)C(OH)Me2
LiAlH4
CH3
CH3—C(NH2)C(OH)Me2
HNO2
CH3—C—C(OH)Me2 OH
pinacole pinacolone rearrangement
(CH3)3CCOCH3
H
(+)
(+)
(+)
5. (b)
CH2COOH
N3H H2SO4
CH2NH2
HNO2
CH2
H2O
(+)
OH
OH2
6. (a) X is semicarbazide which gives condensation with (b),(c) & (d) but not with (a) (carbinol is MeOH) H 2O Heat 7. (d) RCOCl + NaN3 → RCON3 → RNCO → RNH 2
8. (b) 4th N is least basic as its lone pair of electrons are highly delocalized with two carbonyl groups. Lone pair of electrons present on 3rd N are also delocalized with one carbonyl thus, it is less basic than 1st and 2nd. 9. (b) This is Ritcher reaction.
334
Problems in Organic Chemistry
10. (d) Reduction of an organic compound by H2 / Ni is an example of molecular addition. CH3CN + 2H – H → CH3CH 2 NH 2 In isocyanide molecular addition is not possible. Me
Me
11. (c) Me— CH—CH2—NH2 A
Me (+)
NaNO2+HCl
Me— CH—CH2
Me— CH—CH3 (+)
Me negative Vector maeyer test
Me— CH—CH3 OH
OH
NaNO2+HCl
H2O
C
blue colour in Victor maeyer test
CH2NH2
12. (d) When primary amine reacts with sodium nitrite & HCl carbocation forms. If it is attacked by chloride ion alkyl halide forms & alcohol forms if water attacks on carbocation. If H+ ion is removed from carbocation alkene is formed. 13. (a) Lone pair of electron present on nitrogen is highly delocalised. 14. (b) It contains 4 chiral carbon atom. 3rd nitrogen is more basic than 4th because lone pair of electrons present on 4th nitrogen is delocalized. O O H2N
•• CH2OC—NH2 * *
OMe *
N
O
NH *
15. (a) Boiling point ∝ surface area Boiling point of alcohol is greater than amine due to strong H– bonding . Since O is more electronegative than N thus, alcohol forms stronger H– bond in comparison to amine. 16. (a) Lone pair of electron present on nitrogen are less delocalised in comparison to other.
17. (d)
18. (c) ClCOCH2CH2CH2NH2
(+)
LAH
OHCH2CH2CH2CH2NH2
H (+)OH2
(+)
–H (+)
(+)
N
N
H
H
H
•• NH2
335
Amines
O
O O
19. (d)
O
O
(+)
NH2
•• NH2
NH
O O(–)
[Y]
O OH
(+)
H3O 20. (d) MeCN + CH3MgBr → MeCOMe
Me2CHNH2 i-PrNH 2
KMnO4
(+)
H3O
Me2C=NH
NaNO2+HCl
MeCOMe Cu,heat
CH3CHOHMe
21. (b) Because lone pair of electron of nitrogen in pentagonal ring is not localized
22. (c) MeO
*
N H
N H
H
*
* H
*
*
O
*
OMe
O—C—
MeO2C OMe
OMe
(Reserpine)
23. (b) OH
(+) (–)
N
N A
OMe
NaO
2 moles of NaOH
N
N
H
H HO2C
(+) (–)
OH
NaO2C
OH
OH
OH
Passage - II (24 to 28) CH3I
Ag2O, H2O (+)
N CH3
N P
I
(–)
H2SO5
Heat (+)
N
(+)
(–)
NMe2
OH
N
R
(–)
O
Q heat
Me2NOH
+ T
S
336
Problems in Organic Chemistry
Passage - III (29 to 31)
N
I
Heat
OH
(+)
A
CH3
O2H2 ∆
(+)
N
N
MeNH
(–)
N
N
H
Me
B Heat
C
(+)
N
N
Me
N
N
OH
Me
(–)
O
Passage - IV (32 to 40) Compound A & B may be nitro compound because they can react with Fe + HCl & the product formed after reaction can react with nitrous acid. E & F do not show haloform test hence both does not contain MeCO linkage or both may be 30 alcohol. Since H decolourises bromine water hence it should have C-C multiple bond. NO2
NH2
Fe + HCl
A
NaNO2+ H2SO4
OH
C NO2
E
Fe + HCl
B
D NH2
CH=NH
KMnO4
C
C F E
NaNO2 + H2SO4
NH2
CHO
NaBH4
OH
NaOH
(+)
F SOCl2
CH2OH ONa
H3O
G
CHO
CH2Cl
G (+)
H
H3O E
OH
40. (d) All are optically inactive because no one have chiral carbon atom. 41. (b) 2nd on treatment with nitrous acid will produce optically inactive alcohol. CH3 |
CH3 NaNO 2 + H 2SO 4
|
CH3 |
CH3 |
→ CD3CHCH 2( + ) → CD3 —C — CH3 → CD3 —C — CH3 CD3 CHCH 2 NH 2 42. (b) See victor mayer & haloform tests in your text book. 43. (b) 2° amine is formed.
(+ )
H 2O
|
OH
337
Amines
44. (a) It on tautomerism becomes aromatic O
OH
HN
NH
N
N
tautomerism
A O
O
HO
OH
Since A has 3 OH groups thus, it can react with 3 moles of Ac2O. 45. (b) Sterically hindered tertiary amine & can show amine inversion
(+)
NaNO2+HCl
46. (a)
NH2
ozonolysis
H Heat
OH
MeCOMe + EtCHO
+ve haloform but –ve Tollen's test
–ve haloform but +ve Tollens test
47. (b) 48. (c) Factual question N ≡ C — OH 49. (b) O = C = N — H (–) ••
50. (a)
••
NCO
Ambidentate nucleophile
In AgCNO, nitrogen site is not free to attack because it is a covalent compound (Ag-NCO) Thus reaction occurs asKOH
MeI + AgNCO → MeONC → not a primary amine 51. (b) Factual question 52. (d) Migrating tendency of groups follow the following order n butyl > n propyl > ethyl > methyl 53. (a) In acetic acid two H bonds are present between two molecules. In nitro methane there is one H bond in its tautomeric structure. H O
O N (CH2)2CN
54 (a)
(CH2)3NH2
H2/Pd Heat
Heat –H2O
3
3
D
D
1 14
2
C O
55. (a)
CONH2
2 14
KBrO
CN
H O / OH (–)
2 2 →
NH2
H4
S configuration
|
1
C O
H4
56. (c) CN
N H2 / Pd
R configuration
CONH 2 |
CONH2
KBrO
→
NH 2 |
NH2
N H H 2SO 4
3 (X) ←
COOH |
NH2
H O( + )
3 ←
CN
| ( C) NH2
338
Problems in Organic Chemistry
57. (a) Aliphatic diazonium salts are not stabilized by resonance however resonance occurs in aliphatic diazonium salts. (+)
(–)
(+)
(–)
(+)
(–)
(+)
(–)
N
N
N
N
N
N
N
N
(+)
(+)
(+)
58. (b) For the preparation of aromatic amines pthalimide has to react with aromatic halide which is not favourable reaction because in aromatic halide C-X bond acquires double bond character due to resonance, which is difficult to break. 59. (a) Due to resonance NH2 group acquires positive charge and oxygen acquires negative charge so when acid is added it attacks on oxygen and when base attacks it removes water from NH2 (–)
O
O
Me—C—NH2
Me—C (+)
OH
Me—C
(–)
(+)
NH2
(–)
OH base
O Me—C
NH
H (+)
NH2
60. (c) C–H bond is less polar than N–H bond thus, base removes H from N–H bond & not from C–H bond. 61. (c) After first acylation N is attached with COMe group so lone pair of electron present on N becomes delocalized as it undergoes resonance with CO group and can not attack easily on 2nd RCOCl molecule. 62. (d) Aniline is poor base than methyl amine & can not precipitate FeCl3 (+ )
(–)
→ MeNH3 + OH CH3 NH 2 + H 2O (–)
2Fe+3 + 6OH → 2Fe(OH)3 ↓ Brown ppt More than one may correct:1. 2.
b, c, d (b),(c) & (d) contain hydrophobic ‘Ph’ group which make them insoluble in water. b, d CH3CONH2
P2O5/∆
CH3CN
(+)
H3O
3. 4.
CH3COOH + NH3
a, b, c Hoffman mustard oil reaction is the characteristic property of primary amines a, b In both the cases chiral centers are present while (c) & (d) are not resolvable due to amine inversion.
5.
a, b, d
6.
a, b, d RCOOH has more boiling point than amine because two intramolecular hydrogen bonds are present between two molecules of RCOOH
7.
a, d
NaNO2 + HCl
LAH
NO2
NH2 Optically inactive
OH Optically active
339
Amines
8.
a, b, d
9.
c, d
H
10.
H KOBr
CONH 2 →
a, b CH3
NH2 CH3
Product is optically inactive as it contains plane of symmetry 11.
a, d product of this reaction is O
* * NH2 CH3
It is optically active and bears two chiral carbon atom. 12.
b, c
LEVEL -II
Question No. 6 to Question No. 8 OH
OH
CH3CHCONH2
KOBr
H–Bonding
CH3—CH—NH2
–NH3
CH3CHO (A)
(Unstable)
KOBr
NH 2CONH 2 → NH 2 NH 2 (Hydrazine) (B) NaNO2 + HCl
N
N
H (2°amine)
(yellow liq)
NO
CH3CHOHCH2CONH2
CH3COCH3 (D)
(C)
∆
CrO3
CH3COCH2CONH2
CH2COCH2COOH
NaNO2 + HCl
340
Problems in Organic Chemistry wolf Kishner reduction
9. (d) A
→ Ethane
B
→ Propane (F)
(E)
wolf Kishner reduction
E & F are homologues Question No. 10 to Question No. 12 KHSO4 ∆
Glycerine
CH2=CH—CHO (A)
OH
H O
(+)
•• N
N
H
N
PhNHCH2CH2CHO (B)
(–)
O
(E)
PhNH2
(C)
H
N H
–H2O
(O) –H2O
(D)
N H
Only E is highly aromatic Q.10 (c) C is highly basic Q.11 (b)
12
Biomolecules
Main Features REACTION CHART FOR BIOMOLECULE Modern definition of carbohydrates: - These are polyhydroxy aldehyde or ketone. Carbohydrates on hydrolysis produce poly hydroxyl aldehyde & ketone. In modern classification carbohydrates are also called as saccharides Carbohydrates
Mono Saccharides
Test
Oligo Saccharides (Di, tritetra , Saccharides...........etc) Mono Saccharides
Poly Saccharides
Di Saccharides
Poly Saccharides
(1) Mutarotation
Show mutarotation
Except sucrose rest all show mutaroation
Don’t show mutarotation
(2) Tollen’s & Fehling’s test
+ve
Except sucrose rest all show Tollen’s & Fehling test
–ve
(3) Br2 water test
Except ketose rest all show Br2 water test
Except sucrose rest all show Br2 water test
–ve
Molish test: - Carbohydrates give violet colour on treatment with conc. H2SO4 & α - naphthol. Epimers: - A pair of diastereomers that differ only in the configuration about of a single carbon atom are said to be epimers for eq. CHO
CHO H
OH
H
HO
H
H
OH
HO
H
H
OH
H
HO
H
OH
CH2OH
OH CH2OH
342
Problems in Organic Chemistry
Anomers: - The two sugars that differs in configuration only for the carbon that was the carbonyl carbon in the open form is called as anomers H
OH
HO
H
OH
H
HO
H OH
HO
H
H
H
OH
H
OH
H
O
H
O
CH2OH
CH2OH
Mutarotation: • α & β – D glucose have specific rotation +112° & 18.7° respectively. When either of these forms are dissolved in water specific rotaion of solution is found to be + 52.7°. Thus, specific rotation of α & β – D glucose decrease and increase respectively. This change of optical rotation with time is called mutarotation. • If a carbohydrate show mutarotaion indicates that it exists in two anomeric forms α & β . Protein: (a) Biuret test: - Alkaline solution of protein treated with a drop of aq. CuSO4, bluish – violet colour is developed (b) Ninhydrin test: - Protein on treatment with ninhydrin give a solution having colour ranging from deep blue to violet pink.
LEVEL - I Multiple Choice Questions 1. HO2CCH2CHNH2COOH is called aspartic acid If pKa of α – COOH is 1.88 pKa of α – NH3+ is 9.60 & pKa of side chain is 3.65 then isoelectric point of aspartic acid will be: (a) 5.74 (b) 2.77 (c) 6.13 (d) 3.65 2. (+) H3N (CH2)4— CH COOH β α NH3 (+) γ pKa of the sites α, β & γ are 10.79, 2.18 & 8.95 Iso electric point will be: (a) 9.87 (b) 5.07 (c) 6.5 3. Which is not a reducing sugar? CH2OH CH2OH O H H O H H H H O (a) OH (b) H OH H OH OH
OH H CH2OH H
H (c) OH OH H
O H
OH
H
CH2OH
CH2OH O
H O CH2
H H
OH
(d) 2.18
H
CH2OH OH
H
OH
O
H
H (d) OH OH H
H O
H
H OH
O H
OH
H OH
H
H
OH
343
Biomolecules
4. 5. 6. 7. 8. 9.
Which is incorrect about fructose? (a) It gives Bendict’s test (b) It gives CO2 with HIO4 (c) It forms Glucosazone when treated with Phenyl hydrazine (d) It reacts with bromine water Which one of the following is optically inactive? (a) Leucine (b) Proline (c) Arginine (d) Glycine Fructose gives Fehling test because:(a) Ketonic group present in it can be oxidized by fehling solution (b) Ketonic group converts in to CHO gp by enolization (c) Ketal form of fructose changes in to open chain structure which is sensitive towards base present in fehling solution (d) Fructose does not give fehling test Iso electric point of alanine is 6. At what pH value, maximum concentration of zwitter ion will be present (a) greater than 6 (b) lesser than 6 (c) equal to 6 (d) pH = 8 One mole of sucrose will consume (a) Six moles of AcCl (b) Eight moles of AcCl (c) Five moles of AcCl (d) Ten moles of AcCl Consider the following four carbohydrate molecules CHO CHO CHO CHO H HO
OH H
H H
H
OH
H
OH
HO H
CH2OH
10. 11. 12. 13. 14.
OH OH
HO HO
H H
HO H
H
H HO
OH H
H
H
OH
OH
H
OH
CH2OH
CH2OH
OH
CH2OH
Epimeric pair is:(a) (A) & (B) (b) (A) & (D) (c) (C) & (B) (d) (C) & (A) If a compound shows mutarotation then:(a) It confirms that it will exist in two anomeric forms (b) It will always provide α – D glucose units on hydrolysis (c) Compound will always contain aldehyde as well as ketone group (d) All of these Which is not dextro rotatory compound? (a) Glucose (b) mannose (c) Fructose (d) Sucrose Which statement is correct reagarding the difference between glucose & fructose? (a) Glucose & fructose forms different - osazones (b) Glucose on reduction gives sorbitol while fructose gives a mixture of sorbitol and mannitol. (c) Fructose does not undergo oxidation by periodate while glucose is oxidized by it (d) Glucose exhibits muta rotation but fructose does not. The reagent by which we can prove that ketonic group is present at second location in fructose is:(a) Zymase (b) LiAIH4 (c) HNO3/∆ (d) H2 / Ni Which is D(+) mannose ? If mannose is epimer of D(+) Glucose. CHO CHO CHO CHO (A)
H H (a) H H
(B)
(C)
H H (b) OH HO OH H OH OH
CH2OH
(D)
HO HO (c) H H OH OH OH
CH2OH
H
H
HO
H H (d) H OH OH
CH2OH
H
H OH OH OH CH2OH
344
Problems in Organic Chemistry
15. The reagent by which we can prove that glucose contains straight 'C' chain of six C atom is: (a) Na – Hg / alcohol (b) IO4– (c) Br2 water (d) Red P + HI 16. NaOH CaO pass in Amino acid → gas → 0.1999 kg (salt) ∆ lime water (1mole) evolved Amino acid has: (a) Two NH2 groups (b) One COOH group (c) Two COOH groups (d) Three COOH groups 17. HNO 2 NaOH / CaO Amino acid → (A) → Ethyl alcohol, ∆
18. 19.
Amino acid is:(a) Glutamic acid (b) Glycine The carbohydrate with β - Glycosidic linkage is:(a) Lactose (b) Maltose Structure of amylose is:CH2OH CH2OH H H O O H H (a) O O OH H OH H H
(b)
OH
CH2OH O H OH H H H OH
H H O
H
OH
(d) Aspartic acid
(c) Amylopectin
(d) Sucrose
(b)
O n
CH2OH OH H OH H
CH2OH O H OH H
O
O
H
n
OH
CH2OH O H OH H
(c) Lysine
H
OH
CH2OH O H (d) OH H H H OH
H
CH2OH O H OH H
O
n
OH
O
H H
OH
20. The compound whose trinitrate is called gun - cotton is: (a) Sucrose (b) Gulose (c) Cellulose (d) 21. Compound Property (A) amylopectin (1) Milk sugar (B) Lactose (2) dextrorotatory (C) Sucrose (3) 1, 4 & 1, 6 glycosidic linkage (D) Alanine (4) L - Configuration Correct matching is: (a) (A)→3; (B)→4 ; (C)→1 ; (D)→ 2 (b) (A)→3; (B)→1 ; (C)→2; (D)→ 4 (c) (A)→2; (B)→3 ; (C)→2 ; (D)→ 4 (d) (A)→3; (B)→4; (C)→2 ; (D)→ 1 22. Glycine can be converted in to methane by: (a) Heat, H3O+, Zn/Hg + HCl (b) H2SO4 + NaNO2, KMnO4, NaOH/CaO (c) LiAlH4, NaIO4 (d) HNO2, NaOH/CaO / heat, Red P + HI 23. Correct Structure of β – D – fructose is:HOH2C O OH HOH2C O CH2OH OH HO H HO (a) (b) H CH2OH H H H
n
Lactose
H
OH H H O CH2OH HO (c) HOH C H OH 2
24
OH H OH H Which among the following direct the synthesis of protein as well as responsible for the transfer of genetic information? (a) Amino acids (b) Nucleic acids (c) Pyruvic acid (d) Carbohydrates specially cellulose
O OH H HO H CH2OH
HOH2C
(d)
345
Biomolecules
25. Which statement is correct regarding the difference between DNA and RNA? (a) RNA contains Adenine guanine, Thymine and uracil while DNA contains Adenine, uracil Thymine & guanine (b) RNA contains deoxy ribose while DNA contains oxyribose (c) RNA contains Adenine cytosine, uracil and guanine while DNA contains Adenine cytosine, guanine and Thymine (d) RNA contains phosphate while DNA does not 26. Vitamin Source (A) H (p) Butter (B) K (q) Yeast (C) D (r) Green leafy vegetable (D) A (s) Cod liver oil Correct matching is: (a) (A)→2; (B)→3; (C)→ 4; (D)→1 (b) (A)→3 ; (B)→2; (C)→1; (D)→4 (c) (A)→2; (B)→4; (C)→ 3; (D)→1 (d) (A)→1; (B)→3; (C)→ 2; (D)→4 27. Which statement is correct? (a) Hormones not produced by body & may be stored in the body to fight out the diseases (b) Vitamins may be stored in the body to fight out diseases (c) Vitamins are not stored in the body but are continuously produced (d) Both hormones and vitamins are stored in the body 28. Molecular formula of pentasaccharide would be: (a) C30H52O26 (b) C30H72O36 (b) C5H10O5 (d) C30H60O30 29. Nucleotide contains (a) Base + Sugar (b) Base + Sugar + Phosphate (c) Sugar + Phosphate (d) Base + Phosphate 30. Glucose contains 5 – OH groups. This fact can be proved by treating glucose with (a) Br2 water (b) Ac2O/H+ (c) Phenyl hydrazine (d) Zymase n conc 31. (A) HCl
Glucose concn H2SO4
∆
(B)
(A) and (B) are : (a) Pyruvic acid & black mass (c) Pyruvic acid & Ethanol 32. concn (A)
(b) leavulic acid & Ethanol (d) leavulic acid & black mass
HCl
Glucose concn H2SO4
∆
HCl & H2SO4 are:(a) Oxidizing agent
33.
HNO3 Fructose → oxalic (O)
LiAlH
(B)
(b) Reducing agent acid +
(c) Dehydrating agent
[X] optically active
(i) A (ii) B
4 → [X] → Glycerine
A & B are respectively:-
(a) Br2 water & NaBH4
(b) HCN & H3O+
(c) HNO3 & NaOH / CaO,∆
(d) All of these
(d) Polymerizing agent
346
Problems in Organic Chemistry
34. Consider the following three carbohydrates:CHO CHO H
OH
HO H H
H
CH2OH
OH
H
HO
H
OH OH
HO H
OH
H
OH
CH2OH
O
H
OH
H
CH2OH
(C)
B
(d) All have D configuration
It is reducing non sugar It is reducing sugar
Sucrose
(iv) hemi acetal of CH3CHO
i, ii & iii
(d) iii & v
D→ Red P + Br2 E→ NH3
(i) D
→ Glycine (a) AcOH (ii) B (iii) C
(b) H2N—CH
H OH
(A) (B) The carbohydrate which does not have D configuration is: (a) C (b) A & B (c) 35. Which is correct about sucrose? (a) It is non reducing non sugar (b) (c) It is non reducing sugar (d) 36. Honey contains: (a) Glucose + sucrose + water (b) Glucose + Lactose + water (c) Glucose + fructose + water (d) Glucose + galactose + water 37. Which will not react with Tollen’s reagent? (i) Glucose (ii) Fructose (iii) (v) hemi acetal of propanone (a) iv & v (b) iii, iv & v (c) 38. Consider the following reagents A→PCl5 B→ KCN C→H3O+ Which reaction is correct?
OH
H OH
(c) H2N—CH (d)
CH2OH
(i) A (ii) B (iii) C
→ Aspartic acid
CH2OH CH2OH OH
(i) A AcOH → (ii) E
(i) A (ii) B (iii) C
→ Aspartic acid
Glycine
39. Which among the following will not produce D - glucose on hydrolysis? (a) Amylose (b) Cellulose (c) Sucrose 40. β – D – Glucopyrarose is:OHCH2
(a)
H HO H H
OH
C
CH2OH OH H (b)
OH O CH2OH
41. Isoelectric point is: (a) Potential difference (c) pH
CH2OH
OH
C
O OH
H H OH
H
OH
H H
(d) Galactose
OH
H (c) HO HO H
CH2OH OH H H
O CH2OH
O OH
H (d)
H OH
H
H
OH
OH
OH
(b) dissociation constant (d) Amount of current passed in the solution.
347
Biomolecules
Passage - I Base of one nucleotide undergoes H - bonding with base of other nucleotide to give rise to formation of primary structure of DNA. Guanine forms 3 - H bonds with cytosine while adinine forms two H - bonds with Thymine. Answer the question from 42 to 44 42. What will be the sequence of bases on the strand of DNA that would be complementary to strand having the following sequences of bases? T T A G G A T A? (a) A A C T T C A T (b) A A T U U T A T (c) A A T C C T A T (d) T T A G G A T A 43. We have two samples of DNA Sample – 1 M. P. = 340 K Sample – 2 M. P. = 300 K Which statement is correct? (a) G C contents in sample – 1 are more than that of sample – 2 (b) A T contents in sample – 1 are more than that of sample – 2 (c) G C contents in sample – 1 are lesser than that of sample – 2 (d) A T contents in sample – 1 are lesser than that of sample – 2 44. Suppose RNA has double helix structure like DNA. What will be the sequence of bases on the strand of RNA that would be complimentary to strand having the following sequence of base? AA C C G AA G G (a) A A G G C T T C A (b) U U G G C U U C C (c) T T G G C T T C C (d) A A G G C U U C A
Passage - II CHO HO H
H HCN / H3O( + ) (i) Heat →(A) + (B) →(C) + (D) OH (ii) NaBH 4
CH2OH D - Threose Answers the questions from 45 to 47 45. Which is not correct? (a) A & B are Epimers (b) C & D are Epimers (c) C & D are anomers (d) A & B are hydroxy carboxylic acids 46. Which is correct? (a) (A) is aldo pentose (b) (C) is aldo tetrose (c) (D) is aldo pentose (d) (B) is aldo tetrose 47. The reaction D – Threose → (A) + (B) is: (a) Stereo Selective (b) Stereo Specific (c) Rearrangement reaction (d) Racemisation 48. A carbohydrate should contains at least (a) 3 – carbon atoms (b) 4 – carbon atoms (c) 5 – carbon atoms (d) 6 – carbon atoms 49. When a carbohydrate reacts with HIO4 it gives one glyoxal two formic acid and one molecule of formaldehyde. The carbohydrate is: (a) keto hexose (b) Aldo pentose (c) Aldo hexose (d) keto pentose
348
Problems in Organic Chemistry
z 50. C x (H 2 O) y → xC + yH 2 O if x = y = 6 then z would be: (a) Zymase (b) H2SO4 51. The compound which does not exist as a & b forms are: (a) Glucose (b) Fructose 52. Which among the following can not show mutarotaion? CH2OH CH2OH
H
H (a) OH OH H
O H
H O
H O
H
H OH
(c) HNO3
(d) Na – Hg / alcohol
(c) Maltose
(d) Sucrose
H (b)
OH OH
OH
(c)
H
(d)
53. Which is water insoluble: (a) amylose (b) gulose (c) galactose 54. The carbohydrate which does not contain b- anomer of any mono saccharide is: (a) Cellulose (b) Lactose (c) Sucrose 55. Match the following Compound Configuration
(A) α – L – Glucose
HO HO HO (1) H H
(d) amylopectin (d) Maltose
H H H OH O CH2OH H OH
(B) β – D – Glucose
(C) Epimer of β – D – glucose
(2)
H HO H O
(3)
H HO HO H O
OH H OH H CH2OH OH H H OH H CH2OH
HO H HO H H
H OH H OH O CH2OH
(D) Epimer of α – L – glucose
(4)
(a) A→3; B→ 2; C→1; D→4 (c) A→2; B→ 4; C→1;D→3
(b) A→2; B→ 1; C→4; D→3 (d) A→3; B→ 4; C→1; D→2
349
Biomolecules
Answer Key 1. 11. 21. 31. 41. 51.
(b) (c) (c) (d) (c) (d)
2. 12. 22. 32. 42. 52.
(a) (b) (d) (c) (c) (a)
3. 13. 23. 33. 43. 53.
(a) (c) (d) (c) (a) (d)
4. 14. 24. 34. 44. 54.
(d) (c) (b) (d) (b) (d)
5. 15. 25. 35. 45. 55.
(d) (d) (c) (c) (c) (c)
6. 16. 26. 36. 46.
(b) (c) (a) (c) (c)
7. 17. 27. 37. 47.
(c) (d) (b) (d) (d)
8. 18. 28. 38. 48.
(b) (a) (a) (c) (a)
9. 19. 29. 39. 49.
(d) (a) (b) (d) (b)
LEVEL - II Comprehension Consider the following compounds H CH2OH OHO HO CH2OH (A) H H HO OH
(B)
HO
(C)
OH
H H
HO
H
H
OH
CH2OH O
HO
CH2OH OHO
H
HO
CH2OH O
O HO
OH
OH OH
1. 2. 3.
Which can reduce tollen’s reagent? (a) Only A (b) Only B Which can not exist in anomeric form? (a) A & C (c) A & B Which can not decolourise Br2 water? (a) A (b) B
4.
Ag 2O dim ethyl sulphate C → D → E, D & E. are (ii) acidification aq NaOH
H
CH2OH O H
HO (a) HO
(c) C
CH2OH
OH
H
&
H
HO
CH2OH O H OH H
OMe
MeO
H
MeO CH2OH
O
CH2OH OH
H O
OH
(d) None of these
&
OH
CH2OH
MeO
OH
H COOH
OH OH
CH2OMe O H
COOH
CH2OMe
O
OMe
OMe
OMe COOMe
MeO CH2OMe OMe
OH COOH
HO
OH
OMe
H
HO HO (c) HO
CH2OH O
H
MeO
H HO (b)
CH2OH O H
MeO
OH
OH
(d) A & C
respectively
COOH
OH
(d) A, B & C
(b) B & C (d) All can exist in anomeric forms.
H
H O
H
(c) A & B
&
MeO O MeO
OMe MeO
CH2OMe OMe H COOH OMe
10. 20. 30. 40. 50.
(a) (c) (b) (b) (b)
350
Problems in Organic Chemistry
Multiple Choice Questions ach question contains statement given in two columns which have to be matched. Statement (A, B, C , D) in column I have to be E matched with statements (p, q, r, s) in column II. The answers to these questions have to be appropriately bubbled as illustrated in the following examples. If the correct match are A - p, A - s, B - r, B - r, B - q, C - q, D - S, then the correctly bubbled 4 x 4 matrix should be as follows. A
p
q
r
s
B
p
q
r
s
C
p
q
r
s
D
p
q
r
s
5.
arbohydrate C (A) Amylose (p) (B) Glucose (q) (C) Sucrose (r) (D) Fructose (s)
Phenomenon / Reaction observed Blue colouration with I2 decolourise Br2 water Can reduce tollen’s reagent Can not Exist in anomeric form
6.
Carbohydrate Structure CH2OH O H H OH H OH OH H OH
H
(A) β – D – Glucose
(p)
CH2OH O OH H OH H H OH H OH
H
(B) Epimer of α – D – mannose
(q)
(r)
CH2OH O OH H OH HO H OH H H
(s)
OH H—C H OH HO H OH H O H CH2OH
H
(C) α – D – Glucose
(D) β – D – Mannose
Answer Key 1. (d)
2. (d)
3. (a)
5. A – p, B – q, r, C – s, D – r
4. (b) 6. (A) – q, (B) – p, s, (C) – p, s, (D) – r
351
Biomolecules
SOLUTION 1. (b) Isoelectric point is the average of pKa of similar groups . Iso electric point = (1.88 + 3.65) / 2 = 2.77 2. (a) Iso electric point = (10.79 + 8.95) / 2 = 9.87 3. (a) Because none of the hemiactal linkage is free 4. (d) Keto group can not be oxidized by bromine water 5. (d) NH2CH2COOH glycine 6. (b) Fructose in presence of OH(–) ion gives mixture of glucose, fructose & mannose. This is called Lobry – de –bryun –van– ekenstein rearrangement. 7. (c) At pH = 6 zwitter ion is present because pH = 6 is isoelectric point and amino acid does not move towards any electrode 8. (b) Because it has 8 OH groups. CH2OH O
H
H OH
alpha- D- glucose
H H
OH H HOH2C beta- D- fructrose
H
9. (d)
OH O O
α–β glycosidic linkage
H OH CH2OH OH H
CHO
CHO
Epimeric carbon
H H
HO
HO
OH H
H
OH
H
OH
H
OH
H
OH
H
C
HO
C
CH2OH A
CH2OH
10. ( a) Muta rotation is the phenomenon in which anomeric forms (α & β forms) change their optical rotation spontaneously when dissolved in water. For e.g. specific rotation of pure α –D glucose is +113° & that of pure β-D glucose is –19° but when either of these forms are dissolved in water specific rotation is found to be + 53° thus, mutarotation indicates that carbohydrate exists in α & β forms 11. (c) Fructose is laevo rotatory compound. 12. (b)
CHO H HO H H
CH2OH OH H OH OH
CH2OH D-glucose
[H]
H HO H H
OH H OH OH CH2OH sorbitol
CH2OH +
OH HO H H
H H OH OH CH2OH mannitol
CH2OH [H]
O H OH OH
HO H H
CH2OH D-fructose
352
Problems in Organic Chemistry
13. (c) Fructose breaks in to two acids (oxalic & tartaric acids) in which ratio of carbon is 2:4.this indicates that keto group is present at 2nd location. [O]
CH2OHCOCHOCHHOHCHOHCH2OH → COOH—COOH + HOOC—CHOHCHOH—COOH 14. (c) Mannose is the epimer of α – D– glucose 15. (d) Glucose on reduction with Red P + HI gives n–hexane 16. (c) 0.1999 kg CaCO3 = 200 g CaCO3
= 2 mol CaCO3
Thus, amino acid contains 2 moles of COOH group. COOH
17. (d) HOOC—CH2—CH
COOH
HNO2
COOH—CH2CH
NH2
18. (a)
19. (a)
decarboxylation
EtOH
OH
20. (c)
21. (b) Lactose is called milk sugar as it is found in the milk of all mammals. Alanine is amino acid & except glycine all amino acids are found in L configuration. HNO 2 decarboxylation [H] 22. (d) H 2 NCH 2COOH → H 2OCH 2COOH → CH3OH → CH 4
23. (d)
24. (b)
25. (c)
26. (a)
27. (b) 28. (a) [Glucose] × 5 – 4 H2O = C30H52O26
29. (b) Base + Sugar + Phosphate
30. (b) One mole of glucose consumes 5 moles of Ac2O. 31. (d) Concentrated H2SO4 is a good dehydrating agent. concn H SO
2 4 → 6C + 6H O 2 C6 H12O6
32. (c) Dehydrating agent
concn HCl
→ leavulic acid C6 H12O6
33. (c) Fructose on oxidation gives oxalic as well as tartaric acid
CH2OHCOCHOHCHOHCHOHCH2OH
[O]
fructose
COOH CHOH CHOH
LAH
COOH [X]
CH2OH CHOH CHOH CH2OH
COOH–COOH + HOOC–CHOHCHOH–COOH Tartaric acid [X]
COOH HNO3
CHOH CHOH
CH2OH
CH2OH Sodalime Heat
CHOH CH2OH glycerine
34. (d) All have D configuration because at 2nd last carbon atom H is on the left. 35. (c) Since it is sweet in taste hence it is sugar but it can not reduce tollen’s reagent thus, it is non reducing. 36. (c) Glucose + fructose + water
353
Biomolecules
37. (d) 38. (c) H2N—CH
CH2OH
PCl5
CH2Cl
H2N—CH
OH
KCN
Cl
H2N—CH
CH2CN
H2N—CH
CN
(+)
CH2COOH
H3O
COOH
39. (d) Galactose is monosaccharide.
40. (b)
41. (c) It is that pH at which amino acid does not migrate towards any electrode. 42. (c) A A T C C T A T (Thymine forms H bonds with Adenine & Cytosine with Guanine) 43. (a) Melting point ∝ extent of H - bond. 44. (b) A A G G C T T C A (in RNA in place of thymine, uracil is present)
HO H
H OH
CN
CN
CHO Passage-II (45 to 47)
OH H
H HO H
HCN
OH
CH2OH
HO HO H
H H OH
CH2OH
CH2OH (+)
H3O
A
H HO H
COOH OH H OH
HO HO H
COOH H B H OH CH2OH
CH2OH release of H2O Heat
H HO H
CO
CO OH H O
OH HO H
CH2OH
CO
CO H HO H
OH H O
OH + HO H
CH2OH
H H
NaBH4
O CH2OH
CHO * OH H H HO H OH C
+
CH2OH
Starred carbons are epimeric carbons. 45. (c) C & D are epimers but not anomers. 46. (c) (D) is aldo pentose 48. (a) Cx (H2O)y
47. (d) Both A & B are formed in equal amounts
Here x = 3 to 8 (not for poly saccharides)
CHO OH H * H HO H OH D
CH2OH
H H O CH2OH
354
Problems in Organic Chemistry
49. (b)
CHO OH OH OH
CHO
HIO4
+ 2HCOOH + HCHO CHO
CH2OH
50. (b) Dehydration of glucose
51. (d) Sucrose
52. (a)
53. (d) Starch contains two parts. One part is water soluble (amylose) & 2nd part is water insoluble (amylopectin). Gulose & galactose are mono saccharides & soluble in water 54. (d) It contains two units of α – D– glucose.
55. (c)
LEVEL -II 1. (d) A is fructose B is glucose while C is Maltose. 2. (d) All can show muta rotation thus all can exist in anomeric forms i.e. α & β forms. 3. (a) In basic medium fructose converts in to glucose & exhibits tollen’s and fehling test but it can't show Br2 water test because Br2 water cannot convert fructose in to glucose. H
4. (b) HO
CH2OH O H
HO
H
OH
Ag2O acidification
(C)
Hemiacetal but enguaged in bonding H CH2OH Hemiacetal site & shows tollen’s O O test OH HO OH H CH2OH O HO H CH2OH O OH OH HO H
HO H MeO
(D)
CH2OMe O O
MeO
OMe MeO
CH2OMe OMe OMe COOMe
OH COOH
Methylating agent
13
Aromatic Chemistry
Main Features Preparation Properties R NaOH / CaO Ph heat COOH (+) (–)
3Cl2 / hv boil
H3PO2 or C2H5OH,∆
PhN2Cl R
H3O
Ph
KMnO4 boil
(+)
or R
PhCOOH (For C6H5R)
3Cl2 / hv
SO3H Zn dust ∆
Ph — OH
CrO2Cl2 (+)
Fe ∆
Acetylene (3-moles) n-hexane
V2O5/O2 500°C
Anh. AlCl3
C6H6 + RCl
D / D2O
Al2O3 / Cr2O3 600°C, 10 –15 atm
ArN2X
C6H6Cl6 (For benzene) CHCO O CHCO (For benzene) PhCHO (For toluene) D
PhN2Ar
PhMgX + H2O
Aromatic electrophilic substitution PhCCl3 (For toluene)
Preparation Properties PhCH3
(i) CrO2Cl2 in CS2
+ve tollen’s test
+
PhCH2OH PhH+ CO + HCl PhCHCl2 PhCN
(ii) H3O PCC or Cu /∆ ZnCl2 H2O/∆
CuSO4 NaOH
CHO
NaOH LAH
PhCOONa + PhCH2OH PhCH2OH ARSE at meta location
DI BAL–H KCN KMnO4
–ve Fehhing’s & Bendict’s test
Ac2O / AcONa H3O(+)
PhCH — COPh OH PhCOOH
Benzoin
PhCH = CHCOOH [Perkin’s reaction]
356
Problems in Organic Chemistry
Preparation Properties aq NaOH
FeCl3
PhH + X2 [X= Cl,Br]
PhOH
(+)
673K, 200 atm, H Cu or CuCl
PhN2X + HX [X= Cl,Br]
Cu /∆
Ph – Ph [Ullman reaction] (For iodobenzene)
X PhN2Cl + KI [For iodo benzene]
ARSE at ortho and para locations ∆
PhN2Cl + HBF4 [For fluoro benzene]
Na/Ether
X=Cl, Br F&I
RX / Na Ether
Catalyst
PhH + O2 + HCl [Raschig’s process]
Ph - Ph [Fittig reaction] Ph – Ph + Ph – R + R – R [Wurtz-Fittig reaction]
Mg/Ether
PhMgX
Preparation Properties (+)
H /Na2Cr2O7
673K, 200 atm
PhX + NaOH
(+)
O
O
H
Br2 water
Boil
PhN2Cl + H2O
OH
(+)
PhN2Cl Pyridine
NaNO2+HCl t > 8°C
PhN2Cl
White ppt of 2,4,6 tri bromo phenol OH
PhN2 (Orange)
OH COOH
NaOH / CaO ∆
PHENOL
(+)
FeCl3
NaOH /∆ (+) H
PhSO3Na
HCHO/H
3
[Fe(OPh)6] –[violet colouration] OH
dil HNO3
PhCHMe2
Bakelite plastic
o & p derivatives
Ph
air H3O+/∆
NO2 concn HNO3 Zn /∆
C6H6 [Reduction]
PCl5
Picric acid [2,4,6-tri nitro phenol]
C6H6Cl + POCl3 + HCl
Preparation Properties PhCH2R PhCH2OH or PHCHO PhCOCH3 PhCN PhCCl3
hot KMnO4 hot KMnO4 hot KMnO4 H3O(+) (i) KOH (+) (ii) H
NaOH/CaO
∆
C6H6 ARSE at meta position
COOH
NH3 / heat Na N3H H2SO4 NaHCO3
PhCONH2 PhCOONa + 1/2H2 PhNH2 PhCOONa + CO2 + H2O
357
Aromatic Chemistry (i) NaOH/∆ (+) (ii) H
PhOH
SO3Na
SO3H oleum ∆
aq NaOH
NaCN/∆
NaCl/∆ H3O(+) ∆
PhCN PhCl PhH
Preparation Properties conc HNO3 conc H2SO4
Sn+HCl or Fe+HCl or H2/Ni or L.A.H.
NO2
PhNH2
EtONO2
ASRE at meta location Fuming HNO3
Strongly acidic medium pass electric current
(+) (–) NO2 BF4
weakly acidic medium pass electric current
acetyl nitrate
Zn + NaOH CH3OH
p–OH–C6H5–NH2 Ph–NH2 PhN2Ph Azo benzene
Zn + NaOH
PhNH–NH–Ph Hydrazo benzene O
Na2AsO3/NaOH
PhN=N–Ph Azoxy benzene
Preparation Properties PhNO2
Sn+HCl or Fe + HCl or H2/Ni or L.A.H.
NH2
PhCOCl
PhNHCOPh
PhSO2Cl KOH
PhSO2NPh
Na NH3 / Cu2O heat
PhCl PhOH
NH3 + ZnCl2 heat
(–)
PhNHNa + 1/2 H2
CHCl3 / OH(–)
ANILINE
PhNC (carbyl amine)
NaNO2 + HCl 0-5°C
PhN2Cl (Diazotisation)
PhCONH2
KOBr
PhMgBr
NH2Cl
NaNO2 + HCl t > 8°C
PhOH
(+) H / Na2Cr2O7 Br2 in CCl4 Br2 water (i) AcCl (ii) HNO3 + H2SO4 (iii) Hydrolysis
(Hinsberg test)
(–) (+)
H2PtCl6
O
O
o & p bromo aniline white ppt of 2,4,6-tri bromo aniline o & p nitro aniline
[PhNH3]2 [PtCl6]
358
Problems in Organic Chemistry
Preparation Properties KI solid
PhI + N2 + HCl
Cu or CuCl HX
N2Cl PhNH2
PhX + N2 + HCl ( X= Cl ,Br)
HBF4
NaNO2 + HCl 0-5°C
PhF + N2 + BF3
boil with water
PhOH
EtOH or H3PO2 + H2O
benzene
PhNH2, H pH
+
p–NH2–C6H4–N2Ph [coupling reaction]
4-5
(yellow dye) PhOH, OH(–) pH 9-10
p–OH–C6H4–N2Ph [coupling reaction] (orange dye)
PhH + NaOH
Zn + HCl
Ph-Ph [Gomberg reaction] PhNHNH2
Na2SO3 / H2O KCN / CuCN (i) HBF4
(ii) NaNO2 / Cu
Zn + HCl
PhNH2
PhNHNH2 PhCN PhNO2 + N2 + NaBF4
LEVEL - I Multiple Choice Questions NH2
NO2
1. NH2
NH2
For this conversion sequence of reagents required is:(a) (i) KMnO4 (ii) Na2S (c) (i) Caro’s acid/HNO3/H2O2 (ii) Sn + HCl
NaNH 2 2. ortho – Bromo toluene →
(b) (i) H+/K2Cr2O7 (ii) NH4HS (d) (i) CF3CO3H (ii) Na2S
(A)
NaNH
2→ ortho – Bromo anisole (B) (A) & (B) are:OCH3 CH3 CH3 NH2 NH2 & & NH2 (a) (b)
OCH3
CH3 &
(c)
NH2
Br
OCH3
CH3 NH2
NH2
NH2
&
(d)
NH2
359
Aromatic Chemistry (i) NaOH / CO
Br
2 → X → 2 Y,Compound Y is : −–z 3. Phenol (+) Water
(ii) H
OH
OH Br
COOH
COOH
(a)
(b)
OH Br (c)
Br
Br 4. Identify the correct reaction. NH2 Cl (a) + H2O
(d)
Br
COOH
Br
NH2
Cl
OH + HCl
OH + H2O
(b) O2N
Warm
NH2
OH Br
NO2
+ HCl
Warm
O2N
NO2
NH2 Cl
OH
O2N (c)
O2N
NO2
NO2
HCl +
+ H2O Warm
NO2
NO2
(d) None, because Cl present on Benzene nucleus can not be removed by H2O. I I
5.
(A)
Br Br (B)
6.
(A) And (B) are:(a) Both are Cu / ∆ (c) A = Zn + Et2O B = Cu /∆ Idenify the product of the following reaction. Product
(b) A = Cu /∆ B = Zn + Et2O (d) Both are Fe + HCl
NO2 (i) KMnO / ∆ (ii) NaOH + CaO
4 → Product
(a) Nitrobenzene 7. Select the incorrect reaction.
(b) Benzene
(d) o – nitro benzoic acid
NH2
(i) 3 moles of CH Cl
Me
O Me
3 → (a) PhNH 2 (ii) Heat (iii) NaOH
+
Na 2Cr2O7 / H (b) PhNH 2 →
Me NH2
NH2
(c) CO2 + H2O + NO2
O NO2
conc HNO3 (c) PhNH 2 →
+
conc H 2SO 4
NO2
NaNO 2 + HCl PhNH 2 (d) PhNH 2 →[X] → yellow ppt. 0°C
360
Problems in Organic Chemistry
8. Consider the following reactions.
FeCl NaNO 2 CH3 NH 2 → CH3 N 2Cl .......(1) CCl4 + 4Ph — H → Ph 4C + 4HCl
3 PhNH 2 + CH3Cl → o − Me − C6 H 4 — NH 2 .......(3)
Reactions with wrong products are: (a) 1 & 3 (b) 1, 2, 3 & 4
HCl
(i) KMnO (ii) NaOH / CaO, heat
4 Anthracene → C6 H 6 .......(4)
AlCl
CH Cl
(c) 1, 2 & 3
dil H SO
(d) 4, 1 & 3
Br / D
3 2 4 2 → → → [X] Anh. AlCl3 Fe D Major
CH2
9.
C NH O
[X] would be:Br CH2
(a) NH2
Br CH2 (b) C OH
C OH CH3 O
NH2
O
CH3
Br CH2
(c)
CH3
NH2 NO2
10.
NO2
H 2S/ NH3 →(A) ,
NO2
(A) & (B) are:NH2
CH3 NH2
(a)
SO3H
11.
NaOH
NH2 NH2
(d)
NH2
NH2
NH2
CH3
NO2 NaOH H (+)
(C)
NaCN H3O(+)
(B)
NH2
&
NH2
NH2
&
NO2
(b)
NH2
NO2
(c)
NO2
CH3
NH2
CH3 NH2
&
NO2
NO2
& NH2
→ [X] (A)
H S/ NH 50°C
2 3 →(B)
NO2
(d) None of these
CH3
NH2
Correct sequence of acid strength is:(a) A > B > C (b) B > A > C
.......(2)
(c) C > A > B
(d) A > C > B
361
Aromatic Chemistry
NH2 CHCl KOH
12.
dil HCl Heat
X,
3 → →
X would be
Cl
(+)(–)
(a) Cl
NH2Cl (b) Cl
∆ V 2O 2 HO 2 2∆
13.
NH2 (c) Cl
Cl CN (d)
NC
(A) (B)
NaNH2 CH3OH
(C)
Cl2 FeCl3
(D)
Which is correct about A, B, C & D ?
(a) All are non aromatic
(b) ‘C’ is a conjugated diene
(c) A & B are non aromatic while (C) & (D) are aromatic
(d) A, B & C are non aromatic but (D) is aromatic
14. What product will be obtained when 2 – methyl – 2 – phenyl propane is subjected to oxidation by MnO4– / H+ followed by decarboxylation by NaOH / CaO / heat ?
(a) Benzene
(b) Propane
(c) Iso butane
(d) Methane
(c) µ2 > µ1 > µ3
(d) µ3 > µ2 > µ1
15. Consider the following three derivatives of toluene CH3
CH3
CH3
NO2 NO2
(1)
NO2 (3)
(2)
Their dipole moments will follow the sequence.
(a) µ1 > µ2 > µ3
(b) µ2 > µ3 > µ1
16. Aniline can be converted in to m-methyl nitro benzene by using:
(a) NaNO2 + HCl (0°C), NaNO2 / Cu, CH3Cl / AlCl3
(b) CF3COOOH, CH3Cl / AlCl3
(c) NaNO2 + HCl (0°C), CH3Cl / AlCl3, NaNO2 / Cu
(d) All of these
17. Which among the following is / are meta directing ( + ) (–)
– N 2 Cl,
(a) All of these
– CCl3,
– NHAc,
(b) 1, 2 & 3
(c) 1 & 2
OH CO
18.
O
+ CO
Product [X] will be:-
(i) H SO (ii) NaOH
2 4→
(X)
–CHCl2 (d) 1, 2 & 4
362
Problems in Organic Chemistry
OH O
OH
C C
(a)
C
OH COO(–)
(b) C
O
O
OH
OH CO O
(c)
C
(d) None of these
OH
OH
19. How many mono nitro derivatives will be obtained by the nitration of [X] ? OH (i) Br (aq)
2 → (X) (ii) Zn, heat
(a) 1
(b) 2 (+)
H 20. Phenol + Benzoic acid →
(c) 3 (+)
(d) 4
(–)
NO 2 BF4 (X) → (Y) major, (Y) would be:-
COOPh COOPh
(a)
COOPh
COOPh
(b)
(c) NO2
NO2
NO2
(d) NO2
MeO OMe (i) H3O,( + ) rearrangement OMe → (X), (X) would be:(ii) Br2 (aq) OMe
21.
OH
OH Br (b)
Br
(a) OH
HO
OH Br (c) OH
OH Br
Br
(d) Br OH
O
Br Br 22. o - cresol is treated with allyl chloride and product obtained is exposed to sun light. What would be the end product ? OH
CH2—CH=CH2 CH3
CH2—CH=CH2
H3C
(a)
(b) OCH3 CH3
OH CH3
(c)
CH2—CH=CH2
(d)
363
Aromatic Chemistry
23. Which product is unexpected in the following reaction ? OAc OCOPh
AlCl3
+ Et
Products
Me OH
OH Ac
(a)
COPh (c)
(b) Et
Et
Hint: - Fries rearrangement is intermolecular as well as intramolecular
→ Products + (CN)2 C = C(CN)2 ∆
24.
(d) All products are possible
Product of this reaction is:NC CN CN CN CN
(a)
CN
(b)
(d) No reaction occurs
CN CN
CN CN
(c)
CN CN
OH
OH Zn(CN) & HCl
H O
2 2 → →
25.
OH
OH CHO
Electrophile of this reaction would be:-
(a) HOC Cl
(+ )
(+ )
(+ )
(b) CHO (c) Zn(CN CH = NH (d)
26. 3 – Hydroxy phenol (Resorcinol) in basic medium reacts with benzene diazonium chloride to give [X], [X] on acidification gives [Y], [Y] OH OH OH OH N2Ph N2 (a) (b) (c) (d) N2Ph OH HO ON2Ph OH 27. Bismark brown (Y) is an azo dye which can be prepared as follows:
NaNO 2 + HCl / 0°C Meta phenylenediamine →[X]
[X] + metaphenylenediamine →[Y] 1
:
2
364
Problems in Organic Chemistry
Identify bismark brown:-
NH2
NH2 (a)
NH2 N=N (b) H2N
N=N
NH2
NH2
NH2 N=N
(c)
NH2
N=N
NH2
NH2
NH2
28. Reaction with wrong product is:SO3H
N2
(a) Sulphanilic acid
(i) NaNO + HCl / 0°C (ii) α− Naphthyl amine
2 →
NH2 ONa
(i) NaOH
(b) Beta - Naphthol ( + ) (–) →
(ii) Ph N2 Cl
N2Ph NH2
(c) Sulphanilic acid
SO3H
N2
(i) NaNO + HCl / 0°C (ii) α− Naphthyl amine
2 →
N2Ph ONa
(i) NaOH
(d) Beta - Naphthol ( + ) (–) →
(ii) Ph N2 Cl
NH2
OH
Hints:-
Alpha- Naphthyl amine CO 2
beta-Naphthol
H( + )
→ → [X] 29. PhOH Base
H2N
NaNO 2 + HCl NH2 →
S
0°C
Pyridine X (1-mole) + Y) → Anthracene yellow will be;(–)
[ Y]
Anthracene yellow (–)
COO
OOC
HO (a)
N2
S
N2
NH2
N=N
(d) H2N
H 2N
N=N
OH
365
Aromatic Chemistry
OH
OH
(–)
OOC (b)
(–)
S
N2
COO
N2
OH
OH CHO
OHC S
N2
(c)
N2
(–)
(–) (d) OOC
S
N2
COO
N2
O EtONO
30.
2 → Products
O
O
Product of this reaction will be:NO2
O
O
(a)
OEt
O
(b) O
O
O
Br
NO2 (d)
(c) O
O
O
CHO
MeO
OH
MeO
OH
31.
This conversion can be best performed by:- (a) CH3Br / Na / Et2O, H3O+, CrO2Cl2 + (c) H3O , CHOCl / AlCl3 NO2
(b) Mg/Et2O, CH2O, H3O+, Cu /∆ (d) Zn(CN)2 / HCl / H3O+
CH3 1 mol EtSNa
→ Product D
32. Cl Cl
The product of above reaction would be:NO2
NO2
NO2
CH3
CH3
(a)
(b) Cl
SEt
(c) EtS
SEt
Cl NH EtOH
NO2 CH3
Cl
CH3
(d) Cl
SEt SEt
O Me2S
3 →(B) 3 → CH COCH CHO 33. (A) + Na 3 2
Compounds (A) is:(a) o-xylene
(b) m-xylene
(c) p-xylene
(d) mesitylene
366
Problems in Organic Chemistry
34. Identify the product [X] of the reaction given below:∆ OD + D2SO4 ( in heavy water) → [X]
OD
OD
OD
D D (b) D
(a) D D
D
(c) D
D
(d) All of these
D
D C H NH
KHSO ∆
6 5 2 →(B) , Compound (B) is:4 →(A) 35. Glycerine (+)
H
, heat
(a) CH2 = CH — CH = NPh
(b) N H
(c) CH2 = CH — CH = O
NH—CH—CH2CH3
(d)
OH
36. Which intermediate will not form in the course of the reaction given below? OH O
|
KCN
||
2PhCHO → Ph — CH — C— Ph O(–)
OH |
|
(–)
|
8
4
8
9
CN 3 4
1
2
7
2
6
3
6
3
2
5
N1
S
4 1 10 5 H Thiophene Pyrrole Anthracene Which statement is correct regarding the nitration of these compounds ? (a) In naphthalene nitration occurs at 1st position (b) In anthracene nitration occurs at 9th position (c) In pyrrole nitration occurs at 3rd position (d) In thiophene nitration occurs at 4th position
N
(a)
HNO H 2SO 4
3→
HNO3 → H 2SO 4
(X), (X) would be:NO2
(b) N (c) NO2 S S
S
N
(d) All of these
AcONO
CO N
2 → Product
Major product of this reaction would be:CO
(a) O2N
S
N
(b)
CO S
N
NO2 NO2
O2N
3
2
5
5 4 Naphthalene
S
1
7
38.
39.
|
|
|
CN
37. Consider the following four compounds.
|
||
(a) Ph — C — N (b) Ph —C— CN (c) Ph — C — CH — Ph Ph — C — C— Ph (d) CN
OH O(–)
O
(–)
(c)
CO S
N
(d) CO S
N
367
Aromatic Chemistry
40. Match the following Reaction Colour
FeCl3 (A) Phenol →
Ph N 2 Cl (B) Phenol (–) →
2 → (C) Phenol
(3) Violet
K 2S2O8 (D) Phenol →
(4) Colourless
Correct matching is:(a) A→3 B→ 4 (b) A→3 B→ 1 (c) A→3 B→ 4 (d) A→3 B→ 1
+
(1) Orange
–
(2) Red
OH
NaNO H 2SO 4
C→ 2 C→ 4 C→ 1 C→ 2
D→ 1 D→ 2 D→ 2 D→ 4
Note: - Phenol on oxidation with K2S2O8 gives a mixture of catechol & quinol this reaction is called Elbs persulphate oxidation:41. Which reaction will yield naphthalene as an end product? (i) Bayers reagent
heat → Product (a) Benzyne + 1, 3 – butadiene (X) → (X) (+)
3 →(X) → Product (b) Ph(CH2)3COCl (ii) LiAlH
(c) 2 moles of
(d) All of these
(ii) H
H( + ) / ∆
(i) AlCl
4
I Cu / ∆
I
Product
→
42. Identify the product of following oxidation reaction
hot KMnO4 → product
(a) HOOC
COOH
(b)
(c)
CHO
(d)
CHO
HOOC
COOH
HOOC
COOH
HO
OH
HO
OH
43. In the reaction of p-chloro toluene with sodamide, the major product is (a) o-Toluidine (b) m-toluidine (c) p-toluidine
(d) p-chloro aniline
44. Which will not give 1, 3 5 - T.N.B. as a product ? (i) KMnO
(i) conc HNO
4 (a) 2, 4, 6, T.N.T. → Product (ii) NaOH / CaO / ∆
3 3 3 2 4→ → Product (d) m - nitro benzylalc (c) meta phenylenediamine Product HNO + H SO , heat (ii) KMnO
CF CO H
3
2
4
3→ (b) Phenol Product (ii) Zn & heat
(i) HNO / H SO 4
(iii) N3H / H 2SO 4
368
Problems in Organic Chemistry KOH
→ Product 45. T.N.T. + benzaldehyde heat Product of this reaction would be:NO2
(a) No reaction takes place
(b) O2N
CN = CH
+ H2O
NO2
(c) O2N
CHO
NO2
NO2
CHO
CH2
CH2
(d) O2N
NO2
NO2
46. When aniline is heated with H2SO4 at 453 K, compound [X] forms which on treatment with (i) Ac2O/Br2 (ii) H3O+ at 100°C / NaOH gives [Y], What is [Y] NH2 NH2 NH2 NH2 Br Br (a) (b) (c) (d) Br SO3H Br SO3H SO3H 47. Aniline can be distinguished from phenol by: (a) Br2 water (b) Zn dust / ∆
PCl5
48. Ph-COOH
[B] Ph-COOOH
[A] ZnCl2
[Y] would be:(a) PhOH
(c) C6H5N2Cl / NaOH
(d) All of these
[C]
[Y]
(b) PhCOONa
(c) PhCOOOH
(d) PhCOOH
49. We would like to convert Benzaldehyde in to 3 – bromo – 4 – nitro toluene. Which sequence of reagent is suitable for this transformation ? (a) Zn / Hg+HCl, Br2 /Fe, HNO3 / H2SO4 (b) Br2 / Fe, HNO3 / H2SO4, Zn / Hg + HCl (c) Br2 / Fe, Zn / Hg + HCl, HNO3 / H2SO4 (d) Zn /Hg+HCl, HNO3 / H2SO4, Fe / Br2 50. Would like to convert Benzoic acid in to 3, 4, 5 tri bromo benzaldehyde which sequence of reagent is suitable for this transformation. (a) LiAlH4, Br2 / Fe, Cu /∆ , Br2 / Fe, Br2 / Fe (b) Br2 / Fe, LiAlH4, Br2 / Fe, Br2 / Fe, Cu /∆ (c) LiAlH4, Cu /∆ , Br2 / Fe, Br2 / Fe, Br2 / Fe (d) Br2 / Fe, Br2 / Fe, Br2 / Fe, LiAlH4, Cu / ∆ NaNO + H SO 273K
2 2 4 → (A) 51. Aniline
(i) dil H SO
2 4 →(B) CH2(OEt)2 (ii) HCl, A
(iii) Boil with water
Compound (B) would be:OH OH CH2OH NH2 OH (b) (c) (d)
(a)
CH2OH CHO
CH2OH
369
Aromatic Chemistry
52.
Nitro benzene can be converted in to cyclo hexan – 1, 2, 3, 4, 5 pentadiol by:(a) HNO3 / H2SO4 / t > 180°C, LiAlH4, NaNO2 + HCl (t > 8°C) (b) Fuming HNO3 / H2SO4, heat, Fe + HCl, NaNO2 + H2SO4 (0°C), H2O / boil (c) NaNH2 / CH3OH, LiAlH4, HONO, KMnO4 (0°C) (d) HNO3 / H2SO4 / t > 180°C, LiAlH4, H2 / NI, NaNO2 + HCl (t > 180°C) Na / NH (l)
H (1 mol)
3 → 2 →(X) 53. Toluene Ni / ∆
CH3OH
Which is the most correct structure of [X] ?
(a)
(b) (c) (d)
CH3 (i) NaNH / MeOH
54.
(X), (X) would be:-
2 →
SO3H
(ii) KMnO 4 (0°C)
OH
Me HO
(b) HO SO3H
H SO3H
H
OH
55. Consider the following compounds. SO3H
(c) H H
SO3H
CH3
CH3
D
OH
SO3H
D
(1) Correct order of rate of nitration will be:- (a) 3 > 2 > 4 > 1 (b) 2 = 3 > 1 > 4 (i) H O( + ) (ii) HCl / rerrangement
3 →
O
OH
O
Cl (b) Cl H OH
Cl
(a) H
OH
H
D
D (2)
56.
OH
Me
HO H (d) SO3H H OH
HO
H
OH
OH OH
OH
OH
(a) H
Me
OH
Me
(3)
(c) 2 > 3 > 4 > 1
D (4)
(d) 3 = 2 > 1 = 4
(X), (X) would be:OH Cl
OH
Cl (c) (d) Cl OH
OCH2CH=CH2 ∆ → (X), (X) would be:-
57.
OH
OH CH2—CH=CH2
(a)
(b)
OCH2—CH=CH2
(c) CH2CH=CH2
OH
OH
(d) CH2CH=CH2
370
Problems in Organic Chemistry
58. Benzene ring can be halogenated by using interhalogens. Identify the product of the following halogenation reaction. O
O
ICl → Product
O
O
(a)
(b) I
O
O
O
O
I (c)
O
O
(d)
Cl
I
59. C6H6 C6D6 (1) (2) If K1 & K2 are the rate of aromatic electrophilic substitution in 1 & 2 then select the correct statement (a) In case of nitration K1 > K2 (b) In case of sulphonation K1 = K2 (c) In case of sulphonation K1 > K2 (d) In case of nitration K1 < K2
Passage - I Consider the following molecules OH
OH OH
(A)
(B)
(C)
CH2OH
OH
(D)
(E)
Answer the questions from 60 to 69 60. Which is not aromatic alcohol ? (a) Only B (b) (B), (C), (D) & (E)
(c) (B), (C) & (D)
(d) (B) & (D)
61. Which will release CO2 on treatment with sodium bi carbonate ? (a) (A), (C) & (D) (b) (C), (D) & (E) (c) Only (A)
(d) None
62. (B) & (C) can be distinguished by: (a) Neutral FeCl3 (b) Br2 water
(c) NaOCl
(d) Anhydrous ZnCl2 + HCl
63. (D) and (E) can be distinguished by: (a) Neutral FeCl3 (b) NaOCl
(c) Victor maeyer test
(d) All of these
64. The alcohol which will not undergo acid catalysed dehydration to give hydrocarbon is / are: (a) (A) & (E) (b) (A), (E) & (D) (c) (A), (C), (D) & (E) (d) Only (A) 65. Alcohol that can able to produces violet colouration with neutral FeCl3 is: (a) Only (A) (b) (A), (C), (D) & (E) (c) (A) & (E)
(d) (A) & (D)
66. Order of stability of carbocations formed by given alcohols when treated with H+ is: (a) A > B > C > D> E (b) D > C > B > E (c) E > C > D > B
(d) D > C > E > B > A
67. Benzene ring with highest electron density is: (a) (A) (b) (E)
(c) (C)
(d) (D)
68. Correct order of acid strength is: (a) A > E > B > C > D (b) A > E > C > D > B
(c) A > C > D > E > B
(d) A > E > D > C > B
371
Aromatic Chemistry
69. Which will provide highly stable alkene on acid catalysed dehydration ? (a) (D) (b) (E) (c) (B)
(d) (C)
Passage - II Na 2S
Dinitrobenzene (A)
NaNO2 (281K) H2SO4
(B)
(G)
NaNO2 + HCl (273 K) HCl Cu
(D)
KNO2
(C)
(E)
H2O/heat
(F)
Answer the questions from 70 to 74 70. Compound (A) is: (a) 1, 2-di nitro benzene
(b) 1, 3-di nitro benzene
71. In this passage two compounds, which are identical are: (a) A & E (b) G & F
(c) 1, 4-di nitro benzene
(d) All of these
(c) B & E
(d) A & E as well as G & F
72. Compound (B) is:
(a) 2-nitro aniline
(b) 3-nitro aniline
(c) H2N
NH2 (d) SH
73. Which among the following will produce highly explosive compound on mono nitration ? (a) F (b) E (c) G 74. Compound (F) can be converted in to(B) by using: (a) Zn / heat, EtNO2, CF3COOH (c) EtNO2, Zn / heat, LAH
(d) C
(b) Zn / heat, EtNO2, (NH4)2S (d) All of these
Passage - III Alumina, 600°C Oxide of Cr
C6H14(A)
(B)
(does not decolourise Br2 water)
(E)
+
does not give haloform test
(F)
(i)Cl2 / Fe (ii) Mg,Ether
Propanol /H (+)
H3O
(D)
air
(G)
(iii) D2O (+)
(C)
gives haloform test
Answer the questions from 75 to 81 75. Compound (A) is:
(a) iso hexane
(b) n - hexane
(c) neo hexane
(d)
NH2
372
Problems in Organic Chemistry
76. Compound (B) is:-
(a) Benzene
(b)
(c)
(d) Toluene
77. Compound (C) is: (a) Styrene
(b) Cumene
(c) Stilbene
(d) Ethyl benzene
78. Compound (D) is: (a) Ph — CH2O2H
(b) PhCMe2O2H
(c) PhCH2CH2CH2OH
(d) Ph(CH2)3O2H
79. Compound (E) is: (a) Phenol
(b) 1 – Propanol
(c) Acetone
(d) Propanal
80. Compound (F) is: (a) Phenol
(b) Propanol-2-ol
(c) Acetone
(d) Ethanal
81. Compound (G) is:Cl OD
OH
(a) C6H5D
(b)
(c) (d)
Passage - IV An organic compound (A) C7H8O, is insoluble in aq NaHCO3 but soluble in NaOH. (A) On treatment with Br2 water gives (B) (tribromo derivative) Answer the question from 82 to 84 82. Compound (A) is:OH OCH3
(a)
OH CH3
CH3 (b) (c)
83. Compound (B) is:OCH3 Br Br
(d) All of these
OH Br
OH Br
Br
OH Br
Br
Br CH3
(a)
Br
(b) Br
(c)
(d) Br
84. What will be (A) if it does not dissolve in NaOH but shows reactions given in the passage? OCH3 OH
OH
OH CH3
(a)
(b) (c) (d) CH3 CH3
Passage -V Pyrrole on treatment with alkaline chloroform followed by heating with H+ gives (B) & (C). (B) can react with 2, 4 DNP but (C) can not. (B) on reaction with NaOH form (E) & (F). (E) on reaction with H2SO4 & soda lime gives (X). (F) can be converted in to (B) by (Ag/heat). (C) on reaction with Mg/ ether followed by treatment with H2O gives (Y).
373
Aromatic Chemistry
Answer the questions from 85 to 90 85. Compound (B) would be:CHO
(a) N
CHO (b) N
H
(c)
Cl
(d) N
H
86. Compound (C) would be:
(a) N
CHO (b) CH2OH N
H
(c)
Cl
(d) Cl N
(d)
N
H
87. Compound (F) would be:CH2OH
(a) N
CH2OH (b) N
(c) N
H
H
N
CH2OH
88. Compound (E) would be:(–)
COO
COOH
(a) N
COOH (b) COO N
H
(–)
(c)
H
89. Compound (Y) would be: (a) Pyridine
(b) Pyrrole
90. In Passage- V identical compounds are: (a) X & Y (b) X, Y and Pyrrole
(d)
N
N
H
H
(c) Pyrolidine
(d) Piperidine
(c) Pyrrole & X
(d) Pyrrole & Y
Passage - VI Seven compounds are present in seven buckets named A, B, C, D, E, F & G.
PhOH
NaNO2+ HCl
AgNO2
PhNH2
(A)
(B)
(C)
(D)
i–Pr–I
n–Pr–I
NaOH
(E)
(F)
(G)
A cotton shirt is immersed in these containers in the order as given in questions. Identify the colour developed on shirt in each case. Answer the questions from 91 to 94 91. D→ B (ice cold)→ A→ G (a) Yellow
(b) Orange
(c) Red
(d) Blue
92. D→ B (ice cold)→ D→ G (a) Yellow
(b) Orange
(c) Red
(d) Blue
374
Problems in Organic Chemistry
93. E→C→B→G (a) Red
(b) Blue
(c) Yellow
(d) Orange
94. F → C → B→ G (a) Red
(b) Blue
(c) Green
(d) Orange
PCl
95.
5 o − (COOH) − C6 H 4 − CH 2 – Ph →
[C] would be
ZnCl
N H
2 2 4 [A] →[B] →[C] NaOH in glycol / D
O
(a)
(b)
O
(c)
(d) O
96. Match the following: Compound Nature (A) Picric acid (1) Highly acidic (B) m-Toludene (2) Acidic (C) p-Toludene (3) More basic (D) o-Cresol (4) Less basic
Correct option is:(a) A–1B–2C–3D–4
(b) A–1B–3C–4D–2
(c) A–1B–4C– 3D–2
(d) A–2B–1C–4D–3
97. Match the following: Reaction Product (A) Raschig’s (1) Chloro benzene (B) Schotten Bauman (2) Salicylic acid (C) Kolbe’s (3) Cinnamic acid (D) Reimer Tieman’s by CCl4 (4) Benzanilide (E) Perkin
Correct option is:(a) A-1 B-1 C-2 D-4 E-2 (c) A-4 B-2 C- 2 D-3 E-1
98. Match the following:- Product of reaction
(b) A-1 B-4 C-2 D-2 E-3 (d) A-4 B-1 C-2 D-3 E-2 Comparison of Acidic Nature of Products
(i) fuming HNO , Conc H SO (ii) KMnO 4
3 2 4 (A) Ph – CH3 → (A)
Dil HNO
3 →(B) Ph – OH (B)
Conc HNO
3 (C) Ph – OH → (C)
CrO Cl
(1) Highly acidic (2) Good acidic (3) Acidic
2 2 →(D) Ph – CH3 (D)
(4) Not acidic
(c) A–1 B–2 C–3 D–4
(a) A–2 B–3 C–1 D–4
(b) A–2–B–4–C–1 D–3
99. Consider the following reaction. (+)
H 2 ,Ni (i) aq NaOH (i) H , Heat (W) → (X) →(Y) →(Z) Heat SO3H (ii) fuse with NaOH, H (+) (ii) H3O( + )
(d) A–1 B–3 C–2 D–4
375
Aromatic Chemistry
Now match the following
Compound
Test
W
1. Red Colour with FeCl3
X
2. No Colour in Victor Maeyer
Y
3. Blue Colour in Victor Maeyer
Z
4. Violet colour with FeCl3
Correct option is:-
(a) W – 1 X – 2 Y – 3 Z–4
(b) W – 1 X – 4 Y – 3 Z – 2
(c) W – 4 X – 2 Y – 4 Z – 3
(d) W – 4 X – 4 Y – 2 Z – 3
100. Assertion: - Friedel craft reaction of Aniline by CH3Cl / AlCl3 is not possible.
Reason: - Lone pair of electron present on nitrogen is not localised.
(a) Assertion is True, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is True 101. Assertion: -Benzaldehyde does not give Fehling test.
Reason: - It undergoes Cannizaro reaction.
(a) Assertion is True, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true 102. Assertion:- Nitration of benzene & C6D6 occur with the different rates.
Reason: - Breaking of C–H & C–D bonds do not occur in rate determining step.
(a) Assertion is True, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true 103. Assertion:-When aniline is subjected to nitration by conc HNO3 & H2SO4 meta nitro aniline is formed in considerable amount.
Reason: - NH2 is o/p directing but ring deactivating group.
(a) Assertion is True, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true 104. Assertion:-Benzene does not undergo addition reaction easily.
Reason: -Benzene losses it’s aromaticity after addition reaction & becomes less stable.
(a) Assertion is True, Reason is true: Reason is a correct explanation for Assertion. (b) Assertion is true Reason is true: Reason is not a correct explanation for Assertion (c) Assertion is True, Reason is False (d) Assertion is False, Reason is true
376
Problems in Organic Chemistry
Answer Key 2. (d)
3. (c)
4. (c)
5. (b)
6. (a)
7. (c)
8. (c)
9. (b)
10. (c)
11. (d) 12. (b)
13. (d)
14. (c)
15. (d)
16. (d)
17. (c)
18. (b)
19. (a)
20. (c)
21. (c) 22. (a)
23. (d)
24. (b)
25. (c)
26. (c)
27. (b)
28. (b)
29. (a)
30. (c)
31. (b) 32. (b)
33. (c)
34. (b)
35. (b)
36. (c)
37. (b)
38. (c)
39. (a)
40. (d)
41. (a) 42. (a)
43. (b)
44. (d)
45. (b)
46. (c)
47. (c)
48. (a)
49. (c)
50. (a)
51. (d) 52. (c)
53. (c)
54. (b)
55. (d)
56. (c)
57. (a)
58. (d)
59. (c)
60. (b)
61. (d) 62. (c)
63. (c)
64. (a)
65. (a)
66. (d)
67. (a)
68. (b)
69. (a)
70. (b)
71. (d) 72. (b)
73. (b)
74. (b)
75. (b)
76. (a)
77. (b)
78. (b)
79. (a)
80. (c)
81. (a) 82. (a)
83. (d)
84. (a)
85. (b)
86. (c)
87. (a)
88. (b)
89. (a)
90. (c)
91. (b) 92. (a)
93. (b)
94. (a)
95. (a)
96. (c)
97. (b)
98. (d)
99. (b)
100. (b)
101. (a) 102. (d)
103. (c)
1. (d)
104. (a)
Multiple Choice Questions (More Than One May Correct) 1. Phenol can not exhibit: (a) Liberman’s nitroso test H ( + ) / Na Cr O
2 2 7 2. [X] →
[X] may be:(a) Aniline
(b) Neutral FeCl3 test
O
(c) Bromine water test
(d) Biuret test
(c) m- hydroxy phenol
(d) p-amino phenol
(c) Victor maeyer test
(d) Coupling reaction
(c) o-hydroxy aniline
(d) m-hydroxy aniline
O
(b) Phenol
3. Phenol & aniline can not be differentiated by: (a) Neutral FeCl3 test (b) Bromine water test NHOH
4.
H
(+)
(a) Aniline
[X] X may be :-
(b) p-Hydroxy aniline
5. Both phenol & methyl alcohol can show: (a) Reaction with sodium bi carbonate (c) Acylation reaction
(b) Reaction with sodium (d) Reduction with zinc dust
6. Which of the following will undergo diazotization reaction? NHPh
(a) 7.
CH2CH2NH2
CH2NH2
(b) (c)
CH3
Phenol can be prepared by:(a) The hydrolysis of benzene diazonium salt (b) Decarboxylation of salicylic acid (c) Hydrolysis of chloro benzene by aq NaOH/25°C (d) Fusing sodium benzene sulphonate with NaOH pellets followed by acidifications
(d) None of these
377
Aromatic Chemistry
8. Correct order of basic character is:H
O
(a)
N
N (b) N
H
H
> N H
N
(c) N
H
H
>
>
(d) None of these
N H
9. The product of acid hydrolysis of A & B can be distinguished by :
[A] CH3 — C — OCOPh ||
[B] CH3—CH = CH OCOPh
CH2
(a) Cu+2 / NaOH
(b) [Ag(NH3)2]OH
(c) NaCN / HCl
(d) Victor Maeyer test
→ RNH 10. The appropriate reagent for the following transformation is:- RCOOH 2 (a) Ammonia / heat (b) (i) NH3 / heat (ii) KBrO / heat (c) Hydrazoic acid(in acidic medium) (d) (i) LAH (ii) Ammonia / Al2O3 11. In the following transformation A & B are:CH3
Li / liquid NH MeOH
1 mole of H Ni / heat
3 → A 2 →B
CH3
(a) B is
(b) B is
CH3
CH3
(c) A is
(d) A is
CH3
12. In following scheme of reactions Z would be:*
Cl alc. KOH
[X] *
[X] ZnCl2
→[Y] →[Z]
(a)
*
(b)
*
*
*
(c) * *
(d)
Answer Key 1. (d)
2. (a), (b)
3. (b), (c)
4. (b), (c)
5. (b), (c)
6. (d)
7. (a), (b), (d)
8. (a), (b), (c)
9. (a), (b)
10. (b), (c)
11. (b), (d)
12. (b), (c)
LEVEL - II 1. Select the reaction with correct product AlCl
3 (a) Ph — H + n — Pr — Cl → PhCH2CH2CH3
3 2 → (c) Ph — H + nPr — Cl PhCH2CH2CH3
AlCl ,H O
FeCl
3 (b) Ph — H + n — Pr — Cl → PhCH2CH2CH3
(d) None of these
378
Problems in Organic Chemistry
2. You have following four compounds O
O
O
O
O
O
O
O
S
S
(2) (3) If R1, R2, R3 & R4 are the rate of nitration of 1, 2, 3 & 4 respectively then:(a) R1 > R2 > R4 > R3 (b) R1 > R4 > R3 > R2 (c) R1 > R4 > R2 > R3 (1)
O
O
O
(4)
(d) R1 > R2 > R3 > R4
3. Select the reaction which is most likely to occur:-.
(a) Me3CCOCl + PhH + Anhydrous AlCl3 → Ph — CMe3
∆ (b) PhNO2 + CH3COCl + Anhydrous AlCl3 → m — NO2 — C6H4COCH3
3 (c) PhOH + CH3Cl → p — CH3 — C6H4 — OH (d) All are possible
AlCl
Comprehension Comprehension - I Nitration of benzene nucleus is carried out by HNO3 & H2SO4 as:+ HNO3
NO2
conc. H2SO4
Identify P1 to P7 OH
4.
+ concn HNO3 →
P1 + P2 (Major) (Minor)
OH
5.
+ concn HNO3
+ concn H2SO4 → P3
NH2 NO2BF4 low temp
6.
NO2BF4 High temp
P4 P5
Cl
7.
8.
N O
2 5 → P6
NO PF
••
N H
5 6 → P (Thermodynamically controlled product). 7
O
379
Aromatic Chemistry
Comprehension - II Since benzene ring is electron rich hence it does not favour the attack of nucleophile. When electron with drawing groups are present on benzene ring attack of nucleophile on ring becomes easy as intermediate anion is stabilized by electron with drawing groups. Cl
OH aq NaOH,H ( + ) → High + T&P
OH
Cl O2N
NO2
+
H 2O
(warm)
O2N
NO2
→ NO2
NO2
Identify P1 to P4 Cl Cl MeSNa
→ P1 in DMF
9. CN
(–)
O
F
10.
→P
+
2
NO2 NO2 (–)
11.
+ N H 2 → P3 +
P4 (Major) (Minor)
Cl
12. Consider the following reactions NO2 (–)
NO2
NO2
+ CH2=CH—O
F
OCH = CH2
Cl
OCH = CH2
NO2
NO2
NO2
NO2
(–)
+ CH2=CH—O
Br
(–)
.... R1
+ CH2=CH—O
NO2
OCH = CH2
(–)
.... R3
..... R4
+ CH2=CH—O I
If R1, R2, R3 & R4 are the rates of reaction then arrange them in decreasing order.
.... R2
OCH = CH2
380
Problems in Organic Chemistry
13. Consider the following reactions NO2
NO2
NO2
+ MeSNa F SMe
NO2
NO2
+ MeSNa Br
+ MeSNa
................ R1
Cl
F
Cl
SMe
Cl
Cl
SMe
NO2
NO2
................ R3
+ MeSNa I
Br
Cl
NO2
Cl
I
................ R2
................ R4
SMe
If R1, R2, R3 & R4 are the rates of reaction then arrange then in decreasing orders.
Matrix Match Questions Each question contains statement given in two columns which have to be matched. Statement (A, B, C, D) in column I have to be matched with statements (p, q, r, s) in column II. The answer to these questions has to be appropriately bubbled as illustrated in the following examples. If the correct match are A - p, A - s, B - r, B - q, C - q, D - S, then the correctly bubbled 4 x 4 matrix should be as follows.
14.
Pair of compound OH
OH
(A)
Reagent for distinction
&
(p) KOH / I2
NH2 CONH2
(B)
&
(q) PhN2Cl with Pyridine
NH2 NH2
(C)
&
(r) neutral FeCl3
OH CHOHCH3
(D)
&
(s) Br2 water
381
Aromatic Chemistry
15.
List - I CH3
(A)
List - II CH3
NaNH
2→
Cl
(p) Aryne Mechanism
NH2
SO3H
SO3H Cl
(B)
KNH
2→
(q) Anhydrous AlCl3
NH2 CH3
CH3
(C)
(CH ) CCl
3 3 →
(r) Nucleophilic Substitution
CMe3
(D)
→
16.
List - I NHCOC6H5
(A)
List - II
CHMe2
(B)
(s) Intermediate carbocation is form
(p) Gives Phenol or air oxidation followed by hydrolysis.
(q) Benzanilide
OCOPh
(C)
CCl3
(D) 17.
List - I NO2
(A)
(r) Formed by schotten Bauman reaction
(s) Shows reverse hyperconjugation List - II
(p) Gives p – benzoquinone with H+/Na2Cr2O7
NH2
(B) (q) In a strongly acidic medium when current is passed it gives p – hydroxyl aniline OH
(C)
(r) Gives coupling reaction but not react with neutral FeCl3
COCH3
(D)
(s) Can not show coupling reaction but reacts with NaHSO3
382
Problems in Organic Chemistry
18. Match the products W, X, Y and Z in list - I with list - II
List - I
List - II
CrO3 (A) PhCH2OH3 → W
(p) Shows addition reaction with Cl2/CCl4
CrO
3 → X (B) PhCHOHCH3
LiNH
2→ (C) PhNH2 CH OH
Y
3
NO2 (D) O2N
(q) Negative fehling test
Z
(r) Can react with hydroxyl amine
(s) Contains two C = C bonds
NO2
19.
List - I
List - II
Ph Ph OH
(A)
(B) OH OH (C)
Ph NH2
(D) 20.
(p) Gives ketone on oxidation (q) Gives epoxide on dehydration with H2SO4
(r) Can show Diels Alder reaction with
(s) Turns brown on reaction with H2SO5
List - I
List - II
(A) Corbolic acid (B) Oil of winter green (C) Given white ppt with Br2 water (D) Paracetamol
(p) Phenol (q) Phenyl salicylate (r) p - Hydroxy benzanilide (s) Anisole
Answer Key 1. b,
2. c,
O
NO2 5.
+ O
P2
P1
HO
O
NO2 Cl
NO2
O
NO2
4.
3. a,
6. P4 P3
Cl NO2
7.
&
8. N
NO2
For 9, 10, 11, 12 &13 See Solution.
H
NO2
NO2 P5
383
Aromatic Chemistry
Answers matrix match 14. (A) – q, r, s, (B) – q, s, (C) – q, s, (D) – p, q, r, s 16. (A) – q, r, (B) – p, (C) – r, (D) – s 18. (A) – q, r (B) – q, r, (C) – p, q, s (D) – p, q 20. (A) – p, (B) – q, (C) – p, s, (D) – r
15. (A) – p, r (B) – p,r, (C) – q, s, (D) – q, s 17. (A) – q, (B) – p, r, (C) – p, (D) – s 19. (A) – p, r, (B) – p, (C) –p, q (D) – p, s
SOLUTION NH2
NO2
NO2 Na2S
CF3CO3H
1. (d) NH2
NO2 CH3
Br
2. (d)
(–) NH2
CH3
CH3
..
CH3
NH2
NH3
–NH3, –Br (–)
NH3
(+)
or (–)
NH3
More stable
less stable due to +I effect of CH3 group
OCH3
OCH3
OCH3 Br
(–)
NH2
–NH3, –Br (–)
(+)
(–)
NH3
NH2
OCH3
(+)
NH3
(–)
CH3
or
(–) (+)
NH3
More stable due to -I effect of oxygen
OCH3
NH3 3. (c) X is salicylic acid which is obtained by kolbe reaction of phenol. Since COOH is electron withdrawing group hence on reaction with Br2 water IPSO attack occurs & COOH is replaced by Br. OH OH COOH Br Br Br2 water
(X) (Y) Br Br of bromine water can replace –R groups. 4. (c) Nucleophile does not attack on benzene ring because benzene nucleus is electron rich. So as substitutions with –R groups increases on benzene ring, benzene ring becomes electron deficient and favours the attack of nucleophile. Water is a poor nucleophile hence it can’t replace Cl from benzene through necleophilic substitution reaction. H2O can do it only when all ortho & para positions are substituted by electron with drawing groups like NO2. 5. (b) A = Cu (Ullman reaction) B = Zn + Et2O (Fitting reaction) NO2
electron rich
NO2
NO2 COOH
6. (a)
KMnO4, heat
NaOH, CaO heat
COOH electron dificient KMnO4 attacks on electron rich benzene nucleus.
384
Problems in Organic Chemistry
7. (c) Along with o & p products meta derivative also forms with 47% 8. (c) 1st is wrong because aliphatic amine can not form stable di azo salt 2nd is wrong because single carbon atom can not bear 4 phenyl groups due to steric hindrance. 3rd is wrong because benzene ring is deactivated as aniline reacts with AlCl3 to (+ )
produce Ph NH 2
(–)
AlCl3 Br
9. (b)
CH3Cl → AlCl3
CH2 C
CH2 C
NH
NH
O H OH
O
Br CH2 CH3 NH2 COOH 10. (c) NH3 + H2S is used for partial reduction of aromatic nitro compounds 11. (d) (B) is phenol (C) is benzoic acid & (A) is benzene sulphonic acid. The acid strength of these compounds will follow the following order A > C > B
12. (b)
Cl
13. (d) A =
Cl CHCO CHCO
dil HCl hydroloysis
Cl
NH2 Cl
O C=
B=
D=
Ph
14. (c)
oxidation
COOH
CH3
decarboxylation
CH3
CH3
NO2
15. (d) θ =120°
θ =60°
NO2
NO2 θ = 0°
Dipole moment ∝ 1 / θ 16. (d) 17. (c) CCl3 is meta directing group due to reverse hyperconjugation 18. (b) X is phenolphthalein .In basic medium (NaOH) it converts it self in to (b) Br
19. (a) X
Br
is
Br Here all positions are identical hence, only one product will be obtained afer mono nitration.
385
Aromatic Chemistry
20. (c) Benzoic acid & phenol on esterification produces phenyl benzoate (X) OCOPh
OCOPh
NO2(+)
OCOPh +
NO2
NO2 Majo due to less steric hinderence
MeO
HO
OMe
21. (c)
–2H2O
OH OH
OH
OH
Br
Br
OH
H3O(+)
OMe OMe
O Tautomerism
Br2 water
O
OH
OH Br
OCH2CH = CH2
OH CH3
22. (a)
CH3
– HCl
+ CH2 = CH – CH2Cl CH2 O CH3
O
CH
H CH2—CH = CH2
CH3
CH2
OH CH3
CH2CH = CH2
(+) (+) 23. (d) In it two electrophiles are generated by AlCl3,CH3CO and PhCO . Hence both the electrophiles can attack on both the benzene nucleus consequently all products are possible. 24. (b) 9th & 10th positions of anthracenes are more reactive hence Diels – Alder reaction will occurat 9th & 10th position. → ZnCl + 2HCN 25. (c) Zn(CN)2 + 2HCl 2 → CH(+) = NH H — CN + H(+) 26. (c) N2Ph can not enter at 2nd position due to steric hinderence. N2Ph will attach at 6th position because at this location as it is less sterically hindred & both the OH groups have same agreement. OH 1
HO
6
2
less Chances
More Chances
5
3 4
NH2
N2Cl diazotisation
27. (b)
N2Cl
NH2 NH2
(+)
(+)
N2
N2
H2N Also see previous question. OH
28. (b)
(RESORCINOL)
more sensitive position for electrophilic attack
NH2
NH2
386
Problems in Organic Chemistry
OH 2
29. (a) (X) is salicylic acid
COOH
1
3
both OH & COOH groups have common agreement at 5th position hence coupling
6
4 5
will occur at COOH
5th
position. COOH (+)
(+)
N2
HO
N2
S
OH
Oa 3 4
2
30. (c)
b O
1
cO
5 6
a is – R group however b & c are +R groups all these three have common agreement at 2nd position
hence nitration will occur at 2nd position. 31. (b) 32. (c) See aromatic nucleophilic substitution in your text book. CH3
CH3 Birch reduction
33. (c)
ozonolysis
CH3
CH3COCH2CHO
CH3
34. (b) OD is ortho para directing group thus, D+ attacks at ortho para locations. 35. (b) A is acrolein CH2 = CH — CHO (–)
O
O = C—H
C
CH
..
NH2
NH (+) 2
CH2
H CH
H HO
CH2
NH
(+) ARSE
(+)
H
H
H
NH
N
tautomerism
O
OH
OH
NH
H (+)
H , heat
N
36. (c) It is benzoin condensation
H (–)
O
OH
CN
Ph—C—H
PhC—CN
OH
CN (–) O
O (–)
Ph–C–H
+
(–)
O
Ph–C–H + PhC—CN
(–)
OH O
OH
Ph—CH—C—CN
Ph—CH—C—CN
(–)
OH
Ph
Ph
BenzoinPh—CH—COCN
(–) –CN
387
Aromatic Chemistry
37. (b) In thiophene and pyrrole ARSE occurs at 2nd location. See q. No. 11 Chapter -04 topic electrophilic substitution reaction. 4
38. (c) 5
3
S
N 2 nitration can occur at 2nd & 5th positions but at 2nd position H is absent hence substitution is not possible at 2nd
1
position consequently nitration occurs at 5th position. 39. (a) Since electronegativity of S is lesser than N so extent of resonance by lone pair with double bond is maximum in thiophene part hence electrophile will attack on thiophene part. 40. (d) See test of phenol 41. (a)
OH
Cold KMnO4
D–Alder
+
OH H(+), heat
(Naphthanlene)
42. (a) 43. (b) NO2 group is o/p directing for nuclceophilic substitution. 45. (b) It is an example of Cross aldol condensation. (-) CH3 CH2 O2N NO2 O2N NO2 (–) + OH H2O + NO2
NO2 NO2
NO2
O
(–)
CH2 + Ph—C—H
O2N
NO2
NO2 CH = CH—Ph
46. (c)
(–) OH, heat –H2O
NO2
NH2
(+)
NH2
(–)
NH3HSO4
Br
453 K –H2O
NH2
(+)
H2O
NO2
H2SO4
NH2
OH
CH2—CH—Ph
O2N
NO2
(–)
O
CH2 — CH —Ph
O2N
NO2
O2N
44. (d)
Br
H3O, NaOH 100°C
SO4H Ac2O + Br2
SO3H
47. (c) C6H5N2Cl on coupling reaction with phenol gives orange coloured dye while yellow coloured dye with aniline. 48. (a)
PhCOOH
PCl5
PhCOCl (A)
PhCOOPh (C)
Baeyel villiger PhCOPh Oxidation (B)
PhCOCl + PhOH → PhCOOPh (A) [Y]
C6H6 AlCl3
388
Problems in Organic Chemistry
49. (c) Benzaldehyde on bromination gives meta bromo benzaldehyde which on reduction with zinc amalgam & HCl gives meta bromo toluene which finally gives 3 – bromo – 4 – nitro toluene on nitration CHO
50. (a)
CH2OH
CH2OH Br2 Fe
LAH
CHO
(+)
Br2 Fe
Br
Br
Br
Br
Br
Br
(–)
N2Cl NaNO2 + H2SO4
51. (d)
CHO
Br2 Fe
Cu ∆
Br NH2
CHO
meta directing group
OEt dil H2SO4 –2EtOH
CH2 OEt
A OH
OH –H2O
CH2
HCl
CH2O
OH
H—C—H (+)
Formaldehyde
OH (+) N
2Cl
(–)
A
boil with water
HOCH2
HOCH2 NH2
NO2
NO2
LiAlH4
Birch reduction
52. (c)
OH
OH HO
OH Baeyers reagent
HO
HONO
OH 1
Birch reduction
53. (c)
(A) 2
Since 2nd is less substituted alkene hence it is less stable & undergoes reduction easily in comparison to 1st . (A)
H (1 mol) Ni, heat
2 →
CH3
54. (b)
Birch reduction
SO3H
OH
Me
CH3 Cold KMnO4 Synaddition
H OH
HO
SO3H
SO3H
H
OH H
55. (d) Rate of reaction will be least in 1 & 4 as SO3H is ring deactivating group. While rate of reaction will be large in 2 & 3 as methyl group is ring activating group. Since isotopic effect is not observed in nitration thus rate of reaction will follow the order 3 = 2 > 1 = 4 O OH (+) H3O 56. (c) O O O O O OH –H2O OH HCl
Cl OH
See Q.No. 35, topic electrophilic addition, chapter 04
389
Aromatic Chemistry
57. (a) See Question Number 22 58. (d) –R group
O
O
+R group Activated ring
59. (c) Isotopic effect is observed only in case of sulphonation.
Passage - I (60 to 69) 60. (b) For aromatic alcohol OH group should be directly linked to the benzene nucleus. 61. (d) This is the property of carboxylic acid. Phenol & aliphatic alcohols do not release carbon dioxide with NaHCO3 62. (c) (C) will give haloform test with NaOCl. 63. (c) (D) is 3° alcohol and will not give any colouration in victor maeyer test while ‘E’ is primary alcohol & produce red colouration in victor maeyer test. 64. (a) A & E can not form alkene by acid catalyzed dehydration. 65. (a) Aromatic alcohols give violet colouration with neutral FeCl3. 66. (d) Phenol can not form cation on reaction with H(+) hence 1st option is not correct. Carbocation formed by (D) is highly stable due to resonance as well as hyperconjugation. (+) (+)
(+)
(+)
From (D) Resonance & more hyperconjugation
From (C) Resonance & hyperconjugation
From (E) Only Stabilized by resonance
From (B) not Stabilized by resonance
67. (a) Due to +R effect to OH group. 68. (b) 69. (a)
Passage - II (70 to 74) NO2
NO2
NO2 NaNO2 + HCl
Na2S
t > 8°C
NO2
NH2
(A)
OH
(B)
(C) NaNO2 + HCl, 0°C
NO2
NO2
NO2 KNO2, Cu
HCl Cu
NO2
N2Cl
Cl (D)
(C)
(E) NO2
H2O, boil
NO2 (F) Compound ‘E’ on mono nitration will form tri nitro benzene which is highly explosive [T.N.B]
390
Problems in Organic Chemistry
Passage - III (75 to 81) Cl Cl2
aromatisation
n – Hexane (A)
Fe CH3CH2CH2OH, H –H2O
CH3 O–O–H CH3
CH3
(+)
D2O
D CH3
air
(G)
(C) OH
(+)
H3O
CH3COCH3 + (E)
(F)
(doesnot gives haloform test)
(gives haloform test)
Passage - IV (82 to 84) OH
ONa NaOH
(A) CH3
CH3
OH Br
Br
Br2, water
CH3 Br (B)
When (A) is insoluble in NaOH then it will be anisole. Br OCH3 OCH 3
Br water
2 →
Br
Br
Passage - I (85 to 90) Cl
CHCl3, NaOH H
N
(+)
CHO
N H
MgCl
Mg, Ether
+
H (A)
N (C)
(B)
N H2O
NaOH
H2SO4 sodalime
H
(–) + COO
N
N (X)
MgCl
Mg, Ether
H
(E)
(Y) N CH2OH
N H
(F)
391
Aromatic Chemistry
91. (b) 92. (a) 93. (b) 94. (a) 95. (a) 96. (c) 97. (b) 98. (d)
Coupling reaction Coupling reaction Victor maeyer test of aliphatic alcohol Victor maeyer test of aliphatic alcohol A–1 B–4 C– 3 D–2 A–1 B–4 C–2 D–2 E–3 Compounds A, B, C & D are 2,4,6-tri nitro benzoic acid, para or ortho nitro phenol 2,4,6-tri nitro phenol & benzaldehyde respectively.
99. (b)
100. (b) AlCl3 reacts with aniline (Lewis acid - base reaction) thus, benzene ring gets deactivated. (–)
AlCl3 + PhNH 2 → PhNH 2 → AlCl3 Base (Acid) 101. (a) 2PhCHO +
NaOH
(from fehling solution)
→ PhCH 2OH + PhCOONa
102. (d) Nitration of C6H6 & C6D6 occur with same rate because breaking of C— H & C — D bonds is not the part of rate determining step. 103. (c) Due to protonation by nitrating mixture (nitric acid & sulphuric acid) NH2 group becomes NH3(+) which is electron with drawing and meta directing group. 104. (a) Benzene losses it’s aromaticity after addition reaction & becomes less stable More than one may correct:1. 2.
(d) Biuret test is shown by the compounds containing CONH2 linkage. (a, b) 3. b, c (+)
(b, c)
NH
(+)
(+)
4.
NH
NH—OH2
NHOH H
–H2O
NH2
H2O
NH2
H2O
(+)
HO
OH (b, c) (d) Diazotisation reaction is shown by aromatic amine & not by aliphatic amine as aromatic diazonium salts are stabilized by resonance. 7. (a, b, d ) Chlorobenzene can not be hydrolysed by aqueous NaOH at room temperature. Fusion of sodium benzene sulphonate with NaOH pellets gives sodium phenoxide which on acidification gives phenol
5. 6.
392
8.
Problems in Organic Chemistry
(a, b, c) In (a) first compound is more basic than 2nd because it becomes aromatic after protonation. (+)
O
OH
OH
(+)
H
Aromatic
N H
9.
H O( + )
2 → (a, b) [A]
N
N (+)
H
H
MeCOMe
(–ve tollen's & fehling 's test)
+ PhCOOH
H O( + )
3 [B] → EtCHO + PhCOOH (–ve tollen's & fehling 's test)
10.
(b, c)
11. *
12.
Cl
(b, c)
CH3
[B]
* alc. KOH
[Y]
[X] [X]
CH3
(b, d) [A]
* alc. ZnCl2
(+)
(+)
[Y]
*
[Y]
*
* *
*
LEVEL -II (+ )
(–) 1. (b) [FeCl4 ] [CH3CH 2 CH 2 ] is formed, This is weak ion pair thus, rearrangement does not occur because these ions do not get separated. 2. (c) In 1 & 4 benzene rings are activated and moderately activated respectively. While in 2 & 3 benzene rings are moderately deactivated. (+ )
AlCl
3 3. (a) Me3COCl → Me3CO( + ) → CO + Me3 C → Ph – CMe3
OH
4.
O
NH2
NO2
+ O
5.
O2N
+ conc.HNO3
OH
Ph –H
O nitrating mixture
O
P3
P2
HO
P1 NO2
393
Aromatic Chemistry
NO2
6.
P4 NO2 P5
7. N2O5 is a nitrating agent. Cl
Cl
Cl NO2
N2O5
&
NO2
nitration
8. N
N
H
H
Cl
SMe
Cl
(–)
Cl
MeS
9.
NO2
CN
CN
10. No reaction because base is sterically hindered. 11. Here benzyne mechanism takes place. NO2
NO2 (–) NH2
Cl
– NH3
NH3 NO2
NH3 NO2
(+)
NH3
(–) (+)
(–)
NH3 (More Stable)
NO2
(Less Stable)
NO2 NH2 NH2
(Major)
(Major)
394
Problems in Organic Chemistry
X
12. Here following type of intermediate is formed will follow the following order. R1 > R2 > R3 > R4 > R5 X Cl 13.
Base (–)
- I effect of X stabilizes the intermediate hence rate of reactions
NO2
Base (–)
It is more stable than others because -ve charge can go to empty d orbital or chlorine thus order is:-
NO2
R2 > R1 > R3 > R4
Practical Organic Chemistry
14
Main Features Detection of elements (Lassaigne test): - Organic compound is fused with sodium metal and then fusion mixture is mixed in boiled distilled water. Boiled solution is now filtered. This filtered solution is called sodium extract (S.E.). Elements present in organic compound get converted in to their sodium salts as follows:
→ NaCN Na + C + N
→ Na2S 2Na + S
→ NaX Na + X
→ NaCNS Na + C + S + N
[ X = Cl , Br ,I , F]
Test of Nitrogen:
S.E. + FeSO4 + conc H2SO4 boil & cool now add FeCl3 + conc HCl (Prussian blue colouration)
→ Fe(OH)2 + Na2SO4 FeSO4 + 2NaOH
→ Na4[Fe(CN)6] + 2NaOH Fe(OH)2 + 6NaCN
→ Fe [Fe(CN) ] + 3NaCl Na4[Fe(CN)6] + FeCl3 4 6 3 (Pr ussian blue)
Test of Sulphur:
(i)
S.E. + sod. Nitro prusside (deep violet colouration)
→ Na4[Fe(CN)5NOS] Na2S + Na2[Fe(CN)5NO]
(ii)
S.E. + dil CH3COOH + lead acetate (black ppt)
→ PbS ↓ +2CH3COONa Na2S + (CH3COO)2Pb (black)
When both sulphur & nitrogen are present
S.E + FeSO4 + NaOH + boil & cool + FeCl 3 + conc HCl ( blood red colouration)
→ [Fe(CNS)]Cl + NaCl NaCNS + FeCl3 2 (Blood red)
Test of Halogen
S.E + dil HNO3 + AgNO3
(Precipitates)
→ AgCl + NaNO3 NaCl + AgNO3 White ppt
396
Problems in Organic Chemistry
→ AgBr + NaNO3 NaBr + AgNO3 (Dirty yellow ppt)
→ AgI + NaNO3 NaI + AgNO3 (Bright yellow)
Precipitates of AgCl & AgBr are soluble in liquid. NH3 but precipitates of AgI are insoluble in liquid. NH3.
LEVEL - I Multiple Choice Questions 1.
For the detection of nitrogen in urea (Lassaigne test) some times we add naphthalene for better result because
(a) Naphthalene breaks urea easily.
(b) In urea % of C is small so naphthalene helps in the formation of more cyanide ions.
(c) Naphthalene forms nitro naphthalene which can show test of nitrogen easily by aq. FeSO4
(d) Naphthalene reduces melting point of urea
2. X + resorcinol + conc H2SO4 Red-green fluoroscence
∆
solution
pour in a solution containing aq NaOH
X would be:-
(a) Phthalic acid
(b) p - nitro toluene
(c) Salicylic acid
(d) Phenolphthalein
3.
Phenol and acetic acid can be distinguished by:-
(a) Haloform
(b) Neutral FeCl3
(c) Na
(d) (a), (b) & (c)
4.
Choose the answer that has the following compounds located correctly in the separation scheme.
Citric acid - A, A + B + C water
Compound (–3)
Benzophenone - B,
Benzoic acid - C
water soluble + Insoluble Portion Solution (–1)
dil HCl
saturated NaHCO3
Soluble Portion + Insoluble
InNaHCO3(–2)
1 2 3 1 2 3
(a) A
B
C
(b) C
B
A
(c) B
C
A
(d) A
C
D
5.
Choose the answer that has the following compounds located correctly in the separation scheme.
Succinic acid (A), A+B+C
water
dil HCl
(3)
Anthracene (B), water soluble + Insoluble Solution (–1)
Salicylic acid (C) NaHCO3
Soluble Portion + Insoluble Portion(–2)
1 2 3 1 2 3
(a) A
B
C
(b) C
B
A
(c) C
A
B
(d) A
C
B
397
Practical Organic Chemistry
6.
Lassaigne test for the detection of nitrogen does not hold well for:-
(a) NH2OH
7.
Lassaigne test for the detection of nitrogen will fail in case of:-
(a) Guanidine
8.
Formic acid and acetic acid can be differentiated by:-
(a) Calomel
9.
Formaldehyde and acetaldehyde can be differentiated by:-
(a) Phenol / dil NaOH
(b) NH2NH2
(c) N3H
(b) 2, 4 - DNP (b) NaHCO3
(c) Carbyl amine
(d) Tilden reagent
(c) Na
(d) CH3OH / H(+)
(b) [Ag(NH3)2]OH
(c) NH2NH2 in glycol / NaOH OH OH 10. & can be differentiated by:-
(d) All of these
(c) Na
(a) KOBr
(d) All of these
(b) NaNO2 + H2SO4
(d) Cu+2 / OH(–)
11. PhN2Cl some times does not show positive lassaigne test for Nitrogen because.
(a) It dissociates PhN2(+) & Cl(–) ion so Nitrogen does not come in solution in form of NaCN
(b) On heating it decomposes to give N2 gas.
(c) PhN2Cl is stabilized by resonance and thus Nitrogen does not come in the solution easily.
(d) It sublimes on heating thus can not form fusion extract.
12. A + AgNO3 → AgCl
A would be:-
(a) PhN2Cl
white ppt
(b) Me3N(+)Cl(–)
(c) PhCH2Cl
13. Match the following
Column - I
Column - II
(A) Acetaldehyde
1. heat/NaOH/CuSO4
(B) Glucose
2. NaHSO3
(C) HCOOH
3. Molish test
(D) Urea
4. HgCl2
A
B
C
D
A
B
C
D
(a) 1
2
3
4
(b) 2
3
4
1
(c) 4
1
3
2
(d) 3
2
4
1
14. Match the following
Column - I
Column - II
(A) Glucose
1. HgCl2
(B) Phenol
2. concn H2SO4
(C) CH3NH2
3. concn H2SO4 + NaNO2
(D) HCOOH
4. HgCl2/CS2
A
B
C
D
A
B
C
D
(a) 4
2
1
3
(b) 2
3
4
1
(c) 2
3
1
4
(d) 4
3
2
1
(d) All of these
398
Problems in Organic Chemistry
Passage - I Mixture (A + B + C) Dissolved in water & filter
Insoluble (B) & (C)
Soluble (A)
wash with water dried and then add saturated solution of NaHCO3
Soluble (C)
Insoluble (B)
Answer the question from 15 to 17 15. Compound A would be: (a) Benzophenone (b) Aniline (c) Pthalic acid (d) NH2CSNH2 16. Compound B when heated with conc sulphuric acid gives yellow coloured solution. Compound B is : (a) Acetic acid (b) Benzene (c) Toluene (d) Benzophenone 17. Compound C on on heating produces another compound which produces a pink coloured solution on reaction with (i) Phenol / H2SO4 (ii) NaOH COOH OH
(a)
NH2 COOH
(b)
OH
(c)
COOH
(d) All
Passage - II (A + B + C) water
water soluble A
water insoluble B + C Saturated NaHCO3
Insoluble - B
Soluble - C dil-HCl
C
399
Practical Organic Chemistry
Answer the question from 18 to 20
18. Compound C is fairly soluble in hot water but not is cold water. Compound C is:COOH (a) CH3COOH (b) (c) para di chloro benzene O2N
(d) Citric acid
19. Compound B has zero dipole moment. COOH
(a) para di chloro benzene (b)
(c) PhNH2
(d) meta di chloro benzene
(c) aniline
(d) Phthalic acid
HOOC 20. A gives foam with NaOH. A would be:-
(a) Citric acid
(b) meta di chloro benzene
Answer Key 1. (b)
2. ( a)
3. (b)
4. (a)
5. (a)
6. (d)
7. (d)
8. (a)
9. (a)
10. (b)
11. (b)
12. (d)
13. (b)
14. (b)
15. (d)
16. (d)
17. (c)
18. (b)
19. (a)
20. (a)
SOLUTION LEVEL -I 1.
(b) In urea % C is low so nitrogen of urea does not convert in to cyanide easily. For this either we have to use two or more ignition tubes for fusion mixture or we have to use naphthalene along with urea in ignition tube.
2.
(a) HO
OH
OH HO
H
O
HO
OH
H2SO4
H
C
O
O
C O
C O
C Phthalic anhydride
NaOH
Red-green fluoroscence
3.
(b) Phenol gives violet while acetic acid gives blood red colouration on treatment with neutral FeCl3
4.
(a) Citric acid is soluble in water (1) Benzoic acid is insoluble in water due to the presence of hydrophobic benzene ring but forms PhCOONa with sod. bi sulphite white NaHCO3 (2)
5.
(a) Same as above.
6.
(d) Carbon is absent in all compounds.
7.
(d) Carbon is absent in NOCl.
8.
(a) Formic acid shows reducing properties.
→ 2HCl + CO2 + 2Hg↓ Hg2Cl2 + 2HCOOH 9.
(grey)
(a) Formaldehyde form a very hard plastic (Bakelite) with Phenol & H(+) ion.
10. (b) Phenol shows liberman’s nitroso test with NaNO2 + H2SO4. 11. (b) On heating PhN2Cl evolves N2 so cyanide ion formation does not occur.
400
Problems in Organic Chemistry
12. (d) All of these. 13. (b) Acetaldehyde gives addition reaction with sod. bi sulphite & forms white crystals. Urea on heating gives biuret which turns violet by NaOH & aq. CuSO4. Formic acid can reduce Hg2Cl2. (sec. Q.No. 8) 14. (b)
→ Carbon black Glucose + H2SO4 (concn)
→ liberman’s nitroso test Phenol + H2SO4 + NaNO2
→ Mustard oil reaction CH3NH2 + HgCl2 + CS2
→ 2HCl + CO2 + Hg2Cl2 HCOOH + 2HgCl2
(grey)
15. (d) Thiourea is water soluble due to H bonding. 16. (d) Benzophenone on sulphonation gives meta acetyl benzene sulphonic acid which is yellow in colour 17. (c) Pthalic acid on heating produces phallic anhydride which further produces phenolphthalene on reaction with PhOH & sulphuric acid. Phenolphthalene gives pink colouration in basic medium. 18. (b)
19. (a)
20. (a) Since it gives foam with NaOH thus it must have COOH group. Hence it is citric acid.
Physical Properties of Organic Compounds
15
Main Features There are two types of forces (a) Primary forces: Ionic and covalent bonds. Former is stronger than later. (b) Secondary forces:
H – Bonding > dipole – dipole attraction > other van-der waal forces
Van-der waal forces a Molecular weight Melting Point: Melting point of ionic compound is greater than covalent compound. NaCl = 801°C CH3CH2CH2CHO = – 90°C M.P. of ionic compound a Lattice energy In case of covalent molecules, following factors are responsible M.P. a Packing a Molecular weight a Dipole moment a Intermolecular forces of attractions Boiling Point: B.P. of ionic compound is greater than covalent compound. In case of covalent molecules, following factors are responsible B.P. a Mol wt. a Intermolecular forces of attractions a Surface area B.P. of halides a Polarisability (I > Br > Cl > F) B.P. of chlorides, bromides and iodides increases as number of halogen atom in organic compound increases but b.p. of fluorides decreases when number of fluorine atoms increases. Solubility in water: a 1/Mol. wt. a Dipole – dipole or ion – dipole attraction a Dipole moment In general for organic compound having comparable molecular weight solubility order is : RCOOH > RNH2 > ROH > RCOR > RCHO > ROR Solubility of chain isomers a 1/surface area CH3CH2CH2CH2CHO < CH3 — CH — CH2CHO
CH3
402
Problems in Organic Chemistry
Problems (Multiple Choice) 1. Compound with largest melting point is : (a) CH3(CH2)5Cl (b) LiCl
(c) CH3COOH
(d) CH3OH
2. Least boiling point will be exhibited by (a) CH3Cl (b) CH3Br
(c) CH3I
(d) CH3F
3. Which is insoluble in water? (a) Urea
(c) Chloroform
(d) Both (b) and (c)
(c) CH3COONa and T.H.F.
(d) HF + NaCl
(b) Methyl bromide
4. In which case dipole – dipole attraction is present (a) CHCl3 and Acetone (b) CHCl3 and H2
5. Correct order of dipole moment of the following organic compounds is:CH3
CH3
CH3
NO2 NO2 (1)
(a) 3 > 2 > 1 > 4
NO2 (3)
(2)
(b) 3 > 2 > 4 > 1
(c) 1 > 2 > 3 > 4
(4)
(d) 1 > 2 > 4 > 3
6. If µ1, µ2, µ3 and µ4 are the dipole moments of alkenes as shown below then select correct statements regarding following four alkenes CH3 CH3 CH3 H H H Cl Cl C=C C=C C=C C=C H Cl H Cl H H H CH3 (1) (2) (3) (4)
(a) µ1 > µ2
(b) µ4 > µ3
(c) µ1 = µ2 = µ3 = 0
(d) µ1 = µ2
7. Which among the following will have largest melting point? (a) CH3COONa (b) CH3CH2COONa
(c) HCOONa
(d) CH3CH2OH
8. Compounds having infinite solubility in water is (a) Iso propyl alc (b) Phenol
(c) Acetone
(d) Both (a) and (c)
9. Organic Compound A. n – Pentane B. Butanal C. Butan – 1 – ol D. Sod. Acetate
Melting Point (1) 324°C (2) –96°C (3) –90°C (4) –130°C
Correct matching is : (a) A → 4, B → 3, C → 2, D → 1 (c) A → 3, B → 2, C → 1, D → 4
(b) A → 4, B → 2, C → 1, D → 3 (d) A → 4, B → 3, C → 1, D → 2
10. Consider the following two carbonyl compounds where l1 and l2 are the bond lengths of ‘C – O’ bond O Ketone
O l2 Aldehyde
l1 C
CH3
C CH3
Et
H
(a) l1 > l2
(b) Ketone is more polar than aldehyde
(c) Aldehyde has more boiling point than ketone
(d) Ketone has less % enol content than aldehyde
403
Physical Properties of Organic Compounds
11. Out of CH3CH2CH2OH, CH3COCH3, CH3OCH2CH3 and CH3Cl the compound which is completely insoluble in water is : (a) CH3CH2CH2OH (b) CH3COCH3 (c) CH3OEt (d) CH3Cl 12. The compound which makes partially miscible solution when added in to water is OH
NH2 (b) CH3CH2OH (c) (d)
(a) CH3SO3H
13. Which pair of organic compound is completely insoluble in water? (a) PhNH2 and CH3F (b) CH3CH2OCH2CH3 and CH3F
(c)
and PhOH
(d)
OH
and CH3F OH
14. Among the following compound the compound with highest M.P. is : CH3CH2CH2COONa NaCl CH3CH2COOH (I) (a) I
(II) (b) II
OH OH (IV)
(III) (c) III
(d) IV
15. Boiling points of CH3CH2F, CH3CF3 and CF3 – CF3 are respectively – (a) –32°C, –47°C and –78°C (b) –78°C, –47°C and –32°C (c) –47°C, –32°C and –78°C (d) –78°C, –78°C and –32°C 16. Select correct statement OH
(a)
has more m.p. than
(b) 3° alcohol has more m.p. than 1° and 2° alcohols of same molar mass (c) 1° amines are more soluble in water in comparison to alcohols having comparable molar mass (d) All are correct
17. CH3CH2OH is more soluble in water in comparison to CH3OCH3 because (a) Alcohols are more polar than ethers
(b) Alcohol can form hydrogen bonds with water while ether can not form hydrogen bond with water (c) Alcohol dissociates in water while dissociation of ether is not possible (d) Alcohol forms more hydrogen bonds with water in comparison to ether
18. Which is an example of negative deviation?
(a) C2H5OH + H2O
(b) CHCl3 + CH3COCH3
(c) C2H5OH + CH3OH
(d)
OH +
19. Arrange the following in order of boiling point :
CH3F CH3Br CH3I
F (1) (2) (3) (4) (a) 4 > 3 > 1 > 2 (b) 4 > 3 > 2 > 1 (c) 1 > 4 > 3 > 2 (d) 1 > 2 > 3 > 4 20. Compound with largest boiling point is :
(a) CH3COOC2H5
(b) CH3CHO
(c) CH3COOH (d)
404
Problems in Organic Chemistry
21. Compound with maximum solubility in water is :
(a)
(c)
N
(b)
O
(d) All are insoluble in water
22. Out of hexane, pentane, butane and propane, the alkane having second highest melting point is : (a) Butane (b) Hexane (c) Propane (d) Pentane 23. But-2-en 1, 4 dioic acids exists in two forms as shown below CH – COOH
HOOC – CH
CH – COOH
CH – COOH
Maleic acid
Fumaric acid
Select correct statement (a) pKa of maleic acid is lesser than that of fumaric acid (c) Fumaric acid has more M.P. than maleic acid
(b) Maleic acid is more soluble in water than fumaric acid (d) All are correct
24. Select correct statement for following halides Cl
Cl
(I)
(a) (b) (c) (d)
Br
(II)
I
(III)
(IV)
B.P. of IV is lesser than III Dipole moment of III is lesser than IV Dipole moment of II is greater than I Except II rest all halides undergo SN reaction with aq NaOH frequently
25. Which will have maximum solubility in water?
(a) CH3OCH3
(b)
(c) CH3CHF2
(d)
OH NH2 NH2
Answers Key 1. (b)
2. (d)
3. (d)
4. (a)
5. (a)
6. (d)
7. (c)
8. (d)
9. (a)
10. (b)
11. (d)
12. (d)
13. (a)
14. (b)
15. (a)
16. (d)
17. (d)
18. (b)
19. (c)
20. (c)
21. (d)
22. (a)
23. (d)
24. (d)
25. (a)