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JEE ADVANCED LEARNING MODULE
CHEMISTRY 1. CARBONYL COMPOUNDS 2. CARBOXYLIC ACIDS & ACID DERIVATIVES
IDEAL 21ST CENTURY COMPETITIONS www.21stideal.com
CONTENTS CARBONYL COMPOUNDS (Page 1 - 106) TOPIC
PAGE NO.
Theory
03 - 30
EXERCISE-I Subjective Questions
31 - 34
Objective Questions
35 - 42
EXERCISE-II Subjective Questions
42 - 44
Objective Questions
45 - 53
EXERCISE-III Mixed Type Problems
54 - 61
EXERCISE-IV Past Year JEE/AIEEE Problems IIT - JEE Problems (Previous Years)
62 - 67
AIEEE Problems (Previous Years)
67 - 68
ANSWRE KEY
69 - 76
MISSCELLANEOUS QUESTION BANK
77 - 99
ANSWRE KEY
100 - 106
IDEAL 21ST CENTURY COMPETITIONS www.21stideal.com
CARBOXYLIC ACIDS & ACID DERIVATIVES (Page 107 - 181) TOPIC
PAGE NO.
Theory
109 - 138
EXERCISE-I Subjective Questions
139 - 140
Objective Questions
141 - 147
EXERCISE-II Subjective Questions
148
Objective Questions
149 - 156
EXERCISE-III Mixed Type Problems
157 - 162
EXERCISE-IV Past Year JEE/AIEEE Problems IT - JEE Problems (Previous Years)
163 - 165
AIEEE Problems (Previous Years)
165 - 166
ANSWRE KEY
167 - 170
MISSCELLANEOUS QUESTION BANK
171 - 179
ANSWRE KEY
180 - 181
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CARBONYL COMPOUNDS | 1
CARBONYL COMPOUNDS
IDEAL 21ST CENTURY COMPETITIONS www.21stideal.com
CARBONYL COMPOUNDS | 31
PART - I : SUBJECTIVE QUESTIONS Section (A) : Nucleophilic addition reactions A-1.
Why aldehyde are more reactive than ketones (among isomers) towards nucleophilic addition reaction.
A-2.
Arrange the following compounds in decreasing order of rate of nucleophilic addition with RMgBr. (I) Ph–CO–CH3
A-3.
(II) Ph–CHO
(III) CH3CHO
(IV) CCl3CHO
Arrange the following compounds in decreasing orders of nucleophilic addition with semicarbazide ..
..
NH2NHCONH2 i.e., NH2 Z :
(I)
A-4.
A-5.
(II)
(IV)
Arrange the following compounds in decreasing orders of rate of addition of HCN. (a)
(b)
(c)
(d) CH3CHO
Arrange the following compounds in decreasing orders of Keq for hydrate formation.
(a)
A-6.
(III)
(b)
(c)
(d)
Show how you would synthesis the following ? (a) Acetophenone Acetophenone cyanohydrin
(b)
A-7.
Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not. Explain why ?
A-8.
Give the structure of the carbonyl compound and amine used to form the following imines.
(a)
(b)
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CARBONYL COMPOUNDS | 32 A-9.
Show which alcohol and carbonyl compound react to give each of the following product.
(a)
A-10.
(b)
Write the product of following reaction : Ph (a)
H C
CH3 + NaHSO3
(b)
O
18
D2O (c) CH3 C CH3
4 2
OH
(d)
|| O16
H SO
C=N C2H5
+
H
Section (B) : Condensation reactions B-1.
B-2.
Predict the product of following aldol condensation reaction :
(a)
OH / CH3–CH2–CHO
(b)
OH / Ph–CH2–CHO
(c)
OH / Ph–CO–CH3
Indicate the starting aldehyde or ketone from which each of the following compounds are formed by an aldol condensation reaction. (a) 2–Ethyl – 3–hydroxy hexanal ;
(b) 4–Hydroxy–4–methyl–2–pentanone
B-3.
When acetone is treated with excess of benzaldehyde in the presence of base, the crossed condensation add two equivalents of benzaldehyde and expels two equivalent of water and forms [X]. Identify the structure of [X] when [X] reacts with NH2OH how many stereoisomers are formed.
B-4.
Predict the products of following cross condensation reaction,
(a)
B-5.
CH3 | OH / CH3 — C — C H + CH3 – C – H || | || CH3 O O
(b)
OH / Ph – C H + CH3 – CH2 – C – H || || O O
(c)
Ph – C H + || O
OH / C – CH3 || O
What is the principal product of the follwoing reaction ?
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CARBONYL COMPOUNDS | 33 B-6.
Show which esters would undergo claisen condensation to give the following –ketoester.
(a)
(b)
(c)
Section (C) : Cannizzaro's reactions C-1.
Identify the products in the following disproportionation reaction and also mention rate determining step.
+
C-2.
+
OH CH3 C H + H C H Product || || O O
excess
Section (D) : Redox reactions D-1.
Write the product of following reaction,
LiAlH
4 (p) (q) + (r)
(a)
O Per acid
(b)
D-2.
Write the product of following reaction : CHO
CHO
(a)
+ Ag2O
(b)
+ Ag(NH3)2 OH
(d)
HO
HO
CHO
(c)
+
|| O
+ K2Cr 2O 7 / H2SO 4
CHO + KMnO4 (cold, dil.)
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CARBONYL COMPOUNDS | 34 D-3.
Give the structure of compound A to D. CH2=CH–CH2–OH
A(C3H4O)
B(C5H10O2)
C(C5H12O4)
D
Section (E) : -Halogenation, haloform, -deuteration reactions
E-1.
Ph
O ||
C6H5
Br
2 / OH Products and mention stereochemical relation
CH3
E-2.
Write the product of following reaction, (a)
KOH 2 / CH3–CH2–OH I
(c)
KOH 2 / (Me)3C–CO–CH3 I
(b)
KOH 2 / Ph–CH2–CO–CH3 I
Section (F) : Miscellanious reactions F-1.
Formaldehyde and malonic ester reacts in presence of ethoxide ion to give K (C8H12O4) ; what is the structure of K ? How can K be converted into ‘L’ (C2H5 OOC)2 CH – CH2 – CH (COOC2H5)2 what would you get if L was subjected to hydrolysis and heat ?
F-2.
3 2 CH3I + Ph3P 2 2 2
F-3. F-4.
CH CH CH CH Li
Ph C O
benzene
Product. Compound (X) with molecular formula C9H10O forms a semicarbazone and gives negative Tollen’s and Iodoform tests. Upon reduction it gives n-propyl benzene. Deduce the structure of X.
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CARBONYL COMPOUNDS | 35
PART - II : OBJECTIVE QUESTIONS * Marked Questions are having more than one correct option. Section (A) : Nucleophilic addition reactions A-1.
Arrange the following compounds in decreasing orders of rate of exchange of O18 with H2O18 (X) CCl3CHO
A-2.
A-3.
A-4.
(Y) CH3CHO
(Z) CH3COCH3
(W) CF3CHO
(A) W > Z > X > Y (B) W > X > Y > Z (C) W > Y > Z > X (D) W > Z > Y > X Which of the following compound has the largest equilibrium constant for the addition of water ? (A)
(B)
(C)
(D)
Acetaldehyde on reaction with sodium hydrogen sulphite produces
(A)
(B)
(C)
(D)
Write the product of following reaction :
NH NH
2 2
(A)
A-5.
(B)
(C)
(D)
O16 || D2O18 CH3 C H
The intermediate is :
(A)
(B)
(C)
(D)
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CARBONYL COMPOUNDS | 36
A-6.
Aromatic carbonyl compounds having molecular formula C8H8O react with NH2OH how many oximes can be formed : (A) 8
(B) 10
(C) 12
(D) 6
CH2–OH A-7.
H OH OCH3
CH3O H H
[ X]
CH2OH Compound (X) in the above reaction. (A) A-8.
(B)
(C)
(D)
The general order of reactivities of carbonyl compounds towards nucleophilic addition reactions (A) H2C = O > (CH3)2 C = O > Ar2C = O > CH3CHO > ArCHO. (B) H2C = O > CH3CHO > (CH3)2 C = O > ArCHO > Ar2C = O. (C) ArCHO > Ar2C = O > CH3CHO > (CH3)2 C= O > H2C = O. (D) Ar2C = O > (CH3)2C = O > ArCHO > CH3CHO > H2C = O.
A-9.
The correct order of reactivity of PhMgBr with following compounds will be. PhMgBr (C6H5)2CO,
CH3 – CH = O ,
(1) (A) 1 > 2 > 3 A-10.
(CH3)2 C = O
(2)
(3)
(B) 2 > 3 > 1
(C) 3 > 2 > 1
(D) 1 > 3 > 2
Cyanohydrin formation constant will be highest for ? (A) O2N
CHO
(B) CH3 – O
CHO
(D)
CHO
CH3 N
(C)
COCH3
CH3
A-11.
The cyanohydrin of a carbonyl compound on hydrolysis gives lactic acid. The carbonyl compound is (A) HCHO (B) CH3CHO (C) CH3COCH3 (D) CH3COCH2CH3
Section (B) : Condensation reactions B-1.
Which of the following will not undergo aldol condensation ? (A) CH3CHO
B-2.
B-3.
(B) CH3CH2CHO
(C) CD3CHO
(D) PhCHO
(X) is the product of cross aldol condensation between benzaldehyde (C6H5CHO) and acetone What is its structure ?
(A)
(B) C6H5–CH=C–(CH3)2
(C) C6H5–CO–CH2–C=(CH3)2
(D) None of these (buesa ls dksbZ ugha)
In which of the following compounds the methylene hydrogens are the most acidic ? (A) CH3COCH2CH3 (B) CH3CH2COOC2H5 (C) CH3CH2CH(COOC2H5)2 (D) CH3COCH2CN.
IDEAL 21ST CENTURY COMPETITIONS www.21stideal.com
CARBONYL COMPOUNDS | 37 B-4.
(1) CH3COONa A B PhCHO + (CH3CO)2O HBr
( 2) hydrolysis,
The product B is : (A) PhCH = CHCH2Br B-5.
(B)
(D) PhCH = CH – COBr
In the given reaction the product is : H2O
+
B-6.*
(C) PhCH2 CH(Br) COOH
(A)
(B)
(C)
(D)
The compounds that undergo Aldol condensation is :
(A)
(B)
(C)
(D)
Section (C) : Cannizzaro's reactions C-1.
C-2.
Cannizzaro reaction does not take place with (A) (CH3)3CCHO.
(B)
In the reaction,
NaOH A + B. (CH3)3CCHO + HCHO heat
the products (A) and (B) are respectively : (A) (CH3)3CCH2OH and HCOO– Na+. (C) (CH3)3CCH2OH and CH3OH.
(C)
(D) CH3CHO.
(B) (CH3)3CCOONa and CH3OH. (D) (CH3)3COONa and HCOO– Na+.
KOH
+ HCHO (A) + (B)
C-3.
(A)
(C)
(B)
+ CH3OH
(D) Both (A) and (B),
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CARBONYL COMPOUNDS | 38 C-4.
C-5.
In the cannizzaro's reaction the intermediate that will be the best hydride donar ?
(A)
(B)
(C)
(D)
Product of following reaction is
heat + conc. NaOH ?
(A)
C-6.
(B)
(C)
In the given reaction
(D)
Product is
(A)
(B)
(C)
(D)
Section (D) : Redox reactions
D-1.
O O || || ( X) (CH3)2C = CHCH2 CH2CCH3 (CH3 )2 CHCHCH2 CH2CCH3 | OH O O || || (Z) (CH3 )2 CHCCH 2 CH2 CCH 3
(A) (B) (C) (D)
x H2CrO4 B2H6, THF/H2O2,OH B2H6, THF/H2O2,OH H2CrO4
y B2H6, THF/H2O2,OH H2CrO4 / CrO3 OH OH
z OH / OH / H2CrO4 / CrO3 B2H6, THF/H2O2,OH
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CARBONYL COMPOUNDS | 39
O || Ph – CHO
LiAlH 4 [X], Product (X) in this reaction is :
D-2.
OH /
(A)
(B)
(C)
(D)
Cold dil. aq.
(X)
D-3.*
(Y)
KMnO 4 / OH
(Z)
The products of the above reaction is / are :
(A)
D-4.
(B)
(C)
(D)
What will be the product of the following reaction RCO H 3 ?
(A)
D-5.
(C) PhCH(Me)OCOOMe
(D) None of these
2-Methyhlcyclohexanone is allowed to react with metachloroperbenzoic acid. The major product in the reaction is
(A)
D-6.
(B) Ph C —CH || O
(B)
(C)
(D)
In the following conversion
Which of the following regents is suitable ? (A) NH2NH2,KOH, DMSO
(B) NaBH4
(C) Zn-Hg, concentrated H2SO4
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(D) LiAlH4
CARBONYL COMPOUNDS | 40
D-7.
Above conversion can be achieved by (A) NH2–NH2/ NaOH (C) LiAlH4 (A) NH2–NH2/ NaOH (C) LiAlH4
(B) Zn–Hg/HCl (D) NaBH4. (B) Zn–Hg/HCl (D) NaBH4.
Section (E) : -Halogenation, haloform, -deuteration reactions
E-1.
Br / H 2 Major Product is :
(A)
(B)
(C)
( ) optical inactive
(D) A and C both optical active
O
E-3.*
E-4.*
D
D
E-2.
D
The above conversion is carried out
(A) KOD /D2O, H/, LiAlH4 (C) KOD/ D2O, LiAlH4, H/ Which of the following gives haloform reaction
(B) H// KOD, D2O, LiAlH4 (D) LiAlH4, H,KOD/H2O
(A)
(C)
(B)
(D)
In which of the following reaction deuterium exchange is observed ?
(A)
D O / OH 2 excess
(B)
D O / OH 2
(C)
D O / OH 2
(D)
D O / OH 2
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CARBONYL COMPOUNDS | 41 E-5.*
In which of the following reaction deuterium exchange is not observed ?
(A)
(B)
(C)
(D)
Section (F) : Miscellanious reactions F-1.
The major product formed in the reaction. NaOH
(X) C6H5CHO + CH3NO2 heat
F-2.
(A)
(B)
(C) C6H5CH = CH – NO2
(D)
What is the final product of this sequence of reactions ?
(A)
F-3.
.
(B)
(C)
Ph–CH2–COOEt +
(D)
(A)
(B)
Product B is : (A) Ph–CH2–COOH (B) Ph–CH2–COOEt (C)
EtO
(D) None of these
(1) 2 / OH ( 2)H
(P) (Q) (R)
F-4.*
The products of the above reaction is / are :
(A) F-5.
(B)
(C)
(D)
2-pentanone can be distinguished from 3- pentanone by the reagent ? (A) 2, 4- Dinitrophenyl hydrazine (B)Tollen's reagent (C) I2 and dilute NaOH (D) NaHSO3
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CARBONYL COMPOUNDS | 42
F-6.
+ (C6H5)3P = CHCH3
(A) F-7.*
(B)
(C)
(D)
Which of the following will gives iodoform with NaOI ?
(A)
(B)
(C)
(D)
PART - I : SUBJECTIVE QUESTIONS Section (A) : Nucleophilic addition reactions 1.
Write the product of the following reaction
H + HO – NH2
(a)
(c) Ph – CH2 – CH2 – CHO + + Ph – NH – NH2
(d) 2.
3. 4.
H + NH2 – NH2
(b)
When semicarbazide reacts with a ketone (or aldehyde) to form semicarbazone. Only one nitrogen atom of semicarbazide acts as a nucleophile and attack the carbonyl carbon of the ketone. The product of the reaction consequently is R2–C=N–NH–CONH2 rather than R2–C=NCONH–NH2. What factor account for the fact that two nitrogen atom of semicarbazide are relatively non nucleophile ? Cyclopropanone exists as the hydrate in water but 2-hydroxy ethanal does not exist as its hemiacetal expalin why ? [Claiden] Show how would you accomplish the following synthesis
(a)
(b) Br – CH2 – CH2 – C – CH3 HC C – CH2 – CH2 – C – CH3 || O
|| O
O
O CH2 - Br
(c)
CH2 - Ph
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CARBONYL COMPOUNDS | 43 5.
6.
(a) Cis-1, 2-Cyclopentanediol reacts with acetone in the presence of dry HCl to yield compound K, C8H14O2, which is resistant to boiling alkali, but which is readily converted into the starting material by aqueous acids. What is structure of K ? (b) Trans-1, 2-Cyclopentanediol does not form an analogous compound. Explain why ? On the basis of following reaction answer the following questions. H | * Ph C H CH O + NH2 .NH C NH C* CH2 CH3 | || | Me O Me ( d mixture)
(a) (b)
H
(pure optical isomer )
How many stereoisomers will be formed. How many pair(s) of enantiomers are formed ?
Section (B) : Condensation reactions 7.
Predict the product from claisen condensation of the following pair of esters. +
(b)
+
(c)
8.
EtO
(a)
EtO
2. H2O /
Write the components which on crossed claisen condensation could be used to prepare the following esters.
(a)
9.
1. EtO
+
(b)
(c)
Predict the product for each of the following reactions,
(a)
CH3COO Na CHO + CH3 C O C CH3
OHC
|| O
|| O
excess CN (b)
(c)
CH3CHO + CH2
Pyridine /
CN
/ + CH3COCH2COOEt EtONa EtOH
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CARBONYL COMPOUNDS | 44 10.
Predict the product for each of the following reactions.
(a)
OH /
+
(b)
OH /
OO O || || || KOH (d) C 6H5 CCC 6H5 + C 6H5 CH2 CCH2C6H5
OH /
(c)
+
ethanol
EtO
11.
+ CH2 = CH – C – Ph Product (P) is : || O
Section (C) : Cannizzaro's reactions 12.
Write the product of the following reaction, CHO
CDO
(a)
OH /
+
OMe
CHO
(b)
OMe
CHO
OH /
+
NO2
13.
–
–
OH H OH PhCOCHBr2 C A B
The compound 'C' is : 14. 15.
Glyoxal (CHOCHO) on being heated with concentrated NaOH forms. ( i) conc . NaOH, Product is : + CHO (ii ) H3O | COOH
Section (F) : Miscellanious reactions 16.
The compound C4H8CI2 (A) on hydrolysis gives a compound C4H8O, (B). The compound (B) reacts with hydroxylamine and gives a negative test with Tollen’s reagent. What are (A) and (B) Support your answer with proper reasoning and give the equations of reactions.
17.
A ketone (A) which undergoes haloform reaction, gives compound (B) on reduction with LiAlH4. (B) on heating with conc. H2SO4 gives (C) which forms monoozonide (D). (D) on hydrolysis in presence of zinc dust gives only acetaldehyde. Identify (A), (B and (C). Write reactions involved.
18.
Compound ‘X’ (C5H10O) reacts with KCN in dilute H2SO4 followed by heating with dilute acid , results in compound ‘Y’ which is achiral and gives effervescence with NaHCO3 . Reduction of ‘X’ forms ‘Z’ (achiral) which gives positive Lucas test . X responds negatively to iodoform and Fehling’s tests . Identify ‘X’ , ‘Y’ and ‘Z’ .
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CARBONYL COMPOUNDS | 45
PART - II : OBJECTIVE QUESTIONS (without section in class room but store sections in Main Sheet)
Single choice type Section (A) : Nucleophilic addition reactions
1.
Products Products obtained in the reaction is(A) Diastereomer (B) Racemic mixture
2.
(C) Meso compound
(D) Optically pure enantiomer
CHD 2Mg I Conc.H2 SO 4 Y CH2O X H3 O
In the above reaction compound X & Y respectively will be
3.
OH | (A) CHD CH2 OH , CHO – CHO
(B) CHD2 – CH2 – OH , CHO – CHO
(C) CHD2 – CH2 – OH , CD2 = CH2
(D) CHD CH2 OH , CD2 = CH2 | OH
(i) CH CHO
dry ether B. A 3 + Mg
For the given reaction
(ii ) H3O
product B is : (A)
(B)
(C)
(D)
(1)
( 2)
( )
3
4.
CH2 OH (A) (1) | H (2) Mg / CH2 OH CH2 OH (B) (1) | H (2) Mg / CH2 OH
(3) H C O H / H+ || O (3) CH3 C O Et / H+ || O
CH2 OH + (C) (1) | H (2) Mg / THF (3) CH3 C O Et / H || CH2 OH O
CH2 OH
(D) (1) |
CH2 OH
H (2) Mg / THF (3) H C O H / H+ || O
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CARBONYL COMPOUNDS | 46
5.
In the given reaction,
(X)
(Y)
(Y) is
(A)
(B)
(C)
(D)
6.
7.
Product is
(A)
(B)
(C)
(D)
Consider the following sequence of reactions-. HgSO 4 NH2OH A B + C. PhC CH dil.H2SO 4
The products (A), (B) and (C) are respectively,
(A) PhCHO,
and
(B) PhCH2CHO,
and
(C)
(D)
,
,
and
and
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CARBONYL COMPOUNDS | 47
8.
(i) CH3CH2MgBr (ii ) H3 O Cu, (X) (Y) Ph CH CH C H || O
(A) X is 1, 4-addition product ; Y is Ph C CH CH CH2 CH3 || O (B) X is 1, 2-addition product ; Y is Ph CH CH C CH2 CH3 || O (C) X is 1, 4-addition product ; Y is Ph CH CH C CH2 CH3 || O (D) X is 1, 2-addition product ; Y is Ph C CH CH CH2 CH3 || O
(i) KCN / H SO
2 4 Product is
9.
10.
(ii) LiAlH 4
(A) CH3 CH2 CH CH2NH2 | OH
(B) CH3 – CH2 –
– CH2 – NH – CH3
(C) CH3CH2 CH CN | OH
(D) CH3CH2 CH2 CH – NH2 | OH
Consider the following sequence of reactions : H SO
2 4 Heat B. + H2NOH A Heat
The products (A) and (B) are, respectively : (A)
and
(B)
(C)
and
(D)
and
and
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CARBONYL COMPOUNDS | 48 11.
Compound (X) C9H10O gives yellow coloured ppt with 2,4 DNP but does not give red coloured ppt with Fehling’s solution. (X) on treatment with NH2OH/H+ gives compound (Y) C9H11NO. (Y) when treated with PCl5 gives isomeric compound (Z). (Z) on hydrolysis gives propanoic acid and aniline. What will be the correct structure of (X), (Y) and (Z) ? (A)
;
;
(B)
;
;
(C)
;
(D)
12.
;
;
;
+
Product
(A)
(B)
(C)
(D)
Section (B) : Condensation reactions 13.
In the given reaction sequence B is (A)
(A)
14.
(A)
(B)
(B)
(C)
(D)
(B)
(C)
(D)
(B)
The reactant (A) will be :
(A)
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CARBONYL COMPOUNDS | 49
15.
O CH3 || | KOH, H2O Product (C H O) : (mRikn (C H O) CH3 CCH2 CCHO 7 10 7 10 | CH3
(A)
(B)
(C)
(D)
16.
17.
(A)
(B)
(C)
(D)
In the following reaction, A + B
A and B respectively are :
(A)
+ CH3COOC2H5
(B)
+ CH3– C – OC2H5 || O
(C)
+ C2H5O– C –OC2H5 || O
(D)
+ C2H5O– C –CH2– C –OC2H5 || || O O
18.
(X) Identify.
(A)
(B)
(C)
(D)
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CARBONYL COMPOUNDS | 50
CHO
O 19.
+
OH / Product is
COOH
OH
O
(A)
(B) Ph – CH
(C)
Ph – CH
(D)
OH
Section (C) : Cannizzaro's reactions 20.
The suitable reagent for the following reaction is :
(A) LiAlH4
21.
(B) Na / C2H5OH
(C) H2 / Ni
(D) CH2 = O /
Which of the following can be the product/s of following reaction.
Conc. NaOH
(A) I, II, IV
(B) III, IV
(C) II, V
(D) I, V
IDEAL 21ST CENTURY COMPETITIONS www.21stideal.com
+
CARBONYL COMPOUNDS | 51
Section (F) : Miscellanious reactions ( i) CH ( COOC H ) / Pyridine
5 2 2 2 (P) is ( ii) H O,
22.
2
( i) CH ( COOC H ) / fifjMhu
5 2 2 2 (P) gS ( ii) H O, 2
(A)
(B)
(C)
(D)
More than one choice type Section (A) : Nucleophilic addition reactions Which of the following correctly indicate(s) the rate. OH
CN –
CH CHO 3 KCN + HCl OH ( very fast)
Me –
– –
(A) CH3CH
–
23.
– CH
CN
CH CH O
H2O/H+ (B) (hydrate) 3
hydrate
( fast )
HCN / OH CH3MgBr Adduct formation (C) Cyanohydrin C2H5CHO slow fast
(D) All of these
24.
Observe the following reaction
Products.
The correct statement is (A) The product is a mixture of two compounds (B) The product is optically inactive (C) The product is a mixture of two chiral and one achiral stereoisomer (D) The product is a mixture of three stereoisomers.
IDEAL 21ST CENTURY COMPETITIONS www.21stideal.com
CARBONYL COMPOUNDS | 52 25.
The following conversion is/are possible by Ph – CH2 – CH = O
H O ,
KCN / NH Cl
KCN / H O
NH ,
HCN / NaOH
2 NH3 3
3 3 2 (A)
(C) Br / CH COOH
SOCl NH
H O ,
4 3 (B)
H O,
CrO / H
2 3 3 3 (D)
26.
In the given reaction which one of the following statement is correct –
(A) Oxime may be E/Z. (B) Amide on hydrolysis gives a mixture acetic acid, benzoic acid, Aniline and methylamine. (C) Preparation of oxime is nucleophilic addition followed by elimination reaction. (D) Oxime and amides are isomers. 27.
In the given reaction CHCH (A) X is CH3CHO
28.
29.
(X)
(Y)
(B) Y is
(Z)
(C) Z is
(D) Z is CH3COOH
A compound (A) C5H10O forms a phenyl hydrozone and gives negative tollen's and positive iodoform test compound can be : (A)
(B)
(C)
(D)
Which of the following will give 3–pentanone. (1) CH3 – CH2 – MgBr (B) CH3 – CH2 – C N
(A) (CH3 – CH2COO )2 Ca
( 2 ) H3 O
O O || || (1) H O / H (C) CH 3 – C – CH2 – CH 2 – C – OC 2H 5 2 ( 2) NaOH( cao)
(D)
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CARBONYL COMPOUNDS | 53
30.
W
X
Y
Z
Product (Y) and (Z) is / are :
(A)
31.
(B)
(C)
Which of the following reactions is / are correct to get cinnammic acid : (CH CO) O / CH COONa
3 2 3 (A) C6H5-CH=O
CH COOC H / OH –
3 (C) C6H5-CH=O 25
32.
(D)
CH (COOC H )
2 2 5 2 (B) C6H5-CH=O
Pyridine
(i)BrCH2 – COOC 2H5 / Zn /
(D) C6H5-CH=O (ii) H2O / H /
Identify product in the following reaction sequence CH3MgBr OD / D2O SOCl2 (Y) Z CH3I + (W) (X) 1 equivalent (excess)
(A)
(B)
(C)
(D)
IDEAL 21ST CENTURY COMPETITIONS www.21stideal.com
CARBONYL COMPOUNDS | 54
MIXED TYPE PROBLEMS PART - I : MATCH THE COLUMN 1.
Match the column Column I
Column II
(i ) OH (A) MeCO(CH2)4COMe ( ii) H /
(p)
(B) (CH2)4
OEt
(q)
CH COO –
(C) PhCHO +
3
(r) (Ph)2 C — COO – | OH
OH (D) PhCOCOPh
2.
(s)
Column-I
Column-II
Aldol product
Reactant required
(A)
(p)
(B)
(q) HCHO
(C)
(r) Ph–CHO
(D) Ph–CH=CH–CHO
(s) CH3–CHO
3.
Column-I
Column-II
(A)
(p)
1, 4-addition
(B)
(q)
Tautomerism
(r)
AgNO3 / NH4OH
(s)
2, 4 DNP test
(C) (D)
CH3–CH=CH–CH = CH2
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CARBONYL COMPOUNDS | 55
PART - II : COMPREHENSION Read the following passage carefully and answer the questions.
Comprehension # 1 Aldehydes and Ketones reacts with NH2OH to form Aldoximes and Ketoximes respectively. Configuration of these can be determined by Beckmann rearrangement as that group migrates which is anti w.r.t –OH.
R'
..
C
4.
R
..
R
C
N OH
N
H2 O R' C N R R'C N R (II) ( III)
R'
R'C NH R || O
| OH
OH2
H O
H SO4 2 Ph – NH2 + [X] 2
Identify the configuration of [X] compound :
(A) 5.
(B)
. (C)
(D)
Which step is Rate determening step ? (A) I
(B) II
(C) III
(D) IV
(C)
(D)
(1) H SO
2 4 NH2OH (A) (B).
6.
( 2) hydrolysis
The product (B) is :
(A)
(B)
Comprehension # 2 Carbonyl compound which contains –H gives aldol condenation reaction in presence of alkaline medium. The reaction between two molecules of acetaldehyde take place as follows in presence of base.
7.
Aldol condensation reaction is given by (A) C6H5–CHO
(B) CX3 – CHO
(C) O2N
CHO
(D) C6H5–CH2–CHO
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CARBONYL COMPOUNDS | 56
8.
9.
Product (mRikn)
+ (A) Ph – CH = CH – (CH2)5 – CHO
(B) Ph – (CH2)5 – CH = CH – CHO
(C)
(D)
Intramolecular aldol condensation reaction is given by (A) 1,4 diketone
(B) 1,6 diketone
(C) 1,5 and 1,7 diketone
(D) All of these
Comprehension # 3 The conversion of aldehyde having no alpha hydrogen to a mixture of carboxylic acid and primary alcohol is known as cannizzaro reaction. The most important feature of this reaction is the conjugate base of hydrate of aldehyde.
10.
Which step is rate determining step (A) step I
11.
(B) step II
(C) step III
(D) step I and II both
Aldehyde having -Hydrogen does not give cannizzaro reaction because (A) Such aldehyde will form enolate ion (B) Such aldehyde will not form conjugate base of hydrate of aldehyde (C) Such aldehyde may undergo polymerisation (D) All of these
12.
In the given reaction final product is (nh xbZ vfHkfØ;k eas vfUre mRikn gaS )
(A) CH3OH +
(B) HCOONa +
(C) HCOONa +
(D) HCOONa +
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CARBONYL COMPOUNDS | 57
Comprehension # 4 Perkin reaction is the condensation reaction between aromatic aldehyde and aliphatic acid anhydride in the presence of sodium or potasium salt of the acid of the corresponding anhydride to yield , unsaturated aromatic acid. Ph–CHO + (CH3CO)2O
Ph–CH=CH–COOH + CH3–COOH
Mechanism CH3–COONa
CH3–COO + Na
CH3–COO +
CH3–COOH +
H O/
2 Product is
13.
(A)
(B)
(C)
(D)
( i) H O /
2
14.
(ii) H
(A)
(B)
(C)
(D)
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CARBONYL COMPOUNDS | 58
15.
H O/ 2 Product is
+ (CH3–CH2CO)2O
(A)
(B)
(C)
(D)
Comprehension # 5 Ester having -hydrogen on treatment with a strong base eg. C2H5ONa undergoes self condensation to produce -keto esters.This is called claisen condensation. C2H5O– + H – CH2COOC2H5
C2H5OH +
16.
+
(A)
+ C2H5OH
(C)
17.
+ C2H5OH
+
(B)
(D) None of these
product is
(A)
(B)
(C)
(D) Ph – CH2 – COOH
(A)
(B)
(C)
(D)
18.
– COOH
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+ C2H5OH
CARBONYL COMPOUNDS | 59
Comprehension # 6 Study following mechanism of haloform reaction. H | R C C H + HO || | O H
a
R–C–C–H || | O H
X | R C C H (1) HO, (2) I2 || | C O H
I–I b
19.
- d (1) HO (2) I2
Which step is RDS (A) a
20.
X | R C C X || | O X
HOe
X f CHX3 R – C – O R C OH + -CX3 || || O O
(B) b
(C) c
(D) f
Which of the following compounds gives haloform reaction ? (A) CH3 C CH2 C CH3 || || O O
(B) CH3 C CH2 NO 2 || O O C – CH3
(C) CH3 – C – CH2 – CH2 – C || || O O 21.
(D) O
O
Which step produces most acidic compound (A) a
(B) c
(C) d
(D) b
Comprehension # 7 CH2(COOEt)2 + CH3CHO
K (C9H14O4)
(M) (anhydride) 22.
L (C16H26O8)
(C6H10O4)
The product ‘K’ is (A)
(B)
(C)
(D)
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X | RCC X || | O H
CARBONYL COMPOUNDS | 60 23.
24.
The product ‘L’ is (A)
(B)
(C)
(D)
The product ‘M’ is
(A)
(B)
(C)
(D)
PART - III : ASSERTION / REASONING DIRECTIONS : Each question has 5 choices (A), (B), (C), (D) and (E) out of which ONLY ONE is correct. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. (E) Statement-1 and Statement-2 both are False. 25.
Statement-1 : Aldehyde and ketone undergo nucleophilic addition reaction with carbon nucleophile and underog nucleophilic addition-elimination reaction with nitrogen Nucleophile. Statement-2 : Addition of nucleophile on aldehyde and ketone form tetrahedral intermediate, in case of tetrahedral intermediate of nitrogen nucleophile non bonding electrons are present on nitrogen which cause water molecule to eliminate while in case of carbon and hydrogen nucleophile non bonding electron are not present.
26.
Statement-1 : The rate of addition reaction of alcohol on aldehyde can be increased by adding small amount of base. Statement-2 : Addition of alcohols to an aldehyde form acetal.
27.
Statement-1 : Compound II is more reactive towards nucleophilic addition reaction. Statement-2 : Cyclic ketones are more reactive than acyclic ketone due to less steric hinderance and compact structure of cyclic ketone.
28.
Statement-1 : NaHSO3 is used in seperation and purification of aldehydes. Statements-2 : NaHSO3 is reducing agent.
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CARBONYL COMPOUNDS | 61 29.
Statement -1 : Dehydration of aldol takes place by the following mechanism.
Statement -2 : It is due to acidity of – H and stability of conjugated double bond 30.
Statement-1 :
(cinnamaldehyde) fails to undergo aldol condensation
Statement 2: This is due to the fact that cinnamaldehyde does not have acidic – H
PART - IV : TRUE / FALSE 31.
Acid or Base catalysed halogenations of Acetaldehyde follows the reactivity order as Cl2 = Br2 = I2
32.
High Alcohol and low water concentration shifts the aldehyde-hemiacetal and hemiacetal-acetal equilibria towards acetal. OH
33.
Conc. KOH changes Ph–C–C–Ph to Ph–C–COO O O
34.
(CH3)2CuLi changes
Ph
to
PART - V : FILL IN THE BLANKS 35.
Fehling solution (A) consists of aquous solution of copper sulphate, while Fehling’s solution ‘B’ consists of an alkaline solution ............
36.
The structure of the intermediate product, formed by the oxidation of toluene with CrO3 and acetic anhydride, whose hydrolysis gives benzaldehyde is ...........
37.
Ethanol vapour is passed over heated copper and the product is treated with aqeuous NaOH. The final product is ..........
38.
The conversion of carbonyl compound into alkane by reduction is termed as ............
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CARBONYL COMPOUNDS | 62
PART - I : IIT-JEE PROBLEMS (PREVIOUS YEARS) * Marked Questions are having more than one correct option. 1.
Identify (A), (B), (C), (D) and (E) in the following schemes and write their structures : Br / CCl 4 2
[JEE 2001, 5/100]
HgSO / H2 SO 4 NaNH2 A C B 4
C
N H 2NHCONH 2 D
C
N aOD / D 2O ( excess ) E
2.
An alkene A (C16H16) on ozonolysis gives only one product B (C8H8O). Compound B on reaction with NaOH/ I2 yields sodium benzoate. Compound B reacts with KOH/NH2NH2 yielding a hydrocarbon C. (C8H10). Write the structure of compound B and C based on this information two isomeric structures can be proposed for alkene A. Write their structures and identify the isomer which on catalytic hydrogenation (H2/Pd-C) gives a racemic mixture. [JEE 2001, 5/100]
3.
A mixture of benzaldehyde and formaldehyde on heating with aqueous NaOH solution gives : (A) Benzyl alcohol and sodium formate
(B) Sodium benzoate and methyl alcohol
(C) Sodium benzoate and sodium formate (D) Benzyl alcohol and methyl alcohol [JEE 2001, 1/35] 4.
1-Propanol and 2-Propanol can be best distinguished by
[JEE 2001, 1/35]
(A) oxidation with alkaline KMnO4 followed by reaction with Fehling solution (B) oxidation with acidic dichromate followed by reaction with Fehling solution (C) oxidation by heating with copper followed by reaction with Fehling solution (D) oxidation with concentrated H2SO4 followed by reaction with Fehling solution 5.
Write structures of the products A, B, C, D and E in the following scheme. Cl / FeCl
Na Hg / HCl
[JEE 2002, 5/60]
HNO / H SO
2 4 3 B 3 C 2 A
H / Pd / C
2 E D
6.
Five isomeric para-disubstituted aromatic compounds A to E with molecular formula C8H8O2 were given for identification. Based on the following observations, give structures of the compounds. (i) Both A and B form a silver mirror with Tollen’s reagent; also, B gives a positive test with FeCl3 solution. (ii) C gives positive iodoform test. (iii) D is readily extracted in aqueous NaHCO3 solution. (iv) E on acid hydrolysis gives 1, 4–dihydroxybenzene.
7.
[JEE 2002, 5/60]
A biologically compound Bombykol (C16H30O) is obtained from a natural source. The structure of the compound is determined by the following reaction. (a)
On hydrogenation, bombykol gives a compound A (C16H34O). Which react with acetic anhydride to give an ester.
(b)
Bombykol also react with acetic anhydride to give another ester. Which on oxidative ozonolysis (O3/ H2O2) gives a mixture of butanoic acid, oxalic acid and 10 acetoxy decanoic acid.
Determine the number of double bond in Bombykol. Write the structure of compound (A) and Bombykol. How many geometrical isomer are possible for Bombykol. [JEE 2002, 5/60]
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CARBONYL COMPOUNDS | 63 8.
Compound 'A’ (molecular formula C3H8O) is treated with acidified potassium dichromate to form a product 'B' (mol. Formula C3H6O). 'B' forms a shining silver mirror on warming with ammonical AgNO3 . 'B' when treated with an aqueous solution of H2NCONHNH2. HCl & sodium acetate gives a product 'C'. Identify the structure of 'C’. [JEE 2002, 3/90]
(A) CH3CH2CH = NNHCONH2
CH3 | (B) CH3 C N CONHNH2
(C) CH3 – CH = N.NH – CONH2
(D) CH3CH2CH = NCONHNH2
9.
Compound A of molecular formula C9H7O2Cl exists in ketoform and predominantly in enolic form ‘B’. On oxidation with KMnO4, ‘A’ gives m-chlorobenzoic acid. Identify ‘A’ and ‘B’. [JEE 2003, 2/60]
10.
HCl A (C6H12)
[JEE 2003, 4/60]
BC (C 6H13 Cl)
alc KOH
D (isomer of A) B ozonolysis
D E (it gives negative test with Fehling solution but responds to iodoform test). ozonolysis
A F + G (both give positive Tollen’s test but do not give iodoform test.) conc. NaOH
HCOONa + a primary alcohol. F + G Identify to A to G.
(i) NaOH / 100 C
11.
12.
Major Product is
[JEE 2003, 3/84]
(ii) H / H2O
(A)
(B)
(C)
(D)
In the following reaction
[JEE 2005, 3/84] CH COONa heat
3
O || (A) CH2 C O H | Br
13.
O O || || (B) H C C O H
, X can be :
O || (C) CH3 C O H
(D) (CH3CO)2 O
In conversion of 2-butanone to propanoic acid which reagent is used. (A) NaOH, NaI /
(B) Fehling solution
(C) NaOH, I2 /
[JEE 2005, 3/84] (D) Tollen's reagent
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CARBONYL COMPOUNDS | 64 14.
Cyclohexene on ozonolysis followed by reaction with zinc dust and water gives compound E. Compound E on further treatment with aqueous KOH yields compound F. Compound F is : [JEE 2007, 3/162]
CHO
(A) 15.
CHO
(B)
(C)
CO2H
(D)
CO2H
Match the compounds/ions in Column I with their properties/reactions in Column II. [JEE 2007, 8/162] Column I
Column II
(A)
C6H5CHO
(p)
gives precipitate with 2,4 dinitrophenylhydrazine.
(B)
CH3C CH
(q)
gives precipitate with AgNO3 .
(r)
is a nucleophile.
(s)
is involved in cyanohydrin formation.
(C) (D) 16.
COOH
–
CN –
In the following reaction sequence, the correct structures of E, F and G are
[E]
[JEE 2008, 3/163]
[F] + [G]
(*implies 13C labelled carbon)
(A) E =
F=
G = CHI3
(C) E =
F=
G = CHI3
(B) E =
F=
G = CHI3
(D) E =
F=
G = CH3
*
Write Up A tetriary alcohol H upon acid catalysed dehydration gives a product I. Ozonolysis of I leads to compounds J and K. Compound J upon reaction with KOH gives benzyl alcohol and compound L, whereas K on reaction with KOH gives only M.
M= 17.
18.
Compound H is formed by the reaction of
(A)
+ PhMgBr
(B)
(C)
+ PhCH2MgBr
(D)
+ PhCH2MgBr
+
The structure of compound I is
(A) 19.
[JEE 2008, 4/163]
[JEE 2008, 4/163]
(B)
(C)
(D)
The structure of compounds J, K and L respecitvely, are : –
[JEE 2008, 4/163]
(A) PhCOCH3, PhCH2COCH3 and PhCH2COO K (B) PhCHO, PhCH2CHO and PhCOO–K+ (C) PhCOCH3, PhCH2CHO and CH3COO–K+
+
(D) PhCHO, PhCOCH3 and PhCOO–K+
IDEAL 21ST CENTURY COMPETITIONS www.21stideal.com
CARBONYL COMPOUNDS | 65 Write Up A carbonyl compound P, which gives positive iodoform test, undergoes reaction with MeMgBr followed by dehydration to give an olefin Q. Ozonolysis of Q leads to a dicarbonyl compound R, which undergoes intramolecular aldol reaction to give predominantly S. 1. O
1. OH
2. Zn, H2O
2.
1. MeMgBr 3 S R P Q 2. H , H2O 3 . H2SO 4,
20.
The structure of the carbonyl compound P is :
(A) 21.
(B)
,
(C)
(D) [JEE 2009, 4/160]
(B)
,
(D)
,
,
The structure of the product S is :
(A)
23.
(C)
The structures of the products Q and R, respectively, are :
(A)
22.
[JEE 2009, 4/160]
(B)
[JEE 2009, 4/160]
(C)
(D)
Match each of the compounds given in Column I with the reaction(s), that they can undergo, given in column II. [JEE-2009, 8/160] Column I
Column II
(A)
(p)
Nucleophilic substitution
(B)
(q)
Elimination
(C)
(r)
Nucleophilic addition
(D)
(s)
Esterification with acetic anhydride
(t)
Dehydrogenation
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CARBONYL COMPOUNDS | 66 24.
In the scheme given below, the total number of intramolecular aldol condensation products formed from 'Y' is: [JEE 2010, 3/163] 1. O
3 2. Zn, H2O
Y
1. NaOH( aq ) 2. heat
Comprehension Two aliphatic aldehydes P and Q react in the presence of aqueous K2CO3 to give compound R, which upon treatment with HCN provides compound S. On acidification and heating, S gives the product shown below :
25.
The compounds P and Q respectively are :
(A)
(C)
26.
[JEE 2010, 3/163]
and
(B)
and
and
(D)
and
The compound R is :
[JEE 2010, 3/163]
(A)
(B)
(C)
(D)
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CARBONYL COMPOUNDS | 67 27.
The compound S is :
[JEE 2010, 3/163]
(A)
(B)
(C)
(D)
PART - II : AIEEE PROBLEMS (PREVIOUS YEARS) 1.
2.
On vigorous oxidation by permangnate solution (CH3)2C = CHCH2CHO gives (A) (CH3)2CO and OHCCH2CHO
(B) (CH3 )2 C CHCH2CHO | | OH OH
(C) (CH3)2CO and OHCCH2COOH
(D) (CH3)2CO and CH2(COOH)2
Maximum dehydration takes place in : O ||
OH
(A)
O ||
(B)
(C)
6.
(D)
OH
OH
(B) Butan-2-one
(C) Acetamide
(D) Acetic acid
Which one of the following undergoes reaction with 50% sodium hydroxide solution to give the corresponding alcohol and acid ? [AIEEE-2004] (A) Phenol
5.
CH3
Which one of the following is reduced with Zn, Hg and HCl acid to give the corresponding hydrocarbon? [AIEEE-2004] (A) Ethyl acetate
4.
[AIEEE-2002]
O || OH
3.
[AIEEE-2002]
(B) Benzoic acid
(C) Butanal
(D) Benzaldehyde
The best reagent to convert pent-3-en-2-ol into pent-3-ene-2-one is
[AIEEE-2005]
(A) Pyridinium chloro-chromate
(B) Chromic anhydride in glacial acetic acid
(C) Acidic dichromate
(D) Acidic permanganate
The reaction
[AIEEE-2005] O
R – C + NuX
O R–C + X Nu
is fastest when X is : (A) OCOR
(B) OC2H5
(C) NH2
(D) Cl
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CARBONYL COMPOUNDS | 68 7.
Reaction of cyclohexanone with dimethylamine in the presence of catalytic amount of an acid forms a compound if water during the reaction is continuously removed. The compound formed is generally known as [AIEEE-2005] (A) an amine
8.
9.
(B) an imine
(C) an enamine
(D) a Schiff's base
The decreasing order of the ratio of HCN addition to compounds A to D is
[AIEEE-2006]
(a) HCHO
(b) CH3COCH3
(c) PhCOCH3
(d) PhCOPh
(A) d > b > c > a
(B) d > c > b > a
(C) c > d > b > a
(D) a > b > c > d
In the following sequence of reactions,
[AIEEE-2007, 3/120]
Mg
P 2 H O HCHO B A Ether CH3CH2OH C 2 D. The compound ‘D’ is
(A) n-propyl alcohol 10.
(B) propanal
(C) butanal
(D) n-butyl alcohol
In the following sequence of reactions, the alkene affords the compound 'B'
[AIEEE-2008, 3/105]
O3 H 2O CH3CH CHCH3 A B .The compound B is Zn
(A) CH3COCH3 11.
(B) CH3CH2COCH3
(C) CH3CHO
(D) CH3CH2CHO
In Cannizzaro reaction given below
[AIEEE-2009, 4/144]
.. :O H 2Ph CHO PhCH2OH + PhCO2
the slowest step is : (A) the transfer of hydride to the carbonyl group
(B) the abstraction of proton from the carboxylic group
(C) the deprotonation of PhCH2OH
(D) the attack of : O H at the carboxyl group
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CARBONYL COMPOUNDS | 69
EXERCISE - 1 PART - I A-1.
The reactivity of carbonyl group towards nucleophilic reagent depends upon the extent of positive charge present on the carbon. In ketones two alkyl group attach to the carbon reduce the positive charge on the carbonyl carbon due to their electron repelling nature while in aldehyde the positive charge on carbonyl carbon is reduced to lesser extent because only one alkyl group is present. Also, In ketones steric hinderance is more than aldehyde.
A-2.
IV > III > II > I
A-5.
c>b>a>d
A-6.
(a)
A-3.
IV > III > I > II
A-4.
d>b>c>a
HCN / NaOH
or KCN / dil.H2SO4
(b)
(1) HCN / NaOH
( 2) H2O / H
A-7.
Cyanohydrin is formed by nucleophilic attack on carbonyl group (C=O), 2, 2, 6-trimethylcyclohexanone has more steric crowding and three methyl groups in tri methyl cyclohexanone which nearly neutralise the positive charge on carbon hence CN– nucleophile does not react with 2,2,6-trimethylcyclohexanone.
A-8.
(a)
A-9.
(a) CH3–CHO +
A-10.
(a) Ph CH SO3Na ; (b) C 2H5 C NHCH3 ; (c) CH3 C CH3 ; (d) || || | O18 OH O
+ CH3–CHO
(b) H2N – CH2 – CH2–CH2–CH2–CO–CH3
(b)
+
Section (B) : Condensation reactions Ph B-1.
(a) CH3 – CH2 – CH C – CHO | CH3
B-2.
(a) CH3CH2CH2CHO ;
B-3.
[X] =
B-4.
CH3 | (a) CH3 C CH CH – C H ; | || CH3 O
(b) Ph – CH2 – CH C – CHO
(c) Ph–CO–CH=C
| Ph
CH3
(b) CH3COCH3 , Four stereoisomers are formed.
O || (b) Ph CH C C H ; | CH3
(c) Ph CH CH C
|| O
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CARBONYL COMPOUNDS | 70
B-5.
B-6.
(a) CH3CH2CH2COOEt ; (b) PhCH2COOCH3 ; (c) (CH3)2CHCH2COOEt
Section (C) : Cannizzaro's reactions C-1.
(P) =
(Q) = D–CH2–OH
C-2.
CH2 OH | HO CH2 C CH2 OH + HCOO–Na+ | CH2 OH
Section (D) : Redox reactions D-1.
(a)
(p)
, (q)
O
, (r) CH3 – CH2 – CH2OH
O
(b)
COOH | D-2.
(a)
;
(b)
;
|| O COOH |
(c)
D-3.
;
(d)
| OH
OH
(A) CH2=CH–CHO
(B)
(C)
(D)
Section (E) : -Halogenation, haloform, -deuteration reactions
E-1.
Ph Br
E-2.
O ||
C6H5 CH3
+
Br
O ||
Ph
(a)
CHI3 + HCOONa
(c)
CHI3 + (Me)3C–COONa
C6H5
, (Diastereomer).
CH3
(b)
CHI3 + PhCH2COONa
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CARBONYL COMPOUNDS | 71 Section (F) : Miscellanious reactions F-1.
(K)
F-2.
Ph2C = CH2
, Product is HOOC — (CH2)3 — COOH.
F-3.
F-4.
X
.
PART - II Section (A) : Nucleophilic addition reactions A-1. (B) A-2. (C) A-3. (D) A-6. (B) A-7. (B) A-8. (B) A-11. (B) Section (B) : Condensation reactions B-1. (D) B-2. (A) B-3. (D) B-6.* (A, B, D) Section (C) : Cannizzaro's reactions C-1. (D) C-2. (A) C-3. (A) C-6. (B) Section (D) : Redox reactions D-1. (B) D-2. (A) D-3.* (A, B, C) D-6. (A) D-7. (B) Section (E) : -Halogenation, haloform, -deuteration reactions E-1. (A) E-2. (C) E-3.* (A, B, C) Section (F) : Miscellanious reactions F-1. (C) F-2. (B) F-3. (A) F-6. (B) F-7.* (A, B, C)
A-4. A-9.
(A) (B)
A-5. A-10.
(B) (A)
B-4.
(B)
B-5.
(D)
C-4.
(C)
C-5.
(C)
D-4.
(A)
D-5.
(C)
E-4.*
(A, B, C)
E-5.*
(A, C)
F-4.*
(A, C, D)
F-5.
(C)
EXERCISE - 2 PART - I Section (A) : Nucleophilic addition reactions
1.
(a)
(b)
N – NH – Ph || (d) Ph – C – Ph
2.
3.
(c) Two nitrogen atoms of semicarbazide that are adjacent to the (C=O) group have their lone pair present in conjugation so nucleophilicity decreases. H O
2
In cyclopropanone angle strain decreases after addition of water, while cyclic hemiacetal structure of 2-hydroxy ethanal will develope angle strain.
H2O HO–CH2–CHO
Hemiacetal (Not form due to angle strain)
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CARBONYL COMPOUNDS | 72
4.
CH2 OH | H CH2 OH
(a)
(i) LiAlH
4
(ii) H
CH2 OH | H CH2 OH
O
O
(b) Br – CH2 – CH2 – C – CH3 Br – CH2 – CH2 – C – CH3 || O
(i) HCCNa
HC C – CH2 – CH2 – C – CH3 (ii) H || O
O
O CH2 - Br
(c)
5.
(a)
6.
(a) 4
CH2 OH | H CH2 OH
O
O
(i) PhMgBr CH2–Br
CH2 - Ph
(ii) H2O
(b) The – OH groups in the trans isomer are too far apart for the cyclic structure to form.
(b) 2
Section (B) : Condensation reactions 7.
(a)
(b)
8.
(a) Ph CH2 C OCH3 + H C OR || || O O
(c) CH3CH2COOH
(b)
+ Ph CH2 C OCH3 || O
(c) Ph C OR + CH3 CH2 C OC 2H5 || || O O 9.
(a)
HOOC–CH=HC
(b)
CH3–CH = C
CH=CH–COOH CN CN
(c)
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CARBONYL COMPOUNDS | 73
10.
(a)
; (b)
; (c)
; (d)
11.
Section (C) : Cannizzaro's reactions D CH2–OH
CH–OH
12.
(a)
+
(b)
+
OMe
13.
PhCH(OH)COOH
15.
14.
HO – CH2 – COONa.
+
Section (F) : Miscellanious reactions 16.
(i) (B) reacts with hydroxyl amine and thus (B) has carbonyl group, i.e., > C = O group kettonic either or aldehyde (—CH = O ) group. (ii) (B) gives negative test with Tollen’s reagent and thus (B) is ketone. (iii) (B) is obtained by hydrolysis of (A) and thus both CI atoms should be on same carbon atoms, i.e., (A) must be gem dihalide as well as not on terminals. Thus, only possibility of (A) is CH3CCI2CH2CH3.
(i)
(ii)
(iii)
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CARBONYL COMPOUNDS | 74 17.
(A)
(B)
CH3CHOHCH2CH2CH3
(Butan-2-ol )
(C)
CH3 CH = CHCH3
(But -2- ene )
18.
,
OH | C 2H5 C C 2H5 , | COOH
C 2H5 CH C 2H5 | OH
( Y ) (achiral)
( Z ) (achiral)
,
OH | C 2H5 C C 2H5 , | COOH
C 2H5 CH C 2H5 | OH
( Y ) ( vfdjSy)
( Z ) ( vfdjSy)
PART - II Section (A) : Nucleophilic addition reactions 1.
(A)
2.
(C)
3.
(B)
4.
(C)
5.
(B)
6.
(B)
7.
(D)
8.
(B)
9.
(A)
10.
(D)
11.
(B)
12.
(A)
15.
(C)
16.
(A)
17.
(C)
26.
(A, B, C, D)
Section (B) : Condensation reactions 13.
(B)
14.
(B)
18.
(C)
19.
(B)
Section (C) : Cannizzaro's reactions 20.
(D)
21.
(C)
Section (F) : Miscellanious reactions 22.
(A)
Section (A) : Nucleophilic addition reactions 23.
(A, C)
24.
(B, C, D)
25.
(A, B, C)
27.
(A, B, C, D)
28.
(A, D)
29.
(A, B, D)
30.
(B, D)
31.
(Aa, B, D)
32.
(A, B, C)
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CARBONYL COMPOUNDS | 75
EXERCISE - 3 1.
(A) - q ; (B) - p ; (C) - s ; (D) - r
2.
(A) - p,q ; (B) - p,r ; (C) - q,s ; (D) - r,s
3.
(A) - p, q, s ; (B) - p, q, r, s ; (C) - p ; (D) - q, r, s
4.
(A)
5.
(B)
6.
(A)
7.
(D)
8.
(D)
9.
(D)
10.
(B)
11.
(D)
12.
(C)
13.
(A)
14.
(A)
15.
(A)
16.
(A)
17.
(D)
18.
(A)
19.
(A)
20.
(C)
21.
(B)
22.
(C)
23.
(C)
24.
(D)
25.
(A)
26.
(D)
27.
(A)
28.
(B)
29.
(A)
30.
(A)
31.
True
32.
True
33.
True
34.
False
35.
Sodium potassium tarterate
36.
C6H5CH(OCOCH3)2
37.
Aldol
38.
Clemmensen’s reduction or Wolf kishner reduction
EXERCISE - 4 PART - I 1.
2.
(A) =
(B) =
(D) =
(E) =
(C) =
or ;k
A=
B = C 6H5 C CH3 || O
C = C6H5 – CH2 – CH3
Trans isomer give racemic mixture. 3.
(A)
4.
(C)
O || CH2 – CH2– CH3
5.
A = Cl
B= Cl
D=
C=
'A'
E=
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CARBONYL COMPOUNDS | 76
6.
(A)
(B)
(C)
(D)
(E)
7.
Structure of Bombykol is CH3 – CH2 – CH2 – CH = CH – CH = CH(CH2)8CH2OH
8.
(A)
9.
10.
11. 15.
A=
B=
C=
D=
E=
F & G are
(B)
12.
(D)
13.
(C)
14.
(A)
19.
(D)
27.
(D)
(A) – (p, q, s) ; (B) – (q) ; (C) – (q, r, s) ; (D) – (q, r)
16.
(C)
17.
(B)
18.
(A)
20.
(B)
21.
(A)
22.
(B)
23.
(A) - p, q, t ; (B) - p, s, t ; (C) - r, s ; (D) - p
24.
1
25.
(B)
26.
(A)
PART - II 1.
(D)
2.
(B)
3.
(B)
4.
(D)
5.
(A)
6.
(D)
7.
(C)
8.
(D)
9.
(A)
10.
(C)
11.
(A)
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CARBONYL COMPOUNDS | 77
PART - I : OBJECTIVE QUESTIONS Single choice type 1.
Which of the following will be product of following reactions ? CH3 | acid + HO CH2 C CH2 OH | CH3 ( excess )
(A)
(C)
R
2. (A) Zn – Hg/HCl 3.
(B)
(D)
above reaction has reagents ‘R’
(B) N2H4/HO–
(C) H2/Ni
(D) NaBH4
Which of the following compounds on reaction with conc. NaOH followed by
(A)
(B)
(C)
(D)
4.
gives following cyclic ester..
H
Conc . OH y x excess /
The major product (y) in the above reaction :
(A)
(B)
(C)
(D)
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CARBONYL COMPOUNDS | 78
H O 3 A + B
5.
6.
7.
8.
9.
10. 11. 12.
13.
Compound (A) and (B) can be differentiated by : (A) 2, 4-DNP (B) Fehling solution (C) Lucas reagant (D) NaHSO3 The formation of cyanohydrin from a ketone is an example of : [JEE 90] (A) Electrophilic addition (B) Nucleophilic addition (C) Nucleophilic substitution (D) Electrophilic substitution The enolic form of acetone contains : [JEE 90] (A) 9 sigma bonds, 1 pi bond and 2 lone pairs (B) 8 sigma bonds, 2 pi bond and 2 lone pairs (C) 10 sigma bonds, 1 pi bond and 1 lone pairs (D) 9 sigma bonds, 2 pi bond and 1 lone pairs m-chlorobenzaldehyde on reaction with conc. KOH at room temperature gives : [JEE 91] (A) Potassium m-chlorobenzoate and m-hydroxybenzaldehyde (B) m-hydroxybenzaldehyde and m-chlorobenzyl alcohol (C) m-chlorobenzyl and m-hydroxybenzyl alcohol (D) Potassium m-chlorobenzoate and m-chlorobenzyl alcohol. Under Wolf Kishner reduction conditions, the conversions which may be brought about is ? [JEE 92] (A) Benzaldehyde into Benzyl alcohol (B) Cyclohexanol into Cyclohexane (C) Cyclohexanone into Cyclohexanol (D) Benzophenone into Diphenylmethane Hydrogenation of benzoylchloride in presence of Pd and BaSO4 gives [JEE 92] (A) Benzyl alcohol (B) Benzaldehyde (C) Benzoic acid (D) Phenol Among the given compounds, the most susceptible to nucleophilic attack at the carbonyl group is : [JEE 97] (A) MeCOCl (B) MeCHO (C) MeCOOMe (D) MeCOOCOMe An organic compound C3H6O does not give a precipitate with 2, 4-dinitrophenyl hydrazine reagent also does not react with metallic sodium it could be : (A) CH3–CH2–CHO (B) CH3–COCH3 (C) CH2=CH–CH2–OH (D) CH2=CH–OCH3 The product of the reaction (X) will be :
14.
(A) C6H5CH=CH–COOH
(B)
(C)
(D)
In the given reaction sequence Ph–CH2–CHO
(X)
Product (X) will be : (A)
(B)
(C)
(D)
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CARBONYL COMPOUNDS | 79
15.
(B)
+
A and B are :
(A)
,
(B)
(C) Both ‘a’ and ‘b’
(D)
16.
,
In the given reaction
[X]
, [X] is :
(A) 2 Mole 17.
(B)
(D) All of these
Among the following compounds which one will react with acetone to give a product that contains carbonnitrogen double bond ?
(A) C H N HC H 6 5 6 5
18.
(C)
(B) (CH3)3 N
(C)
(B)
(C)
The end product is :
H
(A)
(D)
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(D) C6H5 N H N H2 .
CARBONYL COMPOUNDS | 80
19.
The final product of the following reaction is : O=CH O=CH ..
acid
C 2H 5 N H 2 +
CH3
H3C
CH2
C2H5 | +N ..
(A)
CH
H5C2–N
N ..
(B)
HC
(C)
N–C2H 5 CH
(D)
CH3 CH3
20.
H3C
In the given reaction
CH3
P
P will be :
(A)
21.
(B)
(C)
(D)
A compound (X) on treatment with SOCl2 produces another compound (Y). The latter on hydrolysis yields a mixture of benzoic acid and methylamine. The compound (X) is :
(A)
(B)
(C)
(D)
22.
Above conversion is carried out : (A) X
(C) X
Y
Y
(B) X
Y
(D) (A) and (C) both
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CARBONYL COMPOUNDS | 81 23.
Compound ‘A’ given below can undergo intramolecular aldol condensation reaction.
product formed is :
(A)
24.
(B)
(C)
(D)
Consider following intramolecular aldol condensation reaction.
X X can be : (A)
(B)
(C)
(D)
25.
‘B’ final product ‘B’ is :
(A)
26.
(B)
(C)
(D)
The product of given reaction is : [X]
(A)
(B)
(C)
(D)
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CARBONYL COMPOUNDS | 82 27.
In the given reaction H2O
+
the major product will be
28.
29.
(A)
(B)
(C)
(D) CH3 – CH2 – CHO
Which of the following reaction will give -keto aldehyde as the final product ? (A)
+
(B)
+
(C)
+
(D) All of these (mi;qZä lHkhA)
An organic compound X on treatment with acidified K2Cr2O7 gives compound Y which reacts with I2 and sodium carbonate to form Triodomethane. The compound X can be : O
(A) CH3OH
30.
(D) CH3CH(OH)CH3 .
(B) Hydroxylamine
(C) Fehling solution
(D) Sodium bisulphite.
Which hydrogen of the given compound is least acidic in nature
(A) 1 32.
(C) CH3CHO
C2H5CHO and (CH3)2CO can be distinguished by testing with (A) Phenyl hydrazine
31.
(B) CH3CCH3
(B) 2
(C) 3
(D) 4
In the given reaction sequence
(A)
(B)
Compound (B) is : (A)
(B)
(C)
(D)
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CARBONYL COMPOUNDS | 83 33.
Arrange the following compound in decreasing order of Keq for hydrate formation.
(A) III > II > I > IV
(B) I > II > III > IV
(C) II > III > I > IV
(D) III > IV > II > I
KCN,EtOH Product is : H O
34.
2
35.
(A)
(B)
(C)
(D)
G
H
The products G & H are respectively. (A)
(C) 36.
and
(B)
and
and
(D)
and
In the given reaction , (X) will be :
(A)
(B)
(C)
(D)
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CARBONYL COMPOUNDS | 84
37.
The number of stereoisomeric products formed in the following reaction is
HO
2
(A) 1 38.
(B) 8
(D) 2
What will be the product of the following reaction O O || || H H3 C C (CH2 )2 C CH3
+
(A)
39.
(C) 4
(B)
(C)
(D)
In the Cannizzaro reaction given below :
[JEE 1996]
OH 2Ph – CHO Ph – CH2OH + PhCO2¯
The slowest step is : (A) The attack of
OH at the carbonyl group
(B) The transfer of hydride to the carbonyl group
(C) The abstraction of proton from the carboxylic acid
40.
(D) The disproportionation of Ph – CH2OH
+
(A)
(B)
(C)
(D)
More than one choice type
41.
H O HCN 3 Product
The correct statement about product is (A) The product is optical inactive
(B) The product is meso compound
(C) The product is mixture of two enantiamer
(D) Product exist is two diastereomer forms
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CARBONYL COMPOUNDS | 85 42.
Which of the following reaction correct product is mention :
(A)
(B)
(C)
(D)
43.
NH –NH / OH
2 2
NH –NH / OH
2 2
Zn – Hg / Con.HCl
Which of the following compound gives a positive iodoform test (A) 1 - Phenyl ethanol
44.
Zn – Hg / Con.HCl
(B) 2-pentanone (C) 2- pentanol
(D) Isopropyl alcohol
The sutabile reagent of the following reaction is :
45.
(A) K2Cr2O7 / H
(B) N2H4 / OH
(C) CrO3 / H, Zn-Hg/ Con. HCl
(D) CrO3 / H, N2H4 / OH,
Which of the following aldol reaction product is correctly mention :
(A)
OH + CH3 – CH2 – CHO
OH
(B) PhCHO +
(C) PhCHO + CH3COCH3
(D)
OH
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CARBONYL COMPOUNDS | 86 46.
Observe the following series of reaction. Al (OC H )
2 5 3 Q (CH3)2CHCOO– + (CH3)2CH–CH2OH P Which of the following statement is true (A) The reaction involved are cannizaro reaction followed by acidic hydrolysis (B)The reaction involed are cannizaro reaction followed by alkaline hydrolysis. (C) Compound p is isobutanal and the reaction involved are tischenko reaction followed by alkaline hydrolysis (D) Compound (Q) have fruity smell.
47.
During aldol condensation between excess of formaldehyde and acetaldehyde in basic medium we can get (A) Trihydroxy methyl acetaldehyde (B) Sodium formate (C) CH2= CH – CHO (D) Tetrahydroxymethylmethane
48.
Which of the following is possible combination to prepare 1 -phenyl 1,3 butadiene from Wittig reaction ?
(A)
+ Ph3 P = CH – CH = CH2
(C)
49.
+ Ph3P = CH – CH = CH2
+
(D) Angle strain in carbonyl compound
(B)
(C)
–CHO
(D) –CN
–CHO
Which one of the following -dicarbonyl-compound will not show enolisation ?
(A)
52.
(D)
Which one of the following pair is not correctly matched (A)
51.
+
Stability of hydrates of carbonyl compound depends on (A) Steric hindrance (B) presence of –I group on gemdiol carbon (C) Intramolecular hydrogen bonding
50.
(B)
(B)
(C)
(D)
, unsaturated acids can be prepared by which one of the following reaction ? (A) C6H5–CHO + (CH3CO)2O
(B)
(C)
(D) CH3–CH2–CHO + Br–CH2–COOC2H5
+
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CARBONYL COMPOUNDS | 87 53.
HCCH H / Ni H P 2 Q 2CH2=O R
R is an ether which a good solvent for grignard reagent. Which of the following are correct. (A) P = HO–CH2–CC–CH2OH (B) Q = HO–CH2–CH2–CH2–CH2–OH (C) R =
(D) R =
O
O
54.
The correct statements are : COOH
(A) R =
O O
55.
56.
57.
CH3
(B) S =
O
CH3
(C) Q =
O
(D) Q and T are homologues.
O
Which of the following will undergo aldol condensation : [JEE-98, 3M] (A) Acetaldehyde (B) Propanaldehyde (C) Benzaldehyde (D) Trideuteroacetaldehyde Among the following compounds, which will react with acetone to give a product containing > C = N – : [JEE-98, 3M] (A) C6H5NH2 (B) (CH3)3N (C) C6H5NHC6H5 (D) C6H5NHNH2 The enol form of acetone, after treatment with D2O, gives : [JEE-99, 2M] OD | (A) CH3 C CH2
O || (B) CD3 C CD3
OH | (C) CH2 C CH2D
OD | (D) CD3 C CD3
Assertion-Reason DIRECTIONS :
58. 59.
60. 61. 62.
Each question has 5 choices (A), (B), (C), (D) and (E) out of which ONLY ONE is correct. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. (E) Statement-1 and Statement-2 both are False. Statement-1 : CCl3CHO forms an isolable crystilline hydrate. Statement-2 : Powerfully electron withdrawing chlorine atoms stabilise hydrate by intramolecular H-bonding. Statement-1 : Carbonyl compound are more soluble in water than the corresponding alkanes Statement-2 : H-bonding between carbonyl oxygen and water makes carbonyl compounds more water soluble than hydrocarbon Statement-1 : Acetophenone and benzophenone can be distinguished by Iodoform test. Statement-2 : Acetophenone and benzophenone both are carbonyl compound. Statement-1 : Isobutanal does not give iodoform test. Statement-2 : It does not have –hydrogen. Statement-1 : Acetylene on treatment with alkaline KMnO4 produce acetaldehyde. Statement-2 : Alkaline KMnO4 is a oxidising agent.
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CARBONYL COMPOUNDS | 88 63. 64. 65.
66.
67.
Statement-1 : Acetic acid does not undergo haloform reaction. Statement-2 : Acetic acid has no alpha hydrogen. Statement-1 : Benzaldehyde is more reactive than acetaldehyde towards nucleophilic addition reaction. Statement-2 : In benzaldehyde >C=O group is resonance stabilised by phenyl ring. Statement-1 : Acetal are easily coverted to parent carbonyl compound. This easy interconverision make acetal attractive as protecting group to prevent carbonyl compound. Statement-2 : Acetal are easily hydrolysed in acidic as well as basic medium. Statement-1 : Acetaldehyde react with nitromethane in presence of dil. NaOH to give 1-nitro-2-propanol. Statement-2 : The hydrogen atom of acetaldehyde are more acidic than nitromethane.
Statement-1 : The following conversion
can be done by using NH2–NH2/KOH, and not by Zn-Hg/con. HCl. Statement-2 : Zn.Hg/HCl can affect –OH group and shows substitution reaction.
–
68.
69.
70.
Statement-1 :
Heat
Statement-2 :Polyccarbonyl compound with -H give intramolecular aldol condensation in alkaline medium. Statement-1 : The addition of ammonia derivative to a carbonyl compound is carried out in weakly acidic medium. Statement-2 : In weakly acidic medium attacking nucleophile is also protonated. Statement-1 : HCHO is always oxidized in the crossed cannizzaro reaction. Statement-2 : HCHO is the most reactive aldehyde, it exist in aquous OH solution as the conjugate base of its hydrate
71.
OH
, there is also a statistical factor because HCHO has two aldehydic hydrogen
available for transfer while in other aldehyde hydrate anion has only one such hydrogen atom. Statement-1 : Cannizzaro reaction is given by aldehyde having no -hydrogen atom. Statement-2 : Cannizzaro reaction is redox reaction.
COMPREHENSION Read the following passage carefully and answer the questions. Comprehension # 1 P is an alcohol which on heating with Al2O3 forms an alkene Q. Q on ozonolysis produces R and S. When the mixture of R and S is heated with NaOH, a redox reaction takes place and a mixture of an acid salt and alcohol is formed. 72. The alcohol (P) is obtained by :
(A)
+ CH3MgBr
(B)
(C)
+ CH2 = O
(D)
+ CH3MgBr
+ CH3 MgCl
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CARBONYL COMPOUNDS | 89 73.
74.
The compound (Q) is :
(A)
(B)
(C)
(D)
The compounds R and S are :
(A)
(C)
, CH2 = O
+ CH2 = O
(B)
, CH3 CH = O
(D)
+ CH2 = O
Comprehension # 2 Aldehyde and ketone undergo nucleophilic addition reaction because of polarity betwen >C=O group. The reactivity of carbonyl groups toward nucleophile depends upon the nature of inductive effect of group present at carbonyl carbon. 75. Which among the following is most reactive towards nucleophilic addition reaction. (A) FCH2CHO (B) ClCH2CHO (C) Br CH2CHO (D) I CH2CHO 76. Which of the following is least reactive for nucleophilic addition reaction
(A)
(B)
(C)
(D)
77.
Nucleophilic addition reaction over carbonyl compound is shown by : (A) HCN + dil. NaOH (B) NaHSO3 (C) CH3OH + HCl (D) All of these Comprehension # 3 The conversion of aldehyde having no alpha hydrogen to a mixture of carboxylic acid and primary alcohol is known as cannizzaro reaction. The most important features of this reaction is the conjugate base of hydrate of aldehyde.
78.
Order of the above reaction is : (A) 1 (B) 2
(C) 3
(D) 4
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CARBONYL COMPOUNDS | 90
79.
Write order of best hydride ion donar in cannizzaro reaction
(I)
(II)
(III)
(IV)
(A) III > II > I > IV 80.
(B) II > IV > III > I
(C) II > III > I > IV
(D) III > I > II > IV
Which of the following cannot undergo intramolecular cannizzaro reaction ? (A) H — C — C — H || || O O
(B) H — C — C — Ph || || O O
(C) Ph — C — C — Ph (D) PhC C — C — C — H || || || || O O O O
Comprehension # 4 In presence of excess base and excess halogen a methyl ketone is converted first into a trihalo substituted ketone and then into a carboxylic acid. After the trihalo substituted ketone is formed hydroxide ion attacks the carboxyl carbon because the trihalo methyl ion is the group more easily expelled from the tetrahedral intermediate. The conversion of a methyl ketone to a carboxylic acid is called a haloform reaction because one of the product is haloform (CHCI3) or CHI3 or CHBr3.
O H ( excess ) R – C – CH3 2 ( excess ) || O
81.
R – C – OH + CI3 R – C – O + CH I3 || || O O
Which of the following compound show haloform reaction and racemisation in OD– / D2O.
Me
(A) CH3CH2OH
82.
(B)
Et
O
Me Ph
(C)
(D)
Et | (1) 2 / OH Ph – C – C – C – CH3 product is || | || ( 2 ) H ( 3) O Me O
Me | (A) Ph – C – C – COOH || | O Et
(B) Ph – C – CH – Me || | O Et
(C) Ph – C – O – CH – Et | || Me O
(D) Ph – C – CH – OEt | || O Me
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CARBONYL COMPOUNDS | 91 O Ph O
83.
(1) Na O
product is ( 2 ) H ,
HOOC
(A)
O
Ph O
(B)
(C)
(D)
Comprehension # 5 Following is the mechanism of Benzoin condensation reaction. Answer the following based on it O O | || a Ph C H + CN Ph – C – H | CN
b
H | Ph C C Ph | || OH O
84.
85. 86.
Rate of reaction is given by : (A) r = k [PhCHO]2[CN]1
O || Ph – C – H c
OH | Ph – C | CN
H CN | | Ph – C – C – Ph | | O OH
H CN | | Ph – C – C – Ph | | OH O
e
d
(B) r = k [PhCHO]1[CN]1
(C) r = k [PhCHO]1[CN]0 (D) r = k [PhCHO]2[CN]0 Which step involves Bronsted lowery acid base reaction (A) b, d (B) c, d (C) b, e (D) c, e The benzoin obtained in above reaction when oxidised with Cu / 573 K followed by treatment wth ethanolic KOH gives
(A)
(B) Ph C C OH + Ph C CH2 OH || || || O O O
(C) Ph2CH – COO-
(D) Ph CH CH Ph | | OH OH
Comprehension # 6 Observed the following reaction sequences : Ca(OH)2 LiAlH4 2 / NaOH NaOH / CaO (P) (Q) (R) (chiral carbon) CHI3 + S (acid) (T) CrO Cl
2 2 (U) ( V) 3-phenyl propenoic acid.
CH3COONa
87.
What is the structure of (T) : CH3 (A)
(B)
CH2 – CH3
CHO
CH2 – CHO (C)
(D)
CH3
C – CH3 O
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CARBONYL COMPOUNDS | 92 88.
89.
Which of the following structures is not correct.
(A) (Q) =
(B) (V) = CH3– C –O– C –CH3 || || O O
(C) (S) =
(D) (R) =
What is the structure of (P). CI
CH3
CI
CH – CH2
(A)
(B)
(C)
CCl2
(D)
CH3
CH3
Comprehension # 7 Observe the following reactions and its mechanistic steps and intermediate products.
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CARBONYL COMPOUNDS | 93 Reaction -2 + ACO
+
90.
The intermediate product similar to (IV) following in the reaction-(2) :
(A)
91.
92.
(B)
The leaving group of step (V) in reaction-2 : (A) CH3COO (B) CO2
(C)
(D)
(C) PhCOO
(D) Ph–CH=CH–COOH
If Ph–CH=O18 is used then O18 is traced in reaction -2 : (A) Ph – C – O18 – H || O
(B) CH3 – C – O18 – H || O
(C) Ph – CH CH – C – O18 – H || O
(D) H – O – C – O18 – H || O
PART - II SUBJECTIVE QUESTIONS 1.
Identify A, B, C and D in the following scheme and write their strucute.
2.
Identify A, B, C, D and E in the following scheme and write their strucute.
HgSO
4 B
A
3.
CH3Cl
A
4.
Give the structures from A to D. Give structure for compound A to E.
[5 marks]
dil. H2SO 4
B
C
D. [2 marks]
Cyclohexanol
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CARBONYL COMPOUNDS | 94 5.
Give the structure of compound K to M ? Ethanal
6. 7.
Butanol
Compounds (A) and (B) on reaction in ether medium and subsequent acidififcation and oxidation give 2,5-dimethyl-3-hexanone. What are (A) and (B) ? An organic compound C8H12 decolourised the bromine water and Baeyer’s reagent. On complete bromination, it gave a tetrabromo derivative C8H12Br4 having three asymmetric carbon atoms. On ozonolysis, (A) gives ,
and HCOOH. What is 'A'
8.
9.
10. 11. 12.
(A) on hydration gives (B) which gives positive iodoform test. Also (A) gives white precipitate with Tollen's reagent. (B) is also obtained from acetyl chlori de on treating with benzene in the presence of anhydrous AlCl3 . An alkene C6H12 after ozonolysis yielded two products. One of these gave a positive iodoform reaction but a negative Tollen's test. The other gave a positive Tollen's test but negative iodoform test. The structure of C6H12 is. Compound (A) C5H10O forms phenyl hydrozone and gives negative Tollen’s and Iodoform tests. Compound (A) on reduction gives n-pentane. Give structure of (A). Explain the reactions. The calcium salt of an acid A (C6H10O4) on dry distillation produced B (C5H8O). When B is reacted with Bromine solution in acetic acid, ‘C’ (C5H7OBr) is obtained. ‘C’ is reacted with anhydrous acidic glycol (HOCH2CH2OH) solution and PhMgBr is added to this solution carefully and then its is subjected to dilute HCl solution then compound ‘D’ (C11H12O) is formed. ‘D’ reacted with hydroxylamine hydrochloride to give ‘E’ and ‘F’ having molecular formula (C11H13NO). ‘E’ on heating with small amount of H2SO4 produced G which on hydrolysis produced ‘H’. F under the same treatment yields I and J compounds respectively. E, F, G and I all are isomeric and have molecular formula (C11H13NO). H have two hydrogen atom while J have only one hydrogen atom. Identify compounds A to J. ‘H’ on reaction with SOCl2 followed by treatment with anhydrous AlCl3 gives the following compound ‘X’.
X=
13.
14. 15.
A compound has two isomers (A) and (B) of formula C5H10O. Isomer (A) on treating with NaOH (aq.) gives 2, 2-dimethylpropan-1-ol and 2, 2-dimethylpropanic acid salt. The isomer (B) on treating with NaOH (aq.) gives 3-hydroxy, 2-propylheptanal. What are A and B ? (1) B H
Mg / Et O
(1) Ph CHO HCl, ZnCl2 2 6 2 B D C CH3CH2CH2MgCl A ( 2) H2O2 / OH ¯
( 2) H3O ,
An organic compound (A) C4H9Cl on reaction with aqueous KOH gives (B). (A) on reaction with alcoholic KOH gives (C) which is also formed on passing the vapours of (B) over heated copper. The compound (C) readily decolourise bromine water. Ozonolysis of (C) gives two compounds (D) and (E). Compound (D) reacts with NH2OH to give (F) and compound (E) reacts with NaOH to give an alcohol (G) and sodium salt (H) of an acid. (D) can also be prepared from propyne on treatment with water in presence of Hg+2 and H2SO4. Indentify (A) to (H) with proper reasoning.
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CARBONYL COMPOUNDS | 95 16.
17.
An alkene (A) on passing through Br2 / CCl4 gives a compound (B) which on dehydrobromination in the presence of NaNH2 gives a hydrocarbon (C). Compound (C) yield. (D) when warmed with dil H2SO4 in the presence of HgSO4. (D) gives a yellow ppt. of (E) on treatment with I2 and NaOH and also forms sodium salt of 3, 4-dimethylpentanoic acid. Give structure of (A) to (E) with reason. An organic compound (A) contains 40% carbon and 67% Hydrogen. Its vapour density is 15. (A) on reaction with a concentrated solution of KOH gives two compound (B) and (C). When (B) is oxidised original compound (A) is obtained. When ‘C’ is treated with concentrated HCl, it gives a compound (D) which reduces ammonical. AgNO3 solution and also gives effervescences with sodium bicarbonate solution. Write structure (A), (B), (C) and (D).
18.
(H3O ) HCN Alkaline hydrolysis A B CH3–CH=CH–CHO
19.
OH
CH3 | Conc.H2SO 4 (ii ) H O / H (i) CH3 CHMgBr B CH3CHO 2 A
H O18 / H C 2 D
20.
21.
H SO
1. BH / THF
H O 2 4 OH C 3 D. Me2C = O + HCN B A 3 2. H2O2 , OH ¯
22.
An organic compound (A) C9H10O reacts with NH2OH to give two isomers B & C with same moleculer formula C9H10ON.The compound B and C isomerizes to D and E respectively on treating with small amount of H2SO4. Compound (F) C8H8O2 is formed either by (A) on treating with I2/KOH followed by acidification or by acid hydrolysis of (D). Hydrolysis of (E) gives ortho–methylaniline. identify the structure from (A) to (E).
23.
Compound A to D do not gives tollens test however compound C does. Give the structure of A to D. HO CH2 CH2 OH Mg / Et 2O (C H O Br) (C6H111MgO2Br) 4-bromobutanal HA 6 11 1 2
A
B
CH3 OH,HA (C7H14O2) (C6H12O2)
24. 25.
[Solomon]
D C Acetophenone on reaction with hydroxylamine hydrochloride can produce two isomeric oximes. Write structure of the oximes. [JEE-97, 2M] Complete the following giving the structures of the principal organic products. [JEE-97, 8M] (a)
(c)
+ Ph3P = CH2 E
ClCH2CH2CH2COPh + KOH + MeOH I
(b)
(d)
H3CCOCOC6H5 + NaOH / H3O+ J
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CARBONYL COMPOUNDS | 96 26.
How many asymmetric cabon atoms are created during the complete reduction of benzil (PhCOCOPh) with LiAlH4 ? Also write the number of possible stereoisomer of the possible product. [JEE 97, 3M]
MeMgBr
27.
28. 29.
30.
[JEE 97, 2M]
+ (COOEt)2 + EtONa
[JEE 97, 2M]
An aldehyde (A) (C11H8O), which does not undergo self aldol condensation, gives benzaldehyde and two mole of (B) on ozonolysis. Compound (B), on oxidation with silver ion, gives oxalic acid. Identify the compound (A) and (B). [JEE-98, 2M] Write the intermediate steps for each of the following reactions [JEE-98, 2M] H O
3 C H CH = CHCHO C6H5CH(OH)C CH 6 5
31. 32.
32.
33.
(i) LiAlH
A
4 B
[JEE 98, 2M]
(ii) H ,
The compound X does not show any stereoisomerism. On oxidative ozonolysis it produces Formic acid and Y. With HIO4 oxidation, Y forms only acetic acid. X on reaction with 1 mole of Br2 (with moderate heating) forms most stable major product ‘Z’. Report molecular weight of ‘X’ and report whether ‘Z’ will show stereoisomerism or not. Compound ‘A’ (C8H8O) on treatment with NH2OH . HCl gives ‘B’ and ‘C’ . ‘B’ and ‘C’ rearrange to give ‘D’ and ‘E’ respectively on treatment with acid . ‘B’ , ‘C’ , ‘D’ , and ‘E’ are all isomers of molecular formula C8H9NO . When ‘D’ is boiled with alcoholic KOH , an oil ‘F’ (C6H7N) separates out . ‘F’ reacts rapidly with CH3COCl to give back ‘D’ . ‘E’ on boiling with alkali , followed by acidification gives a white solid ‘G’ (C7H6O2) Identify ‘A’ to ‘G’. [JEE-99, 7M] Carry out the following transformation in not more than three steps : [JEE 99, 2M] O || CH3 – CH2 – C C – H CH3 – CH2 – CH2 – C – CH3
34.
35.
36.
37.
[JEE 99, 3M]
Two different Grignard reagents, (X) and (Y) produce C6H5CH2C(CH3)2OH on reaction with (P) and (Q) respectively. Give structures of (X), (Y), (P) and (Q). [REE 2000, 2M] H A
[JEE 2000, 2M]
Base
[JEE 2000, 1M]
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CARBONYL COMPOUNDS | 97
38.
Identify A, B & C and give their structures :
Br
H
NaOH,
[JEE 2000, 3M]
2 C (C7H12O) A + B
39.
An organic compound (A), C8H4O3, in dry benzene in the presence of anhydrous AlCl3 gives compound (B). The compound (B) on treatment with PCl5, followed by reaction with H2 / Pd(BaSO4) gives compound (C), which on reaction with hydrazine gives a cyclised compound (D) (C14H10N2). Identify (A), (B), (C) and (D). Explain the formation of (D) from (C). [JEE 2000, 5M] 40. An organic compound A, C6H10O, on reaction with CH3MgBr followed by acid treatment gives compound B. The compound B on ozonolysis gives compound C, which in presence of a base gives 1-acetyl cyclopentene D. The compound B on reaction with HBr gives compound E. Write the structures of A, B, C and E. Show how D is formed from C. [JEE 2000, 5M] Matching : 41.
Match the compounds/ ions in column I with their properties/reactions in column II. Column-I Column-II (A) CH3 – CH = O (p) Redox-reaction (B)
(C) (D)
CI– (C2H5O)3 AI
Reaction
42.
43.
(q)
Precipitate with Tollen`s reagent
(r) (s)
Nucleophile condensation
Commonly used Reagent
(A) Perkin reaction
(p) CHCl3 / KOH
(B) Aldol reaction
(q) CH3COOK / (CH3CO)2O
(C) Cannizzaro reaction
(r) dil. KOH
(D) Riemer Tiemann reaction
(s) conc. KOH
Column-I
Column-II
(A)
(p)
Aldol condensation
(B)
(q)
Benzoin condensation
(q)
Perkin reaction
(s)
Claisen schmith reaction
(C) Ph–CH=CH–COOH (D)
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CARBONYL COMPOUNDS | 98
44.
45.
Column – I
Column – II
(A)
(p) Iodoform test
(B) CH3.CHO
(q) Fehling test
(C) HCHO
(r) Tollen's test
(D)
(s) FeCl3 test
46.
Column – I
Column – II
(A)
(p)
Decolourise Br2 water
(B)
(q)
2, 4-DNP
(C)
(r)
NaHCO3
(D)
(s)
FeCl3
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CARBONYL COMPOUNDS | 99 47. (A)
Column – I Reaction HCHO + H+
(p)
Column – II Nature of product or intermediate Carbocation intermediate
(q)
Carbanion intermediate
(r)
Free radical intermediate
(s)
Aldol reaction Column – II
(B)
+ OH Na / alcohol
(C)
CH3COCH3
(D)
PhCOCH3 + O H Column – I
48.
(A)
+ C N
(p)
Nucleophilic addition
(B)
HCl + NH2 –NH2
(q)
Nucleophilic addition elimination
(C)
KOH + NH2 –NH2
(r)
Reduction
(D)
2 + O D Excess
(s)
-substitution
49.
D O
Column – I
Column – II
(A)
(p)
Nucleophilic addtion
(B)
(q)
Aliphatic nucleophilic substitution
(C)
(r)
Elimination
(D)
(s)
Electrophilic addition reaction
(t)
Electrophilic substitution reaction
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CARBONYL COMPOUNDS | 100
MISSCELLANEOUS QUESTION BANK PART - I : OBJECTIVE QUESTIONS 1. 6. 11. 16. 21. 26. 31. 36. 41. 46. 51. 55. 58. 63. 68.
(A) (B) (A) (B) (A) (A) (B) (C) (A, D) (C, D) (B, C, D) (A, B, D) (A) (C) (A)
2. 7. 12. 17. 22. 27. 32. 37. 42. 47. 52. 56. 59. 64. 69.
(A) (A) (D) (D) (A) (D) (A) (D) (A, B, C, D) (A, B, C, D) (A, B, C) (A, D) (A) (D) (C)
3. 8. 13. 18. 23. 28. 33. 38. 43. 48. 53. 57. 60. 65. 70.
(B) (D) (D) (D) (C) (A) (A) (B) (A, B, C, D) (A, B) (A, B, D) (A) (B) (C) (A)
72. 75. 78. 81. 84. 87. 90.
(C) (A) (B) (D) (A) (A) (B)
73. 76. 79. 82. 85. 88. 91.
(D) (A) (A) (B) (A) (A) (C)
74. 77. 80. 83. 86. 89. 92.
(C) (D) (C) (A) (A) (C) (A)
4. 9. 14. 19. 24. 29. 34. 39. 44. 49. 54.
(C) (D) (B) (A) (D) (D) (A) (B) (C, D) (A, B, C, D) (A, B, C, D)
5. 10. 15. 20. 25. 30. 35. 40. 45. 50.
(B) (B) (A) (A) (B) (C) (B) (D) (A,B,C) (B, D)
61. 66. 71.
(C) (C) (B)
62. 67.
(D) (A)
PART - II SUBJECTIVE QUESTIONS
1.
2.
(A)
(B)
(C)
(D)
(A)
(B)
(D)
(E)
(Succinic acid )
(C)
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CARBONYL COMPOUNDS | 101 3.
4.
(A) CH3 CH2 NO 2
(B)
(C)
(D)
(A)
(B)
(C)
(L) CH3–CH=CH–COOEt
(M) CH3–CH=CH–CH2OH
(D)
5.
(E)
(K)
6.
8.
A = Propyne
B=
D = CH3 – C C – CH3
C=
E=
(aldol product)
F=
G = 9.
(A) C 6H5 – C C – H
11.
(1) (2) (3) (4)
(A) (A) (A) (A)
(B)
forms phenyl hydrazone and thus carbonyl compound, i.e., an aldehyde or ketone. give s negative Tollen’s reagent test and thus (A) is ketone. also does not respond Iodoform test and thus it is not methyl ketone. on reduction gives n-pentane and thus, (A) is straight chain compound.
(5) Keeping in view of above facts and the formula of (A), i.e., C5H10O, (A) is
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CARBONYL COMPOUNDS | 102
12.
CH2 CH2 COOH | Dry distillati on CH2 CH2 COOH
(A)
Br / CH COOH
23 (B)
NH OH / HCl
+
2
(D)
(E)
(F)
H SO
HO
4 2
(E)
2 HOOC CH2 CH2 CH2 CH NH2 | Ph (G)
(H)
H SO
HO 2 H2N CH2 CH2 CH2 CH COOH | Ph
4 2
13.
(A)
(C)
(F)
(I)
(J)
(CH3 )3 CCHO 2,2 dim ethylpropa nal
(B) CH3 CH2CH2 CH2CH CH2 CHO || | O CH2 CH2CH3 (Pen tan 1 al )
14.
(A) CH3CH = CH2
(B) CH3CH2CH2OH
(C) CH3CH2CH2Cl
15.
(A)
(B)
(C)
(E) HCHO,
16.
,
,
(F)
,
, (G) CH3OH,
(A)
(B)
(C)
(D)
(D) PhCH = CHCH2CH3
(D)
(H) HCOONa
(E) CHI3 17.
(A) 2HCHO,
(B) CH3OH,
(C) HCOONa,
(D) HCOOH
18.
(A) CH3 CH CH CH CN , (B) CH3 CH CH CH COOH | | OH OH
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CARBONYL COMPOUNDS | 103
21.
OH | (A) CH3 C CN | CH3
22.
(A)
OH | (B) CH3 C COOH | CH3
(C) CH2 C COOH (D) HOCH2 CH COOH | | CH3 CH3
(B)
(C)
CH3
O
(E) CH3 – C – NH
(D)
23.
(A)
(B)
(C)
(D)
HCl H2O
24.
It shows two isomers which are geometrical isomers to each other and represented as follow : (i)
C6H5
C6H5
(ii)
C=N CH3
OH C=N
CH3
OH
Syn-phenyl anti Syn-methyl Ketoxime
Syn-methyl anti-Syn-phenyl Ketoxime
Their configuration may be identified with the help of Backmann’s Rearrangement O || CH3 C NH C 6H5
Presence of acid
or PCl5 or Acid anhydride
Pr esence of acid or PCl 5
N phenyl acetamide
or Acid anhydride
O || C 6H5 NH C CH3 N phenyl acetamide
Therefore both isomers give different product with different configturation.
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CARBONYL COMPOUNDS | 104
25.
(a)
CH2 + Ph3P=O
(b)
(c)
26.
(d)
* * Ph CH CH Ph | | OH OH two chiral C atoms ; total 3 isomers :
27.
29.
28.
Ozonolysis
A =
+
CHO CHO | | COOH COOH Two moles
B =
30.
CHO COOH Ag O | oxalic acid 2 | COOH COOH (B)
OH | H3O C 6H5 CH C CH
H2 O
OH O C6H5 – C H – C CH C6H5 – CH = C = C H
C6H5 – CH = C = CHOH
C6H5 – CH = CH – CHO
Enol-form (less stable)
Keto form (more stable)
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CARBONYL COMPOUNDS | 105
( C 6H5 CHO ) (A ) Base
31.
(i ) LiAlH
4
Rearrangement
32.
O3 HIO 4 HCOOH + Y CH3COOH X
So
Y
=
CH3 – C – C – CH3 O
X
=
O
CH3 – C – C – CH3 CH2 CH2
Z
=
CH 3 – C
C – CH 3 (1,4 Addition)
CH2Br CH2Br
32.
Yes, It will show geometrical Isomerism. M.W. (X) = 82 So compound are A=
B=
C=
O || D = CH3 C NH C6H5
O || E = C 6H5 C NH CH3
33.
F = C6H5 – NH2
O || G = C 6H5 C O H
O || BH3 / THF CH3 X Na CH3CH2C CNa CH3CH2C CH CH3 CH2CH2 CCH3 CH3CH2C CCH3 H2O 2 , OH ¯
O || CH3 CH2CH2 CCH3
CH3 – CH2 – CH2 – CH = O
34.
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CARBONYL COMPOUNDS | 106 35.
(X) C6H5CH2MgX (Y) CH3MgX
(P)
(Q)
CH3 | CH3 C CH2 C 6H5 | OH
+
CH3 | CH3 C CH2 C 6H5 | OH
36.
(A)
(Beckmann's rearrangement)
38.
(A) CHBr3
39.
(A)
40.
A=
41.
(A) – (q, s) ;
(B) – (q) ;
(C) – (q, r) ;
(D) – (p)
42.
(A) - (q) ;
(B) - (r) ;
(C) - (s) ;
(D) - (p)
43. 44.
(A) - (q) ; (A) - s ;
(B) - (s) ; (B) - p ;
(C) - (r) ; (C) - q ;
(D) - (p) (D) - r
45.
(A) - q, r ;
(B) - p, q, r ;
(C) - q, r ;
(D) - s
46.
A p, q
Br
C p, q, s
D p, q, s
47.
Ap
B q, s
C q, r
D q, s
48.
A p,
B q,
Cr
Ds
49.
A p, t
B q, r, t
C q, r, s
D s, t
(B)
37.
(C)
(B)
(C)
(D)
,B=
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 107
CARBOXYLIC ACIDS & ACID DERIVATIVES
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 139
PART - I : SUBJECTIVE QUESTIONS Section (A) : Carboxylic acid A-1. A-2.
Describe the preparation of acetic acid from acetylene. In what way can acetic acid can be distinguished from acetone ?
Section (B) : Acid Halide B-1.
Give the IUPAC names of the following compounds.
(a) Cl C CH2 C Cl || O
B-2.
B-3.
O || (c) Ph CH2 C Cl
(b)
|| O
(d) Ph C Cl || O
Observe the following reactions. O || CH3 – C – F
2
O || CH3 – C – Cl
2
O || CH3 – C – Br
2
O || CH3 – C – I
2
H O r1
H O r2
H O r3 H O r4
Identify the order of the rate of reaction r1, r2, r3 and r4, give reason. Predict the products of the following reactions
(i) O || (ii) Ph C Cl
+
+
O || AlCl3 (iii) CH3 – CH2 – C – Cl + Ph — H
(iv)
H / Pd / C
2 Quinoline fDouksyhu
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 140
Section (C) : Ester C-1. Ans. C-2.
The boiling point of esters are always less than the corresponding carboxylic acids. Why ? Because esters do not form hydrogen bonds Explain the order of the rate of esterification of the following acid with MeOH : MeCH2COOH > Me2CHCOOH > Me3CCOOH > Et3CCOOH >> (i-Pr)2CHCOOH
Section (D) : Amide D-1.
Predict the products of following reaction
NaOH
(a)
(b)
H2O
HO
2 + H2SO4
(c)
heat
(d)
+
D-2.
Write balanced equation, having all the organic product , for the reaction of phenylacetamide with (a) hot HCl (aq) (b) hot NaOH (aq)
D-3.
What products would you expect from acidic & basic hydrolysis of each of the following amides ? (a) N,N - Diethylbenzamide (b)
(c)
Section (E) : Anhydride E-1.
E-2.
Predict the products of the following reaction (a) Phenol + acetic anhydride in presence of lewis acid. (b) Aniline + phthallic anhydride on heating. (c) Anisole + succinic anhydride & aluminium chloride (d) Benzaldehyde + acetic anhydride in presence of acetate ion. Predict the products of the following reactions
(a)
H
+
+ CH3NH2
(b)
(c)
H + CH3–CH2–OH
(d)
+
AlCl
3
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 141
PART - II : OBJECTIVE QUESTIONS * Marked Questions are having more than one correct option. Section (A) : Carboxylic acid A-1.
Which of the following acids has the smallest dissociation constant ? (A)
A-2.*
(B) O2N–CH2–CH2–COOH
(C) Cl–CH2–CH2–COOH (D) NC–CH2–CH2–COOH In the following ester there are three carbon oxygen bonds denoted by x , y and z
Their lengths are in order (A) x = y = z (B) x < y
(C) y < z
(D) x < z
H O, acetic acid
2 Product is :
A-3.
H2SO 4heat
(A)
(B)
(C)
(D)
A-4.
X
Y
(C) Structural isomers
(D) Stereoisomer
A-5.
In the above reaction sequence X & Y are : (A) Identical (B) Homologs End product of this reaction is :
(A)
(C)
(D)
(C)
(D)
(B)
(i) P / Br
2 X, X is :
A-6.
(ii) H2O
(A)
(B)
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 142
A-7.
(A)
(B)
A-8.
A-9.
A gS :
A, A is :
(C)
+
(D)
x + y the organic products x and y are
(A) Enantiomers (B) Diastereomers (C) Metamers For the following acids the rate of decarboxylation on heating would be :
(D) achiral molecules
I.
II.
III. NO2–CH2–COOH (A) III > I > IV > II
IV. HOOC–CH2–COOH (C) III > IV > I > II (D) I > IV > III > II
(B) I > III > IV > II
Section (B) : Acid Halide B-1.*
The correct order/s of decreasing reactivity of the given compound towards hydrolysis under identical condition is/are : (A) CH3COCl > CH3CONH2 (B) CH3COCl > (CH3CO)2O (C) CH3COOCH3 > CH3COCl (D) (CH3CO)2O > CH3CONH2
B-2.
3 Phosgene (COCl2) + 1 mol C2H5OH X, X is
NH
(A) NH2 C NH2 || O
B-3.
(C) NH2 C Cl || O
(D) EtO C NH2 || O
O || H O (CH3 )3 C C Cl + LiAlH4 (excess) (vkf/kD;) 3 X, X is : (X gS :)
+ LiAlH4
B-4.
(B) NH = C = O
X,
O || (A) (CH3 )3 C C H
CH3 | (B) CH3 C CH2 OH | CH3
(C) CH3 CH CH2 CH2 OH | CH3
(D) CH3 – CH2 – CH2 – CH2 – CH2 – OH
For the given reaction
A is gSA (A) CH2=CH–OCH3
(B) CH2=CHOH
(C) (CH2=CH)2CuLi
(D) (CH2=CH)2Cl
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 143
B-5.
O || PhCH2 C Cl CH3 CH2 SH X, X is
O || (A) Ph C S CH CH3 | CH3
O || (B) Ph CH2 C S C 2H5
O || (C) CH3 CH2 CH2 C S Ph
O || (D) CH3 CH C S Ph | CH3
O || + Cl C Cl X, X is gSA
B-6.
NH (A)
(B)
(C)
(D)
O C
C O || O
Section (C) : Ester C-1.
List the following esters in order of decreasing reactivity in the second step of a nucleophilic acyl substitution reaction .
(A) IV > I > III > II C-2.
(B) IV > III > I > II
(C) III > IV > I > II
(D) II > I > III > IV
Which one of the following is least reactive for hydrolysis reaction?
(A)
(B)
(C)
(D)
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 144
C-3.
Which one of the following esters is most reactive for saponification ?
(A)
C-4.*
(B)
(C)
(D)
Which of the following is not the correct order of esterification of following acids with CH3OH : HCOOH , CH3 COOH , CH3 – CH2 – COOH ,
I
II
(A) I = II = III = IV
III (B) I > II > III > IV
IV (C) I < II < III < IV(D) I > IV > III > II
ethanol, H2SO 4
C-5.
X, X is :
ethanol, H2SO 4
X, X gS
C-6.
, is
%
(A)
(B)
(C)
(D)
In the given reaction
, [X] is gSA
(A)
(B)
(C)
(D) No reaction is possible
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 145
¯ OCH
3 X, X is
C-7.
CH3OH
C-8.
O || (A) H3 CO C (CH2 )5 OH
O || (B) H3 CO C CH CH2 CH2 CH2 OH | CH3
O || (C) H3 CO C CH2 CH CH2 CH2 OH | CH3
O || (D) H3 CO C CH2 CH2 CH CH2 OH | CH3
Ease of esterification of following alcohol with HCOOH is CH3 –CH2OH (CH3)2 CH – OH (CH3)3 C – OH I II III (A) III > II > I (B) I > II > III (C) II > I > III CH NH heat
3 2 X, X is
C-9.
C-10.
(D) II > III > I
(A)
(B)
(C)
(D)
In the given reaction
[P] will be
(A)
(B)
(C)
(D)
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 146
Section (D) : Amide (1) LiAlH4 X, X is
D-1.
( 2) H2O
(A) D-2.
(B)
(C)
(D)
Consider the following compounds :
I
II
III
IV
The decreasing order of reactivity towards hydrolysis by aqueous NaOH is : (A) I > II > III > IV (B) III > II > IV > I (C) IV > I > II > III (D) I > IV > II > III O
O
D-3.*
15 OH, Br 2, heat – CH2 – C – NH2
– C – NH2 +
X+Y
Product X &Y can not be : 15 CH2 – NH2
NH2 (A)
&
D-4.
(B)
15 NH2
CH2 – NH2 (C)
15 CH2 – NH2
&
15 NH2
NH2
&
(D)
CH2 – NH2
&
In the given reaction sequence
(A)
(B)
(A) and (B) respectively are :
(A)
and
(B)
(C)
and
(D)
and
and
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 147
Section (E) : Anhydride E-1.*
Which of the following will yield a cyclic compound on heating :
(A)
E-2.
(B)
(C)
(D)
In the following reaction X Y.. X is the lowest molecular weight optically active dicarboxylic acid. What is the structure of Y.
(A)
E-3.
(B)
+
(C)
(D)
X, X is
(A)
(B)
LiAlH
PCC
(C)
(D)
NH – NH
4 2 2 A B C, Product (C) is
E-4.
(A)
(B)
(C)
(D)
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 148
PART - I : SUBJECTIVE QUESTIONS Section (A) : Carboxylic Acid 1.
How will you convert acetaldehyde into but – 2–enoic acid.
2.
14CH I 3
3.
H O
Ca(OH)
,
heat
Z + Na14CN X 3 Y 2 Identify compound X, Y and Z. In case of aldehydes and ketones there is nucleophilic addition but in case of acyl compound there is
Section B: Acid Halide 4.
Show how would you prepare the following compounds with the help of appropriate acid chlorides? (A) N, N–dimethylacetamide (B) Ethyl propanoate (C)
5.
(D) Benzyl benzoate
(a) List three reagents for converting a carboxylic acid to its acyl chloride. (b) Select the most convenient of the three reagents, give a reason for your choice and write a balanced equation for its reaction with R—COOH. (i) PCl (excess)
H / Pd / BaSO
6.
SOCl2 OH ¯ 4 3 (Z) (Y) 2 (X) W + S CH3COOH
7.
write the structure of X,Y,Z,W and S (a) Why cannot HCl convert RCOOH to RCOCl ? (b) Why is it more efficient to prepare an ester by the sequence : acid acyl chloride ester, rather than acid ester ?
(ii) H2O
Section C : Ester 8.
Where would you expect to find the labelled oxygen if you carried out an acid catalyzed hydrolysis of methyl benzoate in 18O–labeled water ? Write detailed mechanism to supports your answer.
9.
A + B, find A and B.
Section D : Amide 10. 11.
12.
The respective boiling points and molecular weights (in/gmol) of the following amides are : MeCONH2, 221ºC and 59; MeCONHMe, 204ºC and 73 ; MeCONMe2, 165ºC and 87. Explain. An organic compound (A) C4H7N does not react with acetyl chloride or nitrous acid. (A) when boiled with acetic acid gave another compound B (C6H11O2N) and on reduction with lithium aluminium hydride formed C (C4H11N). Compound (C) liberates nitrogen with nitrous acid. Identify compound (A) , (B) and (C). Both cis- and trans-1,2-cyclohexanedicarboxylic acids form anhydrides on heating, but the anhydride formed from the cis-1,2-cyclopentanedicarboxylic acid only. explain.
Section E: Anhydride 13. 14.
Complete the folliowing reaction (a) Toluene + phthalic anhydride + AlCl3 (b) Product from (a) + Conc. H2SO4 + Give the product of each of the following reactions : PCl3 MeNH 2 B C, (a) succinic anhydride + MeOH A 300 ºC (b) phthalic acid + NH3 D E ,
15.
(CH3CO2 )O (c) MeCH(CH2COOH)2 F.. An organic compound A, C4H4O3, in dry benzene in the presence of anhydrous AlCl3 gives compound B. The compound B on treatment with PCl5, followed by reaction with H2/Pd(BaSO4) gives compound C, which on reaction with hydrazine gives a cyclised compound D (C10H10N2). Identify A,B,C and D. Explain the formation of D from C.
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 149
PART - II : OBJECTIVE QUESTIONS (without section in class room but store sections in Main Sheet)
Single choice type Section A: Carboxylic Acid 1.
2.
Which compound should have zero dipole moment ?
(A)
(B)
(C)
(D)
When picric acid is treated with aqueous sodium bicarbonate The gas evolved would be : (A) H2 (B) CO2 (C) 14CO2
a gas is liberated with effervesence. (D) NO2
3. (C – O) bond lengths designated by A, B, C are in order : (A) A = C < B = D
(B) A < B < C = D
(C) A < C = D < B
(D) All equal lHkh leku
1.Mg diethyl ether product is :
4.
2.CO2 3 . H3 O
(A)
(B)
(C)
(D)
5.
In the reaction sequence
6.
(A) Compound A will be optically active (B) Product is racemic mixture of -hydroxyacid (C) Product is optically active containing -hydroxy acid. (D) Product is optically active contaning -hydroxy amide. Identify X in the reaction sequence X (A) CH 3– CH3
HOOC – CH2 – CH2 – COOH (B) CH2 = CH2
(C) H – C C – H
(D) CH3– CH = CH – CH3
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 150 7.
The final product of the following reaction sequence is
(A)
8.
9.
10.
(B)
(C)
In the reaction sequence X & Y are
(A) Homologs (B) Identical (C) Structural isomers (D) Enantiomers A racemic mixture of carboxylic acid having one chiral centre is treated with enantiomerically pure amine. the products formed are : (A) diastereomers (both optically active) (B) diastereomers (only one is optically active) (C) Racemic mixture (D) two optically inactive ammonium salts of acid. The product P of the following reaction is CH3 – COOH
11.
(D)
P
(A) CH3 – COOH (B) For the following conversion
(C)
(D)
The effective method is : (A) (B) (C) (D)
12.
Q.
In the above reaction Q is.
(A)
(B)
(C)
(D)
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 151 13.
For the following conversion the correct sequence of reagents is
14.
(A)
(B)
(C)
(D)
In the given reaction CH3COOH
(A) CH3–COOH
15.
(B) HOOC–CH2–CH2–COOH
In the reaction sequence CH3 – CH2 – COOH
(C)
x
(A) x is
(B) y is
(C) z is
(D) all are correct
Conc .HI excess
16.
(D)
y
P the producet P is.
(A)
17.
[X], [X] will be
(B)
(C)
(D)
The correct sequence of reagents for the following conversion.
is (i) KCN
(A)
(ii)H3O
(B)
(C)
(D)
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 152
Section (B) : Acid halide 18.
19.
Nucleophilic substitution at acyl carbon of a carboxylic acid derivative generally proceeds by : (A) Addition-elimination mechanism (B) Elimination-addition mechanism (C) SN1 mechanism (D) SN2 mechanism Which of the following proposed reactions would take place most quickly under mild conditions ? (A) CH3–CO–NH2 + NaCl CH3COCl + NaNH2 (B) C6H5–CO–Cl + CH3NH2 (C) CH3–CO–Cl + CH3NH2 CH3–CO–NH–CH3 (D) CH3COOC2H5 + HOH
20.
In the given reaction sequence : C6H5–CN
A
(A)
21.
CH3COONa + C2H5OH
B
C, (C) is
(B)
(C)
(D)
O || NaOH H2N NH2 Ph C Cl (excess) X, X is O || (B) Ph C NH OH
O || (A) H C NH Ph
22.
O || (C) Ph C NH2
O O || || (D) Ph C NH NH C Ph
Pyridine
X, X is
+
(A)
(B)
(C)
(D)
140C
+ CH3CH2CH2CH2NH2 X,
23.
(A)
(B)
(C)
(D)
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 153
Section (C) : Ester 24.
Consider the following statements for hydrolysis reaction : (I)
is more reactive than C6H5COOC2H5
(II)
(III)
is more reactive than
is more reactive than
Of these the correct statements are (A) I and II (B) I, II and III 25.
(C) II and III
(D) I and III
In the given reaction
H/ H O
2
(A) 26.
(B)
[P], [P] will be :
(C)
(D)
In the given reaction
, [X] is
(A)
(B)
(C)
(D)
EtO ¯ EtOH
X
27.
(A) X is
(B) X is
(C) X is
(D) X is
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 154 28.
The product x in the above reaction is
x
(A)
(B)
(C)
(D)
29.
x , x is
(A)
30.
(C)
(D)
Identify the end product in the following reaction
(A)
31.
(B)
(B)
(C)
(D)
The product X in the following reaction is
acetaldehyde + X.
(A)
(B)
(C)
(D)
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 155
32. x in the above reaction is
(A)
(B)
NaOH /
X, X is HO
33.
(D)
(C)
(D)
gSA
2
(A)
34.
(C)
(B)
NH 3 / KOBr HNO2 B X C (CH3)3 C – OH, X is:
(A) (CH3)3C – COOH
35.
(B) (CH3)3C – CH2O
(D) none is true
CH3 O | || Br2, NaOH X, X is PhCH2 CH CH2 C NH2 CH3 | (A) CH3 CH CH2 NH Ph
CH3 | (C) Ph CH2 CH CH2 NH2
36.
(C) (CH3)3C – CONH2
CH3 | (B) Ph CH2 CH CH NH2 | OH CH3 CH3 | | (D) Ph CH CH C NH2 || O
Urea (H2NCONH2) + hot dilute NaOH X + NH3 , X is (A) NaNH C NH.Na || O
(B) H2N C NH Na || O
(C) Na2CO3
(D) H C NH2 || O
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 156
X, X is :
37.
(A)
(B)
(C)
(D)
More than one choice type Section (A) : Carboxylic acid 38.
Correct water solubility order/s amongst the following pairs is/are
COOH CH3 |
(A) CH3 – CH2 – CH2 – CH2 – OH >
CH3 C OH
(B)
| CH3
Cl
<
OH
Cl Cl
(C)
COOH OH
<
(D)
H
H3C H
C C COOH
>
COOH
H3C H
C
Cl
39.
The correct reagent sequence for the following conversion , is
40.
(A)
(B)
(C)
(D)
The correct order/s for the given pair of isomers is / are (A)
>
(Melting point)
(B)
>
(Dipole moment) (
(C)
>
(Boiling point)
(D)
>
(Water solubility)
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C H
CARBOXYLIC ACIDS & ACID DERIVATIVES | 157
MIXED TYPE PROBLEMS PART - I : MATCH THE COLUMN Match the column 1.
Column-I
Column-II
(A)
(p) Hydrolysis
(B)
(q) Esterification
(C)
(r) Saponification
(D)
(s) Acid base reaction (t) SN2 Th reaction
2.
If compound in column I is heated then select the correct match from column II for product obtained. Column-I Column-II Reaction product formed
(A)
(p) Diastereomers
(B)
(q) Racemic mixture
(C)
(r) Meso compound
(D)
(s) CO2 gas will evolve
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 158
PART - II : COMPREHENSION Read the following passage carefully and answer the questions.
Comprehension - 1 Observe the esterification mechanisms for primary and tertiary alcohols. O Conc. H2SO4
Type.1 CH3–COOH + CH3–CH2–CH2–OH CH3–C–O–CH 2–CH2–CH3 Propyl = Pr Mechanism
O CH3–C OH
H CH3–C Step 1.
+ OH
+ CH3–C
OH
OH
.. O
Pr H Step 2
Step 3
CH3–C–O–CH 2–CH 2–CH 3 CH3–C–OH –H –H2O
+O
OH
Pr
H
CH3
O
Type. 2 CH3–C
O
OH
CH3 Conc. H2SO4 + CH3–C–OH CH3–C–O–C–CH3 CH3 OH CH3 O .. O–H CH3 –C CH3 H O CH3–C + + H2O CH3–C–OH Step 1. CH3 CH
CH3
Mechanism
CH3 CH3–C–O–C–CH3 CH3 O
3
18
3.
Conc. H2SO4 CH3–COOH + C2H5OH (P) 18
Conc. H SO
4 (Q) CH3–COOH + (CH3)3C–OH 2
In the above reaction (P) and (Q) are respectively : O
CH3
O
(A) CH3–C–O–C2H5 , CH3–C–O–C–CH3
CH3
O
O
18
18
(B) CH3–C–O–C2H5 , CH3–C–O–C–CH3
CH3
O 18
O
CH3
(C) C2H5–C–O–CH3 , CH3–C–O–C–CH3
CH3
O
CH3
O 18
18
(D) CH3–C–O–C2H5 , CH –C–O–C–CH 3 3
CH3
CH3
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 159 H 4.
CH3COOH +
OH
D
Conc. H2SO4 (X)
CH3 CH3 CH3–COOH +
Ph–C–OH
Conc. H2SO4 (Y)
C2H5
5.
(A) (X) is optically active while (Y) is optically inactive (B) Both (X) and (Y) are optically active (C) Both (X) and (Y) are optically inactive (D) (X) is optically inactive while (Y) is optically active. (+) Octan-2-ol esterifies with Acetic acid to give optically inactive racemised product. It must have gone by (A) Type I mechanism (B) Type II mechanism (C) Mix type I and type II mechanism (D) More by type I and less by type II mechanism
Comprehension - 2 Observe the following sequence of reaction and answer the questions based on it Phenylacetylene 6.
7.
8.
x
y
z
w
Compound z is (A)
(B)
(C)
(D)
Which of the following statement is not correct (A) y decolourises Br2 /H2O solution (C) w on reaction with NaOI gives yellow ppt
(B) on heating z CO2 is liberated (D) x liberates H2 gas with Na metal
Which of the following compound give benzoic acid on KMnO4 oxidation (A) w (B) y (C) z
(D) all.
Comprehension - 3 Hoffmann rearrangement In the Hofmann rearrangement an unsubstituted amide is treated with sodium hydroxide and bromine to give a primary amine that has one carbon lesser than starting amide. General reaction :
+ NaOH + Br2
R–N=C=O isocyanate
R – NH2
Mechanism :
OH
CO2R– NH
R – NH2 If the migrating group is chiral then its cofiguration is retained. Electron releasing effects in the migrating group increases reactivity of Hofmann rearrangement.
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 160 9.
10.
Which of the following compound(s) cannot give Hoffmann rearrangement :
(A)
(B)
(C)
(D)
Arrange the following amides according to their relative reactivity when reacted with Br2 in excess of strong base
(A) IV > I > II > III
11.
(B) II > I > III > IV
(C) II > IV . III > I
+
(A)
(C)
Products
&
&
(D) II > I > IV > III
(B)
&
(D) B & C both are correct
Comprehension - 4
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 161 12. 13.
Which reagent (X) is used to convert I to II (A) NaOBr (B) Br2/CCl4 Which is true for the above reaction
(A)
(C) Br2/H2O
(D) Na2CO3 with H2SO4
is formed during the reaction
(B) The rate determining step is migration of para tollyl group. (C) Formation of II is R.D.S (D) All are correct
14.
+
Under Hoffmans condition above compounds gives respectively.
(A)
(C)
(B)
+
&
(D)
+
PART - III : ASSERTION / REASONING DIRECTIONS : Each question has 5 choices (A), (B), (C), (D) and (E) out of which ONLY ONE is correct. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. (E) Statement-1 and Statement-2 both are False. 15.
16.
17.
Statement-1 : (CH3)3CCH2COOH is more acidic than (CH3)2 SiCH2COOH. Statement-2 : Si is electropositive element than carbon and therefore has an electron donatingl (+I) inductive effect. Statement-1 : The Reactions of acids are quite different from aldehydes & ketones Statement-2 : In acids one nucleophile replaces another on the acyl carbon atom but aldehyde & ketone give simple nucleophilic addition reaction . Statement-1 :
Statement-2 : Decarboxylation takes place readily at -keto acid.
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 162 18.
19.
Statement-1 : Acyl chloride undergo nucleophilic attack more rapidly than an alkyl chloride. Statement-2 : The transition state of the acid derivative leading to the tetrahedral intermediate is less sterically hindered than transition state in SN2 reaction of alkyl halide. Statement-1 : The reaction of CH3SO2Cl with optically active 2-butanol gives an active methanesulphonate which is treated with aq.NaOH giving back 2-butanol but having opposite configuration. Statement-2 :
20. 21.
attacks with inversion at the chiral carbon displacing
Statement-1 : Unsubstituted or mono substituted amides are water soluble. Statement-2 : The dipole moments of most amides are very high. Statement-1 : Phthallimide has less acidic H attached to ‘N’ as compared to benzamide. Statement-2 : Former has two
22.
24.
groups to stabilize the amide anion.
Statement-1 : The base catalysed hydrolysis of ester order is (1) CH3–COOCH3 > CH3–COOC2H5 > CH3–COOCH(CH3)2 .
(2)
23.
.
>
>
>
Statement-2 : SN2 Th is sterically as well as electronically controlled reaction. Statement-1 : C–O bond length is shorter in an ester as compared with an anhydride. Statement-2 : A degree of cross conjugation exist in the anhydride which decreases the delocalisation to each carbonyl oxygen. Statement-1 : Acid catalysed hydrolysis of ester is reversible while base catalysed hydrolysis is irriversible. Statement-2 : In acid catalysed ester hydrolysis carboxylic acid is formed on which nucleophilic attack of alcohol is possible but in base catalysed ester hydrolysis carboxylate anion is formed on which nucleophilic attack is not possible.
PART - IV : TRUE / FALSE 25._ 26._ 27._ 28._ 29._
The characteristic reaction of acid derivatives is S N2Th. Amides have higher boiling points than corresponding carboxylic acids. Acid catalysed hydrolysis of esters is known as saponification. Carboxylic acids and amides liberate hydrogen gas on reaction with alkali metals. Acid catalysed hydrolysis of esters is a reversible reaction while base catalysed hydrolysis of esters is an irreversible reaction.
PART - V : FILL IN THE BLANKS 30._ 31._ 32._ 33._ 34._
Acetic acid 100% free of water is known as .................. acetic acid. Aromatic carboxylic acids are ........................... acids than aliphatic carboxylic acids. On heating malonic acid forms .................. while succinic acid, forms ........................... Benzoic acid and phenol can be separated by the use of ........................... solvent. Phthalamide is ........................... acidic than benzamide.
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 163
PART - I : IIT-JEE PROBLEMS (PREVIOUS YEARS) * Marked Questions are having more than one correct option. 1.
P and
A CH3 – CH2 – COOH bro mine 1. alcoholic KOH ( excess )
B A
[JEE 95, 2M]
2. H
2.
Which of the following carboxylic acids undergoes decarboxylation easily. Explain briefly. (A) C6H5 – CO – CH2 – COOH (B) C6H5 – CO – COOH (C) C 6H5 CH COOH | OH
[JEE 95, 2M]
(D) C 6H5 CH COOH | NH2
3.
The molecular weight of benzoic acid in benzene as determined by depression in freezing point method corresponds to : [JEE 96] (A) ionization of benzoic acid (B) dimerization of benzoic acid (B) trimerization of the benzoic acid (D) solvation of benzoic acid
4.
A compound D(C8H10O) upon treatment with alkaline solution of iodine gives a yellow precipitate. The filtrate on acidification gives a white solid E (C7H6O2). Write the structures of D.E and explain the formation of E. [JEE 96] Complete the following giving the structures of the principal organic products. [JEE 97] An ester A (C4H8O2), on treatment with excess methylmagnesium chloride followed by acidification, gives an alcohol as the sole organic product. Alcohol B, on oxidation with NaOCl followed by acidification, gives acetic acid. Deduce the structures of A and B. Show the reaction involved. [JEE 98] Following reactions gives two products. Write the structure of the products. Explain briefly the formation of the products giving the structures of the intermediates.
5. 6.
7. 8.
i. OH ¯ / ii. H
9.
10.
11.
O || H3 C CH C OH | CH2 C OH || O
[JEE 99,5/120]
When propionic acid is treated with aqueous sodium bicarbonate, CO2 is liberated. The ‘C’ of CO2 comes from : [JEE 99, 2/80] (A) methyl group (B) carboxylic group (C) methylene group (D) bicarbonate An organic compound A, C8H4O3, in dry benzene in the presence of anhydrous AlCl3 gives compound B. The compound B treatment with PCl5, following by reaction with H2 / Pd(BaSO4) gives compound C, which on the hydrazine gives a cyclised compound D (C14H10N2). Identify A, B, C and D. Explain the formation of D from C. [JEE 2000, 5/100] Which of the following acids has the smallest dissociation constant ? [JEE 2002, 1/35] (A) CH3CHFCOOH (B) FCH2CH2COOH (C) BrCH2CH2COOH (D) CH3CHBrCOOH
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 164 12. 13.
14.
There is a solution of p-hydroxy benzoic acid and p-amino benzoic acid. Discuss one method by which we can separate them and also write down the confirmatory test of the functional groups present. Statement-1 : Acetate ion is more basic than the methoxide ion. Statement-2 : The acetate ion is resonance stabilized. [JEE-2004, 3/84] (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True In conversion of 2-butanone to propanoic acid which reagent is used. [JEE-2005, 3/84] (A) NaOH, NaI /
(B) Fehling solution
(C) NaOH, I2 /
(D) Tollen's reagent
Passage
X
15. 16.
Which reagent (X) is used to convert I to II (A) KBr / NaOH (B) Br2 / NaOH Which step is rate determining step
(C) NaHCO3
[JEE 2006, 5/184] (D) N-Bromo succinamide [JEE 2006, 5/184]
(A) Formation of II
(C) Formation of V
(D) Formation of IV
17.
18.
+
(B) Formation of III
under Hofmann conditions will give :
(A)
+
(B)
+
(C)
+
(D)
+
[JEE 2006, 5/184]
Statement-1 : p-Hydroxybenzoic acid has a lower boiling point that o-hydroxybenzoic acid. Statement-2 : o-Hydroxybenzoic acid has intramolecular hydrogen bonding. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True. [JEE 2007,3/162]
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 165
19.
In the reaction
T, the strucutre of the Product T is :
(A)
(B)
(C)
(D)
PART - II : AIEEE PROBLEMS (PREVIOUS YEARS) 1.
Compound A given below is :
[AIEEE-2002]
OCOCH3 COOH
(A) antiseptic 2.
(B) antibiotic
(C) analgesic
(D) pesticide
End product of the following reaction is :
[AIEEE-2002]
alcoholic KOH
Cl red P
2 CH3CH2COOH
(A) CH3 CHCOOH | OH 3.
(B) CH2CH2 COOH | OH
R––C
[AIEEE-2004]
O + Nu
Z
5.
(D) CH2CHCOOH | | Cl OH
Rate of the reaction is fastest when Z is : O
4.
(C) CH2 = CHCOOH
R––C
+Z Nu
(A) Cl (B) OCOCH3 (C) OC2H5 (D) NH2 Consider the acidity of the carboxylic acids : [AIEEE-2004] (a) PhCOOH (b) o-NO2C6H4COOH (c) p-NO2C6H4COOH (d) m-NO2C6H4COOH which of the following order is correct ? (A) a > b > c > d (B) b > c > d > a (C) b > d > a > c (D) b > d > c > a On mixing ethyl acetate with aqueous sodium chloride, the composition of the resultant solution is : [AIEEE-2004] (A) CH3COOC2H6 + NaCl (B) CH3Cl + C2H5COONa (C) CH3COCl + C2H5OH + NaOH (D) CH3COONa + C2H5OH
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 166 6.
p-cresol reacts with chloroform in alkaline medium to give the compound A which adds hydrogen cyanide to form, the compound B. The latter on acidic hydrolysis gives chiral carboxylic acid. The structure of the carboxylic acid is : [AIEEE-2005] CH3
CH3 CH2COOH
(A)
CH2COOH
(B)
OH
OH
CH3
CH3 CH(OH)COOH
(C)
CH(OH)COOH OH
7.
8.
(D) OH
An organic compound having molecular mass 60 is found to to contain C = 20%, H = 6.67% and N = 46.67% while rest is oxygen. On heating it gives NH3 alongwith a solid resdue. The solid residue give violet colour with alkaline copper sulphate solution. The compound is : [AIEEE-2005] (A) CH3CH2CONH2 (B) (NH2)2CO (C) CH3CONH2 (D) CH3NCO A liquid was mixed with ethanol and a drop of concentrated H2SO4 was added. A compound with a fruity smell was formed. The liquid was : [AIEEE-2009] (A) HCHO (B) CH3COCH3 (C) CH3COOH (D) CH3OH
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 167
Exercise - 1 PART : I
B-3.
(i)
C-1.
Because esters do not form hydrogen bonds
D-1.
(a)
(c)
E-1.
(a)
(ii)
(a)
(c)
(iv)
(b)
+ CH3 NH2
+ CH3COOH
(c)
E-2.
(iii)
(d)
+ C2H5OH
(b)
(d) Ph—CH=CH—COOH + CH3—COOH
+ CH3COOH
O || (b) CH3 — NH — C — CH3 + CH3COOH
(d)
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 168
PART - II A-1. A-6. B-1.* B-6.
(C) (D) (A, B, D) (D)
A-2.* A-7. B-2.
(B, C, D) (A) (D)
A-3. A-8. B-3.
(A) (B) (B)
A-4. A-9. B-4.
(B) (D) (C)
A-5.
(A)
B-5.
(B)
C-1. C-6. D-1. E-1.*
(A) (C) (A) (B, C, D)
C-2. C-7. D-2. E-2.
(B) (A) (B) (C)
C-3. C-8. D-3.* E-3.
(A) (B) (B, C, D) (A)
C-4.* C-9. D-4.
(A, B, D) (C) (B)
C-5. C-10.
(C) (B)
LiAlH
PCC
NH – NH
4 2 2 A B C, Product (C) is
E-4.
(A)
(B)
(C)
(D)
EXERCISE - 2 PART - I 14
14
14
O ||
14
X
6.
X=
9. 11.
(A) = PhCOOH, (B) = (A) C3H7CN (B) C3H7CONHCOCH3
14.
(a) A = HOOCCH2CH2COOMe : B = ClCOCH2CH2COOMe ; C = MeNHCOCH2CH2COOMe
CH3 – CN
CH3 – COOH
,Y=
z
14
2.
(b) D =
y
14
CH3 – C – C H3
, Z = Cl3C C H , W = CHCl3 , S = || O
E=
(C) C3H7CH2NH2
(c)
F=
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 169
PART - II 1. 6. 11. 16. 21. 26. 31. 36.
(C) (B) (C) (B) (D) (C) (C) (C)
2. 7. 12. 17. 22. 27. 32. 37.
(C) (D) (B) (A) (C) (B) (C) (B)
3. 8. 13. 18. 23. 28. 33. 38.
(C) (A) (C) (A) (C) (A) (A) (C, D)
4. 9. 14. 19. 24. 29. 34. 39.
(C) (A) (A) (C) (A) (D) (A) (B, D)
5. 10. 15. 20. 25. 30. 35. 40.
(B) (B) (D) (A) (D) (C) (C) (B, D)
EXERCISE - 3 1. 2. 3.
(A) - (s) ; (B) - (q, s) ; (C) - (p, s) ; (D) - (p, r, s) (A) - (p,s) ; (B) - (q,s) ; (C) - (p, s) ; (D) - (r) (B) 4. (A) 5.
(B)
6.
(C)
7.
(D)
8.
(D)
9.
(B)
10.
(D)
11.
(B)
12.
(A)
13. 18.
(B) (A)
14. 19.
(C) (A)
15. 20.
(A) (B)
16. 21.
(A) (D)
17.
(A)
22.
(A)
23.
(A)
24.
(A)
25._
(T)
26._
(T)
27._
(F)
28._
(T)
29._
(T)
30._
Glacial
31._
Weaker de
32._
Acetic acid, Succinic anhydride
33._
aq. NaHCO3
34._
more
EXERCISE - 4 PART - I 2. 8. 9.
11.
(A) 3. (B) First, anion of Malonic ester is formed which attacks as a nucleophile on the other reagent. When propionic acid is treated with aqueous sodium bicarbonate, CO2 is liberated. The ‘C’ of CO2 comes from : [JEE 99, 2/80] (A) methyl group (B) carboxylic group (C) methylene group (D) bicarbonate (C)
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 170
12.
+
These can be separated by aq. HCl.
Neutral FeCl3 Violet blue ppt.
Test : (1)
NaNO / HCl Naphthol / KOH 2 Red Orange dye.
Test : (2)
13. 18.
(D) (D)
14. 19.
(C) (C)
15.
(B)
16.
(D)
17.
(A)
4.
(B)
5.
(A)
PART - II 1. 6.
(C) (C)
2. 7.
(C) (B)
3. 8.
(A) (C)
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 171
PART - I : OBJECTIVE QUESTIONS Single choice type
1.
OC 2H5 dil. HCl | (catalyst ) H3 C C OC 2H5 H2O X + Y (Alcohol) | OC 2H5 O || (A) X = CH3 C O C 2H5
O || (B) X = CH3 CH2 C O CH3
(C) X = CH3COOH
(D) Y = CH3OH
X,
2.
(A)
3.
(D)
C3H7 | C2H5 (B) CH3 C | OH
C 4H9 | (D) CH3 C CH3 | OH
(C)
In the reaction sequence [X] will be
(A) 5.
(C)
O || H3 O C 2H5 O C OC 2H5 + EtMgBr (large excess) X, X is
C2H5 | (A) C 2H5 C C2H5 | OH
4.
(B)
(B) CH2–CH–COOH
(C)
(D)
Which acid can be oxidised by Fehling solution. (A) Malonic acid (B) Acetic acid (C) Oxalic acid
(D) Formic acid
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 172
7.
PhCH Cl
KOH
2 X, X is,
6.
(A)
(B)
(C)
(D)
O || KOH (1 equiv.) CH3 CH2 C OH
X, X is, gS
(A)
(B)
(C)
(D)
1. LiAlD 2. H2O
4 X, X is gS
8.
(A)
(B)
(C)
(D)
1. THF
+ BrMgCH2CH2CH2CH2MgBr X,
9.
2. H3O
(A)
(B)
(C)
(D)
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 173
CH2OH R?
10.
HO
(A) NaBH4
(B) LiAlH4 H SO ( cat.)
O || (A) Ph CH2 CH O C Ph | CH3
CH3 O | || (B) Ph C O C Ph | CH3
Ph O | || (C) Ph C O C CH3 | CH3
Ph O | || (D) Ph CH CH2 O C CH3
(1) excess PhMgBr X,
12.
( 2) H2O
(A)
(B)
(C)
(D)
O || (1) H3 O , LiAlH4 X, C 2H5 O C (CH2 )2 CN (excess ( 2) ¯OH )
(A)
14.
(D) H2 / Ni
4 2 X,
11.
13.
(C) MPV
(B)
(C)
(D)
1. PhMgBr
H2C = CHCH2CN X, 2. H
Ph | (A) CH3 C CH CHO
Ph | (B) CH2 CH CH CHO
(C) H2C = CHCH2COPh
(D) CH3 – CH = CH – COPh
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 174 15.
In the given reaction
( i ) CH MgBr (1 equiv )
3 P,,
( ii) H 2 O / H
(A)
(B)
(C)
(D)
P O Ca ( OH)2 ,I2 Q 4 10 P R, compound R is MeMgBr
16.
H3O
(A) 17.
(B)
(D)
(C)
(D) All lHkh
Which gives unsaturated acid on heating ? (A)
(B)
H O
18.
19.
(C)
3
Select the correct statement : (A) Alkyl C–O bond breaks and O18 is contained in acid (B) Alkyl C–O bond breaks and O18 is contained in alcohol (C) Acyl C–O bond breaks and O18 is contained in alcohol (D) Acyl C–O bond breaks and O18 is contained in acid An optically active compound ‘X’ has molecular formula C4H8O3. It evolves CO2 with aq. NaHCO3. ‘X’ reacts with LiAlH4 to give achiral compound ‘X’ is (A)
(B)
(C)
(D)
(C)
(D)
BH .THF
I2 / NaOH 3
20.
H2O 2 .OH–
(A)
(B)
H
X,
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 175 21.
Consider following compounds
22.
Which of the above compounds reacts with NaHCO3 giving CO2 : (A) I, II and III (B) I and III (C) II and IIII (D) I and II The carboxylic acid which has maximum solubility in water is : (A) Phthalic acid (B) Succinic acid (C) Malonic acid (D) Salicylic acid Peroxyacetic acid (CH3CO3H) is a weaker acid than acetic acid (CH3CO2H) since :
23.
(A) negative charge in
can’t be delocalised into the carbonyl group
(B) CH3 group in CH3CO3 H shows +I effect (C) both are correct (D) none is correct
24.
A,
(A)
(B)
(C)
(D) None
25.
In case of Hoffmann bromamide degradation reaction, intermdediate RNCO is formed by : (A) intramolecular migration (B) Intermolecular migration (C) hydrolysis is RCONH2 (D) none is correct
26.
When RCOOH reacts with NaHCO3, CO2 is formed. CO2 contains carbon of (A) RCOOH (B) NaHCO3 (C) Both (D) none Consider following reactions
27.
I
H .Pd / BaSO
4 2
,
LiAlH4 II ,
LiAlH( t BuO)3 III
is obtained from RCOCl using (A) I, II and III
28.
. +
(A)
(B) I and II
(i) CH CH O
3 2
(ii) H
(B)
(C) I and III
(D) II and III
(C)
(D)
A,
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 176
29.
Reactivity order of different nucleophiles (NH2– CH3COO–, OH) is in order : (A) NH2– < CHC O O– < OH–
(B) CH3COO– < OH– < NH2–
(C) NH2– < OH– < CH3COO–
(D) CH3COO– < NH2 < OH–
30.
+ KOH + Br2
Intermediates of this reactions is/are expect to be :
(A) R–N=C=O
CH OH
3
31. (A)
32.
(B)
(C)
(D) A, C
(C) both are correct
(D) None is correct
A, A is gSA (B)
Give the structure of the expected product of the following reaction :
+ CH3NH2
(A)
33.
(B)
(C)
(D) None
Consider the following reaction : CH3–NH–CH2–CH2–CH2–OH + (A)
(B)
(C)
(D)
produt in the above reaction
is the only product
is the only product
is a major product and
is a minor product
is a major product and
product.
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is a minor
CARBOXYLIC ACIDS & ACID DERIVATIVES | 177 34.
In the given reaction :
+
CH3OH H (excess )
(A)
35.
[X] is :
(B)
(C)
(D) Mixture of all
In the given reaction : C6H5–CH3
NBS
h
A
NaCN
B
H O
3
C
C is gSA
(A) C6H5–CH2 –Br 36.
(B) C6H5–CH2–CN
(C)
(D) C6H5–CH2–COOH
Consider the following reaction
CH3
CH2–
+ CH3MgBr (1 equiv) Product
The product will be :
(A) CH3
CH2–
(B)
–
–
–CH2–CHO
(C)
–
–
–CH2–
(D)
37.
–CH2–CHO
In the given reaction :
SOCl
(CH – CH ) Cd
2 A 3 22
Pyridine
B
B will be :
(A)
(B)
(C)
(D)
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 178
More than one choice type 38.
Find out the possible products in the following reaction. O
H–O–N
Cl
LiAlH 4 ?
O O
O
O
O
HO
NH2
NH2
NH2
(A)
CH2OH
CH2OH
CH2OH
OH
OH
(B)
O
(C) CH3CH2OH (D) H O
CH2 – OH
OH
CH2 – OH H O
39.
Benzoyl chloride is less reactive than acetyl chloride for nucleophilic acyl substitution reaction because :
(A) group of benzoyl chloride is in conjugation with benzene ring. This makes benzoyl chloride more stable than acyl chloride. (B) CH3 group of acetyl chloride makes C–Cl bond stronger due to the + I effect. (C) C6H5– group of benzoyl chloride makes C–Cl bond weaker due to + R effect. (D) CH3–group makes C–Cl bond weaker due to + I effect and C6H5 group makes C–Cl bond stronger due to – I effect.
PART - II SUBJECTIVE QUESTIONS 1.
Complete the following by supplying (A) to (F) Alcoholic KOH P / Br2 Br2 CH3CH2CH2Br [B] [A]
CCl4
2
KOH (alc .) Hg NH2OH [F] [C] [D] [E] Followed by NaNH2
2.
3.
4.
dil. H2SO 4
H
Bring out the following conversions : (i) Ethanal to 2-Hydroxypropanoic acid (ii) Carbon and Hydrogen to Ethanoic acid (iii) Ethanol to Propanone (iv) Ethanal to Lactic acid (v) Ethanal to 2-hydroxybut-3-enoic acid (vi) Benzamide to Toluene (vii) Benzoic acid to Benzophenone (viii) Acetone to 2-Methylpropan- 2-ol (ix) Malonic acid to Acetic acid. Compound A (C6H12O2) on reaction with LiAlH4 yielded two compounds B and C. The compound B on oxidation gave D, which on treatment with aqueous alkali and subsequent heating furnished E.E on catalytic hydrogenation gave C.D was oxidised to form F which is monobasic carboxylic acid with molecular mass 60. Deduce the structures of A, B, C, D. 30 g of monobasic carboxylic acid A consumed 0.5 moles of NaOH for its neutralisation. A is subjected to following treatments : Monochloro Esterifica tion KCN (alc.) (Ester) with C2H5OH
Cl / P [A] 2 derivative Hydrolysis
Heat
Acid
150 ºC
Dicarboxylic acid [C] A [B]
Give structures of A, B, C providing the sequence of reactions.
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 179 5.
Complete the following by supplying [A] to [F] : [A ] C H6 [B] CH4 CH3Br 6 AlCl3 (anhyd .)
6.
7.
8.
[C]
An organic compound ‘A’ on treatment with ethyl alcohol gives carboxylic acid ‘B’ and compound ‘C’. Hydrolysis of ‘C’ under acidic conditions give ‘B’ and ‘D’. oxidation of ‘D’ with KMnO4 also gives B.B on heating with Ca(OH)2 gives E with molecular formula C3H6O. E does not give Tollen’s test or reduce Fehling solution but forms 2, 4-Dinitrophenyl hydrazone. Identify A, B, C, D, E. When a gas A is passed through dry KOH at low temperature, a deep red coloured compound B and a gas C are obtained. The gas A on reaction with but-2-ene followed by treatement with Zn/H2O yields acetaldehyde. Identify A, B, C. Complete the following sequence of reactions with appropriate structures. ( i) KOH ( alc .)
P / Br2 (A) (B). CH3CH2COOH ( ii) H
9.
An acidic compound (A), C4H6O3 looses its optical activity on strong heating yielding (B), C4H6O2 which reacts readily with KMnO4 and decolourises it. (B) forms a derivative (C) with SOCl2 which on reaction with (CH3)2 NH gives (D). The compound (A) also forms unstable compound (E) on treatment with dilute chromic acid (E) decarboxylate readily to give (F) C3H6O which on treatment with amalgamated zinc and HCl gives hydrocarbon (G). Give structures of (A) to (G).
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 180
MISSCELLANEOUS QUESTION BANK PART - I 1.
(C)
2.
(C)
3.
(A)
4.
(D)
5.
(D)
6.
(D)
7.
(A)
8.
(C)
9.
(C)
10.
(B)
11.
(C)
12.
(C)
13.
(D)
14.
(C)
15.
(B)
16.
(B)
17.
(B)
18.
(C)
19.
(C)
20.
(D)
21.
(C)
22.
(C)
23.
(A)
24.
(C)
25.
(A)
26.
(B)
27.
(C)
28.
(B)
29.
(B)
30.
(B)
31.
(A)
32.
(C)
33.
(C)
34.
(C)
35.
(D)
36.
(A)
37.
(C)
38.
(B, C)
39.
(A, D)
PART - II
1.
3.
(A) = CH3CH2CH2OH ;
(B) = CH3CH = CH2 ;
(D) = CH3C CH ;
O || (E) = CH3 — C — CH3 ;
O || A is CH3CH2CH2C — OC2H5
(Ethyl butanoate)
B is C2H5OH
(Ethanol)
C is CH3CH2CH2CH2OH
(1-Butanol)
D is CH3CHO
(Ethanal)
(A) = CH3COOH ;
5.
[A] = Br2 /hv
[B] =
[C] = Cl2 (g)/hv
[D] =
[E] =
[F] = NH2OH/H+
O O || || (A) = CH3 — C — O — C — CH3 ; O || (C) = C 2H5 OC — CH3 ;
;
(F) = CH3 — C — CH3 || N – OH
4.
6.
(B) =
Br | (C) = CH3CH — CH2Br ;
(C) =
O || (B) = CH 3 C — OH
(D) = CH3COOH + C2H5OH ;
O || (E) = CH3 — C — CH3
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CARBOXYLIC ACIDS & ACID DERIVATIVES | 181 7.
(A) = CH3 CH CHCH3 (B) = 2KO3 + KOH.H2O (C) =
9.
1 O2 2
(A) =
;
(B) =
(C) =
;
(D) =
(E) =
;
(F) =
;
(G) =
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