JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 12

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Multiple Choice Questions

Question 1: In the sequence CH3COOH → (SOCl2) A → (benzene, AlCl3) B → (HCN) C → (H3O+) D, the product D is:
(a) 2-hydroxy-2-phenylpropanoic acid
(b) 3-hydroxy-3-phenylbutanoic acid
(c) 2-cyano-2-phenylpropan-2-ol
(d) 2-hydroxy-2-benzylpropanoic acid
Show Answer & Explanation

Answer: (a) 2-hydroxy-2-phenylpropanoic acid

Explanation:
1. A is acetyl chloride and B is acetophenone (Friedel–Crafts acylation).
2. HCN adds to give the cyanohydrin C6H5C(OH)(CH3)CN.
3. Hydrolysis of CN gives COOH, so D is C6H5C(OH)(CH3)COOH.

Teacher's Note:
1. A cyanohydrin hydrolyses to an α-hydroxy acid.

Question 2: In the sequence CH3CH2COOH → (P/Cl2) A → (alc. KOH) B, the structure of B is:
(a) CH2=CH–COOH
(b) CH3–CHCl–COOH
(c) CH2Cl–CH2–COOH
(d) CH3CH2COCl
Show Answer & Explanation

Answer: (a) CH2=CH–COOH

Explanation:
1. Hell–Volhard–Zelinsky reaction gives 2-chloropropanoic acid (A). Alcoholic KOH eliminates HCl to give prop-2-enoic (acrylic) acid.

Teacher's Note:
1. P/Cl2 chlorinates the α-carbon.

Question 3: Which of the following N-methylbenzamides is the most reactive towards hydrolysis?
(a) 4-methyl-N-methylbenzamide
(b) 4-nitro-N-methylbenzamide
(c) 4-chloro-N-methylbenzamide
(d) N-methylbenzamide
Show Answer & Explanation

Answer: (b) 4-nitro-N-methylbenzamide

Explanation:
1. The –NO2 group withdraws electrons by –M and –I effects and increases the positive character of the carbonyl carbon, so hydrolysis is fastest. The methyl group slows it.

Teacher's Note:
1. Electron-withdrawing groups speed up nucleophilic attack on carbonyl carbon.

Question 4: Which of the following reagents can be used to distinguish phenol from benzoic acid?
(a) Victor Meyer test
(b) neutral FeCl3
(c) aqueous NaOH
(d) Tollen's reagent
Show Answer & Explanation

Answer: (b) neutral FeCl3

Explanation:
1. Phenol gives a violet colour with neutral FeCl3 while benzoic acid gives a buff precipitate; NaOH dissolves both.

Teacher's Note:
1. Neutral FeCl3 is a test for phenols.

Question 5: Acyl chlorides undergo:
(a) nucleophilic addition reactions
(b) nucleophilic substitution reactions
(c) electrophilic substitution reactions
(d) electrophilic addition reactions
Show Answer & Explanation

Answer: (b) nucleophilic substitution reactions

Explanation:
1. The carbonyl carbon is attacked by a nucleophile (addition), and chloride then leaves, giving overall nucleophilic acyl substitution.

Teacher's Note:
1. Acid derivatives show nucleophilic acyl substitution.

Question 6: Reduction of acetamide with LiAlH4 gives:
(a) CH3CH2NH2
(b) (CH3)2CHNH2
(c) (CH3)3CNH2
(d) (CH3CH2)2NH
Show Answer & Explanation

Answer: (a) CH3CH2NH2

Explanation:
1. LiAlH4 reduces the amide C=O to CH2: CH3CONH2 → CH3CH2NH2.

Teacher's Note:
1. Amides give amines with the same number of carbons.

Question 7: Which acid derivative is used in the preparation of aldehydes by the Rosenmund reduction?
(a) ester
(b) acid
(c) acid halide
(d) alcohol
Show Answer & Explanation

Answer: (c) acid halide

Explanation:
1. RCOCl + H2 over Pd/BaSO4 gives RCHO + HCl; barium sulphate poisons the catalyst and stops further reduction.

Teacher's Note:
1. Rosenmund uses an acid chloride.

Question 8: Which of the following is the most reactive towards nucleophilic acyl substitution?
(a) CH3CONH2
(b) CH3COOCH3
(c) CH3COCl
(d) CH3COOC2H5
Show Answer & Explanation

Answer: (c) CH3COCl

Explanation:
1. Reactivity order: acid chloride > anhydride > ester > amide, because chloride is the best leaving group.

Teacher's Note:
1. Better leaving group means faster substitution.

Question 9: Acetamide on heating with P2O5 gives:
(a) CH3COOH
(b) CH3–CN
(c) CH3–CH3
(d) CH3–CHO
Show Answer & Explanation

Answer: (b) CH3–CN

Explanation:
1. P2O5 dehydrates the primary amide: CH3CONH2 → CH3C≡N + H2O.

Teacher's Note:
1. Dehydration of amides gives nitriles.

Question 10: Phthalaldehyde (benzene-1,2-dicarbaldehyde) is treated with NaOH to give A, and A on heating with H+ gives:
(a) 2-(hydroxymethyl)benzoic acid
(b) phthalide (isobenzofuran-1(3H)-one)
(c) indan-2-one
(d) 1,3-dihydroisobenzofuran
Show Answer & Explanation

Answer: (b) phthalide (isobenzofuran-1(3H)-one)

Explanation:
1. With no α-H, intramolecular Cannizzaro reaction gives 2-(hydroxymethyl)benzoate. On acidification and heating, the –OH and –COOH close to the five-membered lactone phthalide.

Teacher's Note:
1. Hydroxy acids that can form five-membered rings lactonise.

Question 11: The correct order of reactivity of the following compounds towards hydrolysis is: (i) benzoyl chloride, (ii) 4-nitrobenzoyl chloride, (iii) 4-methylbenzoyl chloride, (iv) 4-formylbenzoyl chloride.
(a) (i) > (ii) > (iii) > (iv)
(b) (iv) > (ii) > (i) > (iii)
(c) (ii) > (iv) > (i) > (iii)
(d) (ii) > (iv) > (iii) > (i)
Show Answer & Explanation

Answer: (c) (ii) > (iv) > (i) > (iii)

Explanation:
1. Electron-withdrawing groups make the carbonyl carbon more positive and speed up attack by water. The –NO2 group is a stronger electron-withdrawing group than –CHO, while –CH3 donates electrons and slows hydrolysis.
2. Order: nitro > formyl > H > methyl.

Teacher's Note:
1. Substituent effects: −M/−I increase reactivity and +I decreases it.

Question 12: Which one of the following, on treatment with 50 percent aqueous NaOH, yields the corresponding alcohol and acid?
(a) C6H5CHO
(b) CH3CH2CH2CHO
(c) CH3COCH3
(d) CH3CHO
Show Answer & Explanation

Answer: (a) C6H5CHO

Explanation:
1. Only benzaldehyde has no α-hydrogen and undergoes the Cannizzaro reaction to give benzyl alcohol and benzoate. The others have α-hydrogens and undergo aldol reactions or none at all.

Teacher's Note:
1. Cannizzaro needs an aldehyde without α-hydrogen.

Question 13: Acetophenone, when reacted with a base such as C2H5ONa, yields a stable compound with the structure:
(a) 2,3-diphenylbutane-2,3-diol
(b) 1,2-diphenylethane-1,2-diol
(c) 1,3-diphenylbut-2-en-1-one
(d) 1,3-diphenylbutan-1-one
Show Answer & Explanation

Answer: (c) 1,3-diphenylbut-2-en-1-one

Explanation:
1. Two molecules undergo aldol condensation: the enolate of one attacks the carbonyl group of the other, and dehydration gives the conjugated, stable enone C6H5C(CH3)=CHCOC6H5 (dypnone).

Teacher's Note:
1. Aldol condensation gives α,β-unsaturated carbonyl compounds on dehydration.

Question 14: Propionic acid with Br2/P yields a dibromo product (Hell–Volhard–Zelinsky). Its structure is:
(a) CH3–CBr2–COOH
(b) CH2Br–CHBr–COOH
(c) H–CBr2–CH2COOH
(d) CH2Br–CH2–COBr
Show Answer & Explanation

Answer: (a) CH3–CBr2–COOH

Explanation:
1. Only the α-carbon is halogenated in the HVZ reaction. Both α-hydrogens of propanoic acid are replaced, giving 2,2-dibromopropanoic acid.

Teacher's Note:
1. Halogenation in the HVZ reaction is at the α-position only.

Question 15: Phenol → (Zn dust, heat) X → (CH3Cl, anhydrous AlCl3) Y → (KMnO4, heat) Z. The product Z is:
(a) benzene
(b) toluene
(c) benzaldehyde
(d) benzoic acid
Show Answer & Explanation

Answer: (d) benzoic acid

Explanation:
1. Zinc dust reduces phenol to benzene (X).
2. Friedel–Crafts methylation gives toluene (Y).
3. Alkaline or acidic KMnO4 oxidises the side chain completely to benzoic acid (Z).

Teacher's Note:
1. Side-chain oxidation of toluene by KMnO4 gives benzoic acid.

Question 16: In the sequence ethylbenzene → (KMnO4/KOH) B → (Br2, FeBr3) C → (C2H5OH, H+) D, the product D is:
(a) ethyl 3-bromobenzoate
(b) ethyl 2-phenyl-2-bromoacetate
(c) ethyl 2-(3-bromophenyl)acetate
(d) ethyl 2-ethoxybenzoate
Show Answer & Explanation

Answer: (a) ethyl 3-bromobenzoate

Explanation:
1. KMnO4 oxidises the ethyl side chain to benzoic acid (B).
2. The –COOH group is meta directing, so bromination gives 3-bromobenzoic acid (C).
3. Esterification with ethanol gives ethyl 3-bromobenzoate.

Teacher's Note:
1. Meta-directing groups put the incoming bromine at position 3.

Question 17: Which of the following reactions will NOT result in the formation of carbon–carbon bonds?
(a) Friedel–Crafts acylation
(b) Reimer–Tiemann reaction
(c) Cannizzaro reaction
(d) Wurtz reaction
Show Answer & Explanation

Answer: (c) Cannizzaro reaction

Explanation:
1. Friedel–Crafts, Reimer–Tiemann and Wurtz reactions all form C–C bonds. The Cannizzaro reaction is a redox disproportionation (hydride transfer) with no new C–C bond.

Teacher's Note:
1. Count the bonds formed in each reaction.

Question 18: The suitable reagent to convert CH3–CH=CH–CHO into CH3CH2CH2CH2OH is:
(a) NaBH4
(b) LiAlH4
(c) Zn–Hg/HCl
(d) H2/Pd
Show Answer & Explanation

Answer: (d) H2/Pd

Explanation:
1. Both the C=C and the C=O must be reduced. NaBH4 and LiAlH4 reduce only the carbonyl group (giving crotyl alcohol) and Zn–Hg/HCl removes the carbonyl oxygen. Catalytic hydrogenation reduces both.

Teacher's Note:
1. H2/Pd reduces C=C as well as C=O.

Question 19: CH3COCH2COOCH3 is treated with NaBH4. The product is:
(a) CH3CH(OH)CH2COOCH3
(b) CH3CH(OH)CH2CH2OH
(c) CH3COCH2CH2OH
(d) CH3CH(OH)CH2COOH
Show Answer & Explanation

Answer: (a) CH3CH(OH)CH2COOCH3

Explanation:
1. NaBH4 is a mild reductant that reduces the ketone to a secondary alcohol but does not reduce the ester group.

Teacher's Note:
1. LiAlH4 would reduce both groups.
2. NaBH4 reduces aldehydes and ketones only.

Question 20:
Assertion (A): Acetophenone shows aldol condensation.
Reason (R): Acetophenone contains a carbonyl group.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Both statements are true.
2. However, aldol condensation requires α-hydrogen atoms, not just a carbonyl group, so R does not explain A.

Teacher's Note:
1. Check for α-hydrogen.

Question 21: m-Bromobenzoic acid → (SOCl2) B → (NH3) C → (NaOH, Br2) D. The product D is:
(a) 3-bromobenzenesulphonamide
(b) 3-aminobenzoic acid
(c) 3-bromoaniline
(d) 3-bromobenzamide
Show Answer & Explanation

Answer: (c) 3-bromoaniline

Explanation:
1. SOCl2 gives the acid chloride, NH3 gives the amide (3-bromobenzamide), and the Hofmann bromamide reaction gives 3-bromoaniline with one carbon less.

Teacher's Note:
1. Hofmann degradation converts RCONH2 into RNH2.

Question 22: An organic compound A on treatment with NH3 gives B, which on heating gives C. C when treated with Br2 in the presence of KOH produces ethylamine. Compound A is:
(a) CH3CH2COOH
(b) CH3COOH
(c) CH3CH2CH2COOH
(d) (CH3)2CHCOOH
Show Answer & Explanation

Answer: (a) CH3CH2COOH

Explanation:
1. Hofmann bromamide degradation removes one carbon, so C is propanamide, CH3CH2CONH2. B is ammonium propanoate and A is propanoic acid.

Teacher's Note:
1. Count carbons: ethylamine has 2, so the amide has 3.

Question 23: Match the compounds in List I with List II. (a) benzaldehyde, (b) phthalic anhydride, (c) phenyl benzoate, (d) methyl salicylate. (i) phenolphthalein, (ii) benzoin condensation, (iii) oil of wintergreen, (iv) Fries rearrangement.
(a) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
(b) (a)-(iv), (b)-(i), (c)-(iii), (d)-(ii)
(c) (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)
(d) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
Show Answer & Explanation

Answer: (a) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)

Explanation:
1. Benzaldehyde undergoes benzoin condensation, phthalic anhydride with phenol gives phenolphthalein, phenyl esters undergo Fries rearrangement, and methyl salicylate is oil of wintergreen.

Teacher's Note:
1. Know the named reactions and the uses.

Question 24: Decarboxylation occurs with the maximum rate in:
(a) CH3COOH
(b) C6H5COOH
(c) C6H5CH2COOH
(d) CH3COCH2COOH
Show Answer & Explanation

Answer: (d) CH3COCH2COOH

Explanation:
1. β-Keto acids lose CO2 very easily on gentle heating through a cyclic six-membered transition state.

Teacher's Note:
1. A carbonyl group at the β-position assists decarboxylation.

Question 25: In the sequence CH3Br → (KCN) A → (H3O+) B → (LiAlH4, ether) C, the end product C is:
(a) acetaldehyde
(b) ethyl alcohol
(c) acetone
(d) methane
Show Answer & Explanation

Answer: (b) ethyl alcohol

Explanation:
1. CH3CN (A) hydrolyses to acetic acid (B). LiAlH4 reduces the acid to ethanol (C).

Teacher's Note:
1. Carboxylic acids are reduced to primary alcohols by LiAlH4.

Question 26: CH3CHO and C6H5CH2CHO can be distinguished chemically by:
(a) Tollen's reagent test
(b) Fehling solution test
(c) Benedict test
(d) iodoform test
Show Answer & Explanation

Answer: (d) iodoform test

Explanation:
1. Both are aldehydes with α-hydrogens and both give Tollens', Fehling's and Benedict's tests. Only acetaldehyde has the CH3CO– group and gives yellow iodoform.

Teacher's Note:
1. Iodoform test needs a methyl ketone group.

Question 27: Benzoyl chloride is treated with H2 over Pd–BaSO4. The product is:
(a) C6H5COCH3
(b) C6H5Cl
(c) C6H5CHO
(d) C6H5OH
Show Answer & Explanation

Answer: (c) C6H5CHO

Explanation:
1. This is the Rosenmund reduction, which stops at the aldehyde stage because BaSO4 poisons the catalyst.

Teacher's Note:
1. Without poisoning, the aldehyde would be reduced further.

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