JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 06

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Chapter-wise Objective Questions: Aldehydes Ketones and Carboxylic Acids

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Multiple Choice Questions

Question 1: Match: (a) \( \ce{CH3CH2CH2CH2NH2} \), (b) \( \ce{CH3C#CH} \), (c) \( \ce{CH3CH2COOCH3} \), (d) \( \ce{CH3CH(OH)CH3} \) with (i) alkaline hydrolysis, (ii) KOH/alcohol and \( \ce{CHCl3} \) gives a bad smell, (iii) white precipitate with ammoniacal \( \ce{AgNO3} \), (iv) Lucas reagent gives cloudiness after 5 minutes.
(a) (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
(b) (a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)
(c) (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)
(d) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
Show Answer & Explanation

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

Explanation:
1. A primary amine gives the carbylamine test (ii).
2. A terminal alkyne gives a white precipitate with \( \ce{AgNO3} \) (iii).
3. An ester undergoes alkaline hydrolysis (i).
4. A secondary alcohol shows cloudiness with Lucas reagent after about 5 minutes (iv).

Teacher's Note:
1. Match each test to its functional group.

Question 2: The suitable reagent to convert \( \ce{CH3-CH=CH-CHO} \) into \( \ce{CH3CH2CH2CH2OH} \) is:
(a) \( \ce{NaBH4} \)
(b) \( \ce{LiAlH4} \)
(c) \( \ce{Zn-Hg/HCl} \)
(d) \( \ce{H2/Pd} \)
Show Answer & Explanation

Answer: (d) \( \ce{H2/Pd} \)

Explanation:
1. Both the C=C and the C=O must be reduced.
2. Catalytic hydrogenation does both; NaBH4 and LiAlH4 reduce only C=O.

Teacher's Note:
1. Catalytic hydrogenation reduces C=C and C=O.

Question 3: \( \ce{CH3-CO-CH2-COOCH3} \) is treated with \( \ce{NaBH4} \). The product is:
(a) \( \ce{CH3CH(OH)CH2COOCH3} \)
(b) \( \ce{CH3CH(OH)CH2CH2OH} \)
(c) \( \ce{CH3COCH2CH2OH} \)
(d) \( \ce{CH3CH(OH)CH2COOH} \)
Show Answer & Explanation

Answer: (a) \( \ce{CH3CH(OH)CH2COOCH3} \)

Explanation:
1. NaBH4 reduces the ketone to a secondary alcohol but does not reduce the ester.

Teacher's Note:
1. NaBH4 is a mild, selective reducing agent.

Question 4: A series of reactions: 3-methylbutan-2-ol with \( \ce{H+} \)/heat gives A (major) and B (minor); A with HBr (dark, no peroxide) gives C (major) and D (minor). The major products A and C are respectively:
(a) 2-methylbut-1-ene and 2-bromo-3-methylbutane
(b) 2-methylbut-2-ene and 2-bromo-2-methylbutane
(c) 2-methylbut-2-ene and 2-bromo-3-methylbutane
(d) 2-methylbut-1-ene and 2-bromo-2-methylbutane
Show Answer & Explanation

Answer: (b) 2-methylbut-2-ene and 2-bromo-2-methylbutane

Explanation:
1. Dehydration follows Saytzeff's rule to give the more substituted alkene, 2-methylbut-2-ene.
2. HBr adds by Markovnikov's rule, placing bromine on the tertiary carbon.

Teacher's Note:
1. Saytzeff for elimination, Markovnikov for addition.

Question 5: m-Bromobenzoic acid is treated with \( \ce{SOCl2} \) to give B, then \( \ce{NH3} \) to give C, then NaOH/\( \ce{Br2} \) to give D. D is:
(a) 3-bromobenzenesulphonamide
(b) 2-amino-5-bromobenzoic acid
(c) 3-bromoaniline
(d) 3-bromobenzamide
Show Answer & Explanation

Answer: (c) 3-bromoaniline

Explanation:
1. The acid chloride (B) gives the amide (C).
2. Hofmann bromamide degradation removes the carbonyl carbon to give 3-bromoaniline.

Teacher's Note:
1. Hofmann degradation gives an amine with one carbon less.

Question 6: An organic compound A on treatment with \( \ce{NH3} \) gives B, which on heating gives C. C when treated with \( \ce{Br2} \) in the presence of KOH produces ethylamine. Compound A is:
(a) \( \ce{CH3CH2COOH} \)
(b) \( \ce{CH3COOH} \)
(c) \( \ce{CH3CH2CH2COOH} \)
(d) \( \ce{(CH3)2CHCOOH} \)
Show Answer & Explanation

Answer: (a) \( \ce{CH3CH2COOH} \)

Explanation:
1. Ethylamine from Hofmann degradation requires propanamide (C).
2. Propanamide comes from heating the ammonium salt (B) of propanoic acid (A).

Teacher's Note:
1. Hofmann removes one carbon.

Question 7: Match: (a) benzaldehyde, (b) phthalic anhydride, (c) phenyl benzoate, (d) methyl salicylate with (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.
2. Phthalic anhydride with phenol forms phenolphthalein.
3. Phenyl benzoate undergoes Fries rearrangement.
4. Methyl salicylate is oil of wintergreen.

Teacher's Note:
1. Learn the name reactions with examples.

Question 8: Decarboxylation occurs with maximum rate in:
(a) \( \ce{CH3COOH} \)
(b) \( \ce{C6H5COOH} \)
(c) \( \ce{C6H5CH2COOH} \)
(d) \( \ce{CH3COCH2COOH} \)
Show Answer & Explanation

Answer: (d) \( \ce{CH3COCH2COOH} \)

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

Teacher's Note:
1. Beta-keto acid decarboxylation is the key step in acetoacetic ester synthesis.

Question 9: 3-Chlorobenzaldehyde is heated with 50% KOH. The products are:
(a) 3-chlorobenzyl alcohol and 3-chlorobenzoate
(b) 3-hydroxybenzyl alcohol and 3-hydroxybenzoate
(c) 3-chlorophenylethanol and 3-chlorophenylacetate
(d) 3-hydroxybenzyl alcohol and 1,3-dihydroxybenzene
Show Answer & Explanation

Answer: (a) 3-chlorobenzyl alcohol and 3-chlorobenzoate

Explanation:
1. Aromatic aldehydes without alpha-H give Cannizzaro reaction, and the aryl chloride is inert to alkali under these conditions.

Teacher's Note:
1. Aryl halides do not substitute easily.

Question 10: \( \ce{CH3CHO} \) and \( \ce{C6H5CH2CHO} \) can be distinguished chemically by:
(a) Tollens' reagent test
(b) Fehling's solution test
(c) Benedict's test
(d) iodoform test
Show Answer & Explanation

Answer: (d) iodoform test

Explanation:
1. Acetaldehyde has the \( \ce{CH3CO-} \) group and gives the iodoform test; phenylacetaldehyde does not.
2. Both reduce Tollens', Fehling's and Benedict's reagents.

Teacher's Note:
1. The iodoform test needs \( \ce{CH3CO-} \) or \( \ce{CH3CH(OH)-} \).

Question 11: Benzoyl chloride with \( \ce{H2} \) over Pd-BaSO4 gives:
(a) \( \ce{C6H5COCH3} \)
(b) \( \ce{C6H5Cl} \)
(c) \( \ce{C6H5CHO} \)
(d) \( \ce{C6H5OH} \)
Show Answer & Explanation

Answer: (c) \( \ce{C6H5CHO} \)

Explanation:
1. Rosenmund reduction of an acid chloride gives the aldehyde.

Teacher's Note:
1. The poisoned catalyst stops at the aldehyde.

Question 12: Which reagent cannot be used to prepare acetophenone?
(a) \( \ce{C6H5COCl + (CH3)2Cd} \)
(b) \( \ce{C6H5COCl + CH3MgBr} \)
(c) \( \ce{C6H5CN + C2H5MgBr} \)
(d) \( \ce{C6H6 + (CH3CO)2O/AlCl3} \)
Show Answer & Explanation

Answer: (c) \( \ce{C6H5CN + C2H5MgBr} \)

Explanation:
1. Benzonitrile with ethylmagnesium bromide gives propiophenone, not acetophenone.
2. The other three routes give acetophenone (the acid chloride route under controlled low-temperature conditions).

Teacher's Note:
1. Count the carbons that come from the Grignard reagent.

Question 13: An organic compound X (C5H10O) forms a phenylhydrazone, gives negative iodoform and Tollens' tests and reduces to n-pentane. X could be:
(a) pentan-2-one
(b) pentan-3-one
(c) n-pentanol
(d) pentanal
Show Answer & Explanation

Answer: (b) pentan-3-one

Explanation:
1. Phenylhydrazone formation shows a carbonyl group, and negative Tollens' test excludes pentanal.
2. Negative iodoform test excludes pentan-2-one.
3. Reduction to n-pentane shows an unbranched chain, so X is pentan-3-one.

Teacher's Note:
1. Work through each test one by one.

Question 14: 3-Ethylbenzoic acid is treated with \( \ce{KMnO4/KOH} \) heat, then \( \ce{CH3I} \). The product is:
(a) methyl benzoate
(b) a diester with an extra \( \ce{CH2} \) group
(c) dimethyl 1,3-benzenedicarboxylate (dimethyl isophthalate)
(d) methyl 4-methylbenzoate
Show Answer & Explanation

Answer: (c) dimethyl 1,3-benzenedicarboxylate (dimethyl isophthalate)

Explanation:
1. Oxidation of the ethyl side chain gives a second carboxylate, so a 1,3-dicarboxylate.
2. Methyl iodide converts both carboxylate groups into methyl esters.

Teacher's Note:
1. Alkyl side chains are oxidised to -COOH.

Question 15: Cyclohex-2-enone reacts with \( \ce{H2} \) (1 atm), Pd/carbon, ethanol. The product A is:
(a) cyclohex-2-enol
(b) a cyclohexanol cation
(c) cyclohex-3-enol
(d) cyclohexanone
Show Answer & Explanation

Answer: (d) cyclohexanone

Explanation:
1. At 1 atm with Pd/C, the C=C double bond is hydrogenated selectively.
2. The ketone remains, giving cyclohexanone.

Teacher's Note:
1. C=C is hydrogenated more easily than C=O.

Question 16: Toluene is converted with reagent P to benzaldehyde, which on reaction with Zn-Hg/HCl gives Q. P and Q are respectively:
(a) \( \ce{CrO3/H3O+} \), benzyl alcohol
(b) \( \ce{CrO2Cl2/H3O+} \), toluene
(c) \( \ce{NaOCl} \), toluene
(d) \( \ce{KMnO4/H+} \), benzyl alcohol
Show Answer & Explanation

Answer: (b) \( \ce{CrO2Cl2/H3O+} \), toluene

Explanation:
1. The Etard reaction (\( \ce{CrO2Cl2} \)) oxidises toluene to benzaldehyde.
2. Clemmensen reduction of benzaldehyde gives toluene.

Teacher's Note:
1. Strong oxidants would give benzoic acid.

Question 17: The appropriate reagent for converting \( \ce{CH2=CH-CH2-CHO} \) into \( \ce{CH3CH2CH2CH2OH} \) is:
(a) metal and \( \ce{H2} \)
(b) \( \ce{LiAlH4} \)
(c) \( \ce{NaBH4} \)
(d) all of these
Show Answer & Explanation

Answer: (a) metal and \( \ce{H2} \)

Explanation:
1. Both the C=C and C=O groups must be reduced.
2. Only catalytic hydrogenation does both; hydrides leave the C=C.

Teacher's Note:
1. LiAlH4 and NaBH4 do not reduce isolated C=C.

Question 18: Benzoyl chloride is converted to benzaldehyde by reagent P. P is:
(a) \( \ce{H2} \) with Pd-\( \ce{BaSO4} \)
(b) \( \ce{LiAlH4} \)
(c) \( \ce{NaBH4} \)
(d) \( \ce{H2} \) with Pd
Show Answer & Explanation

Answer: (a) \( \ce{H2} \) with Pd-\( \ce{BaSO4} \)

Explanation:
1. The poisoned catalyst in Rosenmund reduction stops at the aldehyde stage.

Teacher's Note:
1. LiAlH4 would give benzyl alcohol.

Question 19: Cyclohex-3-enecarbonitrile is treated with DIBAL-H followed by water. The product is:
(a) cyclohex-3-ene carbaldehyde
(b) cyclohexane carbaldehyde
(c) cyclohex-3-ene carboxamide
(d) cyclohexylamine
Show Answer & Explanation

Answer: (a) cyclohex-3-ene carbaldehyde

Explanation:
1. DIBAL-H partially reduces a nitrile to an imine, which on hydrolysis gives an aldehyde.
2. The C=C bond is untouched.

Teacher's Note:
1. DIBAL-H is also used for esters to aldehydes at low temperature.

Question 20: Benzaldehyde and acetic anhydride react with sodium acetate (Perkin reaction) to give A, which is reduced with \( \ce{LiAlH4} \) then \( \ce{H2O} \) to give B. B is:
(a) \( \ce{Ph-CH2CH2CH2-OH} \)
(b) \( \ce{Ph-CH=CH-CH2-OH} \)
(c) \( \ce{Ph-CH2CH2-COOH} \)
(d) \( \ce{Ph-CH=CH-CHO} \)
Show Answer & Explanation

Answer: (b) \( \ce{Ph-CH=CH-CH2-OH} \)

Explanation:
1. Perkin reaction gives cinnamic acid (A).
2. LiAlH4 reduces the carboxylic acid group to a primary alcohol and leaves the conjugated C=C intact, giving cinnamyl alcohol.

Teacher's Note:
1. LiAlH4 does not reduce isolated or conjugated C=C bonds in this case.

Question 21: 3-Ethyltoluene (1-ethyl-3-methylbenzene) is oxidised with \( \ce{KMnO4} \) and heat. The product is:
(a) benzene-1,3-dicarboxylic acid
(b) 3-methylphenylacetic acid
(c) 3-methylbenzoic acid
(d) 3-ethylbenzoic acid
Show Answer & Explanation

Answer: (a) benzene-1,3-dicarboxylic acid

Explanation:
1. Both alkyl side chains have benzylic hydrogens and are oxidised to carboxylic acid groups.

Teacher's Note:
1. The length of the chain does not matter.

Question 22: The correct order of reactivity of acyl chlorides towards hydrolysis (4-substituents: H, \( \ce{NO2} \), \( \ce{CH3} \), \( \ce{CHO} \)) is:
(a) \( \ce{NO2} \) > \( \ce{CHO} \) > H > \( \ce{CH3} \)
(b) H > \( \ce{NO2} \) > \( \ce{CHO} \) > \( \ce{CH3} \)
(c) \( \ce{CHO} \) > \( \ce{NO2} \) > H > \( \ce{CH3} \)
(d) \( \ce{CH3} \) > H > \( \ce{CHO} \) > \( \ce{NO2} \)
Show Answer & Explanation

Answer: (a) \( \ce{NO2} \) > \( \ce{CHO} \) > H > \( \ce{CH3} \)

Explanation:
1. Electron-withdrawing groups raise the positive charge on the carbonyl carbon and speed up hydrolysis.
2. The order is NO2 > CHO > H > CH3.

Teacher's Note:
1. Electron donors slow nucleophilic acyl substitution.

Question 23:
Assertion (A): A mixture of benzaldehyde and acetaldehyde in hot alkaline medium gives cinnamaldehyde.
Reason (R): Benzaldehyde is a stronger electrophile than acetaldehyde.
(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: (c) A is true, but R is false.

Explanation:
1. Crossed aldol condensation with acetaldehyde enolate gives cinnamaldehyde, so A is true.
2. Benzaldehyde is a weaker electrophile because of resonance with the ring, so R is false.

Teacher's Note:
1. Aromatic rings reduce the electrophilicity of C=O.

Question 24:
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. Acetophenone has alpha-hydrogens in its methyl group, so it gives aldol condensation.
2. It contains a carbonyl group, but the presence of C=O alone does not explain it; alpha-H is needed.

Teacher's Note:
1. Alpha-H is necessary for aldol.

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