JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 33

Multiple Choice Questions (MCQs) for JEE Chemistry: Aldehydes Ketones and Carboxylic Acids

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Multiple Choice Questions

Question 1: Compound A (\( \ce{C3H8O} \)) is oxidised with anhydrous \( \ce{CrO3} \) to B (\( \ce{C3H6O} \)). B gives a silver mirror with Tollens' reagent. B treated with semicarbazide, HCl and sodium acetate gives C. The structure of C is:
(a) \( \ce{(CH3)2C=NNHCONH2} \)
(b) \( \ce{(CH3)2C=NCONHNH2} \)
(c) \( \ce{CH3CH2CH=NCONHNH2} \)
(d) \( \ce{CH3CH2CH=NNHCONH2} \)
Show Answer & Explanation

Answer: (d) \( \ce{CH3CH2CH=NNHCONH2} \)

Explanation:
1. B gives a silver mirror, so it is an aldehyde, propanal; A is propan-1-ol.
2. Propanal reacts with semicarbazide at the terminal \( \ce{NH2} \) of the hydrazine part to give the semicarbazone \( \ce{CH3CH2CH=N-NH-CO-NH2} \).

Teacher's Note:
1. The hydrazine nitrogen, not the amide nitrogen, condenses with C=O.

Question 2: Arrange the following ketones in decreasing order of \( K_{eq} \) for nucleophilic addition: (a) acetophenone; (b) 4-chloroacetophenone; (c) 4-methylacetophenone; (d) 4-nitroacetophenone.
(a) a > b > c > d
(b) c > a > b > d
(c) d > a > b > c
(d) d > b > a > c
Show Answer & Explanation

Answer: (d) d > b > a > c

Explanation:
1. Electron-withdrawing groups on the ring increase the positive charge on the carbonyl carbon and favour addition.
2. NO2 (-M, -I) > Cl (-I) > H > CH3 (+I, hyperconjugation).

Teacher's Note:
1. EWG increases and EDG decreases the electrophilicity of C=O.

Question 3: Cyclohexene is treated with \( \ce{O3} \) and then water (oxidative work-up) to give A, and A is heated with \( \ce{Ca(OH)2} \) to give B. B is:
(a) Cyclopentanone
(b) Cyclohexanone
(c) Cyclopent-2-enone
(d) Cyclohexanol
Show Answer & Explanation

Answer: (a) Cyclopentanone

Explanation:
1. A = hexanedioic (adipic) acid.
2. The calcium salt on heating undergoes intramolecular ketonic decarboxylation, giving cyclopentanone.

Teacher's Note:
1. Pyrolysis of calcium adipate gives cyclopentanone.

Question 4: A carbonyl compound A reacts with HCN to form a cyanohydrin B, which on hydrolysis gives an optically active alpha-hydroxy acid C. A gives a positive iodoform test. A, B and C respectively are:
(a) \( \ce{CH3CHO} \); \( \ce{CH3CH(OH)CN} \); \( \ce{CH3CH(OH)COOH} \)
(b) \( \ce{HCHO} \); \( \ce{HOCH2CN} \); \( \ce{HOCH2COOH} \)
(c) \( \ce{CH3CH2CHO} \); \( \ce{CH3CH2CH(OH)CN} \); \( \ce{CH3CH2CH(OH)COOH} \)
(d) \( \ce{(CH3)2CO} \); \( \ce{(CH3)2C(OH)CN} \); \( \ce{(CH3)2C(OH)COOH} \)
Show Answer & Explanation

Answer: (a) \( \ce{CH3CHO} \); \( \ce{CH3CH(OH)CN} \); \( \ce{CH3CH(OH)COOH} \)

Explanation:
1. The alpha-hydroxy acid is optically active, so it needs a chiral carbon: the aldehyde must be RCHO with R different from H (not formaldehyde) and acetone gives an achiral product.
2. Iodoform test needs a CH3-CO or CH3-CH(OH) group: acetaldehyde qualifies, propanal does not.
3. Acetaldehyde gives lactonitrile and then lactic acid (chiral).

Teacher's Note:
1. Chirality of C and iodoform test together identify acetaldehyde.

Question 5: Which carbonyl group of the compound shown (indane-1,2,3-trione) is the most reactive towards hydration?
(a) Carbon 1
(b) Carbon 3
(c) Carbon 2 (central C=O)
(d) All have equal reactivity
Show Answer & Explanation

Answer: (c) Carbon 2 (central C=O)

Explanation:
1. The central carbonyl is flanked on both sides by electron-withdrawing C=O groups, so its carbon is the most electron-deficient.
2. The hydrate (gem-diol) is also stabilised, which is why ninhydrin exists as a stable hydrate.

Teacher's Note:
1. Adjacent C=O groups raise the electrophilicity of a carbonyl carbon.

Question 6: Cyclohexyl bromide is converted to the Grignard reagent A, A reacts with \( ^{14}\ce{CO2} \) to give B after acidification, and B is treated with \( \ce{NaHCO3} \) to liberate gas C. C is:
(a) \( ^{14}\ce{CO2} \)
(b) CO
(c) a mixture of \( ^{14}\ce{CO2} \) and \( \ce{CO2} \)
(d) \( \ce{CO2} \) (unlabelled)
Show Answer & Explanation

Answer: (d) \( \ce{CO2} \) (unlabelled)

Explanation:
1. B is cyclohexanecarboxylic acid labelled with \( ^{14}\)C at the carboxyl carbon.
2. With \( \ce{NaHCO3} \) the acid forms its sodium salt (the labelled carbon stays in the carboxylate), and \( \ce{CO2} \) is produced from the bicarbonate, so it is unlabelled.

Teacher's Note:
1. The gas comes from bicarbonate, not from the acid.

Question 7: In which group can every member be reduced to an alcohol by \( \ce{NaBH4} \)?
(a) Acetyl chloride, acetic acid and acetamide
(b) Ethanal, propanone and butanone
(c) Ethanal, ethyl acetate and acetyl chloride
(d) Propanone, acetamide and ethanal
Show Answer & Explanation

Answer: (b) Ethanal, propanone and butanone

Explanation:
1. \( \ce{NaBH4} \) is a mild reagent that reduces aldehydes and ketones only.
2. It does not reduce acids, esters or amides.

Teacher's Note:
1. NaBH4 reduces aldehydes and ketones.

Question 8: Which of the following compounds gives a positive iodoform test with \( \ce{I2} \) and NaOH?
(a) 1-Phenylethanol, \( \ce{C6H5CH(OH)CH3} \)
(b) Pentan-3-ol
(c) 3-Phenylpropan-1-ol
(d) 2-Methylpropan-1-ol
Show Answer & Explanation

Answer: (a) 1-Phenylethanol, \( \ce{C6H5CH(OH)CH3} \)

Explanation:
1. The iodoform test needs \( \ce{CH3CO-} \) or \( \ce{CH3CH(OH)-} \) (which is oxidised to the former).
2. Only 1-phenylethanol contains the \( \ce{CH3CH(OH)-} \) unit.

Teacher's Note:
1. Look for a methyl carbinol or methyl ketone group.

Question 9: Which of the following forms a hydrazone and gives the iodoform test but not Tollens' test?
(a) 1-Cyclohexylpropan-1-one
(b) Ethanol
(c) Pentan-3-one
(d) Butan-2-one
Show Answer & Explanation

Answer: (d) Butan-2-one

Explanation:
1. A methyl ketone forms hydrazones and gives the iodoform test, and being a ketone it does not give Tollens' test.
2. 1-Cyclohexylpropan-1-one and pentan-3-one have no methyl ketone group, and ethanol does not form hydrazones.

Teacher's Note:
1. Methyl ketone: hydrazone and iodoform positive, Tollens negative.

Question 10: In the Cannizzaro reaction of benzaldehyde with hydroxide, the slowest (rate-determining) step is:
(a) The attack of \( \ce{OH-} \) at the carbonyl carbon
(b) The abstraction of a proton from the carboxylic group
(c) The deprotonation of benzyl alcohol
(d) The transfer of hydride to the carbonyl group of a second molecule
Show Answer & Explanation

Answer: (d) The transfer of hydride to the carbonyl group of a second molecule

Explanation:
1. The hydroxide adds quickly and reversibly to give the alkoxide anion.
2. Hydride transfer from this anion to another benzaldehyde molecule is the slow step.

Teacher's Note:
1. Hydride transfer is rate determining.

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Multiple Choice Questions (MCQs) for JEE Chemistry Aldehydes Ketones and Carboxylic Acids

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