JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 39

Chemistry Objective Questions and Answers: Aldehydes Ketones and Carboxylic Acids

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Multiple Choice Questions

Question 1: The conversion of cyclopentyl methyl ketone into cyclopentanecarboxylic acid can be carried out by:
(a) (i) Tollens' reagent, (ii) H+
(b) (i) Benzoyl peroxide, (ii) H+
(c) (i) I2/NaOH, (ii) H+
(d) H+/KMnO4, heat
Show Answer & Explanation

Answer: (c) (i) I2/NaOH, (ii) H+

Explanation:
1. The methyl ketone undergoes the haloform reaction with I2/NaOH, giving the carboxylate and CHI3.
2. Acidification liberates the acid with one carbon fewer.
3. Tollens' reagent does not oxidise ketones, and hot KMnO4 would cleave the ring.
4. Answer: (c)

Teacher's Note:
1. Look for alpha hydrogens and the CH3CO– or CH3CH(OH)– unit before applying the named test.

Question 2: Butan-2-one is obtained when which of the following compounds is treated with CH3MgBr and then hydrolysed?
(a) CH3CHO
(b) CH3COCl
(c) CH3C≡N
(d) CH3CH2C≡N
Show Answer & Explanation

Answer: (d) CH3CH2C≡N

Explanation:
1. A Grignard reagent adds once to a nitrile and the imine salt hydrolyses to a ketone.
2. CH3CH2CN + CH3MgBr gives CH3CH2COCH3.
3. CH3CN would give propanone, and the acid chloride would go on to a tertiary alcohol.
4. Answer: (d)

Teacher's Note:
1. Look for alpha hydrogens and the CH3CO– or CH3CH(OH)– unit before applying the named test.

Question 3: Which of the following cannot undergo an intramolecular Cannizzaro reaction?
(a) OHC–CHO
(b) OHC–CO–C6H5
(c) CH3CO–COCH3
(d) OHC–CO–D
Show Answer & Explanation

Answer: (c) CH3CO–COCH3

Explanation:
1. This reaction needs a 1,2-dicarbonyl compound with no alpha hydrogen on the CHO side, and at least one CHO group.
2. Biacetyl, CH3COCOCH3, has no CHO group and has alpha hydrogens, so it does not do it.
3. OHC–CO–D and the other two have a CHO group and no alpha hydrogen on that side, so they react.
4. Answer: (c)

Teacher's Note:
1. Look for alpha hydrogens and the CH3CO– or CH3CH(OH)– unit before applying the named test.

Question 4: Methanal is treated with concentrated NaOH. The products are:
(a) HCOO− and HCHO
(b) HCOO− and HCOO−
(c) HCOO− and CH3OH
(d) HCHO and CH3OH
Show Answer & Explanation

Answer: (c) HCOO− and CH3OH

Explanation:
1. Methanal has no alpha hydrogen, so it undergoes the Cannizzaro reaction.
2. One molecule is oxidised to formate and another is reduced to methanol.
3. Answer: (c)

Teacher's Note:
1. Look for alpha hydrogens and the CH3CO– or CH3CH(OH)– unit before applying the named test.

Question 5: Which of the following reactions finally gives a product that gives a positive iodoform test?
(a) 2-Methylcyclopentanone oxime, H3O+
(b) CH3CH2C(=O)–O–C(CH3)=CHCH3, H3O+
(c) 1-Phenylpropyne, Hg2+/dil. H2SO4
(d) Cyclohexanone dimethyl ketal, H3O+
Show Answer & Explanation

Answer: (b) CH3CH2C(=O)–O–C(CH3)=CHCH3, H3O+

Explanation:
1. The enol ester hydrolyses to propanoic acid and the enol of butan-2-one, which gives CH3COCH2CH3.
2. This has a CH3CO– group, so it gives CHI3.
3. The others give 2-methylcyclopentanone, propiophenone (PhCOCH2CH3) and cyclohexanone, none of which has a CH3CO– group.
4. Answer: (b)

Teacher's Note:
1. Look for alpha hydrogens and the CH3CO– or CH3CH(OH)– unit before applying the named test.

Question 6: In which of the following groups does every member give a positive iodoform test?
(a) Methanol, ethanol, propanone
(b) Ethanol, 2-methylpropanal, methanal
(c) Ethanal, ethanol, propan-2-ol
(d) Propanal, propan-2-ol, propanone
Show Answer & Explanation

Answer: (c) Ethanal, ethanol, propan-2-ol

Explanation:
1. A positive test needs CH3CO– or CH3CH(OH)– attached to H or carbon.
2. Ethanal, ethanol and propan-2-ol all have it.
3. Methanol, methanal, propanal and 2-methylpropanal do not.
4. Answer: (c)

Teacher's Note:
1. Look for alpha hydrogens and the CH3CO– or CH3CH(OH)– unit before applying the named test.

Question 7: The conversion CH3CH(OH)CH2CH2COCH3 → CH3CH(OH)(CH2)3CH3 can be achieved by:
(a) Wolff–Kishner reduction
(b) Clemmensen reduction
(c) LiAlH4
(d) NaBH4
Show Answer & Explanation

Answer: (a) Wolff–Kishner reduction

Explanation:
1. Only the ketone group has to become CH2, while the alcohol stays.
2. Wolff–Kishner is carried out in base and leaves the OH group alone.
3. Clemmensen's strong acid would attack the alcohol, and hydrides would reduce C=O only to CH(OH).
4. Answer: (a)

Teacher's Note:
1. Look for alpha hydrogens and the CH3CO– or CH3CH(OH)– unit before applying the named test.

Question 8: A compound C4H8O reacts with 2,4-DNP but gives a negative haloform test. It is:
(a) Butan-2-one
(b) 2-Methylpropanal
(c) Cyclobutanol
(d) Butan-2-ol
Show Answer & Explanation

Answer: (b) 2-Methylpropanal

Explanation:
1. The 2,4-DNP reaction shows a carbonyl group.
2. Butan-2-one has CH3CO– and gives the haloform test.
3. 2-Methylpropanal is a carbonyl compound with no CH3CO– group, and the two alcohols have no C=O.
4. Answer: (b)

Teacher's Note:
1. Look for alpha hydrogens and the CH3CO– or CH3CH(OH)– unit before applying the named test.

Question 9: Which of the following does not form a stable hydrate on addition of water?
(a) PhCO–CO–COPh
(b) Indane-1,2,3-trione
(c) Cyclopropanone
(d) Cyclohexanone
Show Answer & Explanation

Answer: (d) Cyclohexanone

Explanation:
1. The central C=O of a vicinal tricarbonyl is made very electrophilic by its neighbours, and cyclopropanone gains relief of angle strain on becoming sp3.
2. Both of these hydrates are stable.
3. Cyclohexanone sits at the unfavourable end of the equilibrium and gives almost no hydrate.
4. Answer: (d)

Teacher's Note:
1. Look for alpha hydrogens and the CH3CO– or CH3CH(OH)– unit before applying the named test.

Question 10: PhCH(OH)CH3 is oxidised with PCC to A, and A is treated with semicarbazide (NH2NHCONH2). Product B is:
(a) PhC(CH3)=N–CO–NHNH2
(b) PhC(CH3)=N–NH–CO–NH2
(c) PhCH=N–N(CH3)–CO–NH2
(d) PhCH=N–CO–NH2
Show Answer & Explanation

Answer: (b) PhC(CH3)=N–NH–CO–NH2

Explanation:
1. PCC oxidises the secondary alcohol to acetophenone.
2. The terminal NH2 (attached to N–H) of semicarbazide condenses with the C=O to give the semicarbazone.
3. B is PhC(CH3)=N–NH–CONH2.
4. Answer: (b)

Teacher's Note:
1. Look for alpha hydrogens and the CH3CO– or CH3CH(OH)– unit before applying the named test.

Question 11: Which of the following will not give a positive Tollens' test?
(a) CH3CHO
(b) Tetrahydropyran-2-ol (cyclic hemiacetal)
(c) CH3CH2CH2CH2CH(OH)OCH3
(d) CH3CH(OCH3)2
Show Answer & Explanation

Answer: (d) CH3CH(OCH3)2

Explanation:
1. Hemiacetals are in equilibrium with the open-chain hydroxy aldehyde (or the aldehyde and alcohol), so they reduce Tollens' reagent.
2. An acetal such as CH3CH(OCH3)2 is stable in basic medium and has no free CHO group.
3. Answer: (d)

Teacher's Note:
1. Look for alpha hydrogens and the CH3CO– or CH3CH(OH)– unit before applying the named test.

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Practice MCQs for JEE Chemistry Aldehydes Ketones and Carboxylic Acids

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