JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 32

Answer: (b) butanoic acid

Explanation:
1. A Grignard reagent adds to CO₂ to give a carboxylate salt.
2. Acid hydrolysis gives R–COOH with one more carbon than R: CH₃CH₂CH₂COOH.

Teacher's Note:
1. The new COOH is attached where MgBr was.

Answer: (d) 2-methylpropylmagnesium bromide (isobutylmagnesium bromide)

Explanation:
1. The acid is (CH₃)₂CH–CH₂–COOH, so the carbon chain R attached to COOH is (CH₃)₂CH–CH₂–.
2. Hence A is (CH₃)₂CHCH₂MgBr.

Teacher's Note:
1. Remove the COOH carbon to find R.

Answer: (a) benzoic acid

Explanation:
1. A primary alcohol is oxidised to the aldehyde and then to the carboxylic acid by a strong oxidant.
2. C₆H₅CH₂OH gives C₆H₅COOH.

Teacher's Note:
1. Strong oxidant: acid. PCC: aldehyde.

Answer: (c) cyclohexanecarboxylic acid

Explanation:
1. The primary alcohol CH₂OH group is oxidised to COOH.
2. The ring is not altered.

Teacher's Note:
1. Same oxidation as benzyl alcohol to benzoic acid.

Answer: (d) 2-(4-methylcyclohexyl)propanoic acid

Explanation:
1. CrO₃/H⁺ is a strong oxidant that takes a primary alcohol to the carboxylic acid.
2. Only the CH₂OH group is oxidised; the methyl group on the ring is unaffected.

Teacher's Note:
1. Primary alcohol → acid.

Answer: (b) propanoic acid and methanol

Explanation:
1. Ester hydrolysis splits the C(=O)–O bond: RCOOR' + H₂O ⇌ RCOOH + R'OH.
2. The acyl part C₂H₅CO gives propanoic acid and the OCH₃ part gives methanol.

Teacher's Note:
1. The reaction is reversible in acid.

Answer: (c) ethanoic acid and anilinium ion (C₆H₅NH₃⁺)

Explanation:
1. Amide hydrolysis splits the C–N bond.
2. The acyl part gives CH₃COOH and the nitrogen part gives aniline, which is protonated in acid.

Teacher's Note:
1. In acid the amine product is obtained as its ammonium salt.

Answer: (a) 1 : 1

Explanation:
1. Hydrolysis of an anhydride cleaves the C–O–C link.
2. Each acyl group becomes a carboxylic acid, so one mole of each acid forms.

Teacher's Note:
1. Anhydride + H₂O → two acids.

Answer: (b) 3-methylbutanoic acid

Explanation:
1. A nitrile hydrolyses first to an amide and then to the acid, R–C≡N → R–COOH.
2. The carbon skeleton including the nitrile carbon is unchanged.

Teacher's Note:
1. Nitrile carbon becomes the carboxyl carbon.

Answer: (d) Hell-Volhard-Zelinsky (HVZ) reaction

Explanation:
1. Red phosphorus with a halogen brominates the alpha carbon of a carboxylic acid.
2. This alpha-halogenation is the HVZ reaction.

Teacher's Note:
1. Needs an alpha hydrogen.

Answer: (a) 3-methylbutanoyl chloride

Explanation:
1. SOCl₂ replaces the OH of the carboxylic acid by Cl.
2. The product is the acid chloride, with SO₂ and HCl as gaseous by-products.

Teacher's Note:
1. Acid chloride preparation: SOCl₂, PCl₅ or PCl₃.

Answer: (c) butanamide

Explanation:
1. The ammonium salt formed first loses water on heating.
2. RCOOH + NH₃ → RCOONH₄ → RCONH₂ + H₂O.

Teacher's Note:
1. Carboxylic acids give amides on strong heating with ammonia.

Answer: (d) C₆H₅CH(CH₃)CH(CH₃)Br

Explanation:
1. The silver salt with Br₂ undergoes the Hunsdiecker reaction.
2. COOH is replaced by Br, so the product has one carbon less than the acid.

Teacher's Note:
1. RCOOAg + Br₂ → RBr + CO₂ + AgBr.

Answer: (b) 2-bromobutane

Explanation:
1. This is the Hunsdiecker reaction: the COOH group is replaced by Br.
2. CH₃CH₂CH(CH₃)COOH gives CH₃CH₂CHBrCH₃.

Teacher's Note:
1. Product has one carbon fewer than the acid.

Answer: (c) ethylbenzene

Explanation:
1. Soda lime decarboxylates the acid: RCOOH → RH + CO₂.
2. C₆H₅CH₂CH₂COOH gives C₆H₅CH₂CH₃.

Teacher's Note:
1. Product has one carbon fewer than the acid.

Answer: (a) the sodium sulphonate with evolution of CO₂

Explanation:
1. Sulphonic acids are stronger acids than carbonic acid.
2. They decompose NaHCO₃, forming the sodium salt and CO₂ gas.

Teacher's Note:
1. Effervescence with NaHCO₃ shows a strongly acidic group.

Answer: (b) propanal

Explanation:
1. This is the Rosenmund reduction: the poisoned catalyst stops reduction at the aldehyde.
2. CH₃CH₂COCl gives CH₃CH₂CHO.

Teacher's Note:
1. Without the poison the alcohol would form.

Answer: (d) benzoyl chloride

Explanation:
1. The Rosenmund reduction converts acid chlorides into aldehydes.
2. So C₆H₅COCl gives C₆H₅CHO.

Teacher's Note:
1. Esters and ketones are not reduced under these conditions.

Answer: (a) ethanoyl chloride

Explanation:
1. Y gives Tollens and 2,4-DNP tests, so it is an aldehyde, and the iodoform test makes it ethanal.
2. X must be the acid chloride that gives ethanal in a Rosenmund reduction: CH₃COCl.

Teacher's Note:
1. Only acid chlorides are reduced by Rosenmund reduction.

Answer: (c) 2-methylpropanal

Explanation:
1. PCC oxidises a primary alcohol to the aldehyde and stops there.
2. (CH₃)₂CHCH₂OH gives (CH₃)₂CHCHO.

Teacher's Note:
1. PCC is a mild oxidising agent.

Answer: (d) 1-phenylethanol

Explanation:
1. Y is a methyl ketone: it reacts with 2,4-DNP and gives iodoform, but is not an aldehyde.
2. PCC oxidises the secondary alcohol 1-phenylethanol to acetophenone.

Teacher's Note:
1. Positive iodoform needs CH₃CO– group.

Answer: (b) ethanal

Explanation:
1. This is the Stephen reaction: a nitrile is reduced to an imine and hydrolysed to an aldehyde.
2. CH₃CN gives CH₃CHO.

Teacher's Note:
1. Nitrile carbon becomes the CHO carbon.