Write the structure of the aldehyde, carboxylic acid and ester that would yield 4-methylpent-2-en-1-ol on redu

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Q. Write the structure of the aldehyde, carboxylic acid and ester that would yield 4-methylpent-2-en-1-ol on reduction with LiAlH4.

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Step 1. 4-Methylpent-2-en-1-ol is (CH3)2CH-CH=CH-CH2-OH: a primary allylic alcohol with the double bond between C2 and C3, and a methyl branch at C4.

Step 2. LiAlH4 reduces an ALDEHYDE, R-CHO, to a primary alcohol, R-CH2-OH, with the SAME carbon skeleton -- so the aldehyde precursor is (CH3)2CH-CH=CH-CHO (4-methylpent-2-enal).

Step 3. LiAlH4 also reduces a CARBOXYLIC ACID, R-COOH, all the way to the primary alcohol R-CH2-OH -- so the acid precursor is (CH3)2CH-CH=CH-COOH (4-methylpent-2-enoic acid).

Step 4. LiAlH4 reduces an ESTER, R-COOR', to the primary alcohol R-CH2-OH PLUS the separate alcohol R'-OH -- so the ester precursor is any ester of 4-methylpent-2-enoic acid, e.g. the methyl ester (CH3)2CH-CH=CH-COOCH3, releasing methanol as the second product. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set 56/4/12 marks
$Q. Write the structures of the main products of the following reactions :$
(a) Methyl 3-(2-oxocyclohexyl)propanoate (drawn: a cyclohexanone ring bearing, on the carbon next to the $C{=}O$ , the side chain $-CH_2CH_2-\overset{O}{\underset{\|}{C}}-O-CH_3$ ) $\xrightarrow{NaBH_4}$
(b) $6\,(CH_3)_3C-OH + 2Al \longrightarrow$
›Reveal solutionSolution
The key idea is to identify the functional group that reacts with the given reagent. NaBH₄ selectively reduces the ketone to an alcohol, leaving the ester untouched. Aluminium reacts with tert-butyl alcohol to form aluminium tert-butoxide and hydrogen gas.

(a) Methyl 3-(2-oxocyclohexyl)propanoate + NaBH₄

Concept first: Sodium borohydride (NaBH₄) is a mild reducing agent. It reduces aldehydes and ketones to alcohols, but it does not reduce esters, carboxylic acids, or amides under normal conditions. This selectivity is crucial — you must recognise which carbonyl group will react and which will survive.

The molecule has two carbonyl groups:

A ketone on the cyclohexanone ring (the "2-oxo" part)

An ester in the side chain (the methyl propanoate part)

NaBH₄ will attack only the ketone.

Step-by-step reasoning:

Identify the reactive site. The ketone carbon is electrophilic. NaBH₄ provides a hydride ion (H⁻) that attacks this carbon.

Mechanism in brief: The hydride adds to the carbonyl carbon, forming an alkoxide intermediate. Aqueous workup (protonation) then gives the alcohol.

Stereochemistry note: The ketone is on a cyclohexane ring. Hydride attack can occur from either face, producing a mixture of cis and trans alcohols relative to the side chain. In exam contexts, you usually show the racemic product — often drawn with a wedge/dash or simply as the alcohol without specifying stereochemistry unless asked.

The ester remains unchanged. The ester carbonyl is less electrophilic and is not reduced by NaBH₄.

Watch out

A common mistake is to reduce both carbonyls. Remember: NaBH₄ stops at the aldehyde/ketone level. LiAlH₄ would reduce the ester too, but that's a different reagent. …
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Q.Write the structure of the aldehyde, carboxylic acid and ester that would yield 4-methylpent-2-en-1-ol on reduction with LiAlH4.

Concept understanding — Preparation of Alcohols by Hydroboration and Carbonyl Reduction

(a) Methyl 3-(2-oxocyclohexyl)propanoate + NaBH₄

Q.Write the structure of the aldehyde, carboxylic acid and ester that would yield 4-methylpent-2-en-1-ol on reduction with LiAlH4.

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