The structure of the product of the following reaction is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → methyl 2-(2-hydroxycyclohexyl)ethanoate
The starting compound has two carbonyl groups: a ketone (in the ring) and an ester (–\( \ce{COOCH3} \) on the side chain). Sodium borohydride (NaBH4) is a mild reducing agent: it reduces aldehydes and ketones to alcohols, but it does not reduce esters.
So only the ring C=O is reduced, to a secondary alcohol (CH–OH), and the ester group is left unchanged. The product is methyl 2-(2-hydroxycyclohexyl)ethanoate, structure (2).
\( \text{2-oxocyclohexyl}\ce{-CH2COOCH3} \xrightarrow[\ce{(ii) H3O^+}]{\ce{(i) NaBH4}} \text{2-hydroxycyclohexyl}\ce{-CH2COOCH3} \)
Why the other options are incorrect
Option (1) — Incorrect. It reduces the ester carbonyl and leaves the ketone, the opposite of what \( \ce{NaBH4} \) does.
Option (3) — Incorrect. It turns the ester into an aldehyde (–CHO) and leaves the ketone untouched; \( \ce{NaBH4} \) does neither.
Option (4) — Incorrect. The ketone is correctly reduced, but the ester has also been changed into an aldehyde, which \( \ce{NaBH4} \) does not do.
NCERT (2026-27 edition):
Reduction to Alcohols
