The reduction of cyclohexanone with \( NaBH_4 \) produces
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Cyclohexanol
Sodium borohydride, \( \ce{NaBH4} \), is a source of the hydride ion. The hydride ion adds to the electrophilic carbonyl carbon (nucleophilic addition), and the alkoxide formed takes up a proton on work-up. Aldehydes are reduced to primary alcohols and ketones to secondary alcohols.
Cyclohexanone is a ketone, so it gives the secondary alcohol cyclohexanol:
\( \displaystyle \underset{\text{cyclohexanone}}{\ce{C6H10O}} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{NaBH4}} \underset{\text{cyclohexanol}}{\ce{C6H11OH}} \)
\( \ce{NaBH4} \) is a mild reducing agent: it reduces the C=O group of aldehydes and ketones but not a C=C double bond.
Why the other options are incorrect
Option (2) Cyclohexene — Incorrect. Cyclohexene is obtained by dehydration of cyclohexanol with an acid such as conc. \( \ce{H3PO4} \), not by reduction with \( \ce{NaBH4} \).
Option (3) Cyclohexane-1,2-diol — Incorrect. A 1,2-diol is formed by oxidation of cyclohexene with cold dilute \( \ce{KMnO4} \); reduction of a ketone puts only one –OH group in the molecule.
Option (4) Cyclohexane carboxylic acid — Incorrect. This has one carbon atom more than cyclohexanone and is an oxidation level higher; reduction cannot add a carbon or form an acid. It is made, for example, from cyclohexylmagnesium bromide and \( \ce{CO2} \).
NCERT (2026-27 edition):
Reduction
