Which product is formed when a ketone is reduced with sodium borohydride (NaBH4)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Secondary alcohol
Sodium borohydride adds hydrogen across the C=O group. In a ketone the carbonyl carbon carries two carbon groups, so the carbon that ends up with the –OH group is joined to two carbons: a secondary alcohol. For example, propanone gives propan-2-ol:
\( \displaystyle \ce{CH3COCH3} \xrightarrow{\ce{NaBH4}} \ce{CH3CH(OH)CH3} \)
Why the other options are incorrect
Option (1) Primary alcohol — Incorrect. A primary alcohol comes from the reduction of an aldehyde, whose carbonyl carbon carries only one carbon group.
Option (3) Tertiary alcohol — Incorrect. \( \ce{NaBH4} \) adds only hydrogen and makes no new C–C bond. A ketone gives a tertiary alcohol with a Grignard reagent, not on reduction.
Option (4) Alkane — Incorrect. Reducing the C=O group to a \( \ce{CH2} \) group needs Clemmensen (Zn-Hg and concentrated HCl) or Wolff-Kishner reduction; \( \ce{NaBH4} \) reduces a ketone only to the alcohol.
NCERT (2026-27 edition):
7.4.1 Preparation of Alcohols – “(i) By reduction of aldehydes and ketones: Aldehydes and ketones are reduced to the corresponding alcohols by addition of hydrogen in the presence of catalysts (catalytic hydrogenation). … It is also prepared by treating aldehydes and ketones with sodium borohydride (NaBH4) or lithium aluminium hydride (LiAlH4). Aldehydes yield primary alcohols whereas ketones give secondary alcohols.”