Catalytic hydrogenation of ketones with lithium aluminium hydride gives
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Secondary alcohols
Lithium aluminium hydride, \( \ce{LiAlH4} \), is a strong reducing agent that delivers a hydride ion (\( \ce{H-} \)) to the carbonyl carbon; the oxygen then takes a proton in the aqueous work-up. The same result is obtained with \( \ce{NaBH4} \) or by catalytic hydrogenation over Ni, Pd or Pt. Which alcohol forms depends on the groups already on the carbonyl carbon:
An aldehyde, RCHO, carbon with one R group, gives a primary alcohol, \( \ce{RCH2OH} \).
A ketone, RCOR′, carbon with two R groups, gives a secondary alcohol, RCH(OH)R′.
\( \displaystyle \ce{R-CO-R'} \xrightarrow[\text{(ii) }\ce{H3O+}]{\text{(i) }\ce{LiAlH4}} \ce{R-CH(OH)-R'} \)
For example, acetone gives propan-2-ol:
\( \displaystyle \ce{CH3COCH3} \xrightarrow[\text{(ii) }\ce{H3O+}]{\text{(i) }\ce{LiAlH4}} \ce{CH3CH(OH)CH3} \)
So the reduction of ketones gives secondary alcohols.
Why the other options are incorrect
Option (1) Primary alcohols — Incorrect. These come from the reduction of aldehydes (and of carboxylic acids and esters with \( \ce{LiAlH4} \)); a ketone carbon already has two carbon groups.
Option (3) Tertiary alcohols — Incorrect. These need a third carbon group on the carbinol carbon, which reduction cannot supply; they are made from a ketone and a Grignard reagent: \( \ce{R2C=O} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{R'MgX}} \ce{R2C(OH)R'} \).
Option (4) Esters — Incorrect. Esters are made from a carboxylic acid and an alcohol (esterification); they are not reduction products of ketones.
Note: It seems that there is a mistake in the original NTA paper. The question says “catalytic hydrogenation of ketones with lithium aluminium hydride”; it should be “reduction of ketones with lithium aluminium hydride”, since catalytic hydrogenation uses hydrogen gas over Ni, Pd or Pt, while \( \ce{LiAlH4} \) reduces the ketone by transferring a hydride ion.
NCERT (2026-27 edition):
Reduction
