Which of the following reagents can convert a carboxylic acid into an alcohol?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → LiAlH4/H2O
Lithium aluminium hydride is a strong reducing agent. It reduces the carboxyl group of an acid to a –CH2OH group, and treatment with water then sets free the primary alcohol:
\( \displaystyle \ce{RCOOH} \xrightarrow[\text{(ii) } \ce{H2O}]{\text{(i) } \ce{LiAlH4}} \ce{RCH2OH} \)
Why the other options are incorrect
Option (1) NaBH4/H2O — Incorrect. Sodium borohydride is a milder reducing agent; it reduces aldehydes and ketones but does not reduce the carboxyl group.
Option (2) Pd-C/H2 — Incorrect. Catalytic hydrogenation reduces aldehydes and ketones to alcohols, but not the free carboxyl group. To reduce an acid this way it is first converted to an ester, which is then hydrogenated.
Option (4) Zn/CH3OH — Incorrect. Zinc in methanol is not a hydride reagent and does not reduce a carboxylic acid to an alcohol.
NCERT (2026-27 edition):
(ii) By reduction of carboxylic acids and esters – “Carboxylic acids are reduced to primary alcohols in excellent yields by lithium aluminium hydride, a strong reducing agent.”
\( \displaystyle \ce{RCOOH} \xrightarrow[\text{(ii) } \ce{H2O}]{\text{(i) } \ce{LiAlH4}} \ce{RCH2OH} \)