Ethanoic acid can be converted into ethanol using reagent
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(LiAlH_4\) in \(H_2O\)
The carboxyl group is hard to reduce. Of the common reducing agents, lithium aluminium hydride, a strong hydride donor, reduces carboxylic acids to primary alcohols. The reduction is carried out in dry ether, and water (or dilute acid) is then added to decompose the aluminium alkoxide complex and set the alcohol free:
\( \displaystyle \ce{CH3COOH} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{LiAlH4},\ \text{dry ether}} \ce{CH3CH2OH} \)
So ethanoic acid gives ethanol with \( \ce{LiAlH4} \) followed by water.
Why the other options are incorrect
Option (1) \(Sn + HCl\) — Incorrect. This reagent reduces nitro compounds to amines (nitrobenzene to aniline); it does not reduce the –COOH group.
Option (2) \(Na\) and \(C_2H_5OH\) — Incorrect. Sodium reacts with the acidic hydrogen of –COOH to give sodium ethanoate and hydrogen, \( \ce{2CH3COOH + 2Na -> 2CH3COONa + H2} \); the carboxyl group is not reduced.
Option (4) \(H_2\) and \(Pt\) — Incorrect. Catalytic hydrogenation adds hydrogen to C=C and C≡C bonds and to aldehydes and ketones, but it does not reduce a carboxylic acid under ordinary conditions.
Note: It seems that there is a mistake in the original NTA paper, which prints LiAlH4 in H2O in Option (3) when it should read LiAlH4 in dry ether, followed by H2O.
NCERT (2026-27 edition):
Reduction
