What will be the product of the following reaction?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → R-CH2NH2
Lithium aluminium hydride (\( \ce{LiAlH4} \)) is a strong reducing agent. It reduces the C=O group of carboxylic acid derivatives without removing any carbon atom:
Carboxylic acids and esters give primary alcohols.
Amides give amines: the C=O group becomes \( \ce{CH2} \).
Nitriles give primary amines.
\( \displaystyle \ce{R-CONH2} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{LiAlH4}} \ce{R-CH2-NH2} \)
So the product is R–CH2NH2, which has the same number of carbon atoms as the amide.
Why the other options are incorrect
Option (1) R-NH2 — Incorrect. \( \ce{R-NH2} \) has one carbon atom fewer than the amide. It is the product of the Hofmann bromamide reaction (\( \ce{Br2} \) + NaOH), not of \( \ce{LiAlH4} \) reduction.
Option (3) R-COOH — Incorrect. \( \ce{LiAlH4} \) is a reducing agent and cannot turn an amide into an acid; that change needs hydrolysis.
Option (4) R-CHO — Incorrect. The nitrogen of the amide is kept in the product, so an aldehyde cannot form; \( \ce{LiAlH4} \) also does not stop at the aldehyde stage.
NCERT (2026-27 edition):
Reduction of Amides
