The product in the following reaction is

The correct answer is Option (4) → benzylamine (\( \ce{C6H5CH2NH2} \))

Lithium aluminium hydride reduces the C=O group of an amide to a \( \ce{CH2} \) group and gives a primary amine with the same number of carbon atoms. The hydride ion from \( \ce{LiAlH4} \) is added to the carbonyl carbon, the oxygen is removed, and water in the second step frees the amine:

    \( \displaystyle \ce{R-CO-NH2} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{LiAlH4}} \ce{R-CH2-NH2} \)

For benzamide:

    \( \displaystyle \ce{C6H5-CO-NH2} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{LiAlH4}} \ce{C6H5-CH2-NH2} \)

So the product is benzylamine (phenylmethanamine).

Why the other options are incorrect

• Option (1) ethylbenzene — Incorrect. It has one carbon more than benzamide and no nitrogen; reduction cannot add a carbon atom.

• Option (2) benzaldehyde — Incorrect. \( \ce{LiAlH4} \) does not stop at the aldehyde level, and an amide reduced by it keeps its nitrogen.

• Option (3) benzoic acid — Incorrect. It is formed by the hydrolysis of benzamide with acid or alkali, not by reduction.

NCERT (2026-27 edition):
Reduction of Amides