The product 'Y' in the following reaction sequence is

The correct answer is Option (4) → 2-phenylethanamine (C₆H₅CH₂CH₂NH₂)

Reaction of an alkyl halide with alcoholic KCN, followed by reduction of the nitrile, is a way of making an amine with one carbon atom more than the halide (ascent of series). The cyanide ion replaces the halogen by nucleophilic substitution, and \( \ce{LiAlH4} \) reduces –C≡N to –\( \ce{CH2NH2} \).

First, benzyl chloride is converted into phenylethanenitrile (X):

    \( \ce{C6H5CH2Cl + KCN} \xrightarrow[\Delta]{\text{alc.}} \ce{C6H5CH2CN + KCl} \)

Then \( \ce{LiAlH4} \) reduces the nitrile to the primary amine (Y):

    \( \ce{C6H5CH2CN} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{LiAlH4}} \ce{C6H5CH2CH2NH2} \)

So Y is 2-phenylethanamine, with the CN carbon now part of the side chain.

Why the other options are incorrect

• Option (1) benzylamine (C₆H₅CH₂NH₂) — Incorrect. This has no extra carbon; it is formed when benzyl chloride reacts with ammonia, not through the nitrile.

• Option (2) benzyl alcohol (C₆H₅CH₂OH) — Incorrect. This is the product of hydrolysis of benzyl chloride with aqueous KOH; alcoholic KCN gives the nitrile.

• Option (3) benzoic acid (C₆H₅COOH) — Incorrect. An acid is formed by oxidation (for example, of toluene with \( \ce{KMnO4} \)); \( \ce{LiAlH4} \) is a reducing agent.

NCERT (2026-27 edition):
Reduction of Nitriles