An amine (X) is prepared by ammonolysis of benzyl chloride. On adding p-toluenesulphonyl chloride to it the solution remains clear. Molar mass of the amine (X) formed is \underline\hspace2cm g mol^-1 . (Given molar mass in gmol^-1 C: 12, H: 1, O: 16, ~N: 14 )

Compounds Containing Nitrogen PYQ — JEE Main 2024 · 6 Apr · Shift 2 · Q26 with Solution

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Correct answer: 287

Benzyl chloride is $\mathrm{C_6H_5CH_2Cl}$ .

On ammonolysis, benzyl chloride can give benzylamine first, but since the alkyl halide is reactive, further alkylation occurs:

\mathrm{NH_3 \rightarrow C_6H_5CH_2NH_2 \rightarrow (C_6H_5CH_2)_2NH \rightarrow (C_6H_5CH_2)_3N}

The amine $X$ on treatment with $p$ -toluenesulphonyl chloride gives a clear solution.

In Hinsberg test:

Primary amine gives sulphonamide soluble in alkali.
Secondary amine gives sulphonamide insoluble in alkali, so solution does not remain clear.
Tertiary amine does not react with sulphonyl chloride, hence solution remains clear.

Therefore, $X$ must be a tertiary amine.

So,

X = (\mathrm{C_6H_5CH_2})_3\mathrm{N}

This is tribenzylamine.

One benzyl group is:

\mathrm{C_6H_5CH_2} = \mathrm{C_7H_7}

Thus tribenzylamine is:

(\mathrm{C_7H_7})_3\mathrm{N} = \mathrm{C_{21}H_{21}N}

M = 21(12) + 21(1) + 14

= 252 + 21 + 14 = 287

\boxed{287\ \mathrm{g\ mol^{-1}}}