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(A) Ammonolysis of excess benzyl chloride $(C_6H_5CH_2Cl)$ with ammonia $(NH_3)$ leads to the formation of a tertiary amine,tribenzylamine,$(C_6H_5CH_

Vedclass · Accepted Answer

$287$

Vedclass · Answer

(A) Ammonolysis of excess benzyl chloride $(C_6H_5CH_2Cl)$ with ammonia $(NH_3)$ leads to the formation of a tertiary amine,tribenzylamine,$(C_6H_5CH_2)_3N$.This is a $3^{\circ}$ amine,which does not react with $p-$toluenesulphonyl chloride (Hinsberg's reagent) because it lacks an acidic hydrogen atom attached to the nitrogen.Therefore,the solution remains clear.The chemical formula of the amine $(X)$ is $(C_6H_5CH_2)_3N$,which is $C_{21}H_{21}N$.The molar mass is calculated as: $(21 	imes 12) + (21 	imes 1) + (1 	imes 14) = 252 + 21 + 14 = 287 \ g \ mol^{-1}$.

Column-$I$	Column-$II$
$A$. Ammonolysis	$1$. Amine with lesser number of carbon atoms
$B$. Gabriel phthalimide synthesis	$2$. Detection test for primary amine
$C$. Hoffmann Bromamide reaction	$3$. Reaction of phthalimide with $KOH$ and $R-X$
$D$. Carbylamine reaction	$4$. Reaction of alkyl-halides with $NH_3$

An amine $(X)$ is prepared by ammonolysis of benzyl chloride. On adding $p-$toluenesulphonyl chloride to it,the solution remains clear. The molar mass of the amine $(X)$ formed is $g \ mol^{-1}$. (Given molar mass in $g \ mol^{-1}$: $C=12, H=1, O=16, N=14, S=32, Cl=35.5$)

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