$A$ reaction of $0.1 \ mol$ of Benzylamine with bromomethane gave $23 \ g$ of Benzyl trimethyl ammonium bromide. The number of moles of bromomethane consumed in this reaction are $n \times 10^{-1},$ when $n = \dots$ (Round off to the Nearest Integer).
(Given : Atomic masses: $C : 12.0 \ u, H : 1.0 \ u, N : 14.0 \ u, Br : 80.0 \ u$)

$A$ primary amine can be converted to an alcohol by the action of

Consider the following reaction. The product $X$ is used:

$A$ mixture of ethyl amine,chloroform,and alcoholic $KOH$ on heating gives:

$A$ student has been given a compound $x$ of molecular formula $C_6H_7N$. $x$ is sparingly soluble in water. However, on addition of dilute mineral acid, $x$ becomes soluble in water. $x$ when treated with $CHCl_3$ and $KOH$ (alc.) produces $y$. $y$ has a specific unpleasant smell. On treatment with benzenesulphonyl chloride, $x$ gives a compound $z$ which is soluble in alkali. The number of different $H$ atoms present in $z$ is:

The final product $[B]$ is :