The correct option$(s)$ for the following sequence of reactions is(are):
$(A)$ $Q = KNO_2, W = LiAlH_4$
$(B)$ $R =$ benzenamine,$V = KCN$
$(C)$ $Q = AgNO_2, R =$ phenylmethanamine
$(D)$ $W = LiAlH_4, V = AgCN$

The relative rate of reaction of the following amines with methyl iodide is:

$A$ given nitrogen-containing aromatic compound $A$ reacts with $Sn/HCl,$ followed by $HNO_2$ to give an unstable compound $B$. $B,$ on treatment with phenol,forms a beautiful coloured compound $C$ with the molecular formula $C_{12}H_{10}N_2O.$ The structure of compound $A$ is

$A$ nitrile $X$ is treated with $LiAlH_{4}$ to obtain compound $Y$ $(C_{2}H_{7}N)$. In a separate reaction,$X$ is hydrolyzed in an acidic medium to obtain $Z$. The product obtained after mixing $Y$ and $Z$ will be:

Given below are two statements:
Statement-$I$ : Oxidation of $p$-nitrotoluene with acidic $KMnO_4$ gives an acid that is stronger than benzoic acid.
Statement-$II$ : Reduction of $p$-nitrotoluene with $Sn/HCl$ followed by neutralization gives an amine that is more basic than aniline.
In light of the above statements, choose the most appropriate answer from the options given below.

An organic compound $A$ on reduction gives compound $B$,which on reaction with trichloromethane and caustic potash forms $C$. The compound $C$ on catalytic reduction gives $N$-methylbenzenamine. The compound $A$ is,