An alkyl halide reacts with alcoholic ammonia in a sealed tube,the product formed will be

How many of the transformations given below would result in aromatic amines?

The product $(A)$ of the given reaction is:

Complete the following reactions:
$(i)$ $C_{6}H_{5}NH_{2} + CHCl_{3} +$ alc. $KOH \rightarrow$
$(ii)$ $C_{6}H_{5}N_{2}Cl + H_{3}PO_{2} + H_{2}O \rightarrow$
$(iii)$ $C_{6}H_{5}NH_{2} + H_{2}SO_{4}$ (conc.) $\rightarrow$
$(iv)$ $C_{6}H_{5}N_{2}Cl + C_{2}H_{5}OH \rightarrow$
$(v)$ $C_{6}H_{5}NH_{2} + Br_{2(aq)} \rightarrow$
$(vi)$ $C_{6}H_{5}NH_{2} + (CH_{3}CO)_{2}O \rightarrow$
$(vii)$ $C_{6}H_{5}N_{2}Cl \xrightarrow[(ii) NaNO_{2}/Cu, \Delta ]{(i) HBF_{4}}$

$A$ 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

When $methyl$ $iodide$ is heated with $ammonia$,the product obtained is