$$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Cl} \xrightarrow{\text { Alc. } \mathrm{NH}_3} A \xrightarrow{2 \mathrm{CH}_3 \mathrm{Cl}} B.$$ The product $$B$$ is

The method by which aniline cannot be prepared is

A hydrocarbon $$A\left(\mathrm{C}_4 \mathrm{H}_8\right)$$ on reaction with $$\mathrm{HCl}$$ gives a compound $$\mathrm{B}\left(\mathrm{C}_4 \mathrm{H}_9 \mathrm{Cl}\right)$$ which on reaction with $$1 \mathrm{~mol}$$ of $$\mathrm{NH}_3$$ gives compound $$\mathrm{C}\left(\mathrm{C}_4 \mathrm{H}_{10} \mathrm{N}\right)$$. On reacting with $$\mathrm{NaNO}_2$$ and $$\mathrm{HCl}$$ followed by treatment with water, compound $$C$$ yields an optically active compound $$D$$. The compound $$D$$ is

The compound A (major product) is