$A$ colourless substance $'A'$ $(C_6H_7N)$ is sparingly soluble in water and gives a water-soluble compound $'B'$ on treating with mineral acid. On reacting with $CHCl_3$ and alcoholic potash,$'A'$ produces an obnoxious smell due to the formation of compound $'C'$. Reaction of $'A'$ with benzenesulphonyl chloride gives compound $'D'$ which is soluble in alkali. With $NaNO_2$ and $HCl$,$'A'$ forms compound $'E'$ which reacts with phenol in an alkaline medium to give an orange dye $'F'$. Identify compounds $'A'$ to $'F'$.

(A) The substance $'A'$ is $C_6H_5NH_2$ (Aniline).
$1$. $C_6H_5NH_2 + HCl \rightarrow C_6H_5NH_3^+Cl^-$ (Anilinium chloride,$'B'$).
$2$. $C_6H_5NH_2 + CHCl_3 + 3KOH \rightarrow C_6H_5NC + 3KCl + 3H_2O$ (Phenyl isocyanide,$'C'$).
$3$. $C_6H_5NH_2 + C_6H_5SO_2Cl \rightarrow C_6H_5NHSO_2C_6H_5 + HCl$ ($N$-phenylbenzenesulphonamide,$'D'$).
$4$. $C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278 \ K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$ (Benzenediazonium chloride,$'E'$).
$5$. $C_6H_5N_2^+Cl^- + C_6H_5OH \xrightarrow{OH^-} C_6H_5-N=N-C_6H_4OH$ ($p$-hydroxyazobenzene,orange dye,$'F'$).