Isomer $(P) \Rightarrow$ Can be prepared by Gabriel phthalimide synthesis
Isomer $(Q) \Rightarrow$ Reacts with Hinsberg's reagent to give solid insoluble in $NaOH$
Isomer (R) $\Rightarrow$ Reacts with $HONO$ followed by $\beta$-naphthol in $NaOH$ to give red dye.
Isomers $( P ),( Q )$ and $( R )$ respectively are

$(1)$ $C{H_3}C{H_2}N{H_2}$ $(2)$ $\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_2}C{H_3}}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,}\\
{\,C{H_3}C{H_2}NH\,\,\,\,\,\,\,}
\end{array}$
$(3)$ $\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,C{H_3}}\\
{\,|}\\
{{H_3}C - N - C{H_3}}
\end{array}$ $(4)$ $\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,C{H_3}}\\
{\,|}\\
{Ph - N - H}
\end{array}$
Assertion $(A)$ : Experimental reaction of $CH _{3} Cl$ with aniline and anhydrous $AlCl _{3}$ does not give $o$ and $p-methylaniline$.
Reason $(R)$ : The - $NH _{2}$ group of aniline becomes deactivating because of salt formation with anhydrous $AlCl _{3}$ and hence yields $m$-methyl aniline as the product.
In the light of the above statements, choose the most appropriate answer from the options given below.
(image) $\xrightarrow{{[O]}}A\xrightarrow{{SOC{l_2}}}B\xrightarrow{{Na{N_3}}}C\xrightarrow{{Heat}}D$