A) Nitrobenzene $\underset{H_{2}{SO}_{4},heat}{\overset{Fuming\ {HNO}_{3}}{\rightarrow}}\lbrack\ \rbrack\overset{{NH}_{3}/H_{2}S}{\rightarrow}\lbrack\ \rbrack\underset{2.{HBF}_{4},\Delta}{\overset{1.HONO}{\rightarrow}}$
B) Aniline $\underset{\ heat}{\overset{{\ \ \ \ F}_{2\ }\ \ \ }{\rightarrow}}$
C) Fluorobenzene$\ \underset{H_{2}{SO}_{4},heat}{\overset{{\ \ \ \ \ \ HNO}_{3}\ \ \ \ \ \ }{\rightarrow}}$
D) $m$-$C_{6}H_{4}\left( NH_{2} \right)_{2}\underset{2.{CuNO}_{2},3.{HBF}_{4}}{\overset{{\ \ \ \ \ \ \ \ \ 1.HONO\ \ \ \ \ \ }_{\ }}{\rightarrow}}$
Explanation
The correct answer is Option A: Nitrobenzene $\underset{H_{2}{SO}_{4},heat}{\overset{Fuming\ {HNO}_{3}}{\rightarrow}}\lbrack\ \rbrack\overset{{NH}_{3}/H_{2}S}{\rightarrow}\lbrack\ \rbrack\underset{2.{HBF}_{4},\Delta}{\overset{1.HONO}{\rightarrow}}$.
A detailed explanation for this question is being prepared.