A) Benzene $\overset{Br_{2}/FeBr_{3}/heat\ }{\rightarrow}$
B) Aniline $\overset{Br_{2},H_{2}O}{\rightarrow}\lbrack\ \rbrack\underset{2.CuBr}{\ \overset{1.HONO\ }{\rightarrow}}$
C) Nitrobenzene
$\underset{H_{2}{SO}_{4},\mathrm{\Delta}}{\overset{Fuming\ {HNO}_{3}\ }{\rightarrow}}\lbrack\ \rbrack\underset{C_{2}H_{5}OH,heat}{\overset{\ \ \ \ \ \ Fe/HCl\ \ \ \ \ \ \ \ \ }{\rightarrow}}\lbrack\ \rbrack\underset{2.CuBr}{\ \overset{1.HONO\ }{\rightarrow}}$
D) Bromobenzene
$\underset{H_{2}{SO}_{4}}{\overset{\ {\ \ HNO}_{3}\ \ }{\rightarrow}}\lbrack\ \rbrack\underset{C_{2}H_{5}OH,heat}{\overset{\ \ \ \ \ \ \ \ \ \ \ \ Fe/HCl\ \ \ \ \ \ \ }{\rightarrow}}\lbrack\ \rbrack\underset{2.CuBr}{\overset{\ 1.HONO\ }{\rightarrow}}$
Explanation
The correct answer is Option C: Nitrobenzene
$\underset{H_{2}{SO}_{4},\mathrm{\Delta}}{\overset{Fuming\ {HNO}_{3}\ }{\rightarrow}}\lbrack\ \rbrack\underset{C_{2}H_{5}OH,heat}{\overset{\ \ \ \ \ \ Fe/HCl\ \ \ \ \ \ \ \ \ }{\rightarrow}}\lbrack\ \rbrack\underset{2.CuBr}{\ \overset{1.HONO\ }{\rightarrow}}$.
A detailed explanation for this question is being prepared.