A compound 'A' when treated with $\mathrm{HNO}_{3}$ (in presence of $\mathrm{H}_{2} \mathrm{SO}_{4}$ ) gives…
A compound 'A' when treated with $\mathrm{HNO}_{3}$ (in presence of $\mathrm{H}_{2} \mathrm{SO}_{4}$ ) gives compound 'B', which is then reduced with $\mathrm{Sn}$ and $\mathrm{HCl}$ to aniline? The compound ' $\mathrm{A}$ ' is
toluene
benzene
ethane
acetamide
Solution
$\mathrm{A} \stackrel{\mathrm{HNO}_{3} / \mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow} \mathrm{B} \stackrel{\mathrm{Sn} / \mathrm{HCl}}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}$
This indicates that $\mathrm{B}$ is $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2}$ and hence $\mathrm{A}$ is $\mathrm{C}_{6} \mathrm{H}_{6}$