Question

$C_6{H}_6+{HNO}_3+H_2{SO}_4\stackrel{{50}^0-{60}^0}{\to }A\stackrel{Sn+HCl}{\to }B\stackrel{{CH}_{3}COCl}{\to }C$. Then $C$ is:

• A. aniline
• B. nitro benzene
• C. acetanilide
• D. benzedine aniline

Answer 1

Answer: (C)

acetanilide

Nitration of benzene with a mixture of nitric acid and sulfuric acid gives nitrobenzene (compound $A$). The nitro group is then reduced to amino group by tin and $HCl$ to given aniline (compound $B$). Amino group is then acetylated by treatment with acetyl chloride to form acetanilide (compound $C$).

$\underset{benzene}{C_6{H}_6}+{HNO}_3+H_2{SO}_4\stackrel{{50}^0-{60}^0}{\to }\underset{A,nitrobenzene}{C_6{H}_6-N{O}_2}\stackrel{Sn+HCl}{\to }\underset{B,aniline}{C_6{H}_6-N{H}_2}\stackrel{{CH}_{3}COCl}{\to }\underset{C,acetanilide}{C_6{H}_5-NH-COC{H}_3}$

Hence, the correct option is C.