Question

Consider the following sequence of reactions.
$C_6{H}_{5}N{H}_2\underset{Heat}{\overset{{Ac}_{2}O}{\to }}A\underset{\begin{array}{c}Conc.H_{2}S{O}_4\\ 0-{15}^{o}C\end{array}}{\overset{Conc.HN{O}_3}{\to }}\underset{\left(Minor\right)}{B}\underset{Heat}{\overset{aq.H_{2}S{O}_4}{\to }}\underset{\left(Major\right)}{C}$
The major product $\left(C\right)$ is:

• A. 2, 4, 6-trinitroaniline
• B. p-nitroacetanilide
• C. p-nitroaniline
• D. m-nitroaniline

Answer 1

The correct option is C p-nitroaniline

The major product $\left(C\right)$ is $p$-nitroaniline.

Acetylation of aniline with acetic anhydride gives acetanilide.

This is followed by nitration with a mixture of conc nitric acid and conc sulphuric acid to form N-(4-nitrophenyl)acetamide.

In the final step, the acid hydrolysis removes acetyl protecting group..

Option C is correct.