JEE Chemistry 2025 Question with Solution

Given: Nitrobenzene is subjected to the sequence $\text{(i) Sn/HCl, (ii) Ac}_2\text{O, pyridine, (iii) Br}_2\text{, AcOH, (iv) NaOH(aq)}$.

Find: The major product A.

Step 1: Reduction of nitro group

Sn/HCl reduces the nitro group $(-\text{NO}_2)$ to an amino group $(-\text{NH}_2)$. Thus nitrobenzene is converted to aniline.

Step 2: Acetylation of amino group

Acetic anhydride in pyridine converts aniline into acetanilide, that is, the amino group is protected as $-\text{NHCOCH}_3$. This protection is necessary because a free amino group would strongly activate the ring and can lead to excessive bromination.

Step 3: Bromination

The amide group $(-\text{NHCOCH}_3)$ is an ortho/para directing group. Because it is bulky, bromination occurs predominantly at the para position in acetic acid.

Step 4: Hydrolysis

Treatment with aqueous NaOH hydrolyses the amide group back to the amino group. Therefore the final product is para-brominated aniline.

Therefore, the major product A is p-bromoaniline, so the correct option is B.

The sequence involves reduction, protection, electrophilic substitution, and deprotection.

Nitrobenzene gives aniline.

Aniline gives acetanilide.

Bromination occurs mainly at the para position due to ortho/para directing nature and steric preference.

Hydrolysis removes the acetyl protecting group and gives p-bromoaniline. The solution states Correct Option: (2), which maps to B.