JEE Chemistry 2025 Question with Solution

Given: Bromobenzene undergoes sulfonation with $SO_3/H_2SO_4$ to form major product A, followed by bromination with $Br_2/Fe$ to form major product B.

Find: The correct pair of major products A and B.

In electrophilic aromatic substitution with multiple substituents, the directing effects of the groups already present on the ring determine the major product. Bromine is ortho-para directing though deactivating.

In the first step, bromobenzene on sulfonation gives the para sulfonated product as the major product due to ortho-para direction by bromine and lower steric hindrance at the para position. Thus A is p-bromobenzenesulfonic acid.

In the second step, the ring already contains Br and $SO_3H$. Bromine is ortho-para directing, while $SO_3H$ is a strongly deactivating meta directing group. The position favored by both effects is the position ortho to bromine and meta to $SO_3H$. Therefore the second bromine enters that position.

So B is the dibromo sulfonic acid structure shown in option B.

Therefore, the correct option is B.

Given: A sequence of two electrophilic aromatic substitution reactions on bromobenzene.

Find: Which structures correspond to the major products after each step.

1. First reaction: Sulfonation of bromobenzene.

• Bromine directs incoming electrophiles to the ortho and para positions.
• Between these, the para product is favored because it is less sterically crowded than the ortho product.
• Hence A is the para sulfonated product.

2. Second reaction: Bromination of the para-sulfonated product.

• The existing Br group directs to ortho/para positions.
• The $SO_3H$ group directs meta.
• The common favorable position is the one ortho to Br and meta to $SO_3H$.

Thus the final major product matches the structure given in option B.

Therefore, A and B are represented by option B.