What is the major product when benzoic acid reacts with Br2 and FeBr3?

The correct answer is Option (2) → m-Bromobenzoic acid

Br₂ with FeBr₃ brings about electrophilic substitution on the benzene ring. The –COOH group withdraws electrons from the ring, so it deactivates the ring and directs the incoming bromine to the meta position:

    \( \displaystyle \ce{C6H5COOH + Br2} \xrightarrow{\ce{FeBr3}} \ce{BrC6H4COOH + HBr} \)

The bromine enters at C-3, meta to –COOH, giving m-bromobenzoic acid.

Why the other options are incorrect

Option (1) o-Bromobenzoic acid — Incorrect. The ortho position would be the main site only for an ortho/para-directing group; –COOH is meta-directing.

Option (3) p-Bromobenzoic acid — Incorrect. For the same reason, the para position is not the main site of attack.

Option (4) 2,4,6-Tribromobenzoic acid — Incorrect. Substitution at three ortho/para positions at once is seen with strongly activating groups such as –OH in phenol; the deactivating –COOH group gives a single meta substitution.

NCERT (2026-27 edition):
2. Ring substitution – “Aromatic carboxylic acids undergo electrophilic substitution reactions in which the carboxyl group acts as a deactivating and meta-directing group.”