What will be the major product in the following reaction?

The correct answer is Option (2) → 4-Bromoanisole

In anisole the –\( \ce{OCH3} \) group donates electrons to the ring by resonance, so the ring is activated and substitution goes to the ortho and para positions; the activation is strong enough for bromine in ethanoic acid to react without an \( \ce{FeBr3} \) catalyst. The bulky –\( \ce{OCH3} \) group hinders the two ortho positions, so the para product is major (about 90%):

    \( \displaystyle \ce{C6H5OCH3 + Br2} \xrightarrow{\ce{CH3COOH}} \text{4-bromoanisole (major)} + \text{2-bromoanisole (minor)} + \ce{HBr} \)

Both isomers are formed, but the question asks for the major product, which is the para isomer.

NCERT (2026-27 edition):
Halogenation