Identify the major and minor products obtained in the following reaction:
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (D) only
In anisole (methoxybenzene) the lone pair of the oxygen of –\( \ce{OCH3} \) is delocalised into the ring (+R effect). The electron density rises at the ortho and para positions, so the ring is activated and the –\( \ce{OCH3} \) group is ortho, para-directing. Anisole is brominated by bromine in ethanoic acid even without the \( \ce{FeBr3} \) catalyst.
\( \ce{C6H5OCH3} \xrightarrow[\ce{CH3COOH}]{\ce{Br2}} \text{p-}\ce{BrC6H4OCH3} + \text{o-}\ce{BrC6H4OCH3} + \ce{HBr} \)
(p-bromoanisole, major, about 90%; o-bromoanisole, minor)
The para product is major because the ortho positions are hindered by the methoxy group; the ortho product is minor.
(A) p-bromoanisole — Major product — Correct. Para substitution, about 90% yield.
(B) 2,4-dibromoanisole — Minor product — Incorrect. Under these mild conditions only one bromine enters the ring; a dibromo compound is not a product here.
(C) o-bromoanisole — Major product — Incorrect. The ortho isomer is formed, but as the minor product.
(D) o-bromoanisole — Minor product — Correct. The ortho isomer is the minor product.
(E) 2,6-dibromoanisole — Major product — Incorrect. No dibromination takes place, and the para position is free, so a 2,6-product cannot be major.
NCERT (2026-27 edition):
Halogenation
