The major product of the following reaction is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4-methoxytoluene (p-methylanisole)
Methyl chloride with anhydrous aluminium chloride is a Friedel-Crafts alkylation. The Lewis acid \( \ce{AlCl3} \) pulls the chlorine away and produces the electrophile \( \ce{CH3+} \), which substitutes a hydrogen of the ring.
In anisole the –\( \ce{OCH3} \) group releases electrons into the ring by resonance, so it activates the ring and directs the electrophile to the ortho and para positions. Both isomers are formed, but the methoxy group crowds the ortho positions next to it, so the para product is the major one.
So the major product is 4-methoxytoluene.
Why the other options are incorrect
Option (1) 3-methoxytoluene — Incorrect. The –\( \ce{OCH3} \) group is an ortho/para director; the meta isomer is not obtained.
Option (2) 2-methoxytoluene — Incorrect. It is formed, but only as the minor product, because the ortho position is hindered by the methoxy group.
Option (3) 2,4-dimethylanisole — Incorrect. This needs two methyl groups; one molecule of \( \ce{CH3Cl} \) gives the monomethyl product.
NCERT (2026-27 edition):
Electrophilic Substitution – …
