Chlorobenzene reacts with acetyl chloride in anhydrous \( AlCl_3 \) to give major product as
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4-chloroacetophenone (1-(4-chlorophenyl)ethan-1-one)
Chlorobenzene and acetyl chloride with anhydrous \( \ce{AlCl3} \) undergo Friedel-Crafts acylation, an electrophilic substitution. The Lewis acid \( \ce{AlCl3} \) takes the chloride ion from acetyl chloride and forms the acylium ion, the electrophile:
\( \displaystyle \ce{CH3COCl + AlCl3 -> CH3CO^+ + AlCl4^-} \)
Chlorine is an ortho/para director; the bulky acetyl group enters mainly at the para position, since the ortho positions are crowded by the chlorine atom.
So the major product is 4-chloroacetophenone, the structure in Option (3).
Why the other options are incorrect
Option (1) 2-chloroacetophenone — Incorrect. The ortho product is formed, but only as the minor product, because the ortho positions are hindered by the chlorine atom.
Option (2) 3-chloroacetophenone — Incorrect. Chlorine directs to ortho and para positions; the meta product is not obtained.
Option (4) 1-chloro-2,4-diacetylbenzene — Incorrect. Once one acetyl group has entered, this electron-withdrawing group deactivates the ring further, so a second acylation does not take place.
NCERT (2026-27 edition):
Electrophilic Substitution Reactions
