The major product in the following reaction is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4-chloroacetophenone
In Friedel-Crafts acylation, an acyl chloride and anhydrous \( \ce{AlCl3} \) produce the acylium ion, \( \ce{CH3CO^{+}} \). This electrophile attacks the benzene ring and replaces one hydrogen atom by the –\( \ce{COCH3} \) group.
The chlorine atom already on the ring decides where the acyl group goes:
–I effect: chlorine withdraws electrons and makes the ring less reactive than benzene.
+R effect: its lone pairs increase the electron density at the ortho and para positions, so it is an o,p-directing group.
The acyl group is bulky, so attack next to chlorine (ortho) is hindered and the para product is major.
\( \displaystyle \ce{C6H5Cl + CH3COCl} \xrightarrow{\text{anhyd. }\ce{AlCl3}} \ce{ClC6H4COCH3 + HCl} \)
So the major product is 4-chloroacetophenone (1-(4-chlorophenyl)ethan-1-one).
Why the other options are incorrect
Option (1) the diacetyl product — Incorrect. Once an acyl group enters, it strongly deactivates the ring, so a second acylation does not take place.
Option (3) 2-chloroacetophenone — Incorrect. The ortho product forms only as the minor product because of steric hindrance.
Option (4) 3-chloroacetophenone — Incorrect. Chlorine is o,p-directing, so the meta product is not formed.
NCERT (2026-27 edition):
Friedel-Crafts Reaction
