Question

What happens when : Chlorobenzene is treated with $CH_{3}Cl$ in the presence of anhydrous $AlCl_{3}$ ?

Hint:

Halogens are deactivating but ortho/para directing. The bulky chlorine atom makes the para position the preferred site of attack.

Answer 1

Step 1: Role of the Lewis acid catalyst.
Anhydrous $AlCl_3$ accepts a chloride lone pair from $CH_3Cl$, generating the electrophilic methyl carbocation: $CH_3Cl + AlCl_3 \rightarrow CH_3^+ [AlCl_4]^-$. This is Friedel-Crafts alkylation.
Step 2: Effect of Cl on the ring.
The chlorine substituent on chlorobenzene is deactivating (withdraws electrons by $-I$ effect) but ortho/para directing (donates electrons by $+M$ resonance, increasing electron density at ortho and para positions).
Step 3: Attack of the electrophile.
The $CH_3^+$ electrophile preferentially attacks the electron-richer ortho and para positions of the ring, forming a Wheland intermediate (sigma complex). Loss of a proton then restores aromaticity.
Step 4: Products formed.
A mixture of 1-chloro-2-methylbenzene (ortho isomer, minor) and 1-chloro-4-methylbenzene (para isomer, major) is obtained. The para product predominates due to less steric crowding.