What would be the product of alkylation of aniline in the presence of anhydrous AlCl3 as catalyst?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Aniline forms salt with AlCl3.
In a Friedel–Crafts alkylation, the catalyst anhydrous \( \ce{AlCl3} \) is a Lewis acid: it accepts an electron pair from the alkyl halide and produces the electrophile \( \ce{R^{+}} \).
Aniline is a Lewis base, because its nitrogen has a lone pair. So instead of helping the reaction, \( \ce{AlCl3} \) reacts with aniline and forms a salt (an adduct):
\( \displaystyle \ce{C6H5NH2 + AlCl3 -> C6H5NH2^+-AlCl3^-} \)
Why the other options are incorrect
Option (1) p-alkyl aniline — Incorrect. –\( \ce{NH2} \) is an ortho/para director, so this would be expected only if the reaction took place, but it does not.
Option (2) o-alkyl aniline — Incorrect. For the same reason, no ortho product forms; the ring is deactivated by salt formation.
Option (3) both p and o-alkyl aniline — Incorrect. This is the usual result for an activating group, but here no alkylation occurs at all.
NCERT (2026-27 edition):
Chemical Reactions – …
