Why Aniline Does Not Undergo Friedel-Crafts Reaction
Chemistry · Amines · NEET
Aniline does not undergo the Friedel-Crafts reaction because the nitrogen lone pair of the -NH2 group reacts with the Lewis acid catalyst (AlCl3). This makes the nitrogen positively charged. The positive nitrogen pulls electrons away from the ring, so the ring becomes electron-poor and cannot attract the electrophile. Memory hook: "AlCl3 grabs the lone pair, N turns +, ring goes dead."
The nitrogen lone pair of aniline bonds to the AlCl3 catalyst, forming a salt. The nitrogen becomes positive and pulls electrons out of the ring, so the electron-poor ring cannot attract the electrophile and the Friedel-Crafts reaction does not occur.
Your doubts, answered
Why exactly does aniline fail the Friedel-Crafts reaction?
Friedel-Crafts uses a Lewis acid catalyst like anhydrous AlCl3. Aniline's nitrogen has a lone pair and acts as a base. This lone pair reacts with AlCl3 and forms a salt (an acid-base complex). Now the nitrogen carries a positive charge (-NH2·AlCl3 becomes like -N(+)). A positive nitrogen strongly pulls electron density out of the benzene ring by the -I effect. An electron-poor ring cannot attract the electrophile, so no Friedel-Crafts alkylation or acylation happens.
Isn't -NH2 supposed to activate the ring? Why is it deactivating here?
Yes, a free -NH2 group is strongly activating and ortho/para directing because its lone pair pushes electrons into the ring. But in Friedel-Crafts, that same lone pair is no longer free. The AlCl3 catalyst takes the lone pair. Once bonded to AlCl3, the nitrogen becomes positive and now acts like a -NH3(+) type group, which is deactivating and meta directing. So the behaviour flips: helpful lone pair becomes a positive electron-withdrawing centre.
Does the AlCl3 catalyst get used up with aniline?
Yes. Because the basic nitrogen binds the AlCl3, the catalyst is consumed forming the salt. It is no longer available to activate the alkyl halide or acyl halide. This is a second reason the reaction does not proceed properly. So both effects work together: catalyst is trapped, and the ring is deactivated.
Is this the same reason nitration of aniline in strong acid gives some m-nitroaniline?
Yes, the idea is similar. In strong acid, aniline gets protonated to the anilinium ion, C6H5NH3(+). The -NH3(+) group is deactivating and meta directing, so some nitration goes to the meta position. In Friedel-Crafts, AlCl3 plays the role of the acid: it makes the nitrogen positive, giving the same deactivating, meta-directing behaviour.
Do other aromatic amines like N,N-dimethylaniline also fail Friedel-Crafts?
Yes. Any aromatic amine with a basic nitrogen lone pair reacts with the Lewis acid the same way. The nitrogen becomes positive and deactivates the ring. So aniline and its N-alkyl versions all struggle with Friedel-Crafts for the same reason: the basic nitrogen destroys the catalyst and deactivates the ring.
⚠️ The NEET trap ✗ Thinking -NH2 is always activating, so aniline should undergo Friedel-Crafts easily and even faster than benzene. ✓ With AlCl3 present, the lone pair binds the catalyst. The nitrogen turns positive, becomes deactivating and meta directing, and the ring will not react. 🧠 Free -NH2 activates, but -NH2 holding AlCl3 becomes positive and kills the ring.
Real NEET questions
NEET 2024
Given below are two statements. Statement I: Aniline does not undergo Friedel-Crafts alkylation reaction. Statement II: Aniline cannot be prepared through Gabriel synthesis. In the light of the above statements, choose the correct answer.
A · Both Statement I and Statement II are false
B · Statement I is correct but Statement II is false
C · Statement I is incorrect but Statement II is true
D · Both Statement I and Statement II are true ✓
Solution: Statement I is TRUE: In Friedel-Crafts alkylation the Lewis acid AlCl3 reacts with the basic nitrogen lone pair of aniline to form a salt (C6H5N(+)H2·AlCl3 type complex). The positively charged nitrogen strongly deactivates the ring (acts as a meta-directing -I group) and the catalyst is used up, so aniline does not undergo Friedel-Crafts alkylation. Statement II is TRUE: Gabriel synthesis needs an SN2 attack of potassium phthalimide on an alkyl halide. Aryl halides do not undergo SN2, so aromatic amines like aniline cannot be made by Gabriel synthesis. Hence both statements are true, option (D).
NEET 2018
Nitration of aniline in strong acidic medium also gives m-nitroaniline because
A · in absence of substituents the nitro group always goes to the m-position
B · in electrophilic substitution reactions the amino group is meta directing
C · inspite of substituents the nitro group always goes to only the m-position
D · in strong acidic medium aniline is present as the anilinium ion ✓
Solution: In strong acid the basic -NH2 group is protonated to the anilinium ion, C6H5NH3(+). The -NH3(+) group has no lone pair to donate, is electron-withdrawing (deactivating) and therefore meta directing, so some nitration occurs at the meta position. This is the same principle as the Friedel-Crafts case, where AlCl3 (instead of H+) makes the nitrogen positive and deactivating. Option (D) is correct; the free -NH2 group is actually activating and ortho/para directing, so (B) is wrong.
Solved Amines NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
In one line, why does aniline not undergo Friedel-Crafts reaction?
Because the nitrogen lone pair binds the Lewis acid AlCl3, turning nitrogen positive, which deactivates the benzene ring so it cannot react with the electrophile.
Does this apply to both Friedel-Crafts alkylation and acylation?
Yes. Both need the AlCl3 Lewis acid catalyst, and both fail with aniline for the same reason: the basic nitrogen traps the catalyst and deactivates the ring.
How can you still carry out substitution on aniline safely?
Protect the -NH2 group first by acetylation (using acetic anhydride) to form acetanilide. The -NHCOCH3 group is much less basic, so it does not deactivate the ring badly, and controlled substitution becomes possible.
Is aniline a base? Why does that matter here?
Yes, aniline is a base because nitrogen has a lone pair. That basic lone pair is exactly what reacts with the Lewis acid AlCl3, which is what stops the Friedel-Crafts reaction.
Why is this important for NEET?
NEET repeatedly tests aniline reactions in the Amines chapter, including a direct 2024 statement question on Friedel-Crafts. Knowing the lone pair plus AlCl3 salt logic lets you answer many linked questions on directing effects and basicity.
