Electrophilic Substitution in Aniline: Bromination, Nitration and Sulphonation
Chemistry · Amines · NEET
The -NH2 group is a strong activating, ortho/para-directing group, so aniline reacts much faster than benzene. That is why bromine water alone gives the trisubstituted product 2,4,6-tribromoaniline (a white solid), not a single mono-product. The one trap: in strong acid (nitration), aniline turns into the anilinium ion, which is meta-directing, so you also get m-nitroaniline. Memory hook: "Free NH2 = ortho/para; protonated NH3+ = meta."
Free -NH2 activates the ring and directs to ortho/para (bromine water gives 2,4,6-tribromoaniline). In strong acid the group becomes -NH3+, which is deactivating and meta-directing, so nitration also gives m-nitroaniline.
Your doubts, answered
Why does aniline give 2,4,6-tribromoaniline and not just one bromo product?
The -NH2 group donates its lone pair into the ring, so the ring becomes very electron-rich (strongly activated). This makes it react so fast that all three free ortho and para positions get brominated at once. So aniline + bromine water gives a white precipitate of 2,4,6-tribromoaniline, not mono-bromoaniline. Stopping at one bromine is not possible with plain bromine water.
How do you get only para-bromoaniline (mono-substitution) from aniline?
You must first protect the -NH2 group. Acetylate aniline with acetic anhydride to form acetanilide (-NHCOCH3). The lone pair on nitrogen is now shared with the C=O oxygen by resonance, so the ring is less activated and the group becomes bigger. Bromination now gives mainly p-bromoacetanilide. Then hydrolyse (acid or base) to remove the acetyl group and get p-bromoaniline. Steps: acetylate to protect, brominate, then hydrolyse.
Why does nitration of aniline give meta-nitroaniline in strong acid?
Nitration needs a strong acid mixture. In this acid, the basic -NH2 group picks up a proton and becomes the anilinium ion, -NH3+. This positive group has no lone pair to donate, is electron-withdrawing (deactivating), and is meta-directing. So a large fraction of the product is m-nitroaniline. Free aniline that is still un-protonated gives ortho and para. This is exactly what NEET 2018 asked.
Is aniline more reactive than benzene in electrophilic substitution?
Yes. The -NH2 group is a strong activator because its lone pair adds electron density to the ring. So aniline undergoes electrophilic substitution (bromination, nitration, sulphonation) faster and more easily than benzene. Benzene needs a catalyst and a Lewis acid, but aniline reacts even with bromine water at room temperature.
What is the product of sulphonation of aniline?
Aniline with concentrated sulphuric acid first forms anilinium hydrogensulphate. On heating at 453-473 K it rearranges to p-aminobenzenesulphonic acid, commonly called sulphanilic acid, as the major product. Sulphanilic acid exists as a dipolar zwitterion (the -NH2 becomes -NH3+ and the -SO3H becomes -SO3-).
Is the -NH2 group ortho/para or meta directing?
Free -NH2 is ortho/para directing and activating because it donates its lone pair into the ring. But when it is protonated to -NH3+ (in strong acid), it becomes meta directing and deactivating because it no longer has a lone pair to give. So the SAME group can direct differently depending on whether it is protonated. This is a classic NEET catch.
⚠️ The NEET trap ✗ The -NH2 group is always ortho/para directing, so nitration of aniline gives only o- and p-nitroaniline. ✓ In the strong acid used for nitration, aniline is protonated to the anilinium ion (-NH3+), which is deactivating and meta-directing, so a significant amount of m-nitroaniline also forms. 🧠 Acid protonates NH2 to NH3+ -> the meta product appears. Free NH2 = o/p; NH3+ = meta.
Real NEET questions
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: Free -NH2 is a strong activating, ortho/para director because it donates its lone pair into the ring. But nitration uses a strongly acidic mixture, and here the basic amino group is protonated: C6H5NH2 + H+ -> C6H5NH3+. The -NH3+ group carries a full positive charge, has no lone pair to donate, is electron-withdrawing (deactivating) and therefore meta-directing. So a good fraction of nitration goes to the meta position, giving m-nitroaniline alongside the o-/p- products from any un-protonated aniline. Hence (D). Option (B) is wrong because the FREE amino group is not meta directing; (A) and (C) are false generalisations.
NEET 2024
Statement I: Aniline does not undergo Friedel-Crafts alkylation reaction. Statement II: Aniline cannot be prepared through Gabriel synthesis. 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 the Lewis acid AlCl3 reacts with the basic nitrogen lone pair of aniline to form a salt. The nitrogen now carries a positive charge (-NH2.AlCl3 type), which strongly deactivates the ring and behaves like a meta-directing -I group, and the catalyst is used up. So aniline gives no Friedel-Crafts product. Statement II is TRUE: Gabriel synthesis needs an SN2 attack of potassium phthalimide on an alkyl halide, but aryl halides do not undergo this substitution, so aromatic amines like aniline cannot be made this way. Both true, so (D). This links directly to why aniline behaves specially in electrophilic conditions.
Solved Amines NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the product when aniline reacts with bromine water?
A white precipitate of 2,4,6-tribromoaniline. The strong activating -NH2 group makes all three free ortho/para positions brominate at once, even without a catalyst.
Why can't we stop bromination of aniline at one bromine?
Because -NH2 makes the ring so reactive that substitution keeps happening at every available ortho and para spot. To get a mono product like p-bromoaniline you first protect the -NH2 by acetylation, then brominate, then hydrolyse.
What is sulphanilic acid?
It is p-aminobenzenesulphonic acid, the major product of sulphonating aniline with concentrated H2SO4 at 453-473 K. It exists as a zwitterion (dipolar ion) with -NH3+ and -SO3- in the same molecule.
Does aniline undergo Friedel-Crafts reaction?
No. The nitrogen lone pair binds the Lewis acid catalyst (AlCl3) to form a salt, so the nitrogen becomes positive, the ring is deactivated, and the catalyst is consumed. This is asked separately in the next concept, why aniline does not undergo Friedel-Crafts.
Why is nitration of aniline done as acetanilide instead of directly?
Direct nitration in strong acid protonates aniline and gives a mixture including m-nitroaniline, and can also oxidise aniline. Acetylating first (to acetanilide) protects the nitrogen and gives mainly p-nitro product cleanly, which is then hydrolysed.
