Electrophilic substitution in aniline

Pour bromine water into a test tube of aniline and nothing dramatic seems to happen at first glance — until a white solid appears almost instantly, with no heating, no catalyst, nothing like the sluggish reaction benzene needs. That instant precipitate is the signature of a benzene ring that has been supercharged by a single substituent: $-NH_2$ . The nitrogen lone pair feeds electron density straight into the ring, and the ring pays it back by reacting with electrophiles far faster than benzene does, and almost exclusively at the ortho and para positions. This article works through the three ring reactions of aniline named explicitly in the ISC syllabus — bromination, nitration and sulphonation — and the one twist examiners love: nitration cannot be done directly, and needs a protect-react-deprotect workaround. Understanding why the workaround is needed, not just memorising it, is the real test here.

The core insight

The $-NH_2$ group donates its lone pair into the benzene ring by resonance, making aniline strongly activated and ortho/para-directing toward electrophiles, but in strongly acidic nitrating conditions that same nitrogen gets protonated to $-NH_3^+$ , which flips the ring to deactivated and meta-directing, so nitration is done indirectly via acetanilide.

The core idea

Why aniline is hyper-reactive. Nitrogen's lone pair is not stuck on $-NH_2$ ; it delocalises into the $\pi$ -system of the ring by resonance. Draw out the contributing structures and negative charge appears specifically at the carbons ortho and para to the substituent — nowhere else. Those are exactly the carbons an electrophile ( $Br^+$ , $NO_2^+$ , $SO_3$ ) will attack, and because the electron density there is genuinely higher than in benzene, the whole ring reacts faster too. This combination — activating and ortho/para-directing — is what makes aniline such an outlier among aromatic compounds.

Bromination needs no help. Benzene requires a Lewis-acid catalyst such as $FeBr_3$ to generate the electrophile $Br^+$ from $Br_2$ before anything happens. Aniline's ring is so electron-rich that bromine water alone, at room temperature, is enough:

C_6H_5NH_2 + 3Br_2(aq) \rightarrow 2,4,6\text{-}Br_3C_6H_2NH_2\!\downarrow + 3HBr \qquad (1)

where: $C_6H_5NH_2$ is aniline, $Br_2(aq)$ is bromine water (no catalyst added), and the product $2,4,6\text{-}Br_3C_6H_2NH_2$ is 2,4,6-tribromoaniline, a white precipitate; $HBr$ is the by-product released as all three o/p positions react in turn. The reaction cannot easily be stopped at mono-substitution because the ring stays activated (if anything, more so) after each substitution — this instant, catalyst-free white precipitate also doubles as a qualitative test for aniline.

Nitration needs protection, not force. Throwing aniline straight into conc. $HNO_3/H_2SO_4$ backfires twice over: the strongly acidic medium protonates $-NH_2$ to $-NH_3^+$ , which is electron-withdrawing and meta-directing (the opposite of what free aniline does), and any aniline left unprotonated is oxidised by concentrated $HNO_3$ into tarry by-products. The fix is to acetylate first — turning $-NH_2$ into $-NHCOCH_3$ , a group that is still o,p-directing but resists both protonation and oxidation under the reaction conditions — nitrate the acetanilide, then hydrolyse the amide to unmask the free amine. The para isomer dominates.

Sulphonation bakes to the zwitterion. Aniline and concentrated $H_2SO_4$ first give anilinium hydrogen sulphate; heating this further ("baking," around $180\,^\circ C$ ) rearranges it to sulphanilic acid (4-aminobenzenesulphonic acid), a zwitterion where the acidic $-SO_3H$ protonates the basic $-NH_2$ intramolecularly:

C_6H_5NH_2 + H_2SO_4 \xrightarrow{\Delta} {}^{+}H_3N\text{-}C_6H_4\text{-}SO_3^- + H_2O \qquad (2)

where: the product is written in its zwitterion form, para-substituted, with an internal proton transfer from $-SO_3H$ to $-NH_2$ rather than a separate acidic and basic group.

Figure

Resonance donation from the amino lone pair pushes electron density onto the ortho and para carbons of the ring while the meta carbons stay unaffected

Why it works

Nitration of aniline fails with the ordinary $HNO_3/H_2SO_4$ mixture, and understanding exactly why — down to which species is present in solution and what that species does to the ring — is the reasoning examiners test, more than any single equation.

Step 1. Identify what conc. $H_2SO_4$ does to $-NH_2$ before nitration even starts. Aniline is a base ( $-NH_2$ carries a lone pair), and it sits in a strongly acidic medium. It is protonated essentially completely: $C_6H_5NH_2 + H^+ \rightarrow C_6H_5NH_3^+$ . The species actually present for the nitration step is not aniline at all — it is the anilinium ion.

Step 2. Ask what the anilinium ion does to the ring. The nitrogen in $-NH_3^+$ has no lone pair left to donate — it is fully used up bonding to three hydrogens and carries a full positive charge. A positively charged, electron-withdrawing group attached directly to the ring pulls electron density away, deactivating the ring, and by the same resonance logic that made $-NH_2$ ortho/para-directing, an electron-withdrawing group directs meta. So the very act of dissolving aniline in the nitrating mixture destroys the directing behaviour you wanted to exploit.

Step 3. Account for the second failure mode, oxidation. Even the small equilibrium fraction of aniline that remains unprotonated is extremely electron-rich, and concentrated $HNO_3$ is a strong oxidising agent; that fraction gets oxidised to tarry, resinous by-products rather than clean nitro-substitution. Direct nitration therefore gives a poor, messy mixture even setting the meta-directing problem aside.

Step 4. Build the workaround from the diagnosis. Both failures trace back to the free $-NH_2$ lone pair being too available — for protonation and for oxidation. Acetylation converts it to $-NHCOCH_3$ : the lone pair on this nitrogen is partly delocalised into the adjacent carbonyl, so it is a much weaker base (resists protonation under nitrating conditions) and a less easily oxidised group, while still donating enough into the ring to remain o,p-directing. Nitrate the acetanilide cleanly, then hydrolyse to remove the acetyl group and recover the free amine as p-nitroaniline.

Worked example

Problem: Aniline is treated with excess bromine water at room temperature; separately, the same aniline sample is to be converted into p-nitroaniline. (a) Give the product and the reasoning for the bromination. (b) Outline the full route, with equations, from aniline to p-nitroaniline.

Step 1 (bromination). The $-NH_2$ lone pair delocalises into the ring, placing extra electron density at C2, C4 and C6 (ortho and para to $-NH_2$ at C1). These are the sites of highest electron density, so they react preferentially with the electrophile $Br^+$ generated from $Br_2$ — no catalyst needed, because the ring is already activated enough.

Step 2 (bromination, continued). Because the activation is so strong, substitution does not stop after one bromine goes in; it proceeds successively at all three available o/p positions. Product: 2,4,6-tribromoaniline, $Br_3C_6H_2NH_2$ , a white precipitate, with $3HBr$ released.

Step 3 (nitration route). Protect the amine: $C_6H_5NH_2 + (CH_3CO)_2O \rightarrow C_6H_5NHCOCH_3 + CH_3COOH$ (acetanilide). Nitrate the acetanilide, which stays o,p-directing and is not appreciably protonated: acetanilide $+\ HNO_3/H_2SO_4 \rightarrow$ predominantly p-nitroacetanilide, $CH_3CONH\text{-}C_6H_4\text{-}NO_2$ , with a minor ortho fraction separable by their different physical properties.

Step 4 (nitration route, finish). Hydrolyse off the protecting group (dilute acid or base, heat): p-nitroacetanilide $\rightarrow$ p-nitroaniline $+\ CH_3COOH$ . Final answer: (a) 2,4,6-tribromoaniline as a white precipitate; (b) aniline $\rightarrow$ acetanilide $\rightarrow$ p-nitroacetanilide $\rightarrow$ p-nitroaniline via the protect–nitrate–deprotect sequence.

Competitive stretch

If asked to predict the sulphonation product from the same aniline sample: heating aniline with concentrated $H_2SO_4$ first gives anilinium hydrogen sulphate, and further baking near $180\,^\circ C$ rearranges this to sulphanilic acid, the para zwitterion $^+H_3N\text{-}C_6H_4\text{-}SO_3^-$ — the intramolecular acid-base pair formed because $-SO_3H$ is acidic enough to protonate the ring's own $-NH_2$ .

Common traps

Assuming aniline needs a Lewis-acid catalyst for bromination the way benzene does — it reacts with plain bromine water, no catalyst.
Forgetting the bromination product is the tribromo compound (2,4,6-tribromoaniline), not a mono-substituted one, under normal (uncontrolled) conditions.
Trying to nitrate free aniline directly with $HNO_3/H_2SO_4$ — protonation to the meta-directing anilinium ion plus oxidative tar formation both defeat this; acetylation is the syllabus-prescribed fix.
Mixing up directing behaviour after protonation or acylation — free $-NH_2$ is o,p-directing and activating, protonated $-NH_3^+$ is meta-directing and deactivating, but acetamido $-NHCOCH_3$ stays o,p-directing because it resists protonation.

Ask the tutor: want to see the full curved-arrow resonance structures for why $-NH_2$ pushes electron density specifically to ortho and para, or work through why the anilinium ion becomes meta-directing, one resonance form at a time?

Why does aniline react with bromine water without any catalyst, when benzene absolutely needs one?

Benzene's ring has only ordinary $\pi$ -electron density, which is not enough to generate the electrophile $Br^+$ from $Br_2$ on its own — a Lewis acid like $FeBr_3$ has to polarise the $Br-Br$ bond first. Aniline's ring is already far more electron-rich because the $-NH_2$ lone pair is feeding density into it by resonance, concentrated at the ortho and para carbons. That extra electron density is enough to attack $Br_2$ directly and pull out $Br^+$ without any catalytic help, which is also why the reaction is instant at room temperature.

If protonating aniline kills its o,p-directing power, why does acetylating it not do the same thing?

Both protonation and acetylation change what happens to the nitrogen lone pair, but in opposite directions. Protonation removes the lone pair entirely — it is used up forming a bond to $H^+$ — leaving a positively charged, electron-withdrawing $-NH_3^+$ that is meta-directing. Acetylation instead delocalises the lone pair partly onto the adjacent carbonyl oxygen; some donating power is lost, but enough remains conjugated with the ring to keep $-NHCOCH_3$ activating and o,p-directing, while the weaker, less available lone pair also resists picking up a proton under the reaction conditions.

Why does sulphonation of aniline need heating and baking, unlike the instant bromination?

Sulphonation is an equilibrium reaction and the electrophile involved, $SO_3$ (or protonated $SO_3H^+$ ), is generated and reacts more slowly than $Br^+$ from bromine water. The first product, anilinium hydrogen sulphate, is only a salt formed by acid-base reaction, not yet the ring-substituted product; driving it to the actual C-sulphonation product, sulphanilic acid, needs sustained heating (the "baking" step near $180\,^\circ C$ ) to push the slower, reversible sulphonation equilibrium toward the thermodynamically stable zwitterion.