JEE Mains Chemistry · Amines

Electrophilic Substitution in Aniline

The amino group makes the ring so reactive, and so basic, that aniline over-brominates, gives meta product on nitration and fails Friedel–Crafts; acetylation tames it.

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Why this matters

Twenty-four PYQs, two numerical, three from 2026. Eight test bromination, from tribromoaniline yields to the protection and blocking needed for one bromine; ten test nitration and nitrosation, above all why the meta product appears; six test Friedel–Crafts failure, sulphonation to sulphanilic acid and oxidation.

Concept 1 of 3: Bromination of aniline and protection by acetylation

The nitrogen lone pair feeds the ring and makes the ortho and para carbons very rich in electrons. Bromine water therefore substitutes at all three of them at once, and the reaction cannot be stopped at one bromine. To get one bromine, first turn the amine into an amide: the lone pair is then shared with the carbonyl group, the ring is only moderately activated, and the bulky group favours para.

Definition

With bromine water at room temperature aniline gives 2,4,6-tribromoaniline, a white precipitate, in one step.
Only ortho and para attack lets the nitrogen share the positive charge of the σ-complex (an iminium form); a meta σ-complex gets no such help and is not involved.
Protection: acetylate ( $\mathrm{(CH_3CO)_2O}$ , pyridine) to acetanilide, brominate (mainly para), then hydrolyse the amide with acid or base to free the $\mathrm{NH_2}$ .
To get an ortho product, block para first (for example by sulphonation), then remove the blocking group.
Molar masses for yield problems: aniline 93, 2,4,6-tribromoaniline 330 g/mol.

Bromination of aniline with bromine water

\mathrm{C_6H_5NH_2 + 3Br_2 \xrightarrow{H_2O} 2,4,6{-}Br_3C_6H_2NH_2\downarrow + 3HBr}

Worked example

Give a route from aniline to 4-bromoaniline as the main product.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 5 Apr 2024 · Q60Moderate

Example 1 · Amines · Electrophilic Substitution in Aniline

9.3\text{ }g

of pure aniline is treated with bromine water at room temperature to give a white precipitate of the product '

P

'. The mass of product '

P

' obtained is

26.4\text{ }g

. The percentage yield is............

\%

Bromine water cannot stop at one bromine

The amino group activates all three ortho and para positions. With bromine water they are all substituted at once; monobromination needs the amine protected as its acetyl derivative.

The protecting group must come off

Bromination of acetanilide gives 4-bromoacetanilide. The last step, hydrolysis, is what turns it into 4-bromoaniline; a sequence without it ends at the amide.

Drill 7 more on bromination of aniline and protection by acetylation

Concept 2 of 3: Nitration of aniline: the anilinium ion and meta product

The nitrating mixture is a strong acid, and aniline is a base. Much of the aniline is protonated to the anilinium ion,

\mathrm{C_6H_5NH_3^+}

, whose positive nitrogen withdraws electrons and directs meta. So nitration of aniline gives a large share of meta product, besides oxidation tars. Protecting the amine as acetanilide keeps it unprotonated and gives mainly the para product.

Definition

Direct nitration ( $\mathrm{HNO_3/H_2SO_4}$ , 288 K) gives about 51% para, 47% meta and 2% ortho nitroaniline.
In the mixture $\mathrm{H_2SO_4}$ is the acid and $\mathrm{HNO_3}$ accepts a proton (acts as a base) to form $\mathrm{NO_2^+}$ .
The meta product comes from the anilinium ion, not from $\mathrm{NH_2}$ being a meta director.
Route to 4-nitroaniline: acetylate, nitrate, hydrolyse.
In benzanilide, $\mathrm{C_6H_5CONHC_6H_5}$ , the ring on nitrogen is activated and is substituted para to NH; the ring on the carbonyl is deactivated.
N,N-Dimethylaniline with $\mathrm{NaNO_2/HCl}$ at low temperature is nitrosated at the para position by $\mathrm{NO^+}$ .

Product distribution in the direct nitration of aniline

\mathrm{C_6H_5NH_2 \xrightarrow{HNO_3,\ H_2SO_4,\ 288\ K}}\ p\ (51\%) + m\ (47\%) + o\ (2\%)

Worked example

Give a route from aniline to 4-nitroaniline that avoids the meta product.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 28 July 2022 · Q135Moderate

Example 2 · Amines · Electrophilic Substitution in Aniline

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason

R

Assertion A: Aniline on nitration yields ortho, meta & para nitro derivatives of aniline. Reason R: Nitrating mixture is a strong acidic mixture. In the light of the above statements, choose the correct answer from the options given below

NH₂ is not meta-directing

The amino group directs ortho and para. The meta product in nitration comes from the anilinium ion formed in the acid, whose positive nitrogen directs meta.

Para is still slightly ahead of meta

Direct nitration gives about 51% para and 47% meta, with only 2% ortho. A statement that meta exceeds ortho is true; one that meta is the only product is false.

Drill 9 more on nitration of aniline: the anilinium ion and meta product

Concept 3 of 3: Friedel–Crafts failure, sulphonation and oxidation of aniline

Aniline's lone pair makes it a Lewis base as well as an activated ring. A Lewis acid such as

\mathrm{AlCl_3}

bonds to the nitrogen instead of generating an electrophile, and the positive nitrogen then deactivates the ring. Strong acid does the same at first, which is why sulphonation needs a high temperature. The electron-rich ring is also easily oxidised.

Definition

Friedel–Crafts alkylation and acylation fail: aniline forms a salt with $\mathrm{AlCl_3}$ , putting a positive charge on nitrogen, and no ring product forms.
Sulphonation: concentrated $\mathrm{H_2SO_4}$ first gives anilinium hydrogensulphate; heating at 453–473 K gives sulphanilic acid (4-aminobenzenesulphonic acid), which exists as a zwitterion.
Sulphanilic acid contains N and S, so its Lassaigne extract gives a blood-red colour with $\mathrm{Fe^{3+}}$ (thiocyanate).
Oxidation: acidified $\mathrm{K_2Cr_2O_7}$ gives p-benzoquinone; air slowly gives coloured products.

Reaction of aniline	Reagent and conditions	Result	Reason
Friedel–Crafts alkylation or acylation	RCl or RCOCl with anhydrous $\mathrm{AlCl_3}$	No ring substitution; N– $\mathrm{AlCl_3}$ complex	Nitrogen is a Lewis base; the complex deactivates the ring
Sulphonation	Conc. $\mathrm{H_2SO_4}$ , then 453–473 K	Sulphanilic acid, $\mathrm{H_3N^+C_6H_4SO_3^-}$	Anilinium hydrogensulphate rearranges on heating
Oxidation	Acidified $\mathrm{K_2Cr_2O_7}$	p-Benzoquinone	The ring is very electron-rich
Nitration	$\mathrm{HNO_3/H_2SO_4}$ , 288 K	Mixture of para, meta and ortho nitroanilines	Partial protonation to the anilinium ion
Bromination	Bromine water	2,4,6-Tribromoaniline	Strong activation by $\mathrm{NH_2}$

Wherever aniline meets an acid, think of the protonated or complexed nitrogen first.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 29 June 2022 · Q136Moderate

Example 3 · Amines · Electrophilic Substitution in Aniline

In Friedel-Crafts alkylation of aniline, one gets:

Friedel–Crafts on aniline gives no ring product at all

The

\mathrm{AlCl_3}

is tied up by the nitrogen. Do not predict ortho, para or meta alkylanilines; the ring is not alkylated or acylated.

A strongly activated ring is also easily oxidised

Aniline darkens in air and is oxidised to p-benzoquinone by dichromate. Oxidising agents in a sequence can destroy the amine before any intended step.

Drill 5 more on friedel–crafts failure, sulphonation and oxidation of aniline

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Bromination of aniline and protection by acetylation
Bromination of aniline with bromine water
$\mathrm{C_6H_5NH_2 + 3Br_2 \xrightarrow{H_2O} 2,4,6{-}Br_3C_6H_2NH_2\downarrow + 3HBr}$
Nitration of aniline: the anilinium ion and meta product
Product distribution in the direct nitration of aniline
$\mathrm{C_6H_5NH_2 \xrightarrow{HNO_3,\ H_2SO_4,\ 288\ K}}\ p\ (51\%) + m\ (47\%) + o\ (2\%)$

Reference tables (1)

Friedel–Crafts failure, sulphonation and oxidation of aniline5 rows

Reaction of aniline	Reagent and conditions	Result	Reason
Friedel–Crafts alkylation or acylation	RCl or RCOCl with anhydrous $\mathrm{AlCl_3}$	No ring substitution; N– $\mathrm{AlCl_3}$ complex	Nitrogen is a Lewis base; the complex deactivates the ring
Sulphonation	Conc. $\mathrm{H_2SO_4}$ , then 453–473 K	Sulphanilic acid, $\mathrm{H_3N^+C_6H_4SO_3^-}$	Anilinium hydrogensulphate rearranges on heating
Oxidation	Acidified $\mathrm{K_2Cr_2O_7}$	p-Benzoquinone	The ring is very electron-rich
Nitration	$\mathrm{HNO_3/H_2SO_4}$ , 288 K	Mixture of para, meta and ortho nitroanilines	Partial protonation to the anilinium ion
Bromination	Bromine water	2,4,6-Tribromoaniline	Strong activation by $\mathrm{NH_2}$

Wherever aniline meets an acid, think of the protonated or complexed nitrogen first.

Watch out for (6)

Bromine water cannot stop at one bromine
→ Bromination of aniline and protection by acetylation
The protecting group must come off
→ Bromination of aniline and protection by acetylation
NH₂ is not meta-directing
→ Nitration of aniline: the anilinium ion and meta product
Para is still slightly ahead of meta
→ Nitration of aniline: the anilinium ion and meta product
Friedel–Crafts on aniline gives no ring product at all
→ Friedel–Crafts failure, sulphonation and oxidation of aniline
A strongly activated ring is also easily oxidised
→ Friedel–Crafts failure, sulphonation and oxidation of aniline

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