JEE Mains Chemistry · Amines

Coupling Reactions and Azo Dyes

A diazonium ion keeps its nitrogen and attacks a very electron-rich ring, usually para to OH or NH₂, to give a coloured azo compound, Ar–N=N–Ar′.

All Amines notes JEE Mains Chemistry strategy

Why this matters

Nineteen PYQs, four numerical, three from 2026. Twelve ask for the coupling product, the partner or the medium, including the β-naphthol test for aryl amines; seven are dye stoichiometry, percentage of nitrogen and the Griess–Ilosvay test for nitrite.

Concept 1 of 2: Coupling of diazonium salts with phenols and aryl amines

In coupling the diazonium ion does not lose nitrogen; it acts as an electrophile. It is a weak one, so it attacks only rings strongly activated by OH, O⁻,

\mathrm{NH_2}

\mathrm{NR_2}

. It goes para to the activating group, or ortho when para is taken. The

\mathrm{{-}N{=}N{-}}

bridge joins two rings into one long conjugated system, which is why the products are coloured.

Definition

With phenol in mildly alkaline solution (the phenoxide ion is the reactive form): 4-hydroxyazobenzene, orange.
With aniline in mildly acidic solution: 4-aminoazobenzene (aniline yellow), yellow.
With 2-naphthol (β-naphthol) in NaOH: 1-phenylazo-2-naphthol, orange-red. This is the confirmatory test for a primary aromatic amine.
Attack is para to the activating group; if para is blocked, ortho.
Methyl groups ortho to an $\mathrm{N(CH_3)_2}$ group twist it out of the ring plane, cut its resonance with the ring and slow coupling.
Aliphatic diazonium ions decompose before they can couple, so they give no dye.

Coupling of benzenediazonium chloride with phenol

\mathrm{C_6H_5N_2^+Cl^- + C_6H_5OH \xrightarrow{OH^-} C_6H_5{-}N{=}N{-}C_6H_4{-}OH\ (para) + HCl}

Worked example

Benzenediazonium chloride is coupled (a) with 2-naphthol in NaOH and (b) with aniline in mildly acidic solution. Give each product and its colour.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 9 · Q31Moderate

Example 1 · Amines · Coupling Reactions and Azo Dyes

Considering the above reaction,

X

and

Y

respectively are:

Coupling keeps the nitrogen

In replacement reactions

\mathrm{N_2}

leaves; in coupling both nitrogens stay in the product as the

\mathrm{{-}N{=}N{-}}

bridge. An azo dye always contains the two nitrogens of the diazonium ion.

Only strongly activated rings couple

The diazonium ion is a weak electrophile. Benzene, toluene or chlorobenzene do not couple; phenols, naphthols and aryl amines do.

Ortho methyls slow coupling on dimethylanilines

Two methyl groups beside

\mathrm{N(CH_3)_2}

push it out of the ring plane. Its lone pair no longer feeds the ring, so the ring is less activated and couples more slowly.

Drill 11 more on coupling of diazonium salts with phenols and aryl amines

Concept 2 of 2: Azo dye stoichiometry and the Griess–Ilosvay test

Every molecule of dye uses one diazonium ion, so moles of dye equal moles of the amine that was diazotised. If the same amine is also the coupling partner, each dye molecule uses two of it. The Griess–Ilosvay test turns this chemistry into a test for nitrite ion: nitrite diazotises sulphanilic acid, and the salt couples with 1-naphthylamine to give a red dye.

Definition

Moles of dye = moles of diazotised amine (with the partner in excess).
Useful molar masses: aniline 93; 4-aminoazobenzene $\mathrm{C_{12}H_{11}N_3}$ 197; 4-hydroxyazobenzene $\mathrm{C_{12}H_{10}N_2O}$ 198.
% N = (mass of N in one molecule/M) × 100.
Griess–Ilosvay test for $\mathrm{NO_2^-}$ : sulphanilic acid and 1-naphthylamine (α-naphthylamine) in acetic acid; nitrite diazotises the sulphanilic acid and the salt couples para to $\mathrm{NH_2}$ on the naphthylamine, giving a red azo dye.
Diazoaminobenzene, $\mathrm{C_6H_5N{=}N{-}NHC_6H_5}$ , warmed with aniline and a little aniline hydrochloride, rearranges to 4-aminoazobenzene.

Mass of dye from the diazotised amine

n(\text{dye}) = n(\mathrm{ArNH_2}) \qquad m(\text{dye}) = n(\text{dye}) \times M(\text{dye}) \qquad \%\mathrm{N} = \dfrac{\text{mass of N}}{M} \times 100

Worked example

4.65 g of aniline is diazotised and the salt is coupled with excess aniline. What mass of aniline yellow forms, assuming complete conversion?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 6 April 2024 · Q55Moderate

Example 2 · Amines · Coupling Reactions and Azo Dyes

9.3\text{ }g

of pure aniline upon diazotisation followed by coupling with phenol gives an orange dye. The mass of orange dye produced (assume 100% yield/ conversion) is _____ g. (nearest integer)

When aniline plays both roles, it is used twice

To make aniline yellow from aniline alone, one aniline is diazotised and a second is the coupling partner. The moles of dye are half the moles of aniline used in total, but equal to the moles of diazonium salt.

The Griess–Ilosvay colour is red

The azo dye from diazotised sulphanilic acid and 1-naphthylamine is red. The test detects nitrite; it uses the amine chemistry but is not a test for amines.

Drill 6 more on azo dye stoichiometry and the griess–ilosvay test

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)

Coupling of diazonium salts with phenols and aryl amines
Coupling of benzenediazonium chloride with phenol
$\mathrm{C_6H_5N_2^+Cl^- + C_6H_5OH \xrightarrow{OH^-} C_6H_5{-}N{=}N{-}C_6H_4{-}OH\ (para) + HCl}$
Azo dye stoichiometry and the Griess–Ilosvay test
Mass of dye from the diazotised amine
$n(\text{dye}) = n(\mathrm{ArNH_2}) \qquad m(\text{dye}) = n(\text{dye}) \times M(\text{dye}) \qquad \%\mathrm{N} = \dfrac{\text{mass of N}}{M} \times 100$

Watch out for (5)

Coupling keeps the nitrogen
→ Coupling of diazonium salts with phenols and aryl amines
Only strongly activated rings couple
→ Coupling of diazonium salts with phenols and aryl amines
Ortho methyls slow coupling on dimethylanilines
→ Coupling of diazonium salts with phenols and aryl amines
When aniline plays both roles, it is used twice
→ Azo dye stoichiometry and the Griess–Ilosvay test
The Griess–Ilosvay colour is red
→ Azo dye stoichiometry and the Griess–Ilosvay test

Test yourself on Amines

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