Amines • Topic 3 of 3

Diazonium Salts

Diazonium salts have the general formula Ar–N₂⁺ X^– (e.g. benzenediazonium chloride C₆H₅N₂⁺Cl^–). They are the single most useful intermediates in aromatic synthesis: the –N₂⁺ group can be replaced by a wide range of other groups, letting a chemist install substituents that are otherwise hard to attach directly to the ring.

Preparation — Diazotisation

A primary aromatic amine is treated with nitrous acid (NaNO₂ + dilute HCl) at 0–5°C:

C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl^– + NaCl + 2H₂O

The low temperature is essential because the salt decomposes above 5°C (releasing N₂ and giving phenol). This is called diazotisation.

Stability

Aromatic diazonium salts are far more stable than aliphatic ones because the positive charge is delocalised into the benzene ring (resonance). Aliphatic diazonium salts decompose instantly. Even so, arenediazonium salts are stored cold and usually used in solution at once. The benzenediazonium fluoroborate (with BF₄^–) is unusually stable and can be isolated as a solid.

Reactions — two families

(A) Replacement reactions (the N₂ leaves as gas and another group takes its place):

Sandmeyer: with CuCl/HCl, CuBr/HBr or CuCN gives Ar–Cl, Ar–Br, Ar–CN respectively.
Gattermann: with copper powder and HCl/HBr gives Ar–Cl or Ar–Br (cheaper, lower yield).
With H₂O (warm): gives a phenol, Ar–OH.
With KI: gives an aryl iodide, Ar–I (no Cu needed).
With HBF₄ then heat (Balz–Schiemann): gives an aryl fluoride, Ar–F.
With H₃PO₂ (hypophosphorous acid) or ethanol: replaces –N₂⁺ by –H (deamination), giving the parent arene Ar–H.

(B) Retention of the diazo group — coupling reactions: diazonium salts act as weak electrophiles and couple at the para position of activated arenes such as phenol and aniline to give brightly coloured azo compounds (Ar–N=N–Ar′). For example, benzenediazonium chloride + phenol (mild alkali) → p-hydroxyazobenzene (orange). With aniline it gives p-aminoazobenzene (yellow). These azo dyes underpin the textile dye industry.

Why they matter in synthesis

Groups like –F, –I, –CN and –H are difficult to introduce onto a benzene ring by direct substitution, but easy via a diazonium salt. The route nitro → amine → diazonium → product is one of the most flexible in aromatic chemistry, and azo coupling gives the vivid colours of dyes and indicators (e.g. methyl orange).

Solved Examples

Worked Example

Write the equation for the diazotisation of aniline and state two conditions that must be controlled.

Solution

Nitrous acid is generated in situ from NaNO₂ and dilute HCl.
Aniline reacts: C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl^– + NaCl + 2H₂O.
Condition 1: temperature must be kept at 0–5°C, or the salt decomposes to phenol with loss of N₂.
Condition 2: an excess of mineral acid (HCl) is used to keep the medium strongly acidic and the salt in solution.

Answer: C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl^– + NaCl + 2H₂O; keep 0–5°C and excess HCl.

Worked Example

How would you convert benzenediazonium chloride into (i) chlorobenzene, (ii) iodobenzene, (iii) fluorobenzene?

Solution

(i) Chlorobenzene: Sandmeyer reaction with cuprous chloride in HCl — C₆H₅N₂⁺Cl^– ⟶{CuCl/HCl} C₆H₅Cl + N₂.
(ii) Iodobenzene: simply warm with potassium iodide — C₆H₅N₂⁺Cl^– + KI → C₆H₅I + N₂ + KCl (no copper needed).
(iii) Fluorobenzene: Balz–Schiemann — treat with HBF₄ to precipitate the diazonium fluoroborate, then heat: C₆H₅N₂⁺BF₄^– &xrarr{Δ} C₆H₅F + N₂ + BF₃.

Answer: (i) CuCl/HCl (Sandmeyer); (ii) warm KI; (iii) HBF₄ then heat (Balz–Schiemann).

Worked Example

Distinguish the Sandmeyer and Gattermann reactions for preparing chlorobenzene from benzenediazonium chloride.

Solution

Both replace –N₂⁺ by –Cl, releasing N₂.
Sandmeyer uses cuprous chloride (CuCl) dissolved in HCl as the source of the nucleophile/catalyst.
Gattermann uses copper powder with HCl instead of CuCl.
Sandmeyer generally gives better yields; Gattermann is simpler and cheaper but lower yielding.

Answer: Sandmeyer uses CuCl/HCl (higher yield); Gattermann uses Cu powder + HCl (cheaper, lower yield).

Worked Example

Write the reaction of benzenediazonium chloride with phenol in mild alkali. Name the product type and explain its colour.

Solution

This is an azo coupling: the diazonium ion acts as a weak electrophile and attacks the activated phenol ring at the para position.
C₆H₅N₂⁺Cl^– + C₆H₅OH → p-HOC₆H₄–N=N–C₆H₅ (p-hydroxyazobenzene) + HCl.
The product is an azo compound containing the –N=N– chromophore conjugated with both rings.
This extended conjugation absorbs visible light, so the compound is coloured (orange).

Answer: p-Hydroxyazobenzene, an orange azo dye; the conjugated –N=N– chromophore absorbs visible light.

Worked Example

How can aniline be converted to benzene? Why is this route useful?

Solution

First diazotise aniline at 0–5°C: C₆H₅NH₂ ⟶{NaNO₂/HCl} C₆H₅N₂⁺Cl^–.
Then treat the diazonium salt with hypophosphorous acid (H₃PO₂) or ethanol: C₆H₅N₂⁺Cl^– + H₃PO₂ + H₂O → C₆H₆ + N₂ + H₃PO₃ + HCl.
The –N₂⁺ group is replaced by –H; this is called deamination.
It is useful because –NH₂ can first be used as a strong director to place other groups, then removed, allowing substitution patterns not otherwise accessible.

Answer: Diazotise, then treat with H₃PO₂ (deamination) to give benzene; this lets –NH₂ act as a temporary directing group.

Worked Example

Why can aryl fluorides and aryl iodides be made easily from diazonium salts but not by direct halogenation of benzene?

Solution

Direct fluorination of benzene is far too violent to control, and direct iodination is reversible/very slow because I₂ is a poor electrophile and HI reduces the product.
From a diazonium salt, –N₂⁺ is an excellent leaving group (leaves as stable N₂ gas).
Iodobenzene forms simply on warming with KI; fluorobenzene forms by heating the diazonium fluoroborate (Balz–Schiemann).
Thus the diazonium route gives controlled, good-yield access to Ar–F and Ar–I that direct halogenation cannot.

Answer: Direct F/I substitution on benzene is impractical; via diazonium, the easily-lost –N₂⁺ is cleanly replaced by –I (KI) or –F (HBF₄, heat).

Key Points

Diazonium salts Ar–N₂⁺X^– are made by diazotisation: ArNH₂ + NaNO₂ + 2HCl at 0–5°C; the cold temperature stops decomposition to phenol.
Aromatic diazonium salts are stabilised by resonance into the ring (aliphatic ones decompose at once); BF₄^– salts can be isolated as solids.
Replacement: Sandmeyer (CuCl/CuBr/CuCN) → Ar–Cl/Br/CN; Gattermann (Cu powder); H₂O → phenol; KI → Ar–I; HBF₄/Δ → Ar–F; H₃PO₂ → Ar–H (deamination).
Retention/coupling: diazonium ion couples at the para position of phenol/aniline to give coloured azo dyes (Ar–N=N–Ar′), e.g. p-hydroxyazobenzene; basis of dyes and indicators like methyl orange.
The nitro → amine → diazonium → product route installs –F, –I, –CN, –OH and –H that are hard to add by direct substitution.

Quick Check — 4 MCQs

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Q1.Diazotisation of aniline is carried out at:

Explanation: Above 5°C the diazonium salt decomposes to phenol with loss of N₂, so the reaction is done at 0–5°C.

Q2.The Sandmeyer reaction with CuCN converts a diazonium salt to:

Explanation: CuCN replaces –N₂⁺ with –CN, giving an aryl nitrile (benzonitrile from benzenediazonium chloride).

Q3.Treatment of benzenediazonium chloride with H₃PO₂ gives:

Explanation: Hypophosphorous acid replaces –N₂⁺ with –H (deamination), giving benzene.

Q4.Azo coupling of benzenediazonium chloride with phenol occurs mainly at the:

Explanation: The diazonium ion is a weak electrophile and couples at the more accessible para position of the activated phenol ring, giving p-hydroxyazobenzene.

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