Reactions of Diazonium Salts

Two Families of Diazonium Reactions

Diazonium reactions fall into two groups:

1. Replacement of N$_2^+$ — the -N$_2^+$ group leaves as N$_2$ gas and is replaced by another atom or group (-Cl, -Br, -I, -CN, -OH, -F, -H, -NO$_2$). These let you make halo-, cyano-, hydroxy- and other substituted benzenes.
2. Retention of N$_2$ (coupling) — the whole -N=N- unit is kept, joining the aryl group to another aromatic ring to make a brightly coloured azo compound (dye).

Key Point: diazonium reactions either replace -N$_2^+$ (lose N$_2$) with a new group, or retain -N=N- to couple into an azo dye.

Replacement Reactions — Sandmeyer, Gattermann & Others

Sandmeyer reaction: the -N$_2^+$ is replaced by -Cl, -Br or -CN using the cuprous halide/cyanide (CuCl, CuBr, CuCN):

• Ar-N$_2^+$Cl$^-$ + CuCl → Ar-Cl (chlorobenzene) + N$_2$
• with CuBr → Ar-Br; with CuCN → Ar-CN (benzonitrile)

Gattermann reaction: the same replacements by -Cl or -Br can be done with copper powder + HCl/HBr (Cu/HX). The Gattermann method uses copper powder instead of the cuprous salt; the yield is usually lower than Sandmeyer's.

Other replacements:

• -I: simply warming with KI gives iodobenzene (no catalyst needed).
• -F: via the fluoroborate (the Balz-Schiemann reaction, heating Ar-N$_2^+$BF$_4^-$) gives Ar-F.
• -OH: warming the salt with water (H$_2$O/H$^+$) gives a phenol.
• -H: treatment with hypophosphorous acid (H$_3$PO$_2$) or ethanol replaces -N$_2^+$ by -H (removes the group entirely).

Key Point: Sandmeyer (CuCl/CuBr/CuCN) → Ar-Cl/Br/CN; Gattermann (Cu/HX) → Ar-Cl/Br (lower yield); KI → Ar-I; H$_2$O → phenol; H$_3$PO$_2$ → Ar-H.

Coupling Reactions — Azo Dyes

In coupling reactions the -N$_2^+$ group is retained: the diazonium ion acts as a weak electrophile and attacks a highly activated aromatic ring (a phenol or an aromatic amine), joining the two rings through an azo (-N=N-) bridge to give a brightly coloured azo compound.

• With phenol (in mildly alkaline solution) → p-hydroxyazobenzene (an orange dye).
• With aniline (in mildly acidic solution) → p-aminoazobenzene (aniline yellow).

Coupling occurs mainly at the para position (or ortho if para is blocked). The extended -N=N- conjugation between the two rings is what gives azo dyes their intense colours, which is why this reaction is the basis of a whole industry of dyes and pigments.

Key Point: azo coupling — diazonium ion + phenol/aromatic amine → -N=N- linked azo dye (para). The conjugated azo bridge gives the bright colour; basis of the dye industry.

Solved Examples

Example 1: Sandmeyer to chlorobenzene

How is chlorobenzene prepared from aniline?

Solution: Diazotise aniline (NaNO$_2$/HCl, 273–278 K) to C$_6$H$_5$N$_2^+$Cl$^-$, then treat with CuCl (Sandmeyer reaction) → chlorobenzene + N$_2$.

Example 2: Benzonitrile

How would you convert benzene diazonium chloride to benzonitrile?

Solution: Treat it with cuprous cyanide (CuCN) (Sandmeyer): C$_6$H$_5$N$_2^+$Cl$^-$ + CuCN → C$_6$H$_5$CN (benzonitrile) + N$_2$.

Example 3: Iodobenzene

How is iodobenzene made from a diazonium salt?

Solution: Simply warm the diazonium salt with potassium iodide (KI) — no copper catalyst is needed: C$_6$H$_5$N$_2^+$ + KI → C$_6$H$_5$I (iodobenzene) + N$_2$.

Example 4: Diazonium to phenol

Give the product of warming benzene diazonium chloride with water.

Solution: Phenol (the -N$_2^+$ is replaced by -OH): C$_6$H$_5$N$_2^+$Cl$^-$ + H$_2$O → C$_6$H$_5$OH + N$_2$ + HCl.

Example 5: Sandmeyer vs Gattermann

What is the difference between the Sandmeyer and Gattermann reactions for making chlorobenzene?

Solution: Both replace -N$_2^+$ by -Cl. Sandmeyer uses cuprous chloride (CuCl); Gattermann uses copper powder + HCl (Cu/HCl). The Sandmeyer yield is generally better than the Gattermann.

Example 6: Removing the amino group

How can the -NH$_2$ group of aniline be removed entirely (replaced by H)?

Solution: Diazotise to the diazonium salt, then treat with hypophosphorous acid (H$_3$PO$_2$) (or ethanol): the -N$_2^+$ is replaced by -H, giving benzene.

Example 7: Azo coupling with phenol

What forms when benzene diazonium chloride couples with phenol?

Solution: p-Hydroxyazobenzene (an orange azo dye), formed by coupling at the para position of phenol through an -N=N- link.

Example 8: Aniline yellow

Name the azo dye formed when benzene diazonium chloride couples with aniline.

Solution: p-Aminoazobenzene (aniline yellow), formed by para coupling with aniline.

Example 9: Why azo compounds are coloured

Why are azo dyes intensely coloured?

Solution: The -N=N- azo bridge conjugates the two aromatic rings, giving an extended delocalised π system that absorbs visible light — producing the bright colour.

Example 10: Para preference in coupling

Why does coupling occur mainly at the para position of phenol/aniline?

Solution: The diazonium ion is a weak electrophile, so it attacks the most activated, least hindered position. The -OH/-NH$_2$ groups are o/p-directing, and the para position is usually the least hindered, so coupling occurs there (ortho if para is blocked).