MHT-CET Chemistry · Amines

Diazonium Salts and Aromatic Amine Reactions

Aniline with NaNO₂ and HCl at 273–278 K gives benzenediazonium chloride, whose N₂⁺ is replaced by OH (warm water), Cl or Br or CN (Sandmeyer, Cu(I) salts), F (Balz–Schiemann, HBF₄), I (KI) or H (ethanol or H₃PO₂) — and which couples with phenol in mild alkali to give the azo dye p-hydroxyazobenzene.

All Amines notes MHT-CET Chemistry strategy

Why this matters

13 PYQs, none HARD. Ten are diazotisation and replacement — the reagent, the diazonium salt as A and phenol as B in the same two-step sequence four times, Sandmeyer's reagent and what it cannot make (iodobenzene), fluoroboric acid giving Ar–F, ethanol giving benzene; three are azo coupling — which reaction it is, what it makes, and the mild alkaline medium it needs. Two cards.

Concept 1 of 2

Diazotisation and the Replacement Reactions of the Diazonium Ion

Intuition

Nitrous acid, made in place from NaNO₂ and HCl in ice, turns a primary aromatic amine into a diazonium salt that is stable only cold. Its N₂⁺ is the best leaving group there is, so almost anything replaces it: water gives phenol, Cu(I) halides or cyanide give the halo- or cyanoarene (Sandmeyer), KI gives the iodide without copper, HBF₄ then heat gives the fluoride (Balz–Schiemann), and ethanol or hypophosphorous acid simply give benzene.

Definition

Diazotisation: $\text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{NaNO}_2 + \text{HCl},\ 273\text{ K}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-$ (A). Reagent = NaNO₂ + HCl at low temperature.
$+ \text{H}_2\text{O}, \Delta$ → phenol (B) + N₂ + HCl. Benzene → nitrobenzene → aniline → diazonium → phenol is the standard four-step chain (C = phenol).
Sandmeyer: CuCl/HCl → chlorobenzene; CuBr/HBr → bromobenzene; CuCN/KCN → benzonitrile. NOT iodobenzene (use KI, no copper) and NOT fluorobenzene.
Balz–Schiemann: $\text{ArN}_2^+\text{BF}_4^- \xrightarrow{\Delta} \text{Ar-F}$ + BF₃ + N₂ — fluoroboric acid gives Ar–F.
Reduction: $\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_2\text{H}_5\text{OH} \to \text{C}_6\text{H}_6 + \text{CH}_3\text{CHO} + \text{N}_2 + \text{HCl}$ (A = benzene); H₃PO₂ does the same.

Diazotisation and hydrolysis

\text{ArNH}_2 \xrightarrow{\text{NaNO}_2/\text{HCl},\ 273\text{ K}} \text{ArN}_2^+\text{Cl}^- \xrightarrow{\text{H}_2\text{O},\ \Delta} \text{ArOH} + \text{N}_2 + \text{HCl}

Worked example

Convert aniline into (i) iodobenzene, (ii) fluorobenzene and (iii) benzonitrile, naming the reagent after diazotisation in each case.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1AminesMODERATE

Identify product B in the following reaction: Aniline

\\\xrightarrow{NaNO_2+HCl, 273K}

\\\xrightarrow{H_2O, \Delta}

B + N₂↑

[Q78 · 12th May Shift 2 · 2024]

Stopping at the diazonium salt

The salt is A, the intermediate. When the sequence continues with warm water, B is PHENOL — 'benzenediazonium chloride' is offered as option (c) for B every time.

Drill 9 more on diazotisation and the replacement reactions of the diazonium ion

Concept 2 of 2

Azo Coupling: Diazonium Ion Plus Phenol or Aniline

Intuition

The diazonium ion is a weak electrophile, so it attacks only a strongly activated ring — phenol (as phenoxide, in mild alkali) or aniline (in mild acid) — at the para position, keeping both nitrogens as an azo bridge –N=N–. The products are coloured: p-hydroxyazobenzene (orange) from phenol, p-aminoazobenzene (yellow) from aniline. Too much acid protonates the phenoxide; too much base turns the diazonium ion into a diazotate. Mild alkali is the medium.

Definition

$\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{OH}^-,\ 273\text{–}278\text{ K}} \text{C}_6\text{H}_5\text{-N=N-C}_6\text{H}_4\text{-OH}$ (p-hydroxyazobenzene, orange) — this reaction IS azo coupling.
With aniline in mildly acidic solution: p-aminoazobenzene (yellow).
Medium for the phenol coupling: mild alkaline (converts phenol to the more reactive phenoxide); not strongly acidic, not alcoholic.
Not azo coupling: aniline + HNO₂ (diazotisation), diazonium + HBF₄ (Balz–Schiemann), diazonium + Cu/HCl (Gattermann).

Azo coupling

\text{ArN}_2^+ + \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{OH}^-} \text{Ar-N=N-C}_6\text{H}_4\text{-OH}\ (p)

Worked example

Write the product of benzenediazonium chloride with N,N-dimethylaniline in mildly acidic solution, and explain why strong acid stops the reaction.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2AminesMODERATE

Which from following reaction results in azo coupling?

[Q94 · 22 April Shift I · 2025]

Coupling in strong acid

Strongly acidic medium keeps phenol as phenol, too weak a nucleophile for the diazonium ion; strongly basic medium destroys the diazonium ion. The window is mild alkali for phenol, mild acid for aniline.

Drill 2 more on azo coupling: diazonium ion plus phenol or 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)

Diazotisation and the Replacement Reactions of the Diazonium Ion
Diazotisation and hydrolysis
$\text{ArNH}_2 \xrightarrow{\text{NaNO}_2/\text{HCl},\ 273\text{ K}} \text{ArN}_2^+\text{Cl}^- \xrightarrow{\text{H}_2\text{O},\ \Delta} \text{ArOH} + \text{N}_2 + \text{HCl}$
Azo Coupling: Diazonium Ion Plus Phenol or Aniline
Azo coupling
$\text{ArN}_2^+ + \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{OH}^-} \text{Ar-N=N-C}_6\text{H}_4\text{-OH}\ (p)$

Watch out for (2)

Stopping at the diazonium salt
→ Diazotisation and the Replacement Reactions of the Diazonium Ion
Coupling in strong acid
→ Azo Coupling: Diazonium Ion Plus Phenol or Aniline

Drill every past-year question on this subtopic

13 questions from the bank — paginated, with cart and Word-export support.

Drill the 13 questions

Drill every past-year question on this subtopic

Related notes