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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.

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Why this matters

22 PYQs, one HARD. Of the twelve set from 2021 to 2025, nine are diazotisation and replacement — the reagent, the diazonium salt as A and phenol as B in the same two-step sequence three 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

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: C6H5NH2→NaNO2+HCl, 273 KC6H5N2+Cl−\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.
  • +H2O,Δ+ \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: ArN2+BF4−→ΔAr-F\text{ArN}_2^+\text{BF}_4^- \xrightarrow{\Delta} \text{Ar-F} + BF₃ + N₂ — fluoroboric acid gives Ar–F.
  • Reduction: C6H5N2+Cl−+C2H5OH→C6H6+CH3CHO+N2+HCl\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

ArNH2→NaNO2/HCl, 273 KArN2+Cl−→H2O, ΔArOH+N2+HCl\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.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 2 · Q78Moderate

Example 1 · Amines · Diazonium Salts and Aromatic Amine Reactions

Identify product B in the following reaction: Aniline →NaNO2+HCl,273K\xrightarrow{NaNO_2+HCl, 273K} A →H2O,Δ\xrightarrow{H_2O, \Delta} B + N₂↑

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 8 more on diazotisation and the replacement reactions of the diazonium ion

Concept 2 of 2: Azo Coupling: Diazonium Ion Plus Phenol or Aniline

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

  • C6H5N2+Cl−+C6H5OH→OH−, 273–278 KC6H5-N=N-C6H4-OH\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

ArN2++C6H5OH→OH−Ar-N=N-C6H4-OH (p)\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.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 22 April Shift I · Q94Moderate

Example 2 · Amines · Diazonium Salts and Aromatic Amine Reactions

Which from following reaction results in azo coupling?

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

    ArNH2→NaNO2/HCl, 273 KArN2+Cl−→H2O, ΔArOH+N2+HCl\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

    ArN2++C6H5OH→OH−Ar-N=N-C6H4-OH (p)\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)

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