What is a diazonium salt?
A diazonium salt has the general formula R–N₂⁺X⁻, where X⁻ can be Cl⁻, Br⁻, HSO₄⁻ or BF₄⁻. The –N≡N⁺ group is called the diazonium group. Name: add 'diazonium' to the parent name and then the anion: C₆H₅N₂⁺Cl⁻ is benzenediazonium chloride.
Aromatic (aryl) diazonium ions are stable for a short time in the cold because the positive charge is spread into the ring by resonance. Aliphatic ones break down at once, even in the cold.
Preparation (diazotisation)
Aniline is dissolved in dilute HCl and cooled in ice to 273–278 K (0–5 °C). Then sodium nitrite (NaNO₂) solution is added slowly. NaNO₂ + HCl make nitrous acid (HNO₂) right inside the flask.
C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
This is called diazotisation. The salt is not stored; it is used straight away in the next reaction.
Physical and chemical properties
Physical: benzenediazonium chloride is a colourless crystalline solid. It dissolves easily in water and is stable in the cold but reacts with water when warm. It decomposes when dry, so it is kept in solution. Benzenediazonium fluoroborate is insoluble in water and stable at room temperature.
Chemical: its reactions fall in two groups:
- N₂ is displaced (the ring loses N₂ and gains a new group). N₂ is a very stable molecule, so it is an excellent leaving group.
- N₂ is kept (coupling reactions give azo compounds).
Reactions where N₂ is replaced
- By Cl⁻ or Br⁻ (Sandmeyer): ArN₂⁺X⁻ + CuCl/HCl → ArCl + N₂; with CuBr/HBr → ArBr.
- Gattermann: copper powder + HCl or HBr also gives ArCl or ArBr.
- By CN⁻ (Sandmeyer): CuCN/KCN → ArCN (benzonitrile). Hydrolysing ArCN gives benzoic acid, and reducing it gives benzylamine.
- By I⁻: just warm with KI → ArI (iodine cannot be put on the ring well directly).
- By F⁻ (Balz–Schiemann): add HBF₄ → ArN₂⁺BF₄⁻ (stable solid); heat → ArF + BF₃ + N₂.
- By H: hypophosphorous acid H₃PO₂ (or ethanol) removes the group → benzene. Useful to remove an –NH₂ after it has done its directing job.
- By OH: warm the water solution to about 283 K → phenol + N₂ + HCl.
- By NO₂: with HBF₄ then NaNO₂ and Cu on heating → nitrobenzene.
Reactions where N₂ is kept: coupling
The diazonium ion is a weak electrophile. It attacks a very electron-rich ring at the para position:
- With phenol in alkaline solution → p-hydroxyazobenzene (orange dye).
- With aniline in mildly acidic solution → p-aminoazobenzene (yellow dye).
The –N=N– link (azo group) joins the two rings. The long system of alternating single and double bonds absorbs visible light, so the product is coloured. This is how many dyes are made.
Why diazonium salts are so useful in making aromatic compounds
Some groups cannot be put on a benzene ring directly, or give messy mixtures:
- Direct iodination and fluorination do not work well; via diazonium, ArI and ArF are easy.
- –CN cannot be added by direct substitution; the Sandmeyer reaction does it.
- An –NH₂ can steer new groups to places they would not normally go, and then be removed with H₃PO₂. Example: 1,3,5-tribromobenzene from aniline (brominate to 2,4,6-tribromoaniline, then diazotise and replace N₂ by H).
So the diazonium salt is a 'switchboard': one aniline, many products.
Try it: predict, then check
In the 3D free play: before you press Show, say aloud whether N₂ will leave or stay for each reagent. Only phenol and aniline keep it.
At home (no chemicals): look at the ingredient list on a coloured candy or drink: names like 'sunset yellow' or 'tartrazine' are azo dyes. Count how many you can find.
Board exam focus
Expect 1–3 mark questions: write the equation for diazotisation; why it is done at 273–278 K; name reactions (Sandmeyer, Gattermann, coupling); and conversions such as benzene → fluorobenzene, aniline → benzoic acid, nitrobenzene → 1,3,5-tribromobenzene.
Key formulas and definitions
- C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O (273–278 K)
- ArN₂⁺Cl⁻ + CuCl/HCl → ArCl + N₂ (Sandmeyer); CuBr/HBr → ArBr; CuCN/KCN → ArCN
- ArN₂⁺Cl⁻ + Cu/HX → ArX + N₂ (Gattermann)
- ArN₂⁺Cl⁻ + KI → ArI + KCl + N₂
- ArN₂⁺Cl⁻ + HBF₄ → ArN₂⁺BF₄⁻ → (heat) ArF + BF₃ + N₂
- ArN₂⁺Cl⁻ + H₃PO₂ + H₂O → ArH + N₂ + H₃PO₃ + HCl
- ArN₂⁺Cl⁻ + H₂O (warm) → ArOH + N₂ + HCl
- C₆H₅N₂⁺Cl⁻ + C₆H₅OH (OH⁻) → p-HO–C₆H₄–N=N–C₆H₅ (orange)
Worked examples
1. Write the equation to make benzenediazonium chloride and state the condition.
Step 1: Take aniline in dilute HCl. Step 2: Cool to 273–278 K and add NaNO₂ slowly. Step 3: C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O. Condition: keep at 0–5 °C, because the salt breaks down when warm.
2. Convert aniline into chlorobenzene.
Step 1: Diazotise: aniline + NaNO₂ + HCl at 273–278 K → C₆H₅N₂⁺Cl⁻. Step 2: Add CuCl/HCl (Sandmeyer) → C₆H₅Cl + N₂↑.
3. Convert aniline into benzoic acid.
Step 1: Diazotise aniline → C₆H₅N₂⁺Cl⁻. Step 2: CuCN/KCN → C₆H₅CN (benzonitrile) + N₂. Step 3: Hydrolyse with H₃O⁺ (heat) → C₆H₅COOH (benzoic acid).
4. Convert benzene into fluorobenzene.
Step 1: Nitrate: C₆H₆ → C₆H₅NO₂. Step 2: Reduce (Sn/HCl) → C₆H₅NH₂. Step 3: Diazotise at 273–278 K → C₆H₅N₂⁺Cl⁻. Step 4: Add HBF₄ → C₆H₅N₂⁺BF₄⁻. Step 5: Heat → C₆H₅F + BF₃ + N₂.
5. Convert nitrobenzene into 1,3,5-tribromobenzene.
Step 1: Reduce nitrobenzene (Sn/HCl) → aniline. Step 2: Bromine water → 2,4,6-tribromoaniline (–NH₂ sends Br to o and p). Step 3: Diazotise the –NH₂ at 273–278 K. Step 4: H₃PO₂ replaces N₂ by H → 1,3,5-tribromobenzene. The –NH₂ steered the Br atoms, then was removed.
6. How many moles of N₂ gas are given off when 0.2 mol of benzenediazonium chloride is warmed with water? What volume is this at STP (22.4 L/mol)?
Step 1: C₆H₅N₂⁺Cl⁻ + H₂O → C₆H₅OH + N₂ + HCl: 1 mol salt gives 1 mol N₂. Step 2: 0.2 mol salt → 0.2 mol N₂. Step 3: Volume = 0.2 × 22.4 = 4.48 L.
7. What happens when benzenediazonium chloride reacts with phenol in alkaline solution? Why is the product coloured?
Step 1: The diazonium ion attacks the para position of phenol (phenoxide is very electron-rich). Step 2: N₂ is not lost; –N=N– joins the two rings → p-hydroxyazobenzene. Step 3: The long chain of alternating single and double bonds across both rings absorbs visible light, so it looks orange.
8. Starting from 18.6 g aniline (M = 93 g/mol), what is the greatest mass of iodobenzene (M = 204 g/mol) you could make via the diazonium salt?
Step 1: Moles of aniline = 18.6 ÷ 93 = 0.2 mol. Step 2: Aniline → diazonium → iodobenzene, 1 : 1 : 1. Step 3: 0.2 mol iodobenzene × 204 g/mol = 40.8 g (theoretical yield).
Common mistakes
- Doing diazotisation at room temperature. The salt breaks down above about 278 K, so ice is a must.
- Writing that N₂ is lost in coupling reactions. In coupling the N=N stays; only in replacement reactions does N₂ leave.
- Mixing up Sandmeyer (CuCl, CuBr, CuCN salts) and Gattermann (copper powder + HX).
- Thinking aliphatic diazonium salts can be used the same way. They decompose at once, giving N₂ and alcohols.