ChemistryISC Class 12 · Organic Compounds Containing Nitrogen

Diazonium Salts and Their Reactions: ISC Class 12 Chemistry

Updated 15 October 2026 · Fact-checked

Benzenediazonium chloride is made by treating aniline with NaNO₂ and HCl at 273–278 K (diazotisation). Its –N₂⁺ group is easily replaced by Cl, Br, CN (Sandmeyer, Gattermann), I, F, OH or H with loss of nitrogen gas, and it couples with phenols or aromatic amines to give coloured azo dyes.

Cold diazotisation (273–278 K) makes the diazonium salt; its –N₂⁺ is swapped for many groups, or it couples with phenol to make an azo dye.

Illustration drawn for QuestionBuddy.

Understand Diazonium Salts and Their Reactions

Preparation (diazotisation). A primary aromatic amine is dissolved in excess cold dilute HCl and treated with sodium nitrite solution at 273–278 K. Nitrous acid formed in situ converts it into the diazonium salt: $C_{6} H_{5} N H_{2} + N a N O_{2} + 2 H C l \to C_{6} H_{5} N_{2} C l + N a C l + 2 H_{2} O$ . Benzenediazonium chloride, $C_{6} H_{5} N_{2}^{+} C l^{-}$ , is used in solution without isolation, because it is unstable: above about 278 K it decomposes, giving phenol and nitrogen. Arenediazonium salts are much more stable than alkanediazonium salts (which decompose at once even in the cold), because the positive charge is delocalised into the benzene ring by resonance.

A. Reactions in which nitrogen is lost (replacement of –N₂⁺). The N₂ group is an excellent leaving group (N₂ gas escapes), so it can be replaced by many atoms or groups.
• By Cl, Br or CN (Sandmeyer reaction): Cu(I) salts, $C_{6} H_{5} N_{2} C l C u_{2} C l_{2} / H C l C_{6} H_{5} C l + N_{2}$ ; with CuBr/HBr, bromobenzene; with CuCN/KCN, benzonitrile, $C_{6} H_{5} N_{2} C l + C u C N \to C_{6} H_{5} C N + N_{2} + C u C l$ .
• By Cl or Br (Gattermann reaction): copper powder with HCl or HBr: $C_{6} H_{5} N_{2} C l + H B r C u C_{6} H_{5} B r + N_{2} + H C l$ .
• By iodine: warming with aqueous KI (no catalyst), $C_{6} H_{5} N_{2} C l + K I \to C_{6} H_{5} I + K C l + N_{2}$ .
• By fluorine (Balz–Schiemann): fluoroboric acid precipitates the diazonium fluoroborate, which on heating gives fluorobenzene, BF₃ and N₂.
• By H: mild reducing agents, hypophosphorous acid (H₃PO₂) or ethanol, replace –N₂⁺ by H: $C_{6} H_{5} N_{2} C l + C H_{3} C H_{2} O H \to C_{6} H_{6} + N_{2} + C H_{3} C H O + H C l$ .
• By OH: warming the aqueous solution, $C_{6} H_{5} N_{2} C l + H_{2} O \to C_{6} H_{5} O H + N_{2} + H C l$ .
• By NO₂: the fluoroborate heated with aqueous NaNO₂ and copper powder gives nitrobenzene: $C_{6} H_{5} N_{2} B F_{4} + N a N O_{2} \to C_{6} H_{5} N O_{2} + N a B F_{4} + N_{2}$ .

B. Reactions in which nitrogen is retained (coupling). The diazonium ion is a weak electrophile and attacks only strongly activated rings, at the para position (ortho if para is blocked), giving azo compounds Ar–N=N–Ar′, which are brightly coloured because of the extended conjugated system. With phenol in mildly alkaline solution (the phenoxide ion is very reactive), the orange dye 4-hydroxyazobenzene forms: $C_{6} H_{5} N_{2} C l + C_{6} H_{5} O H \to C_{6} H_{5} N = N C_{6} H_{4} O H + H C l$ . With aniline in mildly acidic solution, yellow 4-aminoazobenzene forms. With N,N-dimethylaniline, 4-(dimethylamino)azobenzene, a yellow dye, forms. This coupling is the basis of the azo-dye test for phenols and primary aromatic amines.

Importance in synthesis. Diazonium salts let you put on a benzene ring groups that cannot be introduced directly, or in positions that direct substitution would not give: F and I (direct fluorination is violent, direct iodination reversible), CN and OH, and substituent patterns such as 1,3,5-tribromobenzene or meta-halophenols. They also remove an amino group completely (replacement by H), so an –NH₂ group can be used temporarily to direct substitution and then deleted.

Diazonium Salts and Their Reactions: formulas and key facts

Diazotisation: $C_{6} H_{5} N H_{2} + N a N O_{2} + 2 H C l 273 - 278 K C_{6} H_{5} N_{2} C l + N a C l + 2 H_{2} O$; Keep cold; use the solution at once.
Sandmeyer reaction: $C_{6} H_{5} N_{2} C l C u_{2} C l_{2} / H C l (C u B r / H B r, C u C N / K C N) C_{6} H_{5} X + N_{2}$; X = Cl, Br, CN; copper(I) salts.
Gattermann reaction: $C_{6} H_{5} N_{2} C l C u / H X C_{6} H_{5} X + N_{2}$; Copper powder; X = Cl or Br.
Replacement by H: $C_{6} H_{5} N_{2} C l H_{3} P O_{2} or C H_{3} C H_{2} O H C_{6} H_{6} + N_{2}$; Removes an amino group from the ring.
Coupling with phenol: $C_{6} H_{5} N_{2} C l + C_{6} H_{5} O H O H^{-} C_{6} H_{5} N = N C_{6} H_{4} O H + H C l$; Orange dye; para position; alkaline medium.

How to solve Diazonium Salts and Their Reactions questions

ISC asks diazonium reactions as equations, conversions (often starting from aniline or benzene) and reasons (low temperature, stability, coupling position). Think 'swap the N₂ for X' or 'join two rings with –N=N–'.

1Start every route from a primary aromatic amine: diazotise with NaNO₂ and dil. HCl at 273–278 K.
2For a replacement, choose the reagent: Cu₂Cl₂/HCl or Cu/HCl (Cl), CuBr/HBr or Cu/HBr (Br), CuCN/KCN (CN), KI (I), HBF₄ then heat (F), H₃PO₂ or ethanol (H), warm water (OH), NaNO₂/Cu on the fluoroborate (NO₂).
3For coupling, state the partner, the medium (alkaline for phenols, mildly acidic for amines), the position (para) and the colour.
4When a target has an unusual substitution pattern, use an –NH₂ (or –NO₂ reduced to –NH₂) group to direct substitution, then convert it via the diazonium salt.
5Balance each equation: N₂ is always a product in replacements; HCl in coupling.
6Name the reactions (Sandmeyer, Gattermann, Balz–Schiemann, coupling).

Diazonium Salts and Their Reactions in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Diazotisation and coupling can detect nitrite

Suppose a food-testing laboratory checks cured meat for added nitrite. In the test, the acidified extract is mixed with an aromatic primary amine in the cold, and then with a coupling partner; a pink-red colour shows nitrite, and a deeper colour shows more nitrite.

Explain the chemistry of the test in two steps, and say why the nitrite must be acidified and the mixture kept cold.

Try it first, then show the solutionSolution

In acid, nitrite forms nitrous acid, which diazotises the primary aromatic amine to an arenediazonium salt (as NaNO₂ + HCl does with aniline).
The diazonium ion, an electrophile, couples with the activated aromatic partner to give a coloured azo dye; the amount of dye depends on the amount of nitrite.
Acid is needed to make nitrous acid from nitrite; without it no diazonium salt forms.
Cold conditions keep the diazonium salt from decomposing to a phenol and N₂ before it can couple.

Answer: Nitrite (as HNO₂ in acid) diazotises the amine; the diazonium ion couples to give a coloured azo dye. Acid generates HNO₂; cold prevents decomposition of the diazonium salt.

Memory hook: Nitrite + amine in cold acid = diazonium; diazonium + activated ring = colour.

Tricks to remember Diazonium Salts and Their Reactions

Short memory aids. Say each one aloud once, then use it on a question the same day.

Sandmeyer uses salts, Gattermann uses metal
Sandmeyer: copper(I) Salts (Cu₂Cl₂, CuBr, CuCN). Gattermann: copper powder (metal) with HX.
Use it when: Naming the replacement reaction.
Cold, Coloured, Coupled
Diazotise Cold (273–278 K); couple to get Coloured azo dyes; coupling goes para.
Use it when: Conditions and coupling questions.
KI needs no copper
Iodide replaces N₂ by itself; chloride, bromide and cyanide need copper.
Use it when: Choosing reagents.

Diazonium Salts and Their Reactions: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Removing the diazo group · 1 mark · MCQBoard level

Benzenediazonium chloride is converted into benzene by:
(a) CuCN/KCN
(b) H₃PO₂ and water
(c) KI
(d) warm water

Try it first, then show the solutionSolution

Hypophosphorous acid (or ethanol) is a mild reducing agent that replaces –N₂⁺ by H.
The others give benzonitrile, iodobenzene and phenol.

Answer: (b).

Example 2: Bromobenzene two ways · 2 marks · Short answerBoard level

Write equations for converting benzenediazonium chloride into bromobenzene by (a) the Sandmeyer reaction and (b) the Gattermann reaction.

Try it first, then show the solutionSolution

(a) With CuBr/HBr: $C_{6} H_{5} N_{2} C l + H B r C u B r C_{6} H_{5} B r + N_{2} + H C l$ (1 mark).
(b) With Cu powder and HBr: $C_{6} H_{5} N_{2} C l + H B r C u C_{6} H_{5} B r + N_{2} + H C l$ (1 mark).

Answer: Sandmeyer: CuBr/HBr. Gattermann: Cu powder/HBr. Both give bromobenzene and N₂.

Example 3: Aniline to benzoic acid · 3 marks · Short answerHarder board level

How will you convert aniline into benzoic acid? Give equations for each step.

Try it first, then show the solutionSolution

Diazotise: $C_{6} H_{5} N H_{2} + N a N O_{2} + 2 H C l \to C_{6} H_{5} N_{2} C l + N a C l + 2 H_{2} O$ (273–278 K) (1 mark).
Sandmeyer with CuCN/KCN: $C_{6} H_{5} N_{2} C l + C u C N \to C_{6} H_{5} C N + N_{2} + C u C l$ (1 mark).
Acid hydrolysis of the nitrile: $C_{6} H_{5} C N + 2 H_{2} O + H C l \to C_{6} H_{5} C O O H + N H_{4} C l$ (1 mark).

Answer: Aniline → benzenediazonium chloride → benzonitrile → benzoic acid.

Example 4: Coupling with two partners · 3 marks · Short answerHarder board level

Write equations for the coupling of benzenediazonium chloride with (a) phenol and (b) N,N-dimethylaniline. State the medium used in (a), the position of attack, and why coupling occurs only with such activated rings.

Try it first, then show the solutionSolution

(a) $C_{6} H_{5} N_{2} C l + C_{6} H_{5} O H \to C_{6} H_{5} N = N C_{6} H_{4} O H + H C l$ , in mildly alkaline solution: 4-hydroxyazobenzene, orange (1 mark).
(b) $C_{6} H_{5} N_{2} C l + C_{6} H_{5} N (C H_{3})_{2} \to C_{6} H_{5} N = N C_{6} H_{4} N (C H_{3})_{2} + H C l$ : 4-(dimethylamino)azobenzene, yellow (1 mark).
Attack occurs at the para position. The diazonium ion is a weak electrophile, so it reacts only with rings made very electron-rich by –O⁻/–OH or –NR₂ groups (1 mark).

Answer: Para-coupled azo dyes; alkaline medium for phenol; only strongly activated rings react.

Example 5: Making 3-bromophenol · 5 marks · Long answerHOTS / competency

3-Bromophenol cannot be made by brominating phenol. Starting from nitrobenzene, show how it can be prepared.
(a) Give the sequence of reactions with reagents and conditions. [3]
(b) Explain why bromination is done before reduction, and why the route goes through a diazonium salt. [2]

Try it first, then show the solutionSolution

(a) Brominate nitrobenzene with Br₂/FeBr₃: –NO₂ is meta directing, $C_{6} H_{5} N O_{2} + B r_{2} \to C_{6} H_{4} B r N O_{2} + H B r$ (3-bromonitrobenzene) (1 mark).
Reduce with Sn/HCl: $C_{6} H_{4} B r N O_{2} + 6 [H] \to C_{6} H_{4} B r N H_{2} + 2 H_{2} O$ (3-bromoaniline); diazotise with NaNO₂/HCl at 273–278 K (1 mark).
Warm the diazonium solution with water: $C_{6} H_{4} B r N_{2} C l + H_{2} O \to C_{6} H_{4} B r O H + N_{2} + H C l$ (3-bromophenol) (1 mark).
(b) The –NO₂ group directs the bromine to the meta position; after reduction, –NH₂ would direct ortho/para and over-brominate (1 mark).
–OH cannot be put directly meta to Br; converting –NH₂ via the diazonium salt places –OH exactly where the nitrogen was (1 mark).

Answer: Nitrobenzene → 3-bromonitrobenzene → 3-bromoaniline → diazonium salt → 3-bromophenol.

Where students lose marks in Diazonium Salts and Their Reactions

The slips examiners see most often, and what to write instead.

Mistake: No medium or condition in coupling
ISC examiners noted that candidates did not mention the medium for the coupling reaction (2024).
Do this: Write 'alkaline medium' for phenol and 'mildly acidic medium' for aniline, and give 273–278 K for diazotisation.
Usually costs: 1 mark.
Mistake: Missing by-products
Examiners asked for named reactions with by-products and conditions (2024).
Wrong working
$C_{6} H_{5} N_{2} C l + C_{6} H_{5} O H \to C_{6} H_{5} N = N C_{6} H_{4} O H$
Right working
$C_{6} H_{5} N_{2} C l + C_{6} H_{5} O H \to C_{6} H_{5} N = N C_{6} H_{4} O H + H C l$
Do this: Balance every equation; HCl in coupling, N₂ in replacements.
Usually costs: ½ to 1 mark.
Mistake: Confusing Sandmeyer and Gattermann
Students swap copper(I) salts and copper powder.
Do this: Sandmeyer: Cu₂Cl₂, CuBr or CuCN. Gattermann: Cu powder with HCl or HBr.
Usually costs: 1 mark.
Mistake: Diazotising at room temperature
Students omit the temperature condition.
Do this: Always give 273–278 K (0–5 °C); warmer solutions decompose to phenol.
Usually costs: ½ mark.

Exam tips

Learn a table of replacements: Cl, Br, CN (Sandmeyer/Gattermann), I (KI), F (HBF₄, heat), H (H₃PO₂ or ethanol), OH (warm water), NO₂ (NaNO₂/Cu on the fluoroborate).
In conversion questions starting from benzene, remember the nitration–reduction–diazotisation sequence.
Give the colour of azo dyes (orange with phenol, yellow with aniline) in observation answers.
Explain stability of arenediazonium ions with the word 'resonance'.
When the target has groups meta to each other that direct ortho/para, plan the order of steps carefully.

Practice questions on Diazonium Salts and Their Reactions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Benzenediazonium chloride couples with phenol in mildly alkaline solution to give:
(a) a white precipitate of tribromophenol
(b) an orange dye, 4-hydroxyazobenzene
(c) chlorobenzene and N₂
(d) benzene
Hint
Nitrogen is retained in coupling.
Answer
(b).
Q2 · 2 marks · Short answerBoard level
Why are arenediazonium salts more stable than alkanediazonium salts?
Hint
Think about where the positive charge can go.
Answer
In an arenediazonium ion the positive charge on the –N₂⁺ group is delocalised into the benzene ring by resonance, which stabilises it (it is still used cold). Alkanediazonium ions have no such delocalisation and decompose at once, giving N₂ and a carbocation.
Q3 · 2 marks · Short answerHarder board level
Write the balanced equation for the reaction of benzenediazonium chloride with ethanol, and state what is achieved.
Hint
Ethanol is a mild reducing agent here.
Answer
$C_{6} H_{5} N_{2} C l + C H_{3} C H_{2} O H \to C_{6} H_{6} + N_{2} + C H_{3} C H O + H C l$ . The diazonium group is replaced by hydrogen (the amino group of the original aniline is removed); ethanol is oxidised to ethanal.
Q4 · 3 marks · Short answerHarder board level
Describe a test based on diazotisation and coupling that distinguishes aniline from benzylamine (C₆H₅CH₂NH₂). Explain the difference.
Hint
Only one forms a stable diazonium salt.
Answer
Treat each with NaNO₂/HCl at 273–278 K, then add alkaline phenol. Aniline forms a stable benzenediazonium salt, which couples to give an orange dye. Benzylamine is an aliphatic-type primary amine: its diazonium salt decomposes at once, so N₂ is evolved and no dye forms.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Diazotisation of aniline is carried out at 273–278 K.
Reason (R): At higher temperatures the diazonium salt decomposes, giving phenol and nitrogen.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Warm water converts the diazonium salt into phenol.
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
How can nitrobenzene be prepared from aniline through a diazonium salt? Give the equation for the final step.
Hint
The fluoroborate is used.
Answer
Diazotise aniline (NaNO₂/HCl, 273–278 K), add HBF₄ to precipitate benzenediazonium fluoroborate, then heat it with aqueous NaNO₂ and copper powder: $C_{6} H_{5} N_{2} B F_{4} + N a N O_{2} \to C_{6} H_{5} N O_{2} + N a B F_{4} + N_{2}$ .
Q7 · 3 marks · Short answerHOTS / competency
0.93 g of aniline (M = 93) is diazotised completely and the solution is then warmed until no more gas is evolved. Calculate (a) the volume of nitrogen at STP and (b) the mass of phenol (M = 94) formed.
Hint
1 mol aniline → 1 mol diazonium → 1 mol N₂ and 1 mol phenol.
Answer
Aniline $= 0.010$ mol. (a) N₂ $= 0.010 \times 22.4 = 0.224$ L (224 mL). (b) Phenol $= 0.010 \times 94 = 0.94$ g.
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a chemist needs pure 4-chlorotoluene. Chlorinating toluene with Cl₂/FeCl₃ gives a mixture of 2- and 4-chlorotoluene that is hard to separate. She has pure 4-methylaniline (p-toluidine) available.
(i) Why does chlorination of toluene give a mixture? [1]
(ii) Give the two steps that convert 4-methylaniline into pure 4-chlorotoluene. [1]
(iii) Name the second reaction and write its equation. [1]
Hint
The diazonium group is replaced exactly where the NH₂ was.
Answer
(i) –CH₃ is ortho-para directing, so both positions are substituted. (ii) Diazotise 4-methylaniline with NaNO₂/HCl at 273–278 K; then treat with Cu₂Cl₂/HCl. (iii) Sandmeyer reaction: $C H_{3} C_{6} H_{4} N_{2} C l \to C H_{3} C_{6} H_{4} C l + N_{2}$ ; Cl enters only at the 4-position, where nitrogen was.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Conditions for diazotisation?Show answer
NaNO₂ and dilute HCl at 273–278 K.
2.Reagent to replace –N₂⁺ by iodine?Show answer
Aqueous KI.
3.Product of benzenediazonium chloride with CuCN/KCN?Show answer
Benzonitrile.
4.Product of coupling with phenol?Show answer
4-Hydroxyazobenzene (orange).
5.Product of warming benzenediazonium chloride solution?Show answer
Phenol (with N₂ and HCl).
6.Name the reaction using copper powder and HBr.Show answer
Gattermann reaction.

Diazonium Salts and Their Reactions: frequently asked questions

What is diazotisation?

The conversion of a primary aromatic amine into its diazonium salt by nitrous acid (NaNO₂ + HCl) at 273–278 K.

What is the difference between Sandmeyer and Gattermann reactions?

Both replace –N₂⁺ by Cl or Br. Sandmeyer uses copper(I) salts (and also gives nitriles with CuCN); Gattermann uses copper powder with the halogen acid.

Why are diazonium salts important in synthesis?

They allow F, I, CN, OH, H and other groups to be introduced into the ring in positions that direct substitution cannot reach.

Why does coupling occur at the para position?

–OH and –NH₂ are ortho-para directing, and the para position is less hindered for the bulky diazonium electrophile.

Why is diazotisation done in the cold?

Diazonium salts decompose at higher temperatures to give phenol and nitrogen.