ChemistryCBSE Class 12 · Amines

Diazonium Salts and Their Reactions

Updated 15 October 2026 · Fact-checked

Benzenediazonium chloride is made by diazotisation: aniline with sodium nitrite and HCl at 273–278 K. Its $-N_2^+$ group is easily replaced by Cl, Br, CN (Sandmeyer or Gattermann), I, F, H or OH, with loss of nitrogen gas. It also couples with phenol or aniline to give coloured azo dyes. These reactions add groups that direct substitution cannot.

Understand Diazonium Salts and Their Reactions

What diazonium salts are. Diazonium salts have the general formula $R N_{2}^{+} X^{-}$ , where X may be $C l^{-}$ , $B r^{-}$ , $H S O_{4}^{-}$ or $B F_{4}^{-}$ . The $- N_{2}^{+}$ group is called the diazonium group; $C_{6} H_{5} N_{2}^{+} C l^{-}$ is benzenediazonium chloride. Primary aliphatic amines form very unstable alkyldiazonium salts, but arenediazonium salts are stable for a short time in cold solution, because the positive charge is dispersed over the benzene ring by resonance.

Preparation: diazotisation. Aniline is dissolved in excess HCl and treated with aqueous sodium nitrite at 273–278 K (0–5 °C). Nitrous acid formed in the mixture reacts with the amine: $C_{6} H_{5} N H_{2} + N a N O_{2} + 2 H C l \to C_{6} H_{5} N_{2}^{+} C l^{-} + N a C l + 2 H_{2} O$ . Diazonium salts are not usually stored; the cold solution is used immediately because it decomposes on warming. Benzenediazonium chloride is a colourless crystalline solid that dissolves readily in water. Benzenediazonium fluoroborate is insoluble in water and stable at room temperature.

A. Reactions in which nitrogen is displaced. $N_{2}$ is an excellent leaving group, and its escape as a gas drives these reactions.

*Cl, Br, CN (Sandmeyer reaction):* with $C u_{2} C l_{2}$ /HCl, $C u_{2} B r_{2}$ /HBr or CuCN/KCN, giving chlorobenzene, bromobenzene or benzonitrile.
*Cl, Br (Gattermann reaction):* with copper powder and HCl or HBr. Yields in the Sandmeyer reaction are usually higher.
*I:* simply warm with aqueous KI: $C_{6} H_{5} N_{2}^{+} C l^{-} + K I \to C_{6} H_{5} I + K C l + N_{2}$ . Iodine cannot easily be introduced directly into benzene.
*F:* treat with fluoroboric acid to precipitate arenediazonium fluoroborate, which decomposes on heating to the aryl fluoride: $A r N_{2}^{+} B F_{4}^{-} Δ A r F + B F_{3} + N_{2}$ .
*H (removal of $N H_{2}$ ):* mild reducing agents such as hypophosphorous acid ( $H_{3} P O_{2}$ ) or ethanol replace the group by hydrogen: $A r N_{2}^{+} C l^{-} + H_{3} P O_{2} + H_{2} O \to A r H + N_{2} + H_{3} P O_{3} + H C l$ .
*OH:* warming the aqueous solution to about 283 K and above gives phenol: $A r N_{2}^{+} C l^{-} + H_{2} O \to A r O H + N_{2} + H C l$ .
* $N O_{2}$ :* the fluoroborate heated with aqueous $N a N O_{2}$ and copper gives nitrobenzene.

B. Reactions retaining the diazo group: coupling. The diazonium ion is a weak electrophile and attacks very reactive rings. With phenol in alkaline solution it couples at the para position to give p-hydroxyazobenzene, an orange dye. With aniline in mildly acidic medium it gives p-aminoazobenzene, a yellow dye. These are electrophilic substitution reactions; the $- N = N -$ (azo) group joins two rings, and the extended conjugation makes the products coloured.

Importance in synthesis. Through diazonium salts, groups like F, Cl, Br, I, CN, OH and $N O_{2}$ can be put into the benzene ring, and an amino group can be removed. Many of these (iodobenzene, fluorobenzene, benzonitrile, 1,3,5-tribromobenzene) cannot be made by direct substitution.

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 at once.
Sandmeyer reaction: $A r N_{2}^{+} C l^{-} C u_{2} C l_{2} / H C l A r C l; C u_{2} B r_{2} / H B r A r B r; C u C N / K C N A r C N (+ N_{2})$; Copper(I) salts.
Gattermann reaction: $A r N_{2}^{+} C l^{-} C u / H C l A r C l + N_{2}; C u / H B r A r B r + N_{2}$; Copper powder; usually lower yield than Sandmeyer.
Replacement by I, F, H, OH: $K I A r I; H B F_{4}, Δ A r F; H_{3} P O_{2} or C_{2} H_{5} O H A r H; H_{2} O, warm A r O H$; All with loss of $N_{2}$ .
Coupling with phenol: $C_{6} H_{5} N_{2}^{+} C l^{-} + C_{6} H_{5} O H O H^{-} p - H O C_{6} H_{4} N = N C_{6} H_{5} + H C l$; Orange dye; para coupling.
Coupling with aniline: $C_{6} H_{5} N_{2}^{+} C l^{-} + C_{6} H_{5} N H_{2} H^{+} p - H_{2} N C_{6} H_{4} N = N C_{6} H_{5} + H C l$; Yellow dye.

How to solve Diazonium Salts and Their Reactions questions

Diazonium questions are usually conversions. Treat the diazonium salt as a hub: you reach it from aniline, and from it you reach almost any monosubstituted benzene.

1Work backwards from the target: which group must replace $- N_{2}^{+}$ (Cl, Br, CN, I, F, H, OH, $N O_{2}$ ) or must the azo group be kept?
2Get to the amine first: nitro compound → reduce (Sn/HCl or Fe/HCl) → aniline (or substituted aniline).
3Diazotise: $N a N O_{2}$ + HCl at 273–278 K.
4Choose the replacement reagent: CuCl/CuBr/CuCN (Sandmeyer), Cu/HX (Gattermann), KI, $H B F_{4}$ then heat, $H_{3} P O_{2}$ or ethanol, warm water.
5For acids, go via the nitrile: ArCN, then acid hydrolysis gives ArCOOH.
6For substitution patterns impossible directly (like 1,3,5-tribromobenzene), use the strong o/p-directing $- N H_{2}$ to place groups, then remove it via diazotisation and $H_{3} P O_{2}$ .
7For coupling, state the medium (alkaline for phenol, mildly acidic for aniline) and the position (para).

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.

Coupling reaction makes azo dyes

Suppose a small textile unit makes an orange dye, p-hydroxyazobenzene, by diazotising 0.93 g of aniline at 0–5 °C and adding the cold solution to phenol dissolved in sodium hydroxide.

Write the two reactions involved, explain why the diazotisation is done cold and the coupling in alkali, and calculate the maximum mass of dye. (Aniline 93 g mol⁻¹, dye $C_{12} H_{10} N_{2} O$ 198 g mol⁻¹)

Try it first, then show the solutionSolution

Diazotisation: $C_{6} H_{5} N H_{2} + N a N O_{2} + 2 H C l \to C_{6} H_{5} N_{2}^{+} C l^{-} + N a C l + 2 H_{2} O$ ; kept at 273–278 K because the diazonium salt decomposes to phenol and $N_{2}$ on warming.
Coupling: $C_{6} H_{5} N_{2}^{+} C l^{-} + C_{6} H_{5} O H O H^{-} p - H O C_{6} H_{4} N = N C_{6} H_{5} + H C l$ ; alkali converts phenol into the more reactive phenoxide ion.
Moles of aniline $= 0.93/93 = 0.010$ mol; maximum dye $= 0.010$ mol.
Mass $= 0.010 \times 198 = 1.98$ g.

Answer: Up to about 1.98 g of p-hydroxyazobenzene; cold to keep the diazonium salt intact, alkaline to activate phenol.

Memory hook: Cold to make it, alkaline phenol to couple it: the –N=N– bridge gives the 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: Cu(I) Salts (CuCl, CuBr, CuCN). Gattermann: Copper powder with HX.
Use it when: Naming the reagent in conversions.
H₃PO₂ Hides the amino group
$H_{3} P O_{2}$ (or ethanol) replaces $N_{2}^{+}$ by H: the $- N H_{2}$ is used to steer substitution and then removed.
Use it when: 1,3,5-trisubstituted benzene puzzles.
Phenol couples in base, aniline in mild acid
Phenoxide needs base; aniline must stay unprotonated, so only mildly acidic.
Use it when: Coupling conditions.

Diazonium Salts and Their Reactions: solved examples

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

Example 1: Diazotisation conditions · 1 mark · MCQBoard level

Benzenediazonium chloride is prepared from aniline using $N a N O_{2}$ and HCl at:
(a) 373 K (b) 298 K (c) 273–278 K (d) 323 K

Try it first, then show the solutionSolution

Diazonium salts decompose on warming, so the reaction is carried out at 0–5 °C.

Answer: (c) 273–278 K.

Example 2: Sandmeyer products · 2 marks · Very short answerBoard level

Write the products when benzenediazonium chloride is treated with (a) $C u_{2} C l_{2}$ /HCl and (b) CuCN/KCN. Name the reaction.

Try it first, then show the solutionSolution

(a) $C_{6} H_{5} N_{2}^{+} C l^{-} \to C_{6} H_{5} C l + N_{2}$ : chlorobenzene (1 mark).
(b) $C_{6} H_{5} N_{2}^{+} C l^{-} + C u C N \to C_{6} H_{5} C N + N_{2} + C u C l$ : benzonitrile. Both are Sandmeyer reactions (1 mark).

Answer: (a) Chlorobenzene. (b) Benzonitrile (cyanobenzene). Sandmeyer reaction.

Example 3: Three conversions from aniline · 3 marks · Short answerHarder board level

How will you convert aniline into (a) iodobenzene (b) benzene (c) phenol? Give reagents and equations.

Try it first, then show the solutionSolution

First diazotise: $C_{6} H_{5} N H_{2} + N a N O_{2} + 2 H C l 273 - 278 K C_{6} H_{5} N_{2}^{+} C l^{-} + N a C l + 2 H_{2} O$ (½ mark).
(a) $C_{6} H_{5} N_{2}^{+} C l^{-} + K I \to C_{6} H_{5} I + K C l + N_{2}$ (1 mark).
(b) $C_{6} H_{5} N_{2}^{+} C l^{-} + H_{3} P O_{2} + H_{2} O \to C_{6} H_{6} + N_{2} + H_{3} P O_{3} + H C l$ (or warm with ethanol) (1 mark).
(c) Warm with water: $C_{6} H_{5} N_{2}^{+} C l^{-} + H_{2} O \to C_{6} H_{5} O H + N_{2} + H C l$ (½ mark).

Answer: Diazotise, then (a) KI, (b) $H_{3} P O_{2}$ , (c) warm water.

Example 4: Coupling reactions · 3 marks · Short answerHarder board level

Write the coupling reactions of benzenediazonium chloride with (a) phenol and (b) aniline, giving the medium, the product and its colour. Why does coupling occur at the para position, and what type of reaction is it?

Try it first, then show the solutionSolution

(a) Alkaline medium: $C_{6} H_{5} N_{2}^{+} C l^{-} + C_{6} H_{5} O H \to p - H O C_{6} H_{4} N = N C_{6} H_{5} + H C l$ : p-hydroxyazobenzene, orange (1 mark).
(b) Mildly acidic medium: $C_{6} H_{5} N_{2}^{+} C l^{-} + C_{6} H_{5} N H_{2} \to p - H_{2} N C_{6} H_{4} N = N C_{6} H_{5} + H C l$ : p-aminoazobenzene, yellow (1 mark).
The diazonium ion is an electrophile; –OH and –NH₂ are strong o/p directors, and the para position is less crowded. It is electrophilic aromatic substitution (1 mark).

Answer: (a) p-Hydroxyazobenzene (orange, alkaline). (b) p-Aminoazobenzene (yellow, mildly acidic). Para attack by electrophilic substitution.

Example 5: Two multistep conversions · 5 marks · Long answerHOTS / competency

Show how you would carry out: (a) benzene → 1,3,5-tribromobenzene (b) nitrobenzene → benzoic acid. Give reagents for each step and explain why route (a) is needed.

Try it first, then show the solutionSolution

(a) Benzene → nitrobenzene (conc. $H N O_{3}$ /conc. $H_{2} S O_{4}$ ) → aniline (Sn/HCl or Fe/HCl) (1 mark).
Aniline + bromine water → 2,4,6-tribromoaniline (–NH₂ directs Br to both ortho and the para position) (1 mark).
Diazotise ( $N a N O_{2}$ /HCl, 273–278 K), then $H_{3} P O_{2}$ removes the diazonium group: 1,3,5-tribromobenzene. Direct bromination of benzene cannot place three Br meta to one another, because Br is o/p-directing (1 mark).
(b) Nitrobenzene → aniline (Sn/HCl) → benzenediazonium chloride ( $N a N O_{2}$ /HCl, 273–278 K) → benzonitrile (CuCN/KCN, Sandmeyer) (1 mark).
Benzonitrile + dilute acid, heat (hydrolysis) → benzoic acid: $C_{6} H_{5} C N + 2 H_{2} O H^{+} C_{6} H_{5} C O O H + N H_{3}$ (as $N H_{4}^{+}$ ) (1 mark).

Answer: (a) Nitrate → reduce → brominate (bromine water) → diazotise → $H_{3} P O_{2}$ . (b) Reduce → diazotise → CuCN → hydrolyse.

Where students lose marks in Diazonium Salts and Their Reactions

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

Mistake: Writing diazotisation at room temperature
Do this: Always write 273–278 K (0–5 °C).
Usually costs: ½ mark.
Mistake: Using CuCl for iodobenzene
Do this: Iodide needs only KI; copper salts are used for Cl, Br and CN.
Usually costs: 1 mark.
Mistake: Forgetting $N_{2}$ in displacement reactions
Wrong working
$C_{6} H_{5} N_{2}^{+} C l^{-} + K I \to C_{6} H_{5} I + K C l$
Right working
$C_{6} H_{5} N_{2}^{+} C l^{-} + K I \to C_{6} H_{5} I + K C l + N_{2}$
Do this: Every replacement releases $N_{2}$ ; only coupling keeps it.
Usually costs: ½ mark.
Mistake: Coupling at the meta or ortho position
Do this: Coupling occurs para to –OH or –NH₂ (ortho only if para is blocked).
Usually costs: 1 mark.

Exam tips

Write $N a N O_{2}$ + HCl, 273–278 K over the first arrow in every conversion that goes through a diazonium salt.
Name reactions: Sandmeyer, Gattermann, coupling.
For conversions, number the steps and name every intermediate.
Mention the colour of azo dyes (orange from phenol, yellow from aniline) when asked about coupling.
Use the diazonium route to explain why iodo-, fluoro- and cyano-benzenes are made indirectly.

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 alkaline medium to give:
(a) p-aminoazobenzene (b) p-hydroxyazobenzene (c) azobenzene (d) chlorobenzene
Hint
The OH stays on the coupled ring.
Answer
(b) p-Hydroxyazobenzene (orange dye).
Q2 · 2 marks · Very short answerBoard level
Write the equation for the diazotisation of aniline with its conditions. Why is the diazonium salt not stored?
Hint
Temperature and stability.
Answer
$C_{6} H_{5} N H_{2} + N a N O_{2} + 2 H C l 273 - 278 K C_{6} H_{5} N_{2}^{+} C l^{-} + N a C l + 2 H_{2} O$ . It is unstable and decomposes (to phenol and $N_{2}$ ) as the temperature rises, so it is used immediately in cold solution.
Q3 · 2 marks · Very short answerHarder board level
Distinguish between the Sandmeyer and Gattermann reactions with one example each.
Hint
Reagent type.
Answer
Sandmeyer uses copper(I) salts: $C_{6} H_{5} N_{2}^{+} C l^{-} C u_{2} C l_{2} / H C l C_{6} H_{5} C l + N_{2}$ . Gattermann uses copper powder with the halogen acid: $C_{6} H_{5} N_{2}^{+} C l^{-} C u / H B r C_{6} H_{5} B r + N_{2}$ . Sandmeyer usually gives better yields.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Arenediazonium salts are more stable than alkyldiazonium salts.
Reason (R): In arenediazonium ions the positive charge is dispersed over the benzene ring by resonance.
(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
Delocalisation stabilises ions.
Answer
(a).
Q5 · 3 marks · Short answerHarder board level
Give the products when benzenediazonium chloride reacts with (a) $H_{3} P O_{2}$ and water (b) warm water (c) aqueous KI.
Hint
All release $N_{2}$ .
Answer
(a) Benzene ( $+ N_{2} + H_{3} P O_{3} + H C l$ ). (b) Phenol ( $+ N_{2} + H C l$ ). (c) Iodobenzene ( $+ K C l + N_{2}$ ).
Q6 · 2 marks · Very short answerHarder board level
Why can fluorobenzene not be made conveniently by direct fluorination of benzene? Show how it is made from aniline.
Hint
Fluorine is too reactive.
Answer
Fluorine reacts violently and uncontrollably with benzene. Instead: aniline → $C_{6} H_{5} N_{2}^{+} C l^{-}$ ( $N a N O_{2}$ /HCl, 273–278 K) → add $H B F_{4}$ to precipitate $C_{6} H_{5} N_{2}^{+} B F_{4}^{-}$ → heat: $C_{6} H_{5} F + B F_{3} + N_{2}$ .
Q7 · 3 marks · Short answerHOTS / competency
9.3 g of aniline is converted to chlorobenzene by diazotisation followed by the Sandmeyer reaction. If the overall yield is 80%, what mass of chlorobenzene is obtained? (Aniline 93, chlorobenzene 112.5 g mol⁻¹) Why must excess HCl be used during diazotisation?
Hint
1 : 1 overall.
Answer
Moles of aniline $= 9.3/93 = 0.10$ ; chlorobenzene $= 0.80 \times 0.10 = 0.080$ mol; mass $= 0.080 \times 112.5 = 9.0$ g. Excess HCl generates nitrous acid from $N a N O_{2}$ , keeps the medium acidic and stops the diazonium salt from coupling with unreacted aniline.
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student diazotises aniline correctly but forgets the ice bath, so the solution warms to room temperature before she adds it to an alkaline solution of phenol. She gets a weak orange colour and notices gas bubbles and a smell of phenol.
(a) What gas is evolved and why? Write the equation. (1 mark)
(b) What should the orange compound be, and how is it formed? (1 mark)
(c) Why is the coupling done in alkaline solution? (1 mark)
(d) If she had added the diazonium solution to aniline in mildly acidic medium, what product would form? (1 mark)
Hint
Warming decomposes the salt.
Answer
(a) Nitrogen: the warm diazonium salt reacts with water, $C_{6} H_{5} N_{2}^{+} C l^{-} + H_{2} O \to C_{6} H_{5} O H + N_{2} + H C l$ . (b) p-Hydroxyazobenzene, from electrophilic coupling of the remaining diazonium ion at the para position of phenoxide. (c) Alkali converts phenol into phenoxide, which is much more strongly activated towards the weak diazonium electrophile. (d) p-Aminoazobenzene (yellow).

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.Temperature for diazotisation?Show answer
273–278 K.
2.Reagent to replace $- N_{2}^{+}$ by H?Show answer
Hypophosphorous acid ( $H_{3} P O_{2}$ ) or ethanol.
3.Product of coupling with aniline?Show answer
p-Aminoazobenzene (yellow).
4.Reagent for benzonitrile from benzenediazonium chloride?Show answer
CuCN/KCN (Sandmeyer).
5.Which diazonium salt is stable at room temperature?Show answer
Benzenediazonium fluoroborate.
6.What forms when benzenediazonium chloride is warmed with water?Show answer
Phenol, $N_{2}$ and HCl.

Diazonium Salts and Their Reactions: frequently asked questions

What is diazotisation in Class 12 Chemistry?

The conversion of a primary aromatic amine into a diazonium salt by nitrous acid (from NaNO_2 and HCl) at 273–278 K.

What is the Sandmeyer reaction?

Replacement of the diazonium group by Cl, Br or CN using copper(I) chloride, bromide or cyanide, with loss of nitrogen gas.

Why are diazonium salts important in synthesis?

They allow groups such as F, Cl, Br, I, CN, OH and NO_2 to be introduced into the benzene ring, and the amino group to be removed, often where direct substitution is impossible.

What is a coupling reaction?

An electrophilic substitution in which a diazonium ion attacks a very reactive aromatic ring such as phenol or aniline, usually at the para position, forming a coloured azo compound.

Why are arenediazonium salts more stable than aliphatic ones?

The positive charge on the diazonium group is delocalised into the benzene ring by resonance.