Chemistry Class 12 Part II…
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Diazonium Salts

🎓 Class 12 Chemistry CBSE Theory Ch 9 – Amines ⏱ ~14 min
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Diazonium Salts — The Aromatic Chemist's Universal Joint

There are aromatic compounds you simply cannot make by substituting benzene directly. Aryl fluorides and aryl iodides cannot be prepared by direct halogenation. The cyano group cannot be introduced by nucleophilic substitution of the chlorine in chlorobenzene. Diazonium salts solve all of these problems at once — you place an –NH₂ group where you want the new substituent, convert it to –N₂⁺, and then swap the diazonium group for almost anything.

II. Diazonium Salts

Diazonium salts have the general formula R–N₂⁺X⁻, where R stands for an aryl group and X⁻ may be Cl⁻, Br⁻, HSO₄⁻, BF₄⁻ and so on. They are named by suffixing diazonium to the name of the parent hydrocarbon from which they are formed, followed by the name of the anion. The N₂⁺ group is called the diazonium group.

Naming examples. C₆H₅N₂⁺Cl⁻ is named benzenediazonium chloride, and C₆H₅N₂⁺HSO₄⁻ is benzenediazonium hydrogensulphate.

Why only aromatic diazonium salts are useful

Primary aliphatic amines form highly unstable alkyldiazonium salts, which decompose immediately (as seen in Part 3, giving an alcohol and nitrogen). Primary aromatic amines form arenediazonium salts which are stable for a short time in solution at low temperatures (273–278 K). The stability of the arenediazonium ion is explained on the basis of resonance — the positive charge is delocalised into the benzene ring, which no alkyl group can do.

Alkyldiazonium — no delocalisation R N≡N⁺ charge trapped on N — very unstable decomposes at once → R–OH + N₂↑ Arenediazonium — resonance stabilised N≡N⁺ positive charge shared with the ring stable in solution at 273–278 K This one structural difference is why the whole of diazonium chemistry is aromatic chemistry.
Resonance delocalisation of the positive charge into the ring is what makes arenediazonium salts isolable while alkyldiazonium salts are not.

9.7 Method of Preparation of Diazonium Salts

Benzenediazonium chloride is prepared by the reaction of aniline with nitrous acid at 273–278 K. Nitrous acid is produced in the reaction mixture by the reaction of sodium nitrite with hydrochloric acid. The conversion of primary aromatic amines into diazonium salts is known as diazotisation.

C₆H₅NH₂  +  NaNO₂  +  2HCl  —[273–278 K]→  C₆H₅N₂⁺Cl⁻  +  NaCl  +  2H₂O
Never store it. Due to its instability, the diazonium salt is not generally stored and is used immediately after its preparation. Every synthesis in this part assumes the ice-cold solution is carried straight on to the next step.

9.8 Physical Properties

Benzenediazonium chloride is a colourless crystalline solid. It is readily soluble in water and is stable in cold, but reacts with water when warmed. It decomposes easily in the dry state. Benzenediazonium fluoroborate, by contrast, is water insoluble and stable at room temperature — which is exactly what makes the Balz–Schiemann route to aryl fluorides practicable.

9.9 Chemical Reactions

The reactions of diazonium salts fall into two categories: (A) reactions involving displacement of nitrogen and (B) reactions involving retention of the diazo group.

A. Reactions involving displacement of nitrogen

The diazonium group is a very good leaving group and is substituted by other groups such as Cl⁻, Br⁻, I⁻, CN⁻ and OH⁻, which displace nitrogen from the aromatic ring. The nitrogen formed escapes from the reaction mixture as a gas — an irreversible driving force.

1. Replacement by halide or cyanide ion — Sandmeyer reaction

The Cl⁻, Br⁻ and CN⁻ nucleophiles can easily be introduced into the benzene ring in the presence of Cu(I) ion. This reaction is called the Sandmeyer reaction.

C₆H₅N₂⁺Cl⁻  —[CuCl/HCl]→  C₆H₅Cl  +  N₂↑
C₆H₅N₂⁺Cl⁻  —[CuBr/HBr]→  C₆H₅Br  +  N₂↑
C₆H₅N₂⁺Cl⁻  —[CuCN/KCN]→  C₆H₅CN  +  N₂↑

Alternatively, chlorine or bromine can be introduced by treating the diazonium salt solution with the corresponding halogen acid in the presence of copper powder. This is referred to as the Gattermann reaction.

C₆H₅N₂⁺Cl⁻  +  HCl  —[Cu powder]→  C₆H₅Cl  +  N₂↑
Which is better? The yield in the Sandmeyer reaction is better than in the Gattermann reaction. Remember the distinction by the reagent: Sandmeyer uses the copper(I) salt (CuCl, CuBr, CuCN); Gattermann uses copper powder with the halogen acid.

2. Replacement by iodide ion

Iodine is not easily introduced into the benzene ring directly, but when the diazonium salt solution is treated with potassium iodide, iodobenzene is formed. No copper catalyst is needed.

C₆H₅N₂⁺Cl⁻  +  KI  →  C₆H₅I  +  KCl  +  N₂↑

3. Replacement by fluoride ion — Balz–Schiemann reaction

When arenediazonium chloride is treated with fluoroboric acid (HBF₄), arenediazonium fluoroborate is precipitated, which on heating decomposes to yield aryl fluoride.

C₆H₅N₂⁺Cl⁻  +  HBF₄  →  C₆H₅N₂⁺BF₄⁻↓  —[heat]→  C₆H₅F  +  N₂↑  +  BF₃

4. Replacement by H

Certain mild reducing agents — hypophosphorous acid (phosphinic acid, H₃PO₂) or ethanol — reduce diazonium salts to arenes and are themselves oxidised, to phosphorous acid and ethanal respectively.

C₆H₅N₂⁺Cl⁻  +  H₃PO₂  +  H₂O  →  C₆H₆  +  N₂↑  +  H₃PO₃  +  HCl
C₆H₅N₂⁺Cl⁻  +  CH₃CH₂OH  →  C₆H₆  +  N₂↑  +  CH₃CHO  +  HCl
Why replace a group with nothing? This looks pointless until you see it as a blocking strategy. An –NH₂ group can be used purely to direct an incoming substituent to a particular position, and then removed entirely by diazotisation followed by H₃PO₂. That is how 1,3,5-tribromobenzene is made from aniline — a compound impossible to make by direct bromination of benzene, which gives 1,2- and 1,4-products.

5. Replacement by hydroxyl group

If the temperature of the diazonium salt solution is allowed to rise up to 283 K, the salt gets hydrolysed to phenol.

C₆H₅N₂⁺Cl⁻  +  H₂O  —[283 K]→  C₆H₅OH  +  N₂↑  +  HCl

6. Replacement by –NO₂ group

When diazonium fluoroborate is heated with aqueous sodium nitrite solution in the presence of copper, the diazonium group is replaced by the –NO₂ group.

C₆H₅N₂⁺BF₄⁻  +  NaNO₂  —[Cu, heat]→  C₆H₅NO₂  +  NaBF₄  +  N₂↑
Ar–N₂⁺ the hub Ar–ClCuCl · Sandmeyer Ar–BrCuBr · Sandmeyer Ar–IKI · no Cu needed Ar–FHBF₄ then Δ · Balz–Schiemann Ar–CNCuCN · Sandmeyer Ar–OHwarm H₂O, 283 K Ar–NO₂NaNO₂ / Cu Ar–HH₃PO₂ or C₂H₅OH Every arrow releases N₂ gas — the irreversible driving force behind the whole map.
The diazonium hub. Eight substituents reachable from one intermediate, several of them impossible by direct substitution of benzene.

B. Reactions involving retention of the diazo group — coupling reactions

The azo products obtained have an extended conjugate system in which both aromatic rings are joined through the –N=N– bond. These compounds are often coloured and are used as dyes.

Benzenediazonium chloride reacts with phenol, in which the phenol molecule couples at its para position with the diazonium salt to form p-hydroxyazobenzene. This type of reaction is known as a coupling reaction. Similarly the reaction of a diazonium salt with aniline yields p-aminoazobenzene. This is an example of an electrophilic substitution reaction — the diazonium ion is the electrophile.

C₆H₅N₂⁺Cl⁻  +  C₆H₅OH  →  p-HO–C₆H₄–N=N–C₆H₅  (p-hydroxyazobenzene)  +  HCl
C₆H₅N₂⁺Cl⁻  +  C₆H₅NH₂  →  p-H₂N–C₆H₄–N=N–C₆H₅  (p-aminoazobenzene)  +  HCl
Why the colour? The –N=N– bridge links the two rings into one long conjugated system. Extended conjugation lowers the energy gap between the electronic levels so that absorption moves into the visible region — which is why azo compounds form the largest single class of synthetic dyes.

9.10 Importance of Diazonium Salts in Synthesis of Aromatic Compounds

From the reactions above it is clear that diazonium salts are very good intermediates for the introduction of –F, –Cl, –Br, –I, –CN, –OH and –NO₂ groups into the aromatic ring.

The three things only diazonium chemistry can do.
• Aryl fluorides and iodides cannot be prepared by direct halogenation.
• The cyano group cannot be introduced by nucleophilic substitution of the chlorine in chlorobenzene, but cyanobenzene is easily obtained from the diazonium salt.
• Replacement of the diazo group is the route to substituted aromatic compounds that cannot be prepared by direct substitution in benzene or substituted benzene.
Worked Example 9.5 — Convert 4-nitrotoluene to 2-bromobenzoic acid

Reading the problem. The target has a –COOH where the –CH₃ was, and a –Br where the –NO₂ was, one position round. So the plan is: turn –NO₂ into –NH₂, use it to place –Br via a diazonium salt, and oxidise –CH₃ to –COOH.

Step 1 — reduce the nitro group. 4-Nitrotoluene —[Fe/HCl or Sn/HCl]→ 4-methylaniline (p-toluidine).

Step 2 — diazotise. 4-Methylaniline + NaNO₂ + HCl at 273–278 K → 4-methylbenzenediazonium chloride.

Step 3 — Sandmeyer with CuBr. The diazonium salt with CuBr/HBr gives 4-bromotoluene (the bromine takes the exact position vacated by nitrogen) + N₂↑.

Step 4 — oxidise the methyl group. 4-Bromotoluene —[KMnO₄/KOH, then H₃O⁺]→ 4-bromobenzoic acid. Numbering the product from the –COOH carbon as C-1 makes the bromine C-4; if instead the isomer required is the 2-bromo compound, the same four-step logic is applied starting from 2-nitrotoluene, so that the bromine ends up ortho to the carboxyl group, giving 2-bromobenzoic acid.

The transferable idea: the –NO₂ (then –NH₂, then –N₂⁺) group acts as a positional placeholder. It occupies the site you want, and the diazonium step swaps it for the group you actually want there.

🧪 Activity 9.5 — Why temperature control decides the productL4 Analyse

Diazotisation is one of the few school-level preparations where a few degrees change the product entirely. This activity makes that dependence explicit.

Predict: A student prepares benzenediazonium chloride but lets the ice bath melt so the flask warms to about 285 K before the next reagent is added. What will be in the flask? Write your answer first.
  1. Write the diazotisation equation and note the temperature range specified: 273–278 K.
  2. Note from Section 9.8 that the salt is stable in the cold but reacts with water when warmed.
  3. Write the equation for what happens at 283 K and above.
  4. Predict the consequence for a planned Sandmeyer reaction if the solution is allowed to warm first.
  5. Explain why the salt must never be isolated and dried.

The flask will contain phenol, not the diazonium salt.

At 283 K and above the salt is hydrolysed by the water it is dissolved in: C₆H₅N₂⁺Cl⁻ + H₂O → C₆H₅OH + N₂↑ + HCl. Brisk effervescence of nitrogen is the visible sign that this has happened.

Consequence for the planned synthesis: the Sandmeyer reaction would fail. The CuBr would be added to a solution that no longer contains any diazonium ion, and the student would isolate phenol (or a phenol-contaminated product) instead of bromobenzene. Since the nitrogen has already escaped as gas, the error cannot be reversed by re-cooling — the batch is lost.

Why it must never be dried: Section 9.8 states that benzenediazonium chloride decomposes easily in the dry state. Dry diazonium salts are notoriously shock-sensitive and can decompose explosively, which is why they are always kept in cold solution and used immediately. Benzenediazonium fluoroborate is the deliberate exception — it is water-insoluble and stable at room temperature, which is precisely what allows it to be filtered off and then heated in the Balz–Schiemann synthesis of aryl fluorides.

Intext question

Intext 9.9 — Convert

(i) 3-Methylaniline into 3-nitrotoluene. The –NH₂ must be replaced by –NO₂ at the same position. Diazotise, then use the fluoroborate route: 3-methylaniline + NaNO₂/HCl at 273–278 K → 3-methylbenzenediazonium chloride; add HBF₄ to precipitate the fluoroborate; then heat with aqueous NaNO₂ in the presence of Cu → 3-nitrotoluene + N₂↑.

(ii) Aniline into 1,3,5-tribromobenzene. Direct bromination of benzene cannot give the 1,3,5-pattern, so the –NH₂ is used as a directing group and then removed. Aniline + 3Br₂(aq) at room temperature → 2,4,6-tribromoaniline (white precipitate); diazotise it with NaNO₂/HCl at 273–278 K → 2,4,6-tribromobenzenediazonium chloride; finally reduce with H₃PO₂ (or ethanol) to replace the diazonium group by hydrogen → 1,3,5-tribromobenzene + N₂↑. The amino group has done its directing job and then been deleted.

Competency-Based Questions

A dye works needs to manufacture two products from a single batch of aniline: 4-iodoanisole for a pharmaceutical customer, and an orange azo dye for textiles. The plant chemist has aniline, sodium nitrite, hydrochloric acid, potassium iodide, fluoroboric acid, phenol, copper(I) salts, hypophosphorous acid and an ice plant capable of holding vessels at 273–278 K.

1. Why must the diazotisation be carried out at 273–278 K rather than at room temperature? L2 Understand

Benzenediazonium chloride is stable only in the cold. If the temperature rises to about 283 K or above, the salt reacts with the water it is dissolved in and is hydrolysed to phenol with brisk evolution of nitrogen: C₆H₅N₂⁺Cl⁻ + H₂O → C₆H₅OH + N₂↑ + HCl. The diazonium ion would then no longer be available for the intended substitution, and since the nitrogen escapes as gas the loss is irreversible.

2. Write the equation for preparing iodobenzene from aniline and state why no copper catalyst is required. L3 Apply

C₆H₅NH₂ + NaNO₂ + 2HCl —[273–278 K]→ C₆H₅N₂⁺Cl⁻; then C₆H₅N₂⁺Cl⁻ + KI → C₆H₅I + KCl + N₂↑. Unlike the chloride, bromide and cyanide substitutions, which need Cu(I) as in the Sandmeyer reaction, the iodide ion is a sufficiently good nucleophile to displace nitrogen on its own, so the reaction proceeds simply on treatment with potassium iodide.

3. The customer also asks for fluorobenzene. Explain why it cannot be made by a Sandmeyer reaction and give the correct method. L4 Analyse

There is no Cu(I) fluoride route that works in the Sandmeyer manner, and aryl fluorides cannot be prepared by direct halogenation either. The correct method is the Balz–Schiemann reaction: treat the diazonium chloride with fluoroboric acid so that benzenediazonium fluoroborate precipitates — this salt is water-insoluble and stable at room temperature, so it can be filtered and dried — and then heat it: C₆H₅N₂⁺BF₄⁻ —[Δ]→ C₆H₅F + N₂↑ + BF₃.

4. Describe how the orange azo dye is made, and state which species acts as the electrophile. L3 Apply

Couple the cold diazonium salt with phenol: C₆H₅N₂⁺Cl⁻ + C₆H₅OH → p-hydroxyazobenzene + HCl. Coupling occurs at the para position of the phenol. The diazonium ion C₆H₅N₂⁺ is the electrophile, and the highly activated phenol ring is the nucleophile, so this is an electrophilic substitution. The product is coloured because the –N=N– bridge creates an extended conjugated system spanning both rings, shifting absorption into the visible region. Coupling with aniline instead of phenol gives p-aminoazobenzene.

5. A trainee proposes making 1,3,5-tribromobenzene by treating benzene with excess bromine and FeBr₃. Evaluate this proposal and give a route that works. L5 Evaluate

The proposal cannot give the 1,3,5-isomer. Bromine is an ortho/para director, so the first bromine introduced directs the second to positions 2 and 4, and the product mixture is dominated by 1,2- and 1,4-dibromobenzene and then 1,2,4-tribromobenzene. The 1,3,5-pattern requires meta relationships throughout, which ortho/para direction can never deliver. Working route: start from aniline, whose powerfully activating –NH₂ group forces substitution at all three of its ortho and para positions at once — aniline + 3Br₂(aq) at room temperature gives 2,4,6-tribromoaniline as a white precipitate. Then delete the amino group: diazotise with NaNO₂/HCl at 273–278 K and reduce the diazonium salt with hypophosphorous acid (or ethanol), which replaces –N₂⁺ by –H. The result is 1,3,5-tribromobenzene. The amino group is used purely as a temporary directing device — a strategy available only because diazonium chemistry allows it to be removed cleanly afterwards.

Assertion–Reason Questions

For each pair choose: (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.

Assertion (A): Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

Reason (R): In an arenediazonium ion the positive charge is delocalised into the benzene ring by resonance.

Answer: A. Resonance delocalisation spreads the positive charge over the ring, which no alkyl group can do — hence arenediazonium salts survive in cold solution while alkyldiazonium salts decompose instantly.

Assertion (A): Benzenediazonium fluoroborate can be isolated as a dry solid, whereas benzenediazonium chloride cannot.

Reason (R): Benzenediazonium fluoroborate is water insoluble and stable at room temperature.

Answer: A. The chloride decomposes easily in the dry state and must be kept in cold solution, while the fluoroborate's insolubility and room-temperature stability allow it to be filtered, dried and then heated — which is exactly the Balz–Schiemann procedure.

Assertion (A): Azo compounds obtained by coupling reactions are coloured.

Reason (R): The coupling reaction proceeds by nucleophilic substitution on the phenol ring.

Answer: C. The assertion is true — the –N=N– bridge produces an extended conjugated system spanning both rings, which shifts absorption into the visible region. The reason is false: coupling is an electrophilic substitution, with the diazonium ion acting as the electrophile attacking the electron-rich phenol or aniline ring.
Coming next. Part 6 gathers the chapter summary and works through all fourteen NCERT exercises, from IUPAC naming and chemical tests to multi-step conversions and the reaction sequences of Exercise 9.9.

Frequently Asked Questions

What is diazotisation and at what temperature is it carried out?
Diazotisation is the conversion of a primary aromatic amine into a diazonium salt. Aniline is treated with nitrous acid, generated in the mixture from sodium nitrite and hydrochloric acid, at 273 to 278 K: C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O. The low temperature is essential because the salt is hydrolysed to phenol if warmed to about 283 K, and it decomposes easily in the dry state, so it is used immediately after preparation.
What is the difference between the Sandmeyer and Gattermann reactions?
Both replace the diazonium group by chlorine or bromine. The Sandmeyer reaction uses the corresponding copper(I) halide, CuCl or CuBr, and can also introduce the cyano group using CuCN. The Gattermann reaction uses copper powder together with the halogen acid, HCl or HBr. The yield in the Sandmeyer reaction is better than in the Gattermann reaction.
Why are diazonium salts so important in aromatic synthesis?
They give access to substituents that direct substitution of benzene cannot provide. Aryl fluorides and aryl iodides cannot be prepared by direct halogenation, and the cyano group cannot be introduced by nucleophilic substitution of chlorine in chlorobenzene, yet all three are readily obtained from a diazonium salt. The diazonium group can be replaced by –F, –Cl, –Br, –I, –CN, –OH, –NO₂ or even –H, each time with loss of nitrogen gas.
How are aryl fluorides prepared from diazonium salts?
By the Balz-Schiemann reaction. The arenediazonium chloride is treated with fluoroboric acid, which precipitates arenediazonium fluoroborate. That salt is water insoluble and stable at room temperature, so it can be filtered and dried, and on heating it decomposes to give the aryl fluoride: C₆H₅N₂⁺BF₄⁻ → C₆H₅F + N₂ + BF₃.
What is a coupling reaction and why are azo dyes coloured?
In a coupling reaction the diazonium group is retained rather than displaced. Benzenediazonium chloride reacts with phenol at its para position to give p-hydroxyazobenzene, and with aniline to give p-aminoazobenzene. The diazonium ion acts as the electrophile, so this is an electrophilic substitution. The products are coloured because the –N=N– bridge joins the two aromatic rings into one extended conjugated system, which shifts light absorption into the visible region.
How can 1,3,5-tribromobenzene be prepared when direct bromination of benzene fails?
Bromine is an ortho and para director, so direct bromination can never give the all-meta 1,3,5 pattern. Instead, start from aniline: its powerfully activating –NH₂ group directs bromination to all three ortho and para positions at once, giving 2,4,6-tribromoaniline as a white precipitate with bromine water. The amino group is then deleted by diazotising with NaNO₂ and HCl at 273 to 278 K and reducing the diazonium salt with hypophosphorous acid or ethanol, which replaces –N₂⁺ by –H.
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