Diazonium Salt

Q: What is diazotisation?

Diazotisation is the conversion of a primary aromatic amine into a diazonium salt by nitrous acid, made in the flask from sodium nitrite and a mineral acid, at 273 to 278 K. Aniline gives benzenediazonium chloride and water. The reaction needs an excess of the mineral acid.

Q: Why is diazotisation carried out at 273 to 278 K?

Benzenediazonium chloride is stable only in the cold. Above about 278 K it reacts with the water of the solution, loses nitrogen gas and gives phenol, so the yield falls and the product is impure. An ice bath therefore holds the temperature between 273 and 278 K.

Q: What is the difference between the Sandmeyer and Gattermann reactions?

Both replace the diazonium group by chlorine or bromine. The Sandmeyer reaction uses a copper(I) halide dissolved in the corresponding halogen acid, while the Gattermann reaction uses copper powder with the halogen acid. Gattermann is cheaper and simpler but gives a lower yield.

Q: Why are aromatic diazonium salts more stable than aliphatic ones?

In an aryl diazonium ion the positive charge is delocalised over the two nitrogen atoms and into the benzene ring, which lowers the energy of the ion. An alkyl diazonium ion has no such stabilisation, so it loses nitrogen immediately and leaves a carbocation behind.

Q: What is a coupling reaction?

A coupling reaction joins a diazonium ion to a strongly activated ring, such as that of a phenol or an aryl amine, keeping both nitrogen atoms. The product is an azo compound with an -N=N- bridge, which is coloured and is used as a dye. Coupling takes place at the para position.

Q: Why must coupling be done at a controlled pH?

In strongly alkaline solution the diazonium ion is converted into a diazohydroxide and then a diazotate ion, which do not couple. In strongly acidic solution an amine is protonated, so its ring is no longer activated. Phenols are coupled near pH 9 to 10 and amines near pH 4 to 7.

Q: Why are diazonium salts so important in synthesis?

They let groups be placed on a ring that cannot be introduced directly, such as iodine, fluorine, the cyano group and the hydroxyl group. The amino group can also hold a position while other groups are added and then be removed as nitrogen, which gives isomers direct substitution cannot reach.

Q: What does NEET ask about diazonium salts?

NEET follows NCERT: the preparation of benzenediazonium chloride and the temperature used, the Sandmeyer and Gattermann reactions, replacement of the diazonium group by iodine, fluorine, hydroxyl and hydrogen, and the coupling reactions with phenol and aniline that give the orange and yellow azo dyes. Simple reagent to product matching is the usual format.

Q: What does JEE Main ask about diazonium salts?

JEE Main sets reagent-product and conversion questions: which reagent gives which product from a diazonium salt, multi-step syntheses that use the amino group as a director and then remove it, the pH window for coupling, N-coupling to diazoaminobenzene, and the order of coupling reactivity of substituted diazonium ions.

ChemistryAminesFor NEET aspirants

A diazonium salt is the most useful thing that can be made from a primary aromatic amine. Aniline and nitrous acid at 273-278 K give benzenediazonium chloride, in which the $- N_{2}^{+}$ group can be replaced by $- Cl$ , $- Br$ , $- I$ , $- F$ , $- CN$ , $- OH$ , $- N O_{2}$ or even $- H$ , or kept to build an azo dye. This page covers the preparation, structure and every reaction of the diazonium salt asked in JEE Main and NEET, with mechanisms.

Key Formulas - Quick Reference

Diazotisation: $C_{6} H_{5} N H_{2} + HN O_{2} + HCl 273-278 K C_{6} H_{5} N_{2}^{+} C l^{-} + 2 H_{2} O$
Sandmeyer: $CuCl$ /HCl gives $C_{6} H_{5} Cl$ , $CuBr$ /HBr gives $C_{6} H_{5} Br$ , $CuCN$ /KCN gives $C_{6} H_{5} CN$
Gattermann: Cu powder with HCl or HBr gives $C_{6} H_{5} Cl$ or $C_{6} H_{5} Br$
Iodide needs no catalyst: $C_{6} H_{5} N_{2}^{+} C l^{-} + KI C_{6} H_{5} I + N_{2} + KCl$
Balz-Schiemann: $C_{6} H_{5} N_{2}^{+} B F_{4}^{-} heat C_{6} H_{5} F + B F_{3} + N_{2}$
Warm water gives phenol: $C_{6} H_{5} N_{2}^{+} C l^{-} + H_{2} O above 283 K C_{6} H_{5} OH + N_{2} + HCl$
Deamination: $C_{6} H_{5} N_{2}^{+} C l^{-} + H_{3} P O_{2} + H_{2} O C_{6} H_{6} + N_{2} + H_{3} P O_{3} + HCl$
Coupling with phenol (pH 9-10) gives $p - HO C_{6} H_{4} N = N C_{6} H_{5}$ , orange
Coupling with aniline (pH 4-7) gives $p - H_{2} N C_{6} H_{4} N = N C_{6} H_{5}$ , yellow
Azo compounds are cleaved: $ArN = NA r^{'} + 4 [H] SnC l_{2} /HCl ArN H_{2} + A r^{'} N H_{2}$

1. What Is a Diazonium Salt?

Diazonium salt: a salt of general formula

Ar N_{2}^{+} X^{-}

, in which two nitrogen atoms are joined to an aryl group and carry a positive charge, balanced by an anion such as

C l^{-}

HS O_{4}^{-}

B F_{4}^{-}

. The name comes from "di" (two), "azo" (nitrogen) and "ium" (positive).

The cation is linear at the two nitrogen atoms, which are $sp$ hybridised, and the positive charge is spread over both of them and into the ring:

Ar - N \equiv N^{+} \leftrightarrow Ar - N^{+} = N

Aryl diazonium salts are stable in cold solution because this delocalisation into the ring lowers their energy. Alkyl diazonium salts are not: they lose $N_{2}$ as soon as they form, leaving a carbocation (see Properties of Amines).
Benzenediazonium chloride is a colourless crystalline solid, readily soluble in water and stable in the cold, but it reacts with water when warmed.
Dry diazonium salts decompose violently, so they are never isolated; the solution is used at once.
Benzenediazonium fluoroborate $C_{6} H_{5} N_{2}^{+} B F_{4}^{-}$ is an exception: it is insoluble in water and stable at room temperature, which is what makes the Balz-Schiemann reaction practical.

Figure 1: Delocalisation into the ring is what keeps

Ar N_{2}^{+}

alive at 273-278 K;

R N_{2}^{+}

has no such help and loses

N_{2}

at once.

2. Diazotisation

Treating a primary aromatic amine with nitrous acid, made in the flask from $NaN O_{2}$ and a mineral acid, at 273-278 K is called diazotisation.

NaN O_{2} + HCl HN O_{2} + NaCl

C_{6} H_{5} N H_{2} + HN O_{2} + HCl 273-278 K C_{6} H_{5} N_{2}^{+} C l^{-} + 2 H_{2} O

The electrophile is the nitrosonium ion $N \equiv O^{+}$ . The amine nitrogen attacks it, a proton is lost, the N-nitrosoamine tautomerises to a diazohydroxide, and loss of water gives the diazonium ion (Figure 2).

Figure 2: The electrophile is

N \equiv O^{+}

; aniline attacks it and ends as

C_{6} H_{5} N_{2}^{+}

, which is only stable in the cold.

Why ice-cold? Above about 278 K the diazonium salt reacts with the water of the solution and gives phenol, with brisk evolution of $N_{2}$ .
Why excess mineral acid? It keeps the unreacted amine as its ammonium salt, so the amine cannot couple with the diazonium ion already formed; it also supplies the $H^{+}$ needed to make $HN O_{2}$ and then $N \equiv O^{+}$ .
Secondary and tertiary amines give N-nitroso compounds or ring nitrosation instead, and primary aliphatic amines give nitrogen gas at once, so only primary aromatic amines can be diazotised usefully.

3. Reactions in Which Nitrogen Is Lost

These reactions replace $- N_{2}^{+}$ by another group, with $N_{2}$ gas leaving. They are the reason diazonium salts matter: groups that cannot be introduced directly into a ring are put in this way (Figure 3).

3.1 Replacement by Halogen

Sandmeyer reaction: the diazonium solution is warmed with the copper(I) halide dissolved in the corresponding halogen acid.

C_{6} H_{5} N_{2}^{+} C l^{-} + CuCl HCl C_{6} H_{5} Cl + N_{2} + CuCl

C_{6} H_{5} N_{2}^{+} C l^{-} + CuBr HBr C_{6} H_{5} Br + N_{2} + CuCl

Gattermann reaction: copper powder with the halogen acid does the same job, more cheaply but in lower yield.

C_{6} H_{5} N_{2}^{+} C l^{-} Cu powder/HCl C_{6} H_{5} Cl + N_{2}

Iodide needs no catalyst: simply warming the diazonium solution with potassium iodide gives iodobenzene, which cannot be made by direct iodination.

C_{6} H_{5} N_{2}^{+} C l^{-} + KI C_{6} H_{5} I + N_{2} + KCl

Balz-Schiemann reaction: for fluorine, the diazonium chloride is first converted into the fluoroborate, which is filtered off, dried and heated.

C_{6} H_{5} N_{2}^{+} C l^{-} + HB F_{4} C_{6} H_{5} N_{2}^{+} B F_{4}^{-} + HCl

C_{6} H_{5} N_{2}^{+} B F_{4}^{-} heat C_{6} H_{5} F + B F_{3} + N_{2}

Exam Trick

Cu(I) salt is Sandmeyer, Cu metal is Gattermann, and I and F need neither. Iodine goes in with plain KI; fluorine needs the dry fluoroborate and heat. Trying to make $C_{6} H_{5} F$ by a Sandmeyer reaction with CuF is a favourite wrong option.

3.2 Replacement by Other Groups

By $- OH$ : warming the diazonium solution with water gives phenol. This is also the side reaction that spoils a diazotisation carried out too warm.

C_{6} H_{5} N_{2}^{+} C l^{-} + H_{2} O warm C_{6} H_{5} OH + N_{2} + HCl

By $- H$ (deamination): hypophosphorous acid, or ethanol, reduces the diazonium group away. This removes an amino group that was only there to direct another substituent.

C_{6} H_{5} N_{2}^{+} C l^{-} + H_{3} P O_{2} + H_{2} O C_{6} H_{6} + N_{2} + H_{3} P O_{3} + HCl

C_{6} H_{5} N_{2}^{+} C l^{-} + C_{2} H_{5} OH C_{6} H_{6} + N_{2} + C H_{3} CHO + HCl

By $- CN$ : copper(I) cyanide gives the nitrile, which can be hydrolysed to the acid or reduced to an amine with one extra carbon.

C_{6} H_{5} N_{2}^{+} C l^{-} + CuCN KCN C_{6} H_{5} CN + N_{2} + CuCl

By $- N O_{2}$ : the dry fluoroborate warmed with sodium nitrite and copper powder gives nitrobenzene.

C_{6} H_{5} N_{2}^{+} B F_{4}^{-} + NaN O_{2} Cu C_{6} H_{5} N O_{2} + NaB F_{4} + N_{2}

By an aryl group (Gomberg reaction): the diazonium salt and an arene in alkali give a biaryl.

C_{6} H_{5} N_{2}^{+} C l^{-} + C_{6} H_{6} NaOH biphenyl C_{6} H_{5} - C_{6} H_{5} + N_{2} + HCl

Figure 3: The diazonium group is a doorway: almost any group can replace

- N_{2}^{+}

, which is how substituents that cannot be introduced directly are placed on a ring.

4. Coupling Reactions: Both Nitrogen Atoms Are Kept

A diazonium ion is a weak electrophile, so it attacks only rings that are strongly activated, such as phenols and aryl amines. The product is an azo compound, $Ar - N = N - A r^{'}$ , and the reaction is called coupling. It is an electrophilic aromatic substitution and goes at the para position, or at the ortho position if para is blocked.

C_{6} H_{5} N_{2}^{+} C l^{-} + C_{6} H_{5} OH O H^{-}, pH 9-10 p-hydroxyazobenzene, orange HO C_{6} H_{4} N = N C_{6} H_{5} + HCl

C_{6} H_{5} N_{2}^{+} C l^{-} + C_{6} H_{5} N H_{2} H^{+}, pH 4-7 p-aminoazobenzene, yellow H_{2} N C_{6} H_{4} N = N C_{6} H_{5} + HCl

Why the pH matters. In strongly alkaline solution the diazonium ion adds hydroxide and becomes a diazohydroxide and then a diazotate ion, neither of which couples. In strongly acidic solution an amine is protonated to the aminium ion, whose ring is no longer activated. Coupling therefore works only in a narrow band: mildly alkaline for phenols, because the phenoxide ion couples much faster than phenol, and mildly acidic for amines (Figure 4).

Figure 4: Coupling needs a diazonium ion and an activated ring at the same time, which only happens in a narrow band of pH.

N-coupling. With an amine at low temperature in weakly acidic solution the diazonium ion may attack the nitrogen instead of the ring, giving a diazoamino compound. Warming it with aniline hydrochloride rearranges it to the C-coupled azo dye.

C_{6} H_{5} N_{2}^{+} C l^{-} + C_{6} H_{5} N H_{2} diazoaminobenzene C_{6} H_{5} N = N - NH C_{6} H_{5} + HCl

C_{6} H_{5} N = N - NH C_{6} H_{5} C_{6} H_{5} N H_{3}^{+} C l^{-}, heat p-aminoazobenzene H_{2} N C_{6} H_{4} N = N C_{6} H_{5}

Dyes. The $- N = N -$ bridge joins the two rings into one conjugated system, so azo compounds absorb visible light and are intensely coloured. Methyl orange, an indicator, is made by coupling diazotised sulphanilic acid with N,N-dimethylaniline.

Figure 5: The

- N = N -

bridge extends conjugation across both rings, which is why azo compounds are coloured and are used as dyes and indicators.

Azo compounds as a source of amines. Stannous chloride and HCl (or $N a_{2} S_{2} O_{4}$ ) break the $- N = N -$ bond and give two amines, which is used to identify a dye.

ArN = NA r^{'} + 4 [H] SnC l_{2} /HCl ArN H_{2} + A r^{'} N H_{2}

Reduction of the salt itself. Milder reduction of a diazonium salt, with $SnC l_{2}$ /HCl or sodium sulphite, stops at phenylhydrazine:

C_{6} H_{5} N_{2}^{+} C l^{-} + 4 [H] SnC l_{2} /HCl phenylhydrazine hydrochloride C_{6} H_{5} NHN H_{2} \cdot HCl

JEE Advanced

Coupling is slow because $Ar N_{2}^{+}$ is a weak electrophile, so its rate depends strongly on what the ring carries. Electron-withdrawing groups make the diazonium ion more reactive by increasing the positive charge on the terminal nitrogen, while electron-donating groups make it less reactive. The order of coupling reactivity is therefore $p - M e_{2} N - C_{6} H_{4} N_{2}^{+}$ < $p - C H_{3} O - C_{6} H_{4} N_{2}^{+}$ < $p - C H_{3} - C_{6} H_{4} N_{2}^{+}$ < $p - O_{2} N - C_{6} H_{4} N_{2}^{+}$ . On the partner ring the opposite holds: the more strongly activated the ring, the faster it couples, which is why phenoxide and 2-naphthol react so readily.

Figure 6: Coupling is electrophilic substitution at the para carbon; the weak electrophile

Ar N_{2}^{+}

reacts only with a strongly activated ring such as phenoxide.

5. Diazonium Salts in Synthesis

Two problems in aromatic synthesis are solved by diazonium chemistry.

Groups that cannot be introduced directly. $- I$ , $- F$ , $- CN$ and $- OH$ are all put on a ring through the diazonium salt, and so is $- H$ when an amino group must be removed.
Getting the right isomer. The $- N H_{2}$ group (or its acetyl derivative) directs an incoming group ortho and para, holds that position while other groups are added, and is then removed as $N_{2}$ . This gives substitution patterns that direct methods cannot reach, such as m-bromotoluene (Figure 7) and 1,3,5-tribromobenzene.

Figure 7: The

- N H_{2}

group directs the bromine into position and is then removed as

N_{2}

, giving a substitution pattern that direct methods cannot reach.

Figure 8: Decide which group must replace

- N_{2}^{+}

and the flowchart gives the reagent; the last branch keeps the nitrogen and makes an azo dye.

Figure 9: A diazonium salt either loses

N_{2}

(substitution) or keeps it (coupling); everything on this page is one of the two.

6. Solved Examples

Solved Example 1

Why must the diazotisation of aniline be carried out in ice-cold conditions?

Solution:

Benzenediazonium chloride is stable only between about 273 and 278 K. Above 278 K it reacts with the water of the solution, losing $N_{2}$ and giving phenol, so the yield of the diazonium salt falls and the product is contaminated.

C_{6} H_{5} N_{2}^{+} C l^{-} + H_{2} O above 283 K C_{6} H_{5} OH + N_{2} + HCl

Solved Example 2

Give a test that distinguishes aniline from diethylamine.

Solution:

Add a cold solution of benzenediazonium chloride in weakly acidic medium. Aniline is an aromatic amine with an activated ring, so it couples to give a yellow azo dye, p-aminoazobenzene. Diethylamine is a 2° aliphatic amine: it has no ring to couple and gives only a yellow oily nitrosamine with nitrous acid, not a dye.

Solved Example 3

How would you bring about the conversion of aniline into benzylamine in three steps?

Solution:

Diazotise, put a cyano group in with copper(I) cyanide, then reduce the nitrile. The extra carbon comes from the cyanide.

C_{6} H_{5} N H_{2} NaN O_{2} /HCl, 273-278 K C_{6} H_{5} N_{2}^{+} C l^{-} CuCN/KCN C_{6} H_{5} CN H_{2} /Ni C_{6} H_{5} C H_{2} N H_{2}

Solved Example 4

Convert acetanilide into 1,3-diiodo-5-nitrobenzene.

Solution:

The $- NHCOC H_{3}$ group directs para, so nitrate first, free the amine, put the two iodine atoms ortho to it, and finally remove the amino group as $N_{2}$ .

C_{6} H_{5} NHCOC H_{3} HN O_{3} / H_{2} S O_{4} N O_{2} C_{6} H_{4} NHCOC H_{3} H_{3} O^{+} N O_{2} C_{6} H_{4} N H_{2}

N O_{2} C_{6} H_{4} N H_{2} I_{2}, base N O_{2} C_{6} H_{2} I_{2} N H_{2} (i) NaN O_{2} /HCl (ii) H_{3} P O_{2} N O_{2} C_{6} H_{3} I_{2}

The last step is a deamination: the amino group has done its work of holding the two iodine atoms in the 3- and 5-positions.

Solved Example 5

Write the intermediates in the synthesis of m-bromotoluene from p-toluidine.

Solution:

Bromine must go meta to the methyl group, which direct bromination cannot do. Acetylate, brominate ortho to the amide, hydrolyse, diazotise and deaminate (Figure 7).

C H_{3} C_{6} H_{4} N H_{2} (C H_{3} CO)_{2} O C H_{3} C_{6} H_{4} NHCOC H_{3} B r_{2} C H_{3} C_{6} H_{3} (Br) NHCOC H_{3}

C H_{3} C_{6} H_{3} (Br) NHCOC H_{3} O H^{-}, heat C H_{3} C_{6} H_{3} (Br) N H_{2} NaN O_{2} / H_{2} S O_{4} C H_{3} C_{6} H_{3} (Br) N_{2}^{+}

C H_{3} C_{6} H_{3} (Br) N_{2}^{+} H_{3} P O_{2} C H_{3} C_{6} H_{4} Br

Solved Example 6

How would you prepare (a)

C_{6} H_{5} D

from aniline and (b) optically active sec-butylbenzene, using an amine intermediate?

Solution:

(a) Diazotise aniline and deaminate with deuterated hypophosphorous acid, which delivers D in place of the diazonium group.

C_{6} H_{5} N H_{2} NaN O_{2} /HCl, 273-278 K C_{6} H_{5} N_{2}^{+} C l^{-} D_{3} P O_{2} C_{6} H_{5} D

(b) Friedel-Crafts alkylation of benzene with 2-chlorobutane gives racemic sec-butylbenzene, and a hydrocarbon cannot be resolved. Nitrate it, reduce the nitro group, and the resulting p-amino compound is basic, so it can be resolved with an optically active acid such as tartaric acid. Deaminating the resolved amine through its diazonium salt with $H_{3} P O_{2}$ then gives optically active sec-butylbenzene.

C_{6} H_{6} C H_{3} C H_{2} CHClC H_{3} / AlC l_{3} C_{6} H_{5} CH (C H_{3}) C_{2} H_{5} (i) HN O_{3} / H_{2} S O_{4} (ii) Sn/HCl (iii) O H^{-} p - H_{2} N C_{6} H_{4} CH (C H_{3}) C_{2} H_{5}

Solved Example 7

When aniline is treated with benzenediazonium chloride at low temperature in weakly acidic medium, the product is
(A) 2-aminoazobenzene
(B) 4-aminoazobenzene
(C) diazoaminobenzene,

C_{6} H_{5} N = N - NH C_{6} H_{5}

(D) 3-aminoazobenzene

Solution:

Answer: (C). At low temperature the attack is on the amine nitrogen, which is the faster (kinetic) reaction, giving diazoaminobenzene. Warming it with aniline hydrochloride rearranges it to the thermodynamically favoured C-coupled product, 4-aminoazobenzene.

Solved Example 8

The end product (Z) of the sequence:

C_{6} H_{5} N_{2}^{+} C l^{-}

with Cu/KCN gives (X); (X) with

H^{+}

H_{2} O

gives (Y); (Y) with NaOH and CaO on heating gives (Z). (Z) is
(A) a cyanide
(B) a carboxylic acid
(C) an amine
(D) an arene

Solution:

Answer: (D). (X) is benzonitrile, (Y) is benzoic acid, and soda lime decarboxylates it to benzene, which is an arene.

C_{6} H_{5} N_{2}^{+} C l^{-} Cu/KCN C_{6} H_{5} CN H^{+} / H_{2} O C_{6} H_{5} COOH NaOH + CaO, heat C_{6} H_{6}

Solved Example 9

Arrange these diazonium ions in increasing order of reactivity in azo coupling: (I)

p - M e_{2} N C_{6} H_{4} N_{2}^{+}

, (II)

p - O_{2} N C_{6} H_{4} N_{2}^{+}

, (III)

p - C H_{3} O C_{6} H_{4} N_{2}^{+}

, (IV)

p - C H_{3} C_{6} H_{4} N_{2}^{+}

.
(A) I < IV < II < III
(B) I < III < IV < II
(C) III < I < II < IV
(D) III < I < IV < II

Solution:

Answer: (B). The diazonium ion is the electrophile, so anything that increases the positive charge on the terminal nitrogen speeds up coupling. Electron-donating groups slow it down in the order $- NM e_{2}$ (strongest donor) then $- OC H_{3}$ then $- C H_{3}$ , and the electron-withdrawing $- N O_{2}$ group makes it fastest.

Solved Example 10

Which reagent converts benzenediazonium chloride into benzene?
(A) water
(B) dilute acid
(C) hypophosphorous acid
(D) HCl

Solution:

Answer: (C). $H_{3} P O_{2}$ reduces the diazonium group and replaces it by hydrogen, giving benzene and $H_{3} P O_{3}$ . Ethanol does the same job. Water would give phenol instead.

Solved Example 11

A neutral compound (A),

C_{8} H_{9} ON

, with sodium hypobromite gives an acid-soluble substance (B),

C_{7} H_{9} N

. (B) with aqueous

NaN O_{2}

in dilute HCl at 273-278 K gives an ionic compound (C),

C_{7} H_{7} N_{2} Cl

, which gives a red dye with alkaline

β

-naphthol. (C) with potassium cuprocyanide gives (D),

C_{8} H_{7} N

, whose hydrolysis gives (E),

C_{8} H_{8} O_{2}

; (E) liberates

C O_{2}

from

NaHC O_{3}

and on oxidation gives (F),

C_{8} H_{6} O_{4}

. Nitration of (F) gives two isomeric mononitro derivatives. Identify (A) to (F).

Solution:

(A) is neutral and loses one carbon with hypobromite, so it is an amide; (B) is the amine and gives a diazonium salt (C) that couples to a dye, so (B) is aromatic. (F) has two $- COOH$ groups on the ring, and the fact that nitration gives exactly two mononitro products fixes the substitution as ortho.

(A) = 2-methylbenzamide
(B) = 2-methylaniline (o-toluidine)
(C) = 2-methylbenzenediazonium chloride
(D) = 2-methylbenzonitrile
(E) = 2-methylbenzoic acid
(F) = benzene-1,2-dicarboxylic acid (phthalic acid); its nitration gives the 3-nitro and 4-nitro acids

Solved Example 12

An aromatic compound (A),

C_{7} H_{7} N O_{2}

, dissolves in

NaHC O_{3}

with evolution of

C O_{2}

and, with

NaN O_{2}

/HCl, gives (B),

C_{7} H_{6} O_{3}

. (B) also dissolves in

NaHC O_{3}

, gives a colour with

FeC l_{3}

and can be made from phenol,

CC l_{4}

and NaOH. (B) with excess

HN O_{3}

gives (C),

C_{6} H_{3} N_{3} O_{7}

, and (C) with

PC l_{5}

gives (D), which returns (C) with water. Identify (A) to (D).

Solution:

(B) is made by the Reimer-Tiemann route from phenol, dissolves in $NaHC O_{3}$ and gives a colour with $FeC l_{3}$ , so it carries both $- COOH$ and $- OH$ : it is salicylic acid. Since (A) gives (B) on diazotisation followed by hydrolysis, the $- N H_{2}$ group of (A) has become the $- OH$ of (B).

(A) = 2-aminobenzoic acid (anthranilic acid)
(B) = 2-hydroxybenzoic acid (salicylic acid)
(C) = 2,4,6-trinitrophenol (picric acid); nitration also removes the $- COOH$ group
(D) = picryl chloride, $C_{6} H_{2} Cl N_{3} O_{6}$

Solved Example 13

An organic compound (A),

C_{7} H_{9} N

, with

NaN O_{2}

and HCl at room temperature gives (B),

C_{7} H_{8} O

. Both (A) and (B) give dibromo derivatives with bromine water. (A) with chloroform and alkali gives (C),

C_{8} H_{7} N

. Hydrolysis of (C), then

NaN O_{2}

/HCl in the cold and reaction with CuCN, gives (D), isomeric with (C). Hydrolysis of (D) followed by oxidation gives a dibasic acid that forms only one monohalo derivative. Identify (A) to (D).

Solution:

The dibasic acid giving only one monohalo derivative must be symmetrical: terephthalic acid. Working backwards, every ring is para-substituted.

(A) = 4-methylaniline (p-toluidine); with $HN O_{2}$ at room temperature the diazonium salt is hydrolysed at once
(B) = 4-methylphenol (p-cresol)
(C) = 4-methylphenyl isocyanide (carbylamine reaction)
(D) = 4-methylbenzonitrile, made from the amine through its diazonium salt and CuCN

Hydrolysis of (D) gives 4-methylbenzoic acid, whose oxidation gives terephthalic acid.

Solved Example 14

How would you prepare 1,3,5-tribromobenzene from aniline?

Solution:

Direct bromination of benzene gives the 1,2- and 1,4-isomers, never 1,3,5. Aniline, however, brominates at all three free positions, and the amino group can then be removed.

C_{6} H_{5} N H_{2} 3B r_{2}, water C_{6} H_{2} B r_{3} N H_{2}

C_{6} H_{2} B r_{3} N H_{2} NaN O_{2} /HCl C_{6} H_{2} B r_{3} N_{2}^{+} C l^{-} H_{3} P O_{2} C_{6} H_{3} B r_{3}

Practice Questions

Why is an excess of mineral acid used in the diazotisation of an arylamine?Answer: it keeps the unreacted amine as its salt, so the amine cannot couple with the diazonium ion formed, and it supplies the $H^{+}$ needed to generate $HN O_{2}$ and $N \equiv O^{+}$
Why are aryl diazonium ions more stable than alkyl diazonium ions?Answer: the positive charge of $Ar N_{2}^{+}$ is delocalised into the ring, while an alkyl diazonium ion has no such stabilisation and loses $N_{2}$ at once to give a carbocation
Nitrobenzene is reduced with Sn/HCl to A; A with $HN O_{2}$ /HCl at 273 K gives B; B with $NaN O_{2}$ / $NaHC O_{3}$ and $C u_{2} O$ gives C. Identify A, B and C.Answer: A = aniline; B = benzenediazonium chloride; C = nitrobenzene (the diazonium group is replaced by $- N O_{2}$ )
In the scheme: (A) with $B r_{2}$ / $C H_{3} COOH$ gives (B); (B) with (i) $NaN O_{2}$ / $H_{2} S O_{4}$ (ii) CuBr gives (Z); (X) with $HN O_{3}$ / $H_{2} S O_{4}$ gives (Y); (Y) with $B r_{2}$ /Fe gives (Z); (Z) with Sn/HCl gives 3,4,5-tribromoaniline hydrochloride. Identify A, B, X, Y and Z.Answer: A = 4-nitroaniline; B = 2,6-dibromo-4-nitroaniline; X = 1,2-dibromobenzene; Y = 1,2-dibromo-4-nitrobenzene; Z = 1,2,3-tribromo-5-nitrobenzene
Why can fluorobenzene not be prepared by a Sandmeyer reaction?Answer: there is no useful copper(I) fluoride route; fluorine is introduced by the Balz-Schiemann reaction, in which the dry diazonium fluoroborate is heated
Convert aniline into (a) phenol, (b) benzonitrile, (c) iodobenzene.Answer: diazotise in each case, then (a) warm with water, (b) treat with CuCN/KCN, (c) warm with KI
What happens when benzenediazonium chloride is (a) warmed with water and (b) treated with ethanol?Answer: (a) phenol, $N_{2}$ and HCl; (b) benzene, $N_{2}$ , ethanal and HCl (ethanol acts as the reducing agent)
Why is coupling with phenol carried out in mildly alkaline solution but coupling with aniline in mildly acidic solution?Answer: phenol couples as the more reactive phenoxide ion, which needs alkali, but above pH 10 the diazonium ion becomes a diazotate; an amine needs acid to stay in solution, but below pH 4 it is protonated to the unreactive aminium ion
Give the product of coupling benzenediazonium chloride with p-cresol.Answer: the para position is blocked by $- C H_{3}$ , so coupling goes ortho to the $- OH$ group: 4-methyl-2-(phenyldiazenyl)phenol
Name the two compounds coupled to make methyl orange.Answer: diazotised sulphanilic acid and N,N-dimethylaniline
What is formed when an azo compound $ArN = NA r^{'}$ is reduced with $SnC l_{2}$ /HCl?Answer: the $- N = N -$ bond is cleaved and two amines, $ArN H_{2}$ and $A r^{'} N H_{2}$ , are formed; this is used to identify a dye
Why are diazonium salts never isolated in the dry state?Answer: dry diazonium salts are unstable and decompose explosively; the cold solution is used immediately, except for the fluoroborate, which is stable enough to filter and dry

Common Mistakes to Avoid

Watch out

Diazotising above 278 K. The salt then reacts with water and gives phenol.
Writing a stable diazonium salt from an aliphatic amine, or from a 2° or 3° amine. Only primary aromatic amines give useful diazonium salts.
Using a Sandmeyer reaction for iodine or fluorine. KI works alone, and fluorine needs the Balz-Schiemann route through the fluoroborate.
Coupling in strongly alkaline or strongly acidic solution. The diazotate ion and the aminium ion do not couple.
Expecting benzene or toluene to couple with a diazonium salt. Only strongly activated rings, such as phenols and aryl amines, are reactive enough.
Putting the azo group meta. Coupling goes para, and ortho only when para is blocked.
Forgetting that $N_{2}$ leaves in every substitution reaction, so the product keeps none of the diazonium nitrogen.
Mixing up the Gattermann reaction (Cu powder with HX on a diazonium salt) with the Gattermann-Koch reaction (CO and HCl, which formylates an arene).
Isolating a dry diazonium salt. Only the fluoroborate is safe to filter and dry.

Frequently Asked Questions

What is diazotisation?

Diazotisation is the conversion of a primary aromatic amine into a diazonium salt by nitrous acid, made in the flask from sodium nitrite and a mineral acid, at 273 to 278 K. Aniline gives benzenediazonium chloride and water. The reaction needs an excess of the mineral acid.

Why is diazotisation carried out at 273 to 278 K?

Benzenediazonium chloride is stable only in the cold. Above about 278 K it reacts with the water of the solution, loses nitrogen gas and gives phenol, so the yield falls and the product is impure. An ice bath therefore holds the temperature between 273 and 278 K.

What is the difference between the Sandmeyer and Gattermann reactions?

Both replace the diazonium group by chlorine or bromine. The Sandmeyer reaction uses a copper(I) halide dissolved in the corresponding halogen acid, while the Gattermann reaction uses copper powder with the halogen acid. Gattermann is cheaper and simpler but gives a lower yield.

Why are aromatic diazonium salts more stable than aliphatic ones?

In an aryl diazonium ion the positive charge is delocalised over the two nitrogen atoms and into the benzene ring, which lowers the energy of the ion. An alkyl diazonium ion has no such stabilisation, so it loses nitrogen immediately and leaves a carbocation behind.

What is a coupling reaction?

A coupling reaction joins a diazonium ion to a strongly activated ring, such as that of a phenol or an aryl amine, keeping both nitrogen atoms. The product is an azo compound with an -N=N- bridge, which is coloured and is used as a dye. Coupling takes place at the para position.

Why must coupling be done at a controlled pH?

In strongly alkaline solution the diazonium ion is converted into a diazohydroxide and then a diazotate ion, which do not couple. In strongly acidic solution an amine is protonated, so its ring is no longer activated. Phenols are coupled near pH 9 to 10 and amines near pH 4 to 7.

Why are diazonium salts so important in synthesis?

They let groups be placed on a ring that cannot be introduced directly, such as iodine, fluorine, the cyano group and the hydroxyl group. The amino group can also hold a position while other groups are added and then be removed as nitrogen, which gives isomers direct substitution cannot reach.

What does NEET ask about diazonium salts?

NEET follows NCERT: the preparation of benzenediazonium chloride and the temperature used, the Sandmeyer and Gattermann reactions, replacement of the diazonium group by iodine, fluorine, hydroxyl and hydrogen, and the coupling reactions with phenol and aniline that give the orange and yellow azo dyes. Simple reagent to product matching is the usual format.

What does JEE Main ask about diazonium salts?

JEE Main sets reagent-product and conversion questions: which reagent gives which product from a diazonium salt, multi-step syntheses that use the amino group as a director and then remove it, the pH window for coupling, N-coupling to diazoaminobenzene, and the order of coupling reactivity of substituted diazonium ions.

Previous year questions on Diazonium Salt

16 questions from past papers, each with a step-by-step solution.

Show all 16 questions

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