Sandmeyer Reaction: Diazotisation, Mechanism, Gattermann Reaction and Exam Traps

ChemistryClass 12, AminesFor JEE and NEET aspirants

The Sandmeyer reaction replaces the diazonium group of an arenediazonium salt by Cl, Br or CN with the help of a copper(I) salt: $C u_{2} C l_{2}$ with HCl, $C u_{2} B r_{2}$ with HBr, or CuCN with KCN. Nitrogen gas is given off. The diazonium salt is made first, by diazotisation: a primary aromatic amine such as aniline is treated with sodium nitrite and hydrochloric acid at 273-278 K. In the closely related Gattermann reaction, copper powder with HCl or HBr does the job for Cl and Br. Both are in the Class 12 NCERT chapters on amines and on haloarenes, and JEE Main, JEE Advanced and NEET ask them in reaction sequences that start from aniline.

At a glance

Step 1, diazotisation:
$C_{6} H_{5} N H_{2} + NaN O_{2} + 2HCl 273-278 K C_{6} H_{5} N_{2} Cl + NaCl + 2 H_{2} O$
Sandmeyer, Cl:
$C_{6} H_{5} N_{2} Cl C u_{2} C l_{2} /HCl C_{6} H_{5} Cl + N_{2} ↑$
Sandmeyer, CN:
$C_{6} H_{5} N_{2} Cl CuCN/KCN C_{6} H_{5} CN + N_{2} ↑$
Gattermann:
$Ar N_{2} X Cu/HCl ArCl + N_{2} ↑ + CuX$
Copper: a copper(I) salt in Sandmeyer; copper powder in Gattermann. The Sandmeyer yield is better.
Not Sandmeyer: KI gives $ArI$ with no copper; $HB F_{4}$ then heat gives $ArF$ ; $H_{3} P O_{2}$ gives $ArH$ ; warm water gives $ArOH$ .

1. Step 1: diazotisation

Nitrous acid is unstable, so it is made in the mixture from sodium nitrite and hydrochloric acid. It converts the $N H_{2}$ group of a primary aromatic amine into the diazonium group, $N_{2}^{+}$ ; this is 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} Cl + 2 H_{2} O

Keep it cold. Benzenediazonium chloride is stable in cold aqueous solution but decomposes on warming, giving phenol and nitrogen. It is used at once, in solution, and not stored.
Primary aromatic amines only. The positive charge of an arenediazonium ion is spread over the ring, which keeps it together in the cold. A primary aliphatic amine gives a diazonium ion that loses $N_{2}$ at once (giving mainly an alcohol), so it cannot be used for a Sandmeyer reaction.
The amine is needed. Phenol, nitrobenzene or a halobenzene cannot be diazotised; a nitro group is reduced to $N H_{2}$ first (Sn/HCl or Fe/HCl).

2. Step 2: Sandmeyer and Gattermann

Copper(I) passes one electron to the diazonium ion, which loses nitrogen gas to give an aryl radical; the radical then picks up Cl, Br or CN from the copper, and copper(I) is formed again. The new group takes the exact place of the $N_{2}^{+}$ group on the ring.

Figure 1: The Sandmeyer reaction runs through an aryl radical. Each fish-hook arrow moves one electron; copper switches between +1 and +2 and is regenerated, and nitrogen gas leaves.

Reagent added to the diazonium salt	Product	Name
$C u_{2} C l_{2}$ / HCl	$ArCl$	Sandmeyer
$C u_{2} B r_{2}$ / HBr	$ArBr$	Sandmeyer
CuCN / KCN	$ArCN$	Sandmeyer
Cu powder / HCl	$ArCl$	Gattermann
Cu powder / HBr	$ArBr$	Gattermann
KI	$ArI$	Not Sandmeyer (no copper)
$HB F_{4}$ , then heat	$ArF$	Fluoroborate route (Balz-Schiemann)
$H_{3} P O_{2}$ / $H_{2} O$ , or ethanol	$ArH$	Reduction
$H_{2} O$ , warm	$ArOH$	Hydrolysis

The Sandmeyer reaction gives aryl nitriles, which cannot be made by direct substitution on the ring, and aryl halides with the halogen at a chosen position: the amino group can be used to steer other groups into place before it is replaced. Hydrolysis of the nitrile then gives a benzoic acid.

Exam Trick

"Sandmeyer uses a Salt, Gattermann uses the metal." Copper(I) chloride, bromide or cyanide means Sandmeyer; copper powder with HCl or HBr means Gattermann (Cl and Br only). The Gattermann-Koch reaction is something else: it puts CHO on benzene with CO and HCl (anhydrous $AlC l_{3}$ or CuCl), and there is no diazonium salt.

JEE Advanced

A radical, not an ion. The aryl cation would be very unstable, so the Sandmeyer reaction does not go through it. Copper(I) gives an electron, $N_{2}$ leaves from the neutral radical, and the aryl radical takes the halogen or cyanide from copper(II), which goes back to copper(I).

Why iodide needs no copper. Iodide ion is itself a good enough electron donor and nucleophile, so potassium iodide alone turns the diazonium salt into an aryl iodide.

Planning with the amino group. In a sequence, $N H_{2}$ is a strong ortho/para director that is later removed: diazotisation then $H_{3} P O_{2}$ replaces it by H, and diazotisation then CuBr replaces it by Br. Count the groups at every step. For example, nitrobenzene with $B r_{2}$ / $FeB r_{3}$ (one Br, meta), reduced to 3-bromoaniline, brominated with bromine water (three more Br, at 2, 4 and 6), diazotised and treated with CuBr (the fifth Br takes the place of $N H_{2}$ ) ends with five Br atoms on the ring.

3. Worked examples

Solved Example 1

Aniline is treated with

NaN O_{2}

and HCl at 273-278 K to give A; A with CuCN/KCN gives B; B is boiled with dilute acid to give C. Identify A, B and C.

Solution:

A is benzenediazonium chloride, $C_{6} H_{5} N_{2}^{+} C l^{-}$ . B is benzonitrile, $C_{6} H_{5} CN$ (Sandmeyer). C is benzoic acid, $C_{6} H_{5} COOH$ , from hydrolysis of the nitrile.

Solved Example 2

Benzenediazonium chloride is treated separately with (a)

C u_{2} B r_{2}

HBr

, (b) KI, (c) Cu/HCl, (d) CuCN/KCN, (e)

H_{3} P O_{2}

. Which of these are Sandmeyer reactions?

Solution:

(a) and (d): copper(I) salts giving bromobenzene and benzonitrile. (c) is the Gattermann reaction (chlorobenzene), (b) gives iodobenzene without copper, and (e) gives benzene.

Solved Example 3

Can chlorobenzene be made from phenol with

NaN O_{2}

, HCl and CuCl?

Solution:

No. Phenol has no $N H_{2}$ group, so no diazonium salt forms (nitrous acid attacks the ring of phenol instead). Start from aniline (diazotisation, then $C u_{2} C l_{2}$ / $HCl$ ), or chlorinate benzene with $C l_{2}$ and anhydrous $FeC l_{3}$ .

Solved Example 4

Nitrobenzene is reduced with Sn/HCl, diazotised at 273 K, treated with

C u_{2} C l_{2}

and then with sodium in dry ether. Name the final product.

Solution:

Aniline, then benzenediazonium chloride, then chlorobenzene (Sandmeyer). Sodium in dry ether joins two aryl groups (Fittig reaction): biphenyl, $C_{6} H_{5} - C_{6} H_{5}$ .

4. Exam traps

Watch out

Sandmeyer uses a copper(I) salt; Gattermann uses copper powder with HCl or HBr. The Sandmeyer yield is better.
Gattermann can give $ArCl$ and $ArBr$ , not $ArCN$ ; CuCN/KCN (Sandmeyer) is needed for the nitrile.
The Gattermann reaction (diazonium salt, Cu, HX) is not the Gattermann-Koch reaction (benzene, CO, HCl, anhydrous $AlC l_{3}$ or CuCl).
Aryl iodides need only KI, and aryl fluorides come from the fluoroborate; neither is a Sandmeyer reaction.
Diazotisation is done at 273-278 K; warm water turns the diazonium salt into phenol.
Only primary aromatic amines give diazonium salts that survive; aliphatic ones lose $N_{2}$ at once.
Phenol cannot be diazotised; the starting material must have an $N H_{2}$ group on the ring.
The new group goes exactly where the $N_{2}^{+}$ was; no substituent moves.

Past-paper questions on Sandmeyer reaction

Our past papers: JEE Main 2025-2026, JEE Advanced 2022-2026, NEET 2018-2019, 2022-2026. We found 14 questions in them that test this:

JEE Main 10 (2025-2026)
JEE Advanced 2 (2023, 2025)
NEET 2 (2022-2023)

Each question was read and checked by hand (list checked 2026-10-15). A question that shows its structures only as pictures can be missed by our search. Each opens in a new tab, with the question to try and our step-by-step solution.

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