Chapter 9 · Amines

9.3 Diazonium Salts

Cold nitrous acid turns aniline into benzenediazonium chloride, the switchboard of aromatic chemistry: its –N₂⁺ group can be swapped for Cl, Br, I, F, CN, OH, NO₂ or H, or kept and coupled to phenols and amines to make the azo dyes.

The orange of a fizzy drink, the yellow of a packet of instant noodles, the red of a cotton shirt and the colour change of methyl orange in a titration flask all come from the same kind of molecule: two aromatic rings joined by a $\mathrm{{-}N{=}N{-}}$ bridge, an azo compound. Azo dyes make up more than half of all the synthetic dyes produced in the world. They are made in two steps from aromatic amines, and the key intermediate, discovered by the German chemist Peter Griess in 1858, is a diazonium salt.

Diazonium salts are useful for a second reason. Their $\mathrm{{-}N_{2}^{+}}$ group is an excellent leaving group: it departs as nitrogen gas, one of the most stable molecules there is, and can be replaced by groups that are hard or impossible to put on a benzene ring directly, such as fluorine, iodine, cyanide or a hydroxyl group. Combined with the directing effects of an amino or nitro group, this lets chemists place substituents in patterns that ordinary electrophilic substitution cannot reach.

Diazonium Salts and Diazotisation

📘 Definition 9.8 — Diazonium salts

A diazonium salt has the general formula $\mathrm{R{-}N_{2}^{+}X^{-}}$, where R is an alkyl or aryl group and $\mathrm{X^{-}}$ is an anion such as $\mathrm{Cl^{-}}$, $\mathrm{Br^{-}}$, $\mathrm{HSO_{4}^{-}}$ or $\mathrm{BF_{4}^{-}}$. The $\mathrm{{-}N_{2}^{+}}$ group, $\mathrm{{-}\overset{+}{N}{\equiv}N}$, is the diazonium group. Salts are named by adding "diazonium" to the name of the parent hydrocarbon, followed by the anion: $\mathrm{C_{6}H_{5}N_{2}^{+}Cl^{-}}$ is benzenediazonium chloride, $\mathrm{C_{6}H_{5}N_{2}^{+}HSO_{4}^{-}}$ benzenediazonium hydrogensulphate.

Alkyldiazonium salts are so unstable that they decompose as soon as they are formed (Section 9.2: $\mathrm{N_{2}}$ and an alcohol). Arenediazonium salts survive for a while in cold solution, because the positive charge is spread over the benzene ring by resonance.

📐 Rule 9.11 — Diazotisation and the properties of benzenediazonium chloride

Diazotisation is the conversion of a primary aromatic amine into its diazonium salt by nitrous acid, made in the mixture from sodium nitrite and hydrochloric acid, in an ice bath at $273\text{–}278\ \mathrm{K}$:

$$\mathrm{C_{6}H_{5}NH_{2}+NaNO_{2}+2HCl}$$ $$\mathrm{\xrightarrow{273-278\ K}C_{6}H_{5}N_{2}^{+}Cl^{-}+NaCl+2H_{2}O}$$
  • The temperature must be kept low: above about $278\ \mathrm{K}$ the salt reacts with water, giving phenol and nitrogen. Because the salt is unstable, it is not stored but used at once, in solution.
  • Benzenediazonium chloride is a colourless crystalline solid, readily soluble in water. It is stable in cold water but reacts when warmed, and the dry solid decomposes easily (it can explode).
  • Benzenediazonium fluoroborate, $\mathrm{C_{6}H_{5}N_{2}^{+}BF_{4}^{-}}$, is insoluble in water and stable at room temperature.
✏️ Example 1 — How much sodium nitrite?
A student diazotises $4.65\ \mathrm{g}$ of aniline. (a) Write the equation. (b) What mass of $\mathrm{NaNO_{2}}$ and what volume of $2.00\ \mathrm{mol\,L^{-1}}$ HCl are needed in theory? (c) Why is the beaker kept in ice, and why is the product not dried? Use $M$: aniline $93.13$, $\mathrm{NaNO_{2}}$ $69.00\ \mathrm{g\,mol^{-1}}$.
(a) $\mathrm{C_{6}H_{5}NH_{2}+NaNO_{2}+2HCl}$ $\mathrm{\rightarrow C_{6}H_{5}N_{2}^{+}Cl^{-}+NaCl+2H_{2}O}$, at $273\text{–}278\ \mathrm{K}$.
(b) Moles $n(\text{aniline})=\dfrac{4.65}{93.13}=0.04993\ \mathrm{mol}$; ratio aniline : $\mathrm{NaNO_{2}}$ : HCl $=1:1:2$.
(b) Amounts $m(\mathrm{NaNO_{2}})=0.04993\times69.00=3.45\ \mathrm{g}$; $n(\mathrm{HCl})=0.09986\ \mathrm{mol}$, volume $=\dfrac{0.09986}{2.00}=0.0499\ \mathrm{L}=49.9\ \mathrm{mL}$. (In practice a slight excess of acid is used.)
(c) Above about $278\ \mathrm{K}$ the diazonium salt reacts with water to give phenol and $\mathrm{N_{2}}$; the dry salt is shock-sensitive and can explode. It is used straight away, in its cold solution.

Replacing the Diazonium Group

In the reactions below the $\mathrm{{-}N_{2}^{+}}$ group leaves as nitrogen gas and another atom or group takes its place on the ring. Several of them introduce groups that cannot be put on a benzene ring by electrophilic substitution at all.

📐 Rule 9.12 — Displacement of nitrogen ($\mathrm{ArN_{2}^{+}X^{-}\rightarrow Ar{-}Y+N_{2}}$)
Group introducedReagent and conditionsName
$\mathrm{{-}Cl}$, $\mathrm{{-}Br}$$\mathrm{Cu_{2}Cl_{2}/HCl}$ or $\mathrm{Cu_{2}Br_{2}/HBr}$Sandmeyer reaction
$\mathrm{{-}Cl}$, $\mathrm{{-}Br}$copper powder with HCl or HBrGattermann reaction
$\mathrm{{-}CN}$$\mathrm{CuCN/KCN}$Sandmeyer reaction
$\mathrm{{-}I}$KI(aq), warm—
$\mathrm{{-}F}$$\mathrm{HBF_{4}}$, then heat the dry fluoroborateBalz–Schiemann reaction
$\mathrm{{-}H}$$\mathrm{H_{3}PO_{2}}$ (phosphinic acid) and water, or ethanolreduction
$\mathrm{{-}OH}$warm water (about $283\ \mathrm{K}$ and above)hydrolysis
$\mathrm{{-}NO_{2}}$fluoroborate + $\mathrm{NaNO_{2}}$, copper, heat—

Typical equations (with $\mathrm{Ar}=\mathrm{C_{6}H_{5}}$):

$$\mathrm{ArN_{2}^{+}Cl^{-}\xrightarrow{Cu_{2}Cl_{2}/HCl}ArCl+N_{2}}$$ $$\mathrm{ArN_{2}^{+}Cl^{-}\xrightarrow{CuCN/KCN}ArCN+N_{2}}$$ $$\mathrm{ArN_{2}^{+}Cl^{-}+KI\rightarrow ArI+KCl+N_{2}}$$ $$\mathrm{ArN_{2}^{+}Cl^{-}+HBF_{4}\rightarrow ArN_{2}^{+}BF_{4}^{-}+HCl}$$ $$\mathrm{ArN_{2}^{+}BF_{4}^{-}\xrightarrow{\Delta}ArF+BF_{3}+N_{2}}$$ $$\mathrm{ArN_{2}^{+}Cl^{-}+H_{3}PO_{2}+H_{2}O}$$ $$\mathrm{\rightarrow ArH+N_{2}+H_{3}PO_{3}+HCl}$$ $$\mathrm{ArN_{2}^{+}Cl^{-}+CH_{3}CH_{2}OH}$$ $$\mathrm{\rightarrow ArH+N_{2}+CH_{3}CHO+HCl}$$ $$\mathrm{ArN_{2}^{+}Cl^{-}+H_{2}O\xrightarrow{warm}ArOH+N_{2}+HCl}$$ $$\mathrm{ArN_{2}^{+}BF_{4}^{-}+NaNO_{2}}$$ $$\mathrm{\xrightarrow{Cu,\ \Delta}ArNO_{2}+NaBF_{4}+N_{2}}$$

The Sandmeyer reaction usually gives a better yield than the Gattermann reaction. Iodine is not easily introduced into a benzene ring directly, but iodide reacts readily with the diazonium salt; fluorine cannot be introduced directly at all (fluorine is too reactive), so the Balz–Schiemann route is the standard way to fluoroarenes.

✏️ Example 2 — One salt, many products
Benzenediazonium chloride is treated separately with (a) $\mathrm{Cu_{2}Br_{2}/HBr}$; (b) $\mathrm{CuCN/KCN}$; (c) KI; (d) $\mathrm{HBF_{4}}$, then heat; (e) $\mathrm{H_{3}PO_{2}}$ and water; (f) water at about $300\ \mathrm{K}$. Name each organic product. (g) In (f), $1.00\ \mathrm{g}$ of the salt ($M=140.6\ \mathrm{g\,mol^{-1}}$) is used. What volume of $\mathrm{N_{2}}$ is evolved at $273\ \mathrm{K}$ and $1\ \mathrm{bar}$ ($22.7\ \mathrm{L\,mol^{-1}}$), and what mass of phenol ($M=94.11$) can form?
(a)–(c) (a) bromobenzene (Sandmeyer); (b) benzonitrile, $\mathrm{C_{6}H_{5}CN}$; (c) iodobenzene.
(d)–(f) (d) fluorobenzene (Balz–Schiemann); (e) benzene; (f) phenol.
(g) Moles $n=\dfrac{1.00}{140.6}=7.11\times10^{-3}\ \mathrm{mol}$; one $\mathrm{N_{2}}$ and one phenol per salt.
(g) Answers $V(\mathrm{N_{2}})=7.11\times10^{-3}\times22.7$ $=0.161\ \mathrm{L}=161\ \mathrm{mL}$; $m(\text{phenol})=7.11\times10^{-3}\times94.11$ $=0.669\ \mathrm{g}$.
Check $\mathrm{C_{6}H_{5}N_{2}Cl+H_{2}O\rightarrow C_{6}H_{5}OH+N_{2}+HCl}$ balances: C 6, H 7, N 2, O 1, Cl 1 on each side.
Interactive 1 — The Diazonium Hub: One Salt, Ten Products

Benzenediazonium chloride sits at the centre. Choose a reagent (or click a product box): the reagent travels out along its spoke, a molecule of $\mathrm{N_{2}}$ escapes (except in the two coupling reactions, which keep both nitrogens), and the product lights up. The readouts give the conditions, the name of the reaction and the balanced equation (Ph = $\mathrm{C_{6}H_{5}}$). Press Play to tour the whole hub.

reagent—
product—
reaction—
N₂ lost?—

Coupling Reactions and Azo Dyes

The diazonium ion is a weak electrophile. It cannot attack benzene itself, but it does attack rings made very electron-rich by an $\mathrm{{-}OH}$ (better, $\mathrm{{-}O^{-}}$) or $\mathrm{{-}NH_{2}}$ group. This time the nitrogen atoms are kept, and they end up as a bridge between two rings.

📘 Definition 9.9 — Coupling reactions and azo dyes

In a coupling reaction an arenediazonium ion substitutes the hydrogen at the para position of a phenol or an arylamine (an electrophilic substitution), giving an azo compound, $\mathrm{Ar{-}N{=}N{-}Ar'}$. The extended system of conjugated double bonds across both rings and the $\mathrm{N{=}N}$ group absorbs visible light, so azo compounds are coloured and many are used as azo dyes.

With phenol, in mildly alkaline solution (phenol reacts as the phenoxide ion), the product is the orange dye 4-hydroxyazobenzene ($p$-hydroxyazobenzene; IUPAC 4-(phenyldiazenyl)phenol):

$$\mathrm{C_{6}H_{5}N_{2}^{+}Cl^{-}+C_{6}H_{5}OH\xrightarrow{OH^{-}}}$$ $$\mathrm{4{-}HOC_{6}H_{4}{-}N{=}N{-}C_{6}H_{5}+Cl^{-}+H_{2}O}$$

With aniline, in mildly acidic solution, the product is the yellow dye 4-aminoazobenzene ($p$-aminoazobenzene; 4-(phenyldiazenyl)aniline):

$$\mathrm{C_{6}H_{5}N_{2}^{+}Cl^{-}+C_{6}H_{5}NH_{2}\xrightarrow{H^{+}}}$$ $$\mathrm{4{-}H_{2}NC_{6}H_{4}{-}N{=}N{-}C_{6}H_{5}+HCl}$$

With naphthalen-2-ol (2-naphthol) in alkali, coupling at C-1 gives an orange-red precipitate. This is the usual test for a primary aromatic amine: diazotise it in the cold and add the solution to alkaline naphthalen-2-ol.

Interactive 2 — Azo Coupling: Find the pH Window

Coupling needs two things at the same time: the reactive form of the partner (the phenoxide ion of a phenol, or the free amine, not its ammonium ion) and the diazonium ion itself, which in strongly alkaline solution is converted into non-electrophilic forms. Move the pH: the dashed curves show the fraction of each species, the shaded curve their product, the relative rate of coupling. For aniline, coupling on nitrogen instead of the ring also takes over above about pH 6. This is an illustrative model: the $\mathrm{p}K_{a}$ values of the partners are real, the diazonium and N-coupling switch points are rough. Press Play to sweep the pH.

9.5
reactive partner—
diazonium ion left—
relative rate—
dye—
🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — the mechanism of coupling

In JEE Main, NEET and JEE Advanced 2026 (diazonium salts and electrophilic substitution; JEE Advanced names azo coupling); not required by the CBSE syllabus.

The terminal nitrogen of $\mathrm{Ar{-}\overset{+}{N}{\equiv}N}$ is the electrophilic atom. It attacks the ring carbon para to $\mathrm{{-}O^{-}}$ or $\mathrm{{-}NH_{2}}$, forming a carbocation intermediate (an arenium ion) that is strongly stabilised by the donor group; loss of $\mathrm{H^{+}}$ restores the aromatic ring. The para position is preferred because it is activated and far from the donor group; if it is already occupied, coupling occurs ortho: 4-methylphenol couples at C-2. Benzene, chlorobenzene or nitrobenzene do not couple at all: the diazonium ion is too weak an electrophile for an unactivated ring.

📐 Rule 9.13 — Diazonium salts in synthesis
  • Groups that cannot be introduced directly: $\mathrm{{-}F}$, $\mathrm{{-}I}$ (easily), $\mathrm{{-}CN}$, $\mathrm{{-}OH}$ on a ring that already carries other groups. Benzonitrile can then be hydrolysed to benzoic acid or reduced to phenylmethanamine.
  • Substitution patterns that direct substitution cannot give: use $\mathrm{{-}NH_{2}}$ (strongly o/p-directing) or $\mathrm{{-}NO_{2}}$ (meta-directing, then reduced to $\mathrm{{-}NH_{2}}$) to steer an electrophile, then convert the amino group into the group you want, or remove it altogether with $\mathrm{H_{3}PO_{2}}$.
  • Azo dyes by coupling with phenols, naphthols and arylamines.
🧭 Problem-solving strategy — Planning a route through a diazonium salt
  1. Look at the target. Is there a group (F, I, CN, OH, or a pattern such as 1,3,5) that electrophilic substitution cannot deliver?
  2. Work backwards: that position once carried $\mathrm{{-}NH_{2}}$, which came from $\mathrm{{-}NO_{2}}$ (Section 9.1).
  3. Check the directing effects at each step: $\mathrm{{-}NH_{2}}$ and $\mathrm{{-}NHCOCH_{3}}$ direct o/p, $\mathrm{{-}NO_{2}}$ directs meta.
  4. Diazotise last, at $273\text{–}278\ \mathrm{K}$, and convert straight away.
✏️ Example 3 — Routes through the diazonium salt
Show how to make (a) fluorobenzene from aniline; (b) 1,3,5-tribromobenzene from aniline; (c) 3-bromophenol from benzene; (d) benzoic acid from aniline.
(a) Aniline $\xrightarrow{\mathrm{NaNO_{2}/HCl},\ 273\ \mathrm{K}}$ $\mathrm{C_{6}H_{5}N_{2}^{+}Cl^{-}}$ $\xrightarrow{\mathrm{HBF_{4}}}$ $\mathrm{C_{6}H_{5}N_{2}^{+}BF_{4}^{-}}$ $\xrightarrow{\Delta}$ $\mathrm{C_{6}H_{5}F}$ (Balz–Schiemann).
(b) Bromine water puts Br at 2, 4 and 6 (o/p to $\mathrm{NH_{2}}$): 2,4,6-tribromoaniline. Diazotise ($\mathrm{NaNO_{2}/HCl}$, $273\ \mathrm{K}$), then $\mathrm{H_{3}PO_{2}}$ replaces $\mathrm{N_{2}^{+}}$ by H. The three Br are now at 1, 3, 5, all meta to each other, a pattern that bromination of benzene cannot give.
(c) Benzene $\xrightarrow{\mathrm{HNO_{3}/H_{2}SO_{4}}}$ nitrobenzene $\xrightarrow{\mathrm{Br_{2}/FeBr_{3}}}$ 1-bromo-3-nitrobenzene ($\mathrm{NO_{2}}$ directs meta) $\xrightarrow{\mathrm{Sn/HCl}}$ 3-bromoaniline $\xrightarrow{\mathrm{NaNO_{2}/HCl},\ 273\ \mathrm{K}}$ diazonium salt $\xrightarrow{\mathrm{H_{2}O},\ \text{warm}}$ 3-bromophenol.
(d) Diazotise; $\mathrm{CuCN/KCN}$ gives benzonitrile; hydrolysis ($\mathrm{H_{3}O^{+}}$, heat) gives $\mathrm{C_{6}H_{5}COOH}$.
Check In (c), brominating phenol directly would give 2- and 4-bromophenol (OH directs o/p); the nitro group is used to reach the meta position.
✏️ Example 4 — Making an azo dye
$0.930\ \mathrm{g}$ of aniline is diazotised and the cold solution is poured into a solution of phenol in dilute NaOH. (a) Write the equation for the coupling and name the product. (b) Why must the phenol solution be alkaline, but not strongly so? (c) What mass of dye ($M=198.2\ \mathrm{g\,mol^{-1}}$) forms if the overall yield is $85.0\%$?
(a) $\mathrm{C_{6}H_{5}N_{2}^{+}Cl^{-}+C_{6}H_{5}OH\xrightarrow{OH^{-}}}$ $\mathrm{4{-}HOC_{6}H_{4}N{=}NC_{6}H_{5}+Cl^{-}+H_{2}O}$: 4-hydroxyazobenzene, an orange dye.
(b) The reactive partner is the phenoxide ion, a much more electron-rich ring than phenol, so the solution must be alkaline enough to form it ($\mathrm{p}K_{a}$ of phenol $\approx10$). In strongly alkaline solution, however, the diazonium ion itself is converted into forms that do not couple, so a mildly alkaline solution is best (Interactive 2).
(c) $n=\dfrac{0.930}{93.13}=9.99\times10^{-3}\ \mathrm{mol}$; theoretical mass $=9.99\times10^{-3}\times198.2=1.98\ \mathrm{g}$; at $85.0\%$: $1.68\ \mathrm{g}$.
✏️ Example 5 — Identify the compounds
An aromatic compound A, $\mathrm{C_{6}H_{7}N}$, dissolves in dilute HCl. With $\mathrm{NaNO_{2}}$ and HCl at $273\ \mathrm{K}$ it gives B, which with alkaline naphthalen-2-ol gives an orange-red precipitate. B with $\mathrm{CuCN/KCN}$ gives C, $\mathrm{C_{7}H_{5}N}$, which on hydrolysis gives D, $\mathrm{C_{7}H_{6}O_{2}}$, an acid. B warmed with water gives E, which gives a violet colour with neutral $\mathrm{FeCl_{3}}$. Identify A to E.
A $\mathrm{C_{6}H_{7}N}$, basic, aromatic, gives a diazonium salt in the cold that couples: a primary arylamine, aniline $\mathrm{C_{6}H_{5}NH_{2}}$.
B, C B: benzenediazonium chloride. C: benzonitrile, $\mathrm{C_{6}H_{5}CN}$ ($\mathrm{C_{7}H_{5}N}$ ✓).
D, E D: benzoic acid, $\mathrm{C_{6}H_{5}COOH}$ ($\mathrm{C_{7}H_{6}O_{2}}$ ✓). E: phenol (violet with $\mathrm{FeCl_{3}}$).
⚠️ Common mistakes with diazonium salts

1. Diazotising an aliphatic amine. Ethanamine gives $\mathrm{N_{2}}$ and ethanol at once; only aromatic primary amines give usable diazonium salts.

2. Forgetting the temperature. Diazotisation is done at $273\text{–}278\ \mathrm{K}$; "warm" turns the salt into phenol.

3. Mixing up Sandmeyer and Gattermann. Sandmeyer: copper(I) salt ($\mathrm{Cu_{2}Cl_{2}}$, $\mathrm{Cu_{2}Br_{2}}$, CuCN). Gattermann: copper powder with HX.

4. Losing nitrogen in coupling. Coupling keeps both N atoms ($\mathrm{{-}N{=}N{-}}$); all the replacement reactions lose them as $\mathrm{N_{2}}$.

5. Coupling at the wrong place. The new bond forms para to $\mathrm{{-}OH}$ or $\mathrm{{-}NH_{2}}$, not meta.

Exercises

I
Diazotisation and properties. $M$ ($\mathrm{g\,mol^{-1}}$): 4-methylaniline 107.15, $\mathrm{NaNO_{2}}$ 69.00.
  • 1 Write the equation for the diazotisation of 4-methylaniline and name the product.
  • 2 Why are arenediazonium salts more stable than alkanediazonium salts?
  • 3 Why is diazotisation carried out at $273\text{–}278\ \mathrm{K}$? What happens if the solution is warmed?
  • 4 Name $\mathrm{C_{6}H_{5}N_{2}^{+}HSO_{4}^{-}}$ and $\mathrm{4{-}ClC_{6}H_{4}N_{2}^{+}Cl^{-}}$.
  • 5 What mass of sodium nitrite is needed to diazotise $10.7\ \mathrm{g}$ of 4-methylaniline?
  • 6 Which of these give a diazonium salt that can be used at $273\ \mathrm{K}$: aniline, ethanamine, phenylmethanamine, 4-nitroaniline, $N$-methylaniline?
1 $\mathrm{4{-}CH_{3}C_{6}H_{4}NH_{2}+NaNO_{2}+2HCl}$ $\mathrm{\rightarrow4{-}CH_{3}C_{6}H_{4}N_{2}^{+}Cl^{-}+NaCl+2H_{2}O}$: 4-methylbenzenediazonium chloride.
2 In the arenediazonium ion the positive charge is delocalised over the benzene ring by resonance; an alkyl group cannot do this.
3 The salt is unstable; above about $278\ \mathrm{K}$ it reacts with water, giving phenol and $\mathrm{N_{2}}$.
4 Benzenediazonium hydrogensulphate; 4-chlorobenzenediazonium chloride.
5 $n=\dfrac{10.7}{107.15}=0.0999\ \mathrm{mol}$; $m(\mathrm{NaNO_{2}})=0.0999\times69.00=6.89\ \mathrm{g}$.
6 Only the primary aromatic amines: aniline and 4-nitroaniline. Ethanamine and phenylmethanamine (N on an $sp^{3}$ carbon) give unstable salts that lose $\mathrm{N_{2}}$ at once; $N$-methylaniline is secondary.
II
Replacement reactions. $M$ ($\mathrm{g\,mol^{-1}}$): $\mathrm{C_{6}H_{5}N_{2}Cl}$ 140.6, $\mathrm{C_{6}H_{5}I}$ 204.0; $22.7\ \mathrm{L\,mol^{-1}}$ at $273\ \mathrm{K}$, $1\ \mathrm{bar}$. Give the organic product of benzenediazonium chloride with:
  • 1 $\mathrm{Cu_{2}Cl_{2}}$ in HCl
  • 2 copper powder and HBr
  • 3 aqueous KI
  • 4 $\mathrm{HBF_{4}}$, followed by heating the dry precipitate
  • 5 ethanol
  • 6 water, warmed
  • 7 How do the Sandmeyer and Gattermann reactions differ?
  • 8 Why is iodobenzene made from benzenediazonium chloride rather than by direct iodination of benzene?
  • 9 $2.81\ \mathrm{g}$ of benzenediazonium chloride reacts completely with KI. What volume of $\mathrm{N_{2}}$ forms, and what mass of iodobenzene?
  • 10 Write the equation for the reaction with phosphinic acid ($\mathrm{H_{3}PO_{2}}$) and water, and name the inorganic products.
1, 2 Chlorobenzene (Sandmeyer); bromobenzene (Gattermann).
3, 4 Iodobenzene; fluorobenzene (Balz–Schiemann).
5, 6 Benzene (with ethanal and HCl); phenol.
7 Sandmeyer uses a copper(I) salt, $\mathrm{Cu_{2}X_{2}}$ in HX (or CuCN); Gattermann uses copper powder with HX. Sandmeyer usually gives better yields.
8 Iodine is not easily introduced into benzene directly (the reaction is reversible, and HI reduces the product); iodide reacts readily with the diazonium salt.
9 $n=\dfrac{2.81}{140.6}=0.0200\ \mathrm{mol}$; $V(\mathrm{N_{2}})=0.0200\times22.7=0.454\ \mathrm{L}$; $m(\mathrm{C_{6}H_{5}I})=0.0200\times204.0=4.08\ \mathrm{g}$.
10 $\mathrm{C_{6}H_{5}N_{2}^{+}Cl^{-}+H_{3}PO_{2}+H_{2}O}$ $\mathrm{\rightarrow C_{6}H_{6}+N_{2}+H_{3}PO_{3}+HCl}$; nitrogen, phosphorous acid ($\mathrm{H_{3}PO_{3}}$) and hydrogen chloride.
III
Coupling and azo dyes.
  • 1 Write the equation for the coupling of benzenediazonium chloride with phenol and give the colour of the product.
  • 2 Write the equation for coupling with aniline, and give the conditions and colour.
  • 3 Why does benzenediazonium chloride not couple with benzene?
  • 4 In which position does the diazonium ion attack phenol, and why?
  • 5 Describe a test, based on coupling, that distinguishes aniline from phenylmethanamine.
  • 6 Give the molecular formula and molar mass of 4-aminoazobenzene.
1 $\mathrm{C_{6}H_{5}N_{2}^{+}Cl^{-}+C_{6}H_{5}OH\xrightarrow{OH^{-}}}$ $\mathrm{4{-}HOC_{6}H_{4}N{=}NC_{6}H_{5}+Cl^{-}+H_{2}O}$: 4-hydroxyazobenzene, orange.
2 $\mathrm{C_{6}H_{5}N_{2}^{+}Cl^{-}+C_{6}H_{5}NH_{2}}$ $\mathrm{\rightarrow4{-}H_{2}NC_{6}H_{4}N{=}NC_{6}H_{5}+HCl}$, in mildly acidic solution: 4-aminoazobenzene, yellow.
3 The diazonium ion is a weak electrophile; only rings strongly activated by $\mathrm{{-}OH}$, $\mathrm{{-}O^{-}}$ or $\mathrm{{-}NH_{2}}$ are attacked.
4 Para to OH: the para position is activated by the $+R$ effect of $\mathrm{{-}O^{-}}$, and is less hindered than the ortho positions.
5 Treat each with $\mathrm{NaNO_{2}/HCl}$ at $273\text{–}278\ \mathrm{K}$, then add to alkaline naphthalen-2-ol: aniline gives an orange-red dye; phenylmethanamine evolves $\mathrm{N_{2}}$ and gives no dye.
6 $\mathrm{C_{12}H_{11}N_{3}}$, $197.2\ \mathrm{g\,mol^{-1}}$.
IV
Synthesis and applications. $M$ ($\mathrm{g\,mol^{-1}}$): naphthalen-2-ol 144.17, 1-(phenyldiazenyl)naphthalen-2-ol 248.28, aniline 93.13.
  • 1 Convert benzene into iodobenzene.
  • 2 Convert aniline into 4-bromophenol.
  • 3 Convert nitrobenzene into 1-chloro-3-fluorobenzene.
  • 4 Convert benzene into benzoic acid through a diazonium salt.
  • 5 Challenge. Convert 4-methylaniline into 3-bromotoluene (3-bromo-1-methylbenzene). Explain why you cannot simply brominate toluene.
  • 6 Industry. A dye works makes $1.00\ \mathrm{kg}$ of the orange dye 1-(phenyldiazenyl)naphthalen-2-ol by coupling benzenediazonium chloride with naphthalen-2-ol. The overall yield based on each starting material is $90.0\%$. What masses of aniline and naphthalen-2-ol are needed?
1 Nitrate ($\mathrm{HNO_{3}/H_{2}SO_{4}}$); reduce (Sn/HCl, then NaOH) to aniline; diazotise ($\mathrm{NaNO_{2}/HCl}$, $273\ \mathrm{K}$); add KI.
2 Acetylate; brominate (para); hydrolyse to 4-bromoaniline; diazotise; warm with water.
3 $\mathrm{Cl_{2}/FeCl_{3}}$ (NO₂ directs meta): 1-chloro-3-nitrobenzene; Sn/HCl: 3-chloroaniline; diazotise; $\mathrm{HBF_{4}}$, then heat.
4 Benzene → nitrobenzene → aniline → benzenediazonium chloride → (CuCN/KCN) benzonitrile → ($\mathrm{H_{3}O^{+}}$, heat) benzoic acid.
5 Acetylate the $\mathrm{NH_{2}}$ (to control it); brominate: Br goes ortho to $\mathrm{{-}NHCOCH_{3}}$, i.e. meta to $\mathrm{CH_{3}}$; hydrolyse; diazotise; $\mathrm{H_{3}PO_{2}}$ removes the nitrogen. Bromination of toluene gives mainly 2- and 4-bromotoluene, because $\mathrm{CH_{3}}$ directs o/p.
6 $n(\text{dye})=\dfrac{1000}{248.28}=4.028\ \mathrm{mol}$; needed $\dfrac{4.028}{0.900}=4.475\ \mathrm{mol}$ of each: aniline $4.475\times93.13=417\ \mathrm{g}$; naphthalen-2-ol $4.475\times144.17=645\ \mathrm{g}$.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 2 marks How will you convert aniline into (a) chlorobenzene, (b) phenol? Give the equations and conditions.

Step 1 Diazotise: $\mathrm{C_{6}H_{5}NH_{2}+NaNO_{2}+2HCl}$ $\mathrm{\xrightarrow{273-278\ K}C_{6}H_{5}N_{2}^{+}Cl^{-}+NaCl+2H_{2}O}$.
(a) $\mathrm{C_{6}H_{5}N_{2}^{+}Cl^{-}\xrightarrow{Cu_{2}Cl_{2}/HCl}C_{6}H_{5}Cl+N_{2}}$ (Sandmeyer). (1 mark)
(b) $\mathrm{C_{6}H_{5}N_{2}^{+}Cl^{-}+H_{2}O\xrightarrow{warm}C_{6}H_{5}OH}$ $\mathrm{+N_{2}+HCl}$. (1 mark)

E2. NEET Benzenediazonium chloride is treated with phosphinic acid ($\mathrm{H_{3}PO_{2}}$) in water. The organic product is (A) benzene  (B) phenol  (C) chlorobenzene  (D) aniline

Idea $\mathrm{H_{3}PO_{2}}$ is a mild reducing agent that replaces $\mathrm{{-}N_{2}^{+}}$ by H.
Answer (A) benzene.

E3. JEE Main Numerical value. Benzenediazonium chloride is treated separately with CuCN/KCN; KI; phenol in NaOH(aq); $\mathrm{HBF_{4}}$ then heat; warm water; aniline in dilute acid; $\mathrm{H_{3}PO_{2}}$. In how many of these reactions is $\mathrm{N_{2}}$ evolved?

Idea Replacement reactions lose $\mathrm{N_{2}}$; coupling reactions (with phenol and with aniline) keep both nitrogens as $\mathrm{{-}N{=}N{-}}$.
Answer $7-2=\mathbf{5}$.

E4. JEE Advanced Aniline ($9.31\ \mathrm{g}$) is shaken with excess bromine water to give P. P is diazotised ($\mathrm{NaNO_{2}/HCl}$, $273\ \mathrm{K}$) to give Q, and Q is treated with $\mathrm{H_{3}PO_{2}}$ to give R. (a) Identify P, Q and R. (b) Why is R not obtained by brominating benzene? (c) Find the maximum mass of R. $M$: aniline $93.13$, R $314.8\ \mathrm{g\,mol^{-1}}$.

(a) P: 2,4,6-tribromoaniline; Q: 2,4,6-tribromobenzenediazonium chloride; R: 1,3,5-tribromobenzene, $\mathrm{C_{6}H_{3}Br_{3}}$.
(b) Br is o/p-directing, so successive bromination of benzene gives 1,2- and 1,4-patterns, never three bromines all meta to each other. The $\mathrm{NH_{2}}$ group places them, then is removed.
(c) $n=\dfrac{9.31}{93.13}=0.1000\ \mathrm{mol}$; $m(\mathrm{R})=0.1000\times314.8=31.5\ \mathrm{g}$.

E5. JEE MainNEET Assertion–reason. Assertion (A): The coupling of benzenediazonium chloride with phenol is carried out in mildly alkaline solution. Reason (R): The phenoxide ion is much more strongly activated towards electrophiles than phenol. (A) Both true, R explains A  (B) Both true, R does not explain A  (C) A true, R false  (D) A false, R true

Idea The weakly electrophilic diazonium ion needs a very electron-rich ring; $\mathrm{{-}O^{-}}$ is a stronger donor than $\mathrm{{-}OH}$. (Strong alkali is avoided because it destroys the diazonium ion.)
Answer (A).

E6. Board · Case study Read the passage and answer. Azo dyes contain the group –N=N– joining two aromatic rings. They are made by diazotising a primary aromatic amine in ice-cold acid and adding the solution to a phenol or an amine. Many food colours, such as the orange colour of some soft drinks, are azo dyes carrying sulphonate groups that make them soluble in water. (a) Why are azo compounds coloured? (1 mark) (b) Write the steps, with conditions, for making 4-hydroxyazobenzene from aniline. (2 marks) (c) Why cannot ethanamine be used in place of aniline? (1 mark)

(a) The two rings and the $\mathrm{N{=}N}$ group form an extended conjugated system that absorbs visible light.
(b) Aniline + $\mathrm{NaNO_{2}}$ + HCl at $273\text{–}278\ \mathrm{K}$ → benzenediazonium chloride; add to phenol in dilute NaOH (mildly alkaline) → 4-hydroxyazobenzene (orange).
(c) Ethanediazonium chloride is unstable and decomposes at once to ethanol and $\mathrm{N_{2}}$, so there is nothing left to couple.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How much azo food colour is in an orange soft drink, and how would a food-safety laboratory measure it?

Why it matters. Sunset Yellow FCF (E110), the disodium salt of an azo dye, colours soft drinks, sweets and snacks in many countries. Food regulators set limits for each colour and an acceptable daily intake (ADI); in the European Union foods containing it must carry a warning that it "may have an adverse effect on activity and attention in children". Checking compliance is routine work for food laboratories, and it relies on the strong visible absorption that the azo group gives.

What to do.

  1. Find the structure of Sunset Yellow ($\mathrm{C_{16}H_{10}N_{2}Na_{2}O_{7}S_{2}}$). Mark the azo group and decide which diazonium salt and which coupling partner it is made from. (Hint: the amine is sulphanilic acid, Section 9.2.) Explain the role of the sulphonate groups.
  2. Calculate its molar mass.
  3. (Illustrative data.) A drink diluted ten times has absorbance $A=0.55$ at $480\ \mathrm{nm}$ in a $1.00\ \mathrm{cm}$ cell; take the molar absorption coefficient as $\varepsilon=2.2\times10^{4}\ \mathrm{L\,mol^{-1}\,cm^{-1}}$. Use $A=\varepsilon cl$ to find the concentration of dye in the diluted and in the original drink, in $\mathrm{mol\,L^{-1}}$ and $\mathrm{mg\,L^{-1}}$.
  4. (Illustrative data.) If the ADI is $4\ \mathrm{mg}$ per kg of body mass per day, how many litres of this drink would a $25\ \mathrm{kg}$ child have to drink to reach it? Look up the current official ADI and the legal maximum in drinks in your country, and compare.

Tools: A calculator; a spectrophotometer if your school has one (or a smartphone colorimeter app); food-regulator websites.

Hand in: The annotated structure, your calculations, and a short paragraph on how the regulations differ between two countries.

Going further. Azo dyes that can be split back into certain aromatic amines are banned in textiles in the EU. Find out which amines are of concern and why the $\mathrm{{-}N{=}N{-}}$ bond can be cleaved in the body.

Instructor note (steps 3 and 4 use illustrative values): diazotised sulphanilic acid coupled with 6-hydroxynaphthalene-2-sulphonic acid; the sulphonate groups make the dye water-soluble. $M=452.4\ \mathrm{g\,mol^{-1}}$. Diluted: $c=\dfrac{0.55}{2.2\times10^{4}\times1.00}=2.5\times10^{-5}\ \mathrm{mol\,L^{-1}}$; original $2.5\times10^{-4}\ \mathrm{mol\,L^{-1}}$, i.e. $2.5\times10^{-4}\times452.4=0.113\ \mathrm{g\,L^{-1}}=113\ \mathrm{mg\,L^{-1}}$. ADI for the child $=100\ \mathrm{mg}$ per day, reached with $\tfrac{100}{113}=0.88\ \mathrm{L}$ of drink. (Real drinks are usually well below this level; the numbers here are chosen to make the comparison interesting.)

🧪 Self-check quiz

0/5 correct

Q1. Aniline is diazotised with $\mathrm{NaNO_{2}}$ and HCl at

Q2. Fluorobenzene is best made from benzenediazonium chloride by

Q3. Benzenediazonium chloride and phenol in mildly alkaline solution give

Q4. Which amine does not give a diazonium salt that is stable at $273\ \mathrm{K}$?

Q5. The Gattermann reaction converts $\mathrm{ArN_{2}^{+}Cl^{-}}$ into $\mathrm{ArCl}$ using

← 9.2 Basicity and Chemical Reactions of AminesChapter 10 — Biomolecules →