All subjects · Chemistry · WCH15 (Unit 5)

Making amines and azo dyes

Aliphatic amines are made from halogenoalkanes or nitriles, and phenylamine from nitrobenzene. Phenylamine is the starting point for azo dyes.

Aliphatic amines

From halogenoalkanes: heat with excess ammonia dissolved in ethanol, in a sealed tube: CH3CH2Br + 2NH3 -> CH3CH2NH2 + NH4Br. Excess ammonia makes further substitution (to secondary and tertiary amines) less likely.
By reducing nitriles: LiAlH4 in dry ether: CH3CN + 4[H] -> CH3CH2NH2. This gives only the primary amine.

Phenylamine

Reduce nitrobenzene by heating under reflux with tin and concentrated hydrochloric acid:

C6H5NO2 + 6[H] -> C6H5NH2 + 2H2O

In the acid, the product is the salt C6H5NH3+. Add sodium hydroxide solution to release the free amine, then separate it (for example by steam distillation).

Azo dyes

1. Diazotisation: phenylamine with sodium nitrite and hydrochloric acid (making nitrous acid, HNO2, in situ), below 10 °C: C6H5NH2 + HNO2 + H+ -> C6H5N2+ + 2H2O. The benzenediazonium ion is unstable above about 10 °C, decomposing to phenol and nitrogen gas.

2. Coupling: add the cold diazonium solution to phenol in sodium hydroxide solution. A yellow-orange precipitate of an azo dye forms: C6H5N2+ + C6H5OH -> C6H5N=NC6H4OH + H+. The -N=N- group links two benzene rings, extending the delocalised system so the compound absorbs visible light and is coloured.

Worked example

Outline how to make an azo dye starting from benzene.

Nitrate: conc HNO3 and conc H2SO4, 50 °C, gives nitrobenzene.
Reduce: Sn and conc HCl, reflux, then NaOH, gives phenylamine.
Diazotise: NaNO2 and HCl below 10 °C.
Couple with phenol in NaOH solution.

Answer: Benzene -> nitrobenzene -> phenylamine -> benzenediazonium ion -> azo dye

Key idea: Halogenoalkane + excess ethanolic NH3 (sealed tube), or nitrile + LiAlH4, gives primary amines. Nitrobenzene + Sn/conc HCl, then NaOH, gives phenylamine. NaNO2/HCl below 10 °C gives C6H5N2+, which couples with alkaline phenol to an azo dye.

Check yourself

Why is excess ammonia used to make ethylamine from bromoethane?
- To reduce further substitution to secondary and tertiary amines
- To dissolve the bromoethane
- To act as a catalyst
- To neutralise ethylamine
Show answer
To reduce further substitution to secondary and tertiary amines. With more ammonia, bromoethane is more likely to meet NH3 than ethylamine.
What is formed when propanenitrile, CH3CH2CN, is reduced with LiAlH4 in dry ether?
- Ethylamine
- Propanamide
- Propylamine
- Propanoic acid
Show answer
Propylamine. -CN becomes -CH2NH2, keeping all three carbons: CH3CH2CH2NH2.
Which reagents reduce nitrobenzene to phenylamine?
- Tin and concentrated hydrochloric acid, heated under reflux
- Sodium nitrite and hydrochloric acid below 10 °C
- Hydrogen with iron(III) bromide
- Aqueous sodium hydroxide only
Show answer
Tin and concentrated hydrochloric acid, heated under reflux. Sn/HCl supplies [H].
Why is sodium hydroxide added after reducing nitrobenzene with tin and hydrochloric acid?
- To oxidise the tin
- To nitrate the product
- To make a diazonium salt
- To convert the phenylammonium salt into free phenylamine
Show answer
To convert the phenylammonium salt into free phenylamine. C6H5NH3+ + OH- -> C6H5NH2 + H2O.
Why is diazotisation carried out below 10 °C?
- Sodium nitrite does not dissolve in warm water
- The benzenediazonium ion decomposes at higher temperatures
- The reaction is endothermic
- Phenylamine boils at 10 °C
Show answer
The benzenediazonium ion decomposes at higher temperatures. Above about 10 °C it breaks down to phenol and N2.
Which conditions are used to couple benzenediazonium ions with phenol?
- Phenol in concentrated sulfuric acid, heated
- Phenol with tin and HCl
- Phenol vapour over a platinum catalyst
- Phenol dissolved in sodium hydroxide solution, kept cold
Show answer
Phenol dissolved in sodium hydroxide solution, kept cold. Alkaline phenol, cold, gives the yellow-orange dye.
How many [H] are needed in the equation for reducing one nitrobenzene molecule to phenylamine?
Show answer
6. C6H5NO2 + 6[H] -> C6H5NH2 + 2H2O: 2 H for NH2 and 4 H to make two H2O.

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Last updated 7 November 2026. Spotted a mistake? Tell us.