Part 1 of 5Classifying Amines
Amine
- An amine is derived from ammonia by replacing one or more hydrogen atoms with alkyl or aryl groups.
Primary
- One hydrogen is replaced, so the nitrogen is bonded to one carbon, as in ethylamine CH₃CH₂NH₂.
Secondary
- Two hydrogens are replaced, so nitrogen is bonded to two carbons, as in dimethylamine (CH₃)₂NH.
Tertiary
- All three hydrogens are replaced, giving nitrogen bonded to three carbons, as in trimethylamine (CH₃)₃N.
Aromatic amine
- Nitrogen is attached directly to a benzene ring, as in phenylamine C₆H₅NH₂.
Part 2 of 5Making Aliphatic Primary Amines
1
Heat a halogenoalkane with an excess of ethanolic ammonia in a sealed tube, e.g. CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br.
2
The excess ammonia is needed so the product amine meets few halogenoalkane molecules and less secondary amine forms.
3
Alternatively, reduce a nitrile using LiAlH₄ in dry ether or hydrogen with a nickel catalyst: RCN + 4[H] → RCH₂NH₂.
4
Making the nitrile from a halogenoalkane with KCN adds one carbon atom to the chain, so this route lengthens the carbon skeleton.
Part 3 of 5Making Phenylamine
1
Benzene is nitrated to nitrobenzene with concentrated nitric acid and concentrated sulfuric acid at about 50 °C.
2
Nitrobenzene is reduced by heating under reflux with tin and concentrated hydrochloric acid, forming the salt C₆H₅NH₃⁺Cl⁻.
3
The reduction overall is C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O, with tin acting as the reducing agent.
4
Excess NaOH(aq) is added to liberate the free amine from its salt: C₆H₅NH₃⁺ + OH⁻ → C₆H₅NH₂ + H₂O.
5
The phenylamine is then separated, often by steam distillation, and purified.
| Route | Reagents | Conditions |
|---|---|---|
| Halogenoalkane → amine | excess NH₃ in ethanol | heat in sealed tube |
| Nitrile → amine | LiAlH₄ in dry ether, or H₂ / Ni | room temp / heat |
| Nitrobenzene → phenylamine | Sn + conc. HCl, then NaOH(aq) | reflux |
Part 4 of 5Worked Example
Question
- 12.3 g of nitrobenzene (Mr = 123) is reduced and 7.4 g of phenylamine (Mr = 93) is isolated.
- Calculate the percentage yield.
- Moles of nitrobenzene = 12.3 ÷ 123 = 0.100 mol.
- One mole of nitrobenzene gives one mole of phenylamine, so the maximum is 0.100 mol.
- Maximum mass = 0.100 × 93 = 9.3 g.
- Percentage yield = (7.4 ÷ 9.3) × 100 = 79.6%.
Answer Percentage yield = 79.6%.
Part 5 of 5Exam Checklist & Tips
3.11.1 – Key Points for the Exam
- Primary amine An amine with one alkyl or aryl group attached to nitrogen.
- Nitrile A compound containing the –C≡N group.
- Reduction A reaction in which a compound gains hydrogen or loses oxygen.
- Phenylamine An aromatic primary amine, C₆H₅NH₂.
Common Mistakes
- Using a limited amount of ammonia, which leads to a mixture of secondary and tertiary amines.
- Forgetting the final NaOH step in the phenylamine preparation, so only the salt is obtained.
- Writing nitrile reduction as using a weak reducing agent; LiAlH₄ or H₂/Ni is required.
- Forgetting that the nitrile route adds an extra carbon atom to the chain.