Part 1 of 5Aliphatic Amines from Haloalkanes
A nucleophilic substitution with ammonia replaces the halogen by an amino group.
Reagent
- The haloalkane is heated with excess ethanolic ammonia in a sealed tube so that the volatile ammonia is not lost.
Equation
- CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br; the second mole of ammonia removes the HBr as a salt.
Further reaction
- The amine produced is itself a nucleophile and can attack more haloalkane, forming secondary and tertiary amines.
Control
- Using a large excess of ammonia makes it more likely that the haloalkane meets NH₃ rather than a product amine.
Part 2 of 5Other Routes
Other Routes
Nitriles
- A nitrile is reduced by LiAlH₄ in dry ether, or by H₂ with a nickel catalyst, to a primary amine: RCN + 4[H] → RCH₂NH₂.
Nitrobenzene
- Benzene is nitrated with concentrated HNO₃ and H₂SO₄ at about 50 °C, forming nitrobenzene as the starting material.
Reduction
- Nitrobenzene is heated under reflux with tin and concentrated HCl: C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O.
Final step
- The acid forms phenylammonium ions, so NaOH is added at the end to release the free phenylamine.
| Starting material | Reagents and conditions | Product |
|---|---|---|
| Bromoethane | Excess ethanolic NH₃, heat in sealed tube | Ethylamine |
| Propanenitrile | LiAlH₄ in dry ether | Propylamine |
| Nitrobenzene | Sn / conc. HCl, reflux, then NaOH | Phenylamine |
Part 3 of 5Planning the Preparation of Phenylamine
1
Nitrate benzene with a mixture of concentrated nitric and sulfuric acids, keeping the temperature near 50 °C.
2
Separate the nitrobenzene from the acid mixture and wash it before use.
3
Reflux the nitrobenzene with tin and concentrated hydrochloric acid until reduction is complete.
4
Add excess sodium hydroxide solution to convert the phenylammonium salt into free phenylamine.
5
Isolate the amine, for example by distillation, and purify it before calculating the yield.
Part 4 of 5Worked Example
Question
- 12.3 g of nitrobenzene (M = 123.1 g mol⁻¹) is reduced to phenylamine (M = 93.1 g mol⁻¹).
- 6.50 g of phenylamine is obtained.
- Calculate the percentage yield.
- Moles of nitrobenzene = 12.3 ÷ 123.1 = 0.0999 mol.
- The ratio nitrobenzenephenylamine is 1 : 1, so the theoretical moles of product = 0.0999 mol.
- Theoretical mass = 0.0999 × 93.1 = 9.30 g.
- Percentage yield = 6.50 ÷ 9.30 × 100 = 69.9%.
Answer Percentage yield = 69.9%
Part 5 of 5Exam Checklist & Tips
6.4.2 – Key Points for the Exam
- Nucleophilic substitution A reaction in which a nucleophile replaces a leaving group such as a halide.
- Reflux Heating with a condenser so that vapours return to the flask.
- Nitrile A compound containing the –C≡N group.
- Nitroarene An aromatic compound with a –NO₂ group on the ring.
Common Mistakes
- Using a 1 : 1 ratio of ammonia to haloalkane and expecting a pure primary amine.
- Forgetting the final NaOH step, so the product is left as its salt.
- Writing the nitro reduction without balancing six [H] and two water molecules.
- Using aqueous rather than ethanolic ammonia, which can produce alcohols.