Preparation of aminesAQA A-Level Chemistry: Revision notes
Section 1
Amines and how they are made
Amines are organic compounds derived from ammonia, in which one or more hydrogen atoms are replaced by alkyl or aryl groups. A primary amine has the formula RNH₂, with the nitrogen bonded to one carbon.
There are three preparations to know:
- halogenoalkane + ammonia (primary aliphatic amines)
- reduction of a nitrile (primary aliphatic amines)
- reduction of a nitro compound (aromatic amines)
Section 2
From halogenoalkanes and ammonia
Heat a halogenoalkane with a large excess of ammonia dissolved in ethanol, in a sealed tube (ammonia and many halogenoalkanes are volatile).
CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br
The ammonia acts as a nucleophile, using the lone pair on nitrogen to attack the δ+ carbon. A second ammonia molecule then removes H⁺, which is why two moles are needed.
The excess ammonia is used because the primary amine formed is also a nucleophile and can react with more halogenoalkane, forming secondary and tertiary amines and quaternary ammonium salts. The excess makes this less likely but does not stop it, so the yield of primary amine is low and the product is a mixture.
Do not forget the second mole of ammonia in the equation. It forms NH₄Br with the HBr.
Section 3
From nitriles
A nitrile contains the –C≡N group. It is reduced to a primary amine by adding hydrogen:
CH₃CH₂CN + 4[H] → CH₃CH₂CH₂NH₂
Reagents: LiAlH₄ in dry ether, or hydrogen gas with a nickel catalyst.
Nitriles are made from halogenoalkanes and cyanide ions (KCN in ethanol), so this route adds one carbon to the chain. For example bromoethane → propanenitrile → propylamine.
The advantage over the ammonia route is that no further substitution is possible, so a purer product and a better yield of the primary amine are obtained.
LiAlH₄ reacts violently with water, so it is used in dry ether. Industrially, hydrogen with a nickel catalyst is cheaper and safer.
Heating a nitrile with dilute acid hydrolyses it to a carboxylic acid. It does not form an amine.
Section 4
Aromatic amines from nitro compounds
Aromatic amines such as phenylamine are made by reducing nitro compounds. Nitrobenzene is heated under reflux with tin and concentrated hydrochloric acid:
C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O
In the acidic mixture the amine is protonated and exists as the salt phenylammonium chloride, C₆H₅NH₃⁺Cl⁻. Excess sodium hydroxide is added to remove H⁺ and release the free amine:
C₆H₅NH₃⁺ + OH⁻ → C₆H₅NH₂ + H₂O
Reduction reactions in this topic use [H] in equations. Count the hydrogen: nitro to amine needs 6[H].
Section 5
Uses and a worked yield calculation
Aromatic amines are important in the manufacture of dyes, especially azo dyes, as well as pharmaceuticals.
Worked example. 12.3 g of nitrobenzene (Mr = 123.0) gives 6.51 g of phenylamine (Mr = 93.0).
- Amount of nitrobenzene = 12.3 ÷ 123.0 = 0.100 mol
- 1 : 1 ratio, so the maximum phenylamine is 0.100 mol = 0.100 × 93.0 = 9.30 g
- Percentage yield = 6.51 ÷ 9.30 × 100 = 70.0%
Must know
- Halogenoalkane + excess ethanolic NH₃ in a sealed tube gives a primary amine, but with further substitution
- Nitrile + LiAlH₄ in dry ether (or H₂ with Ni) gives a primary amine with no further substitution
- Nitro compound + Sn and concentrated HCl under reflux, then NaOH, gives an aromatic amine
- Aromatic amines are used to make dyes
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Preparation of amines
- A student prepares butylamine, CH₃CH₂CH₂CH₂NH₂, from 1-bromobutane, CH₃CH₂CH₂CH₂Br. The 1-bromobutane is added to a large excess of ammonia dissolved in ethanol, and the mixture is heated in a sealed tube.Write an equation for the formation of butylamine and explain why two moles of ammonia are needed for each mole of 1-bromobutane.2 marks
- Propylamine, CH₃CH₂CH₂NH₂, is prepared in industry from propanenitrile, CH₃CH₂CN. The propanenitrile is itself made by heating bromoethane with potassium cyanide in ethanol. The nitrile is then reduced to the primary amine.Write an equation for the reduction of propanenitrile using [H] to represent the reducing agent, and suggest one advantage of making a primary amine by this route rather than from a halogenoalkane and ammonia.2 marks
- A student reduces nitrobenzene, C₆H₅NO₂ (Mr = 123.0), to phenylamine, C₆H₅NH₂ (Mr = 93.0), by heating it under reflux with tin and concentrated hydrochloric acid. After the reaction the student adds excess sodium hydroxide solution and extracts the phenylamine. From 12.3 g of nitrobenzene the student isolates 6.51 g of phenylamine.Write an equation for the reduction of nitrobenzene using [H], and explain why sodium hydroxide is added at the end of the reaction.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).