Preparation of Amines Revision Notes | A Level AQA Chemistry

1. Why Amines Matter — and How We Make Them

Amines are nitrogen-containing organic compounds that sit at the heart of pharmaceutical synthesis, dye chemistry, and biochemistry. The $-NH_2$ group makes them nucleophilic and basic — two properties that make them extraordinarily useful and chemically reactive.

At A Level, AQA requires you to know two routes to primary aliphatic amines and one route to aromatic amines. Each route has its own reagents, conditions, and limitations — and examiners test all three with precision. This lesson builds each route from the ground up so you can reproduce and justify them under exam conditions.

2. Route 1 — Halogenoalkane Reacting with Ammonia

The first route to a primary aliphatic amine uses a halogenoalkane and excess ammonia. Think back to nucleophilic substitution from your halogenoalkane work — the same mechanism is operating here, but now the nucleophile is ammonia rather than hydroxide.

Ammonia ($NH_3$) has a lone pair on nitrogen. That lone pair attacks the electrophilic carbon bearing the halogen, displacing the halide ion as a leaving group. The initial product is a primary ammonium salt, which then loses a proton to give the free amine.

The conditions are critical: the reaction is carried out by heating the halogenoalkane with excess concentrated ammonia dissolved in ethanol in a sealed tube. The sealed tube is essential because ammonia is a gas at room temperature and would otherwise escape, reducing yield.

This is where the main limitation of this route becomes clear. Even with excess ammonia, the primary amine product is itself a nucleophile. It can attack a second molecule of halogenoalkane to give a secondary amine, which can react again to give a tertiary amine, and so on. The result is a mixture of primary, secondary, and tertiary amines plus the quaternary ammonium salt. Separation is difficult, so this route gives impure product.

The overall equation for the first substitution step can be written as: $CH_3CH_2Br + 2NH_3 \rightarrow CH_3CH_2NH_2 + NH_4Br$. Notice that two moles of ammonia are consumed: one acts as the nucleophile and one acts as the base to deprotonate the ammonium salt intermediate.

3. Route 2 — Reduction of Nitriles

The second route to a primary aliphatic amine avoids the mixture problem entirely. Reducing a nitrile ($R-C\equiv N$) gives a primary amine ($R-CH_2-NH_2$) as the sole organic product, with no secondary or tertiary amine contamination.

The reducing agent is lithium aluminium hydride (LiAlH$_4$) in dry ether, followed by careful aqueous workup. LiAlH$_4$ is a powerful hydride source — it delivers $H^-$ ions to the electrophilic carbon of the $C\equiv N$ group, ultimately delivering 4[H] to reduce the $C\equiv N$ group — 2H to carbon and 2H to nitrogen — giving $-CH_2-NH_2$.

Worked example: Reduction of Ethanenitrile to Ethylamine

Describe the reagents, conditions, and the change in carbon skeleton when ethanenitrile ($CH_3CN$) is reduced to give a primary amine.

1. $CH_3-C\equiv N$ (ethanenitrile, 2 carbon atoms) — The nitrile carbon is counted as part of the chain. Ethanenitrile has 2 carbons in total.
2. LiAlH$_4$ in dry ether, then aqueous workup — Dry ether is essential because LiAlH4 reacts violently with water. Aqueous workup is performed after reduction is complete to hydrolyse the aluminium complex and release the amine.
3. $CH_3CN \xrightarrow{\text{1. LiAlH}_4\text{/dry ether}} CH_3CH_2NH_2$ — The $C\equiv N$ group is reduced to $-CH_2-NH_2$. Four hydrogen atoms are added across the triple bond.
4. $CH_3CH_2NH_2$ (ethylamine, 2 carbon atoms) — The carbon count stays the same — the nitrile carbon becomes the $-CH_2-$ carbon in the amine. This is a key difference from the halogenoalkane route, where the carbon count is unchanged for a different reason.
5. Product: $CH_3CH_2NH_2$ — a primary amine — Only one organic product forms. No secondary or tertiary amines are produced, making this route cleaner than the halogenoalkane route.

Answer: Ethanenitrile is reduced by LiAlH$_4$ in dry ether (then aqueous workup) to give ethylamine ($CH_3CH_2NH_2$), a primary amine with the same number of carbon atoms as the starting nitrile.

There is an important synthetic consequence here. If you start from a halogenoalkane and want a primary amine with one extra carbon atom, you can first react the halogenoalkane with potassium cyanide (KCN) to form a nitrile — a reaction you will have met in the halogenoalkane topic — and then reduce the nitrile with LiAlH$_4$. This two-step sequence is a powerful chain-extension strategy.

4. Comparing the Two Routes to Primary Aliphatic Amines

Now that both routes are established, it is worth comparing them directly — examiners frequently ask you to evaluate or compare synthetic routes, and this is a prime target.

The nitrile reduction route is generally preferred when a pure primary amine is required, despite the cost and handling difficulties of LiAlH$_4$. The halogenoalkane route is simpler in terms of apparatus but requires careful control of stoichiometry to minimise over-alkylation.

5. Aromatic Amines — Reduction of Nitro Compounds

Aromatic amines are a distinct class: the $-NH_2$ group is attached directly to a benzene ring. The most important example is phenylamine ($C_6H_5NH_2$), also called aniline. You cannot make phenylamine by reacting ammonia with a halogenobenzene under normal conditions — the benzene ring makes the carbon far less susceptible to nucleophilic attack. A completely different route is needed.

Aromatic amines are prepared by the reduction of nitro compounds. Nitrobenzene ($C_6H_5NO_2$) is reduced to phenylamine ($C_6H_5NH_2$) using tin and concentrated hydrochloric acid, followed by addition of sodium hydroxide to liberate the free amine from its salt.

Worked example: Reduction of Nitrobenzene to Phenylamine

Describe the reagents, conditions, and overall equation for the preparation of phenylamine from nitrobenzene.

1. Starting material: nitrobenzene ($C_6H_5NO_2$). Target: phenylamine ($C_6H_5NH_2$). — The $-NO_2$ group must be reduced to $-NH_2$. This requires adding hydrogen atoms and removing oxygen atoms.
2. Tin (Sn) and concentrated hydrochloric acid (HCl); heat under reflux. — Tin acts as the reducing agent in acidic conditions. The reaction is exothermic, so reflux controls the temperature and prevents loss of volatile material.
3. The initial product is a phenylammonium salt: $[C_6H_5NH_3]^+Cl^-$. — In acidic conditions, the amine is immediately protonated. The free amine has not yet been released.
4. Add excess sodium hydroxide (NaOH) solution: $[C_6H_5NH_3]^+Cl^- + NaOH \rightarrow C_6H_5NH_2 + NaCl + H_2O$ — NaOH deprotonates the ammonium salt, releasing phenylamine as a free base. This step is essential — without it, you isolate the salt, not the amine.
5. $C_6H_5NO_2 + 6[H] \rightarrow C_6H_5NH_2 + 2H_2O$ — The $[H]$ notation represents hydrogen from the reducing system (Sn/HCl). Six hydrogen atoms are required to fully reduce $-NO_2$ to $-NH_2$.

Answer: Nitrobenzene is reduced by Sn/conc. HCl under reflux. NaOH is then added to liberate phenylamine from its salt. Overall: $C_6H_5NO_2 + 6[H] \rightarrow C_6H_5NH_2 + 2H_2O$.

The step involving sodium hydroxide is frequently omitted by students under exam pressure. Without it, the product is the phenylammonium chloride salt, not phenylamine itself. Examiners specifically target this omission.

6. Aromatic Amines in the Manufacture of Dyes

The industrial importance of aromatic amines is substantial. Phenylamine and related compounds are the starting materials for the manufacture of azo dyes — a class of brightly coloured compounds that account for the majority of synthetic dyes used in the textile industry.

The synthesis of an azo dye involves two stages. First, phenylamine is converted to a diazonium salt by reaction with nitrous acid (generated in situ from sodium nitrite and hydrochloric acid) at temperatures below 10 °C — a process called diazotisation. Second, the diazonium salt undergoes a coupling reaction with a phenol or another aromatic amine to form the azo compound, which contains the characteristic $-N=N-$ chromophore responsible for the colour.

The colour of an azo dye depends on the extent of delocalisation across the molecule. Different substituents on the aromatic rings shift the wavelength of light absorbed, producing the full spectrum of colours seen in commercial dyes. This is why the reduction of nitrobenzene to phenylamine — a reaction that might seem purely academic — has enormous industrial and economic significance.

7. Exam Technique and Common Mistakes

Preparation of amines questions appear in both short-answer and extended-response formats. The marks are concentrated on conditions, reagents, and the ability to justify choices — not on naming the product.

Before moving into more retrieval, pause and connect the mechanism to the lab method: the reaction conditions determine not just whether the amine forms, but whether you isolate the free amine or an ammonium salt.