Amines as Nucleophiles: Substitution Reactions Revision Notes | A Level AQA Chemistry

1. Why Amines Are Nucleophiles

Picture the nitrogen atom in ammonia or an amine. It has a lone pair of electrons sitting in a non-bonding orbital — and that lone pair is the key to everything in this topic. A nucleophile is any species that can donate an electron pair to an electron-deficient centre, and nitrogen's lone pair makes it an excellent candidate.

In a halogenoalkane such as bromoethane ($\text{CH}_3\text{CH}_2\text{Br}$), the carbon bonded to bromine carries a partial positive charge ($\delta+$) because bromine is more electronegative. That electron-deficient carbon is the electrophilic centre the nitrogen lone pair attacks.

The nucleophilicity of amines is directly linked to the availability of that lone pair. Primary amines ($\text{RNH}_2$) are generally slightly better nucleophiles than ammonia because the alkyl group donates electron density towards nitrogen, making the lone pair marginally more available. This distinction matters when you consider selectivity in reactions.

2. The Core Reaction: Ammonia with a Halogenoalkane

When ammonia reacts with a halogenoalkane, the nitrogen lone pair attacks the $\delta+$ carbon, displacing the halide ion as the leaving group. The initial product is an alkylammonium salt (for example, ethylammonium bromide, $\text{CH}_3\text{CH}_2\text{NH}_3^+\text{Br}^-$) — not a free amine — because the nitrogen has used its lone pair to form a new bond and now carries a positive charge.

To obtain the free primary amine, excess ammonia is used. The excess ammonia acts as a base, deprotonating the ammonium salt to release the primary amine and form an ammonium halide salt as a by-product.

The overall equation for the formation of ethylamine from bromoethane and excess ammonia is:

The reaction is carried out by heating the halogenoalkane with a concentrated aqueous or ethanolic solution of ammonia in a sealed tube. The sealed conditions are necessary to prevent the volatile ammonia from escaping and to maintain the pressure needed for the reaction to proceed at a useful rate.

3. Successive Alkylation: From Primary to Quaternary

Here is where the chemistry becomes genuinely interesting — and where exam questions get more demanding. The primary amine produced in the first reaction still has a lone pair on nitrogen. That means it can act as a nucleophile itself and react with another molecule of halogenoalkane.

This process of successive alkylation continues through four stages: ammonia gives a primary amine, the primary amine gives a secondary amine, the secondary amine gives a tertiary amine, and the tertiary amine gives a quaternary ammonium salt. Each stage is a nucleophilic substitution with the same mechanism.

Notice that the quaternary ammonium salt cannot be deprotonated to give a free amine — nitrogen has no hydrogen atoms left and carries a permanent positive charge. This is a key distinction: quaternary ammonium salts are ionic compounds, not amines.

Worked example: Tracing Successive Alkylation with Bromoethane

Starting from ammonia and excess bromoethane ($\text{CH}_3\text{CH}_2\text{Br}$), write the equation for each stage of alkylation up to the quaternary ammonium salt.

1. $\text{NH}_3 + \text{CH}_3\text{CH}_2\text{Br} \rightarrow [\text{CH}_3\text{CH}_2\text{NH}_3]^+\text{Br}^-$ $[\text{CH}_3\text{CH}_2\text{NH}_3]^+\text{Br}^- + \text{NH}_3 \rightarrow \text{CH}_3\text{CH}_2\text{NH}_2 + \text{NH}_4^+\text{Br}^-$ — Ammonia attacks the electrophilic carbon of bromoethane. The intermediate ethylammonium salt is deprotonated by excess ammonia to give ethylamine (a primary amine).
2. $\text{CH}_3\text{CH}_2\text{NH}_2 + \text{CH}_3\text{CH}_2\text{Br} \rightarrow [\text{(CH}_3\text{CH}_2)_2\text{NH}_2]^+\text{Br}^-$ $[\text{(CH}_3\text{CH}_2)_2\text{NH}_2]^+\text{Br}^- + \text{NH}_3 \rightarrow \text{(CH}_3\text{CH}_2)_2\text{NH} + \text{NH}_4^+\text{Br}^-$ — Ethylamine still has a lone pair on nitrogen and reacts with a second molecule of bromoethane. Deprotonation gives diethylamine (a secondary amine) — nitrogen now carries two ethyl groups.
3. $\text{(CH}_3\text{CH}_2)_2\text{NH} + \text{CH}_3\text{CH}_2\text{Br} \rightarrow [\text{(CH}_3\text{CH}_2)_3\text{NH}]^+\text{Br}^-$ $[\text{(CH}_3\text{CH}_2)_3\text{NH}]^+\text{Br}^- + \text{NH}_3 \rightarrow \text{(CH}_3\text{CH}_2)_3\text{N} + \text{NH}_4^+\text{Br}^-$ — Diethylamine reacts with a third molecule of bromoethane. Deprotonation yields triethylamine (a tertiary amine) — nitrogen now carries three ethyl groups and one lone pair.
4. $\text{(CH}_3\text{CH}_2)_3\text{N} + \text{CH}_3\text{CH}_2\text{Br} \rightarrow [\text{(CH}_3\text{CH}_2)_4\text{N}]^+\text{Br}^-$ — Triethylamine attacks a fourth molecule of bromoethane. The product is tetraethylammonium bromide — a quaternary ammonium salt. No deprotonation is possible because nitrogen has no N–H bonds remaining. This product is ionic and cannot be converted to a free amine.

Answer: The four products are: ethylamine ($\text{CH}_3\text{CH}_2\text{NH}_2$, primary amine), diethylamine ($(\text{CH}_3\text{CH}_2)_2\text{NH}$, secondary amine), triethylamine ($(\text{CH}_3\text{CH}_2)_3\text{N}$, tertiary amine), and tetraethylammonium bromide ($[\text{(CH}_3\text{CH}_2)_4\text{N}]^+\text{Br}^-$, quaternary ammonium salt).

4. The Selectivity Problem and How to Address It

A practical difficulty arises from successive alkylation: because each product is itself a nucleophile, the reaction does not stop cleanly at the primary amine stage. You end up with a mixture of primary, secondary, and tertiary amines alongside the quaternary ammonium salt. This is a real limitation of the halogenoalkane route to amines.

To push the reaction towards the primary amine, a large excess of ammonia is used relative to the halogenoalkane. The statistical excess of ammonia molecules means the halogenoalkane is far more likely to encounter an ammonia molecule than a primary amine molecule, reducing the extent of over-alkylation. Even so, complete selectivity is not achievable, and the mixture must be separated.

Conversely, if the goal is a quaternary ammonium salt, excess halogenoalkane is used to drive the reaction through all four stages. This is a deliberate synthetic strategy rather than an unwanted side reaction.

5. Mechanism in Detail: Curly Arrow Conventions

At A-level, you are expected to draw the mechanism for nucleophilic substitution of a halogenoalkane by an amine or ammonia. The curly arrow conventions are non-negotiable: each full curly arrow represents the movement of an electron pair, and arrows must start from an electron-rich site — either a lone pair or a bond — and point towards the electron-poor site.

Two curly arrows are required in the substitution step: one from the nucleophile's lone pair to the electrophilic carbon, and one from the C–X bond to the halogen. Drawing only one arrow is a common error that loses marks.

6. Exam Technique and Common Mistakes

Exam questions on this topic tend to test three things: mechanism drawing, product identification at each alkylation stage, and the ability to explain selectivity issues. The following guidance addresses the most frequently penalised errors.

7. Consolidation: Comparing Ammonia and Amine Reactions

Now that all four stages of alkylation are clear, it is worth comparing the reactions of ammonia and amines side by side. The mechanism is identical at every stage — nucleophilic substitution — but the structural outcome changes because each successive nitrogen species has fewer N–H bonds and more N–C bonds.

The pattern is consistent: as long as nitrogen retains a lone pair, it can act as a nucleophile and the reaction can continue. The quaternary ammonium salt is the terminus because nitrogen has formed four N–C bonds and has no lone pair remaining; without a lone pair, it cannot act as a nucleophile.

Bringing this topic together: the nucleophilicity of nitrogen — rooted in its lone pair — drives every reaction in this sequence. From ammonia reacting with a halogenoalkane to give a primary amine, through to the formation of a quaternary ammonium salt, the same nucleophilic substitution mechanism operates at each stage, and the selectivity challenge arises precisely because each product retains that nucleophilic character until the quaternary stage is reached.