AQA A-Level Chemistry (7405)

## Learning Outcomes

- Understand the two main AQA routes for preparing primary aliphatic amines
- Explain how halogenoalkanes react with ammonia by nucleophilic substitution
- Describe why excess ethanolic ammonia is used and why mixtures can form
- Identify how nitriles are reduced using H₂/Ni or LiAlH₄ in dry ether followed by hydrolysis
- Apply chain-length reasoning to synthesis routes involving nitriles
- Distinguish clearly between preparation of primary, secondary and tertiary amines and quaternary ammonium salts
- Avoid common errors in equations, conditions and product prediction

## AQA A-Level Chemistry (7405) Syllabus

For the AQA A-Level Chemistry (7405), you are expected to understand the preparation of amines, with a focus on the following syllabus points:

- Know that primary aliphatic amines can be prepared by reacting ammonia with halogenoalkanes
- Know that primary amines can also be prepared by reduction of nitriles
- Identify the conditions for halogenoalkane reactions: excess ethanolic ammonia, heat, and a sealed tube or pressure
- Write the overall equation for preparation from a halogenoalkane, including formation of an ammonium halide
- Explain why excess ammonia is used and why the halogenoalkane route can produce a mixture of products
- Understand that amines are nucleophiles and can react further with halogenoalkanes
- State the reagents for nitrile reduction: H₂/Ni or LiAlH₄ in dry ether followed by hydrolysis
- Explain why the nitrile route is useful for preparing a pure primary amine and for extending the carbon chain by one atom

## Test Your Knowledge

_Attempt these questions before reading this article. If you find some difficult or cannot remember the answers, look more closely at that area during your revision._

1. What are the two main methods for preparing primary aliphatic amines on the AQA specification?
2. Why is excess ethanolic ammonia used when making amines from halogenoalkanes?
3. How does the carbon chain length change when a halogenoalkane is converted to a nitrile and then reduced?
4. Which reagents can reduce a nitrile to a primary amine?
5. Why does the halogenoalkane route often give a mixture of products?

## Introduction

Amines are nitrogen-containing organic compounds based on ammonia. In AQA Chemistry, the preparation of amines matters because it brings together ideas from several parts of organic chemistry: nucleophilic substitution, reduction, reaction conditions, and synthetic planning.

This topic is tested in simple recall questions, mechanism questions, and multi-step synthesis questions. To do well, you need more than memorised equations. You need to know why one route gives mixtures, why another route gives a single primary amine, and how to count carbon atoms correctly through each step.

> **Key Term: amine**
> An organic derivative of ammonia in which one or more hydrogen atoms are replaced by alkyl or aryl groups.

### Preparing primary amines from halogenoalkanes

A primary aliphatic amine can be prepared by reacting a halogenoalkane with ammonia. This is a **nucleophilic substitution** reaction.

The general equation is:

$$
\mathrm{R{-}X + 2NH_3 \rightarrow R{-}NH_2 + NH_4X}
$$

Here, **R–X** is the halogenoalkane and **NH₄X** is the ammonium halide formed.

For example:

$$
\mathrm{CH_3CH_2Br + 2NH_3 \rightarrow CH_3CH_2NH_2 + NH_4Br}
$$

This makes **ethylamine** from **bromoethane**.

> **Key Term: nucleophilic substitution**
> A reaction in which a nucleophile donates a lone pair of electrons to an electron-deficient carbon and replaces another atom or group.

### Why ammonia works as the nucleophile

Ammonia has a lone pair of electrons on nitrogen. That lone pair attacks the carbon atom bonded to the halogen, because that carbon is slightly positive, written as **δ+**. The halogen is more electronegative, so it pulls electron density away from the carbon.

This is the key idea behind the mechanism. The carbon attached to the halogen is the site of attack because the C–X bond is polar.

### Mechanism step by step

Students are often told to "learn the mechanism", but the exam rewards understanding the electron movement.

Step 1:
A molecule of ammonia attacks the carbon attached to the halogen using its lone pair.

Step 2:
As carbon cannot have five bonds, the C–X bond breaks at the same time, and the halide ion leaves.

This forms an **alkylammonium ion**, for example:

$$
\mathrm{CH_3CH_2NH_3^+}
$$

Step 3:
A second ammonia molecule removes a proton from the alkylammonium ion.

Step 4:
This gives the amine and an ammonium ion:

$$
\mathrm{CH_3CH_2NH_2}
$$

and

$$
\mathrm{NH_4^+}
$$

The halide ion then pairs with the ammonium ion to form the ammonium halide, such as **NH₄Br**.

> **Key Term: primary amine**
> An amine in which the nitrogen atom is bonded to one alkyl or aryl group and two hydrogen atoms, with general formula RNH₂.

### Conditions and why they matter

The standard conditions are:

- excess ethanolic ammonia
- heat
- **sealed tube** or **under pressure**

Each part matters.

**Excess ammonia** is used to reduce further substitution. If ammonia is present in high concentration, a halogenoalkane molecule is more likely to collide with ammonia than with the amine product.

**Ethanolic ammonia** is used rather than aqueous ammonia because water could act as a competing nucleophile and lead to alcohol formation.

A **sealed tube** or pressure is needed because ammonia is volatile. If the tube were open, ammonia could escape and the concentration would fall.

> **Key Term: excess ethanolic ammonia**
> Ammonia dissolved in ethanol, used in excess so that formation of the primary amine is favoured over further substitution.
> **Test Tip:**
> If a question asks for conditions, do not just write "ammonia". AQA usually wants **excess ethanolic ammonia** and heating in a **sealed tube** or **under pressure**.

### Why this route often gives a mixture

This is one of the most tested ideas in the topic.

The primary amine formed still contains a lone pair on nitrogen. That means it is also a nucleophile. So it can react with more halogenoalkane.

For example, if ethylamine is formed, it can react further:

$$
\mathrm{CH_3CH_2Br + CH_3CH_2NH_2 \rightarrow (CH_3CH_2)_2NH}
$$

This gives a **secondary amine**.

That secondary amine still has a lone pair and can react again to give a **tertiary amine**.

A tertiary amine can react one more time to give a **quaternary ammonium ion**, which then forms a salt with a halide ion.

So, from one halogenoalkane and ammonia, you may get:

- primary amine
- secondary amine
- tertiary amine
- quaternary ammonium salt

This is why the halogenoalkane route is often poor if you want one pure primary amine.

> **Key Term: quaternary ammonium salt**
> An ionic compound containing a positively charged nitrogen atom bonded to four alkyl groups, together with a negative ion.

### Stepwise substitution sequence

It helps to see the pattern clearly.

Starting with methyl bromide:

$$
\mathrm{CH_3Br + NH_3 \rightarrow CH_3NH_2}
$$

Then:

$$
\mathrm{CH_3Br + CH_3NH_2 \rightarrow (CH_3)_2NH}
$$

Then:

$$
\mathrm{CH_3Br + (CH_3)_2NH \rightarrow (CH_3)_3N}
$$

Then:

$$
\mathrm{CH_3Br + (CH_3)_3N \rightarrow (CH_3)_4N^+Br^-}
$$

A common error is to think only ammonia can be the nucleophile. In fact, ammonia and amines can both act as nucleophiles because both have a lone pair on nitrogen.

> **Test Tip:**
> When explaining why a mixture forms, do not say only "more substitution happens". State the reason: **the amine product also has a lone pair and can act as a nucleophile**.

### Preparing primary amines by reduction of nitriles

The second AQA route is reduction of a nitrile to a primary amine.

The general change is:

$$
\mathrm{R{-}C \equiv N \rightarrow R{-}CH_2NH_2}
$$

This can be written using hydrogen:

$$
\mathrm{R{-}C \equiv N + 2H_2 \rightarrow R{-}CH_2NH_2}
$$

or using reducing agent notation:

$$
\mathrm{R{-}C \equiv N + 4[H] \rightarrow R{-}CH_2NH_2}
$$

For example:

$$
\mathrm{CH_3CN + 2H_2 \rightarrow CH_3CH_2NH_2}
$$

Ethanenitrile is reduced to ethylamine.

> **Key Term: reduction of nitriles**
> The conversion of a nitrile, R–C≡N, into a primary amine, R–CH₂NH₂, by addition of hydrogen.

### Reagents and conditions

AQA expects two possible methods:

- H₂ with a Ni surface
- LiAlH₄ in dry ether, followed by hydrolysis

For hydrogenation, a safe exam answer is **H₂/Ni**. If a question gives extra details such as temperature and pressure, use them, but the core specification point is the reagent and conditions.

For lithium aluminium hydride, the phrase **in dry ether** matters. LiAlH₄ reacts violently with water, so the solvent must be dry. After reduction, the mixture is hydrolysed to give the amine.

> **Key Term: lithium aluminium hydride (LiAlH₄)**
> A strong reducing agent used in dry ether to reduce nitriles to primary amines, followed by hydrolysis.

### Why this route gives a primary amine only

This is the major advantage over the halogenoalkane route.

When a nitrile is reduced, the product is a primary amine. There is no halogenoalkane present for that amine to react with further. So you do not get the same chain of repeated substitution.

That means this route is much better if you want a **pure primary amine**.

Students sometimes write that nitrile reduction gives a mixture because the amine can still act as a nucleophile. That misses the point. An amine can be a nucleophile, but without a suitable electrophile such as a halogenoalkane, there is no further substitution sequence.

### Why synthetic chemists like the nitrile route

This route is also useful because it gives **chain extension by one carbon atom**.

Look at the sequence carefully:

First, make a nitrile from a halogenoalkane:

$$
\mathrm{CH_3CH_2Br + CN^- \rightarrow CH_3CH_2CN + Br^-}
$$

Bromoethane has **2 carbons**.

Propanenitrile has **3 carbons**. The carbon in the **CN** group becomes part of the chain.

Then reduce the nitrile:

$$
\mathrm{CH_3CH_2CN \rightarrow CH_3CH_2CH_2NH_2}
$$

This gives **propylamine**, still with **3 carbons**.

So the overall route from a halogenoalkane to an amine via a nitrile increases the chain length by one carbon atom.

### Worked example: predicting the amine from a nitrile route

Suppose you start with **1-bromopropane** and convert it to a nitrile, then reduce it.

Step 1: Substitution with cyanide

$$
\mathrm{CH_3CH_2CH_2Br \rightarrow CH_3CH_2CH_2CN}
$$

Count the carbons:

- 1-bromopropane = 3 carbons
- butanenitrile = 4 carbons

Step 2: Reduction

$$
\mathrm{CH_3CH_2CH_2CN \rightarrow CH_3CH_2CH_2CH_2NH_2}
$$

Product: **butan-1-amine**

Reasoning:

- the nitrile carbon is retained
- reduction changes **C≡N** into **CH₂NH₂**
- no carbon is lost

This kind of carbon counting is a common synoptic skill.

> **Test Tip:**
> When going through a nitrile intermediate, always count the nitrile carbon. Students often lose one carbon and name the wrong amine.

### Comparing the two routes and avoiding common exam errors

A strong synoptic answer often compares methods, not just states them.

![Primary aliphatic amine synthesis from halogenoalkanes and nitriles, with further substitution to secondary, tertiary and quaternary ammonium products.](https://static.pastpaperhero.com/images/aqa-a-level-chemistry-7405-amines-preparation-of-amines_00.png)

_Preparation routes include ammonia substitution of halogenoalkanes and nitrile reduction, with successive alkylation giving secondary, tertiary and quaternary products._

### Side-by-side comparison

Halogenoalkane + ammonia

- reaction type: nucleophilic substitution
- conditions: excess ethanolic ammonia, heat, sealed tube or pressure
- product outcome: primary amine intended, but mixture often forms
- reason for mixture: amine product is also nucleophilic
- main use: direct preparation from a halogenoalkane when purity is less critical or when conditions are controlled carefully

Nitrile reduction

- reaction type: reduction
- reagents: H₂/Ni or LiAlH₄ in dry ether then hydrolysis
- product outcome: primary amine only
- synthetic advantage: adds one carbon atom if nitrile came from a halogenoalkane
- main use: cleaner route to a primary amine and useful for chain extension

### Worked synthesis comparison

Question: Outline two routes to make **ethylamine**.

Route 1: from a halogenoalkane

$$
\mathrm{CH_3CH_2Br + 2NH_3 \rightarrow CH_3CH_2NH_2 + NH_4Br}
$$

Conditions:

- excess ethanolic ammonia
- heat
- sealed tube

Possible issue:

- further substitution can give diethylamine and other products

Route 2: from a nitrile

$$
\mathrm{CH_3CN + 2H_2 \rightarrow CH_3CH_2NH_2}
$$

or

$$
\mathrm{CH_3CN \xrightarrow[\text{then hydrolysis}]{LiAlH_4\text{ in dry ether}} CH_3CH_2NH_2}
$$

Advantage:

- primary amine only

If asked which route is better for a pure sample, choose **reduction of nitriles** and explain why.

### Common student errors

1. Missing the second ammonia in the overall equation

Students often write:

$$
\mathrm{R{-}X + NH_3 \rightarrow R{-}NH_2 + HX}
$$

This is not the standard overall equation expected. The correct form is:

$$
\mathrm{R{-}X + 2NH_3 \rightarrow R{-}NH_2 + NH_4X}
$$

The second ammonia removes a proton and forms the ammonium halide.

1. Forgetting "excess" and "ethanolic"

Writing just "NH₃" is incomplete for many exam questions. The conditions show you understand how the reaction is controlled.

1. Confusing nitrile reduction with reaction with ammonia

A nitrile is **reduced** to an amine. It is not treated with ammonia to make an amine.

1. Losing the nitrile carbon

In chain-extension questions, students often name propylamine instead of butylamine, or similar one-carbon mistakes.

1. Assuming nitrile reduction gives secondary or tertiary amines

It gives a **primary amine**.

### How to structure an evaluative answer

Some synoptic questions ask which route is more suitable. A good answer should compare both methods directly.

Use this structure:

- state the two possible routes
- give conditions for each
- explain the main advantage and disadvantage of each
- link the choice to the aim of the synthesis

For example, if the aim is a **pure primary amine**, say:

- halogenoalkane + ammonia can produce a mixture because the amine formed reacts further
- nitrile reduction gives only the primary amine
- therefore nitrile reduction is the better route

If the aim is **chain extension**, say:

- going via a nitrile adds one carbon atom
- direct halogenoalkane + ammonia does not increase chain length
- therefore the nitrile route is the better synthetic route

That is stronger than simply saying one route is "better". You must give the chemical reason.

## Key Point Checklist

_This article has covered the following key knowledge points:_

- Amines are derivatives of ammonia in which one or more hydrogens are replaced by alkyl or aryl groups
- Primary aliphatic amines can be prepared from halogenoalkanes by nucleophilic substitution with ammonia
- The general equation is R–X + 2NH₃ → R–NH₂ + NH₄X
- Required conditions for the halogenoalkane route are excess ethanolic ammonia, heat, and a sealed tube or pressure
- Excess ammonia is used to reduce further substitution and favour formation of the primary amine
- The halogenoalkane route often gives a mixture because the amine product also has a lone pair and can act as a nucleophile
- Further substitution can produce secondary amines, tertiary amines and quaternary ammonium salts
- Primary amines can also be prepared by reduction of nitriles
- AQA reagents for nitrile reduction are H₂/Ni or LiAlH₄ in dry ether followed by hydrolysis
- Reduction of a nitrile gives a primary amine only
- Going from a halogenoalkane to a nitrile and then reducing it extends the carbon chain by one carbon atom
- In synthesis questions, correct carbon counting and correct reaction conditions are both essential

## Key Terms and Concepts

- amine
- nucleophilic substitution
- primary amine
- excess ethanolic ammonia
- quaternary ammonium salt
- reduction of nitriles
- lithium aluminium hydride (LiAlH₄)


Amines - Preparation of amines