Free-Radical Chlorination of Methane Revision Notes | A Level AQA Chemistry

1. Why Alkanes React the Way They Do

Alkanes are often described as unreactive — and under normal conditions, that is largely true. Their C–H and C–C bonds are strong, non-polar, and offer no obvious site for attack by ions or polar reagents.

Yet mix methane with chlorine gas and shine UV light on the mixture, and something dramatic happens: a pale yellow gas reacts vigorously, producing a range of chlorinated products. The key is that UV light provides enough energy to break a bond in a very specific way — not into ions, but into radicals.

Understanding this reaction means understanding three things: how radicals form, how they propagate a chain reaction, and how that chain eventually ends. These are the initiation, propagation, and termination steps of free-radical substitution.

2. Homolytic Fission and the Birth of a Radical

Think about what happens when you snap a stick cleanly in half — each half gets an equal share. Homolytic fission works the same way at the electronic level: the two electrons in a bond split equally, one to each atom. The result is two species each carrying an unpaired electron — these are free radicals, denoted with a dot (e.g. Cl•).

Contrast this with heterolytic fission, where both electrons go to one atom, producing ions. Heterolytic fission is what happens in polar mechanisms (nucleophilic substitution, electrophilic addition). Free-radical substitution is different: it is a non-polar, radical chain process driven by UV light, not by charged species.

3. The Three Stages: Initiation, Propagation, Termination

The reaction of methane with chlorine is a chain reaction: once started, it sustains itself through a cycle of radical-generating steps. AQA requires you to know all three stages — initiation, propagation, and termination — and to write equations for each.

3a. Initiation

Initiation is the step that starts the chain. UV light is absorbed by a chlorine molecule, providing enough energy to break the Cl–Cl bond homolytically. This produces two chlorine radicals.

The equation for initiation is: $\text{Cl}_2 \xrightarrow{\text{UV}} 2\text{Cl}\bullet$. Notice that UV light is written above the arrow — it is a condition, not a reactant. Without UV light, the reaction does not start, because the Cl–Cl bond cannot be broken thermally under normal conditions.

3b. Propagation

Propagation is where the chain sustains itself. There are two propagation steps, and they cycle repeatedly. In each step, a radical reacts with a stable molecule to produce a new radical — so the radical count is conserved across the two steps combined.

• Propagation step 1: $\text{Cl}\bullet + \text{CH}_4 \rightarrow \text{CH}_3\bullet + \text{HCl}$ — a chlorine radical abstracts a hydrogen atom from methane, producing a methyl radical and hydrogen chloride.
• Propagation step 2: $\text{CH}_3\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}\bullet$ — the methyl radical reacts with a chlorine molecule, producing chloromethane and regenerating a chlorine radical.

The regenerated Cl• from step 2 feeds straight back into step 1. This is why the reaction is described as a chain — a single initiation event can trigger thousands of propagation cycles before the chain is broken.

3c. Termination

Termination occurs when two radicals collide and combine, removing radicals from the system and ending the chain. Unlike propagation, termination produces no new radicals.

There are three possible termination reactions in this system. Any two radicals present can combine — and since both Cl• and CH3• are present, all three combinations are possible:

• $\text{Cl}\bullet + \text{Cl}\bullet \rightarrow \text{Cl}_2$
• $\text{CH}_3\bullet + \text{Cl}\bullet \rightarrow \text{CH}_3\text{Cl}$
• $\text{CH}_3\bullet + \text{CH}_3\bullet \rightarrow \text{C}_2\text{H}_6$

The third termination step — two methyl radicals combining to form ethane — is why trace amounts of ethane are detected as a by-product. This is a useful piece of evidence that the mechanism genuinely proceeds via methyl radicals.

Worked example: Writing the Full Free-Radical Substitution Mechanism

Write all the equations for the free-radical substitution of methane with chlorine to form chloromethane, labelling each stage correctly.

1. $\text{Cl}_2 \xrightarrow{\text{UV}} 2\text{Cl}\bullet$ — UV light causes homolytic fission of the Cl–Cl bond. Label this 'Initiation'. The UV condition goes above the arrow.
2. $\text{Cl}\bullet + \text{CH}_4 \rightarrow \text{CH}_3\bullet + \text{HCl}$ — Cl• abstracts a hydrogen atom from methane. A methyl radical and HCl are produced. Label this 'Propagation step 1'.
3. $\text{CH}_3\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}\bullet$ — The methyl radical reacts with Cl2. Chloromethane is produced and Cl• is regenerated, sustaining the chain. Label this 'Propagation step 2'.
4. $\text{Cl}\bullet + \text{Cl}\bullet \rightarrow \text{Cl}_2$ $\text{CH}_3\bullet + \text{Cl}\bullet \rightarrow \text{CH}_3\text{Cl}$ $\text{CH}_3\bullet + \text{CH}_3\bullet \rightarrow \text{C}_2\text{H}_6$ — Any two radicals can combine. All three combinations must be stated. Label these 'Termination'. Note that no new radicals are produced in any termination step.
5. Step 1 consumes Cl•, produces CH3•. Step 2 consumes CH3•, produces Cl•. Net radical count: unchanged. — This confirms the chain is self-sustaining. If radicals were consumed without being regenerated, the chain would stop after one cycle.

Answer: Initiation: $\text{Cl}_2 \xrightarrow{\text{UV}} 2\text{Cl}\bullet$. Propagation: $\text{Cl}\bullet + \text{CH}_4 \rightarrow \text{CH}_3\bullet + \text{HCl}$; $\text{CH}_3\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}\bullet$. Termination: three possible radical combination steps.

4. Why a Mixture of Products Forms

Here is where the reaction becomes less tidy. If excess chlorine is present, the chloromethane produced in the first substitution can itself undergo further substitution. The Cl• radical can abstract a hydrogen from CH3Cl, and the resulting radical reacts with Cl2 to give dichloromethane (CH2Cl2).

This process continues: CH2Cl2 can become CHCl3 (trichloromethane), and CHCl3 can become CCl4 (tetrachloromethane). The result is a mixture of all four chlorinated products, along with unreacted methane and trace amounts of ethane from termination.

This lack of selectivity is a significant limitation of free-radical substitution as a synthetic method. You cannot reliably obtain a single pure product without carefully controlling the ratio of methane to chlorine. A large excess of methane favours monosubstitution (CH3Cl), because statistically the Cl• radical is more likely to encounter a CH4 molecule than a CH3Cl molecule.

5. Overall Equation and Reaction Summary

The overall equation for the monosubstitution of methane with chlorine is: $\text{CH}_4(g) + \text{Cl}_2(g) \xrightarrow{\text{UV}} \text{CH}_3\text{Cl}(g) + \text{HCl}(g)$. This is the equation you would write to summarise the net reaction, but the mechanism behind it involves the three-stage radical chain described above.

Notice that the overall equation is the sum of the two propagation steps. Adding them together: $\text{Cl}\bullet + \text{CH}_4 \rightarrow \text{CH}_3\bullet + \text{HCl}$ and $\text{CH}_3\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}\bullet$ gives $\text{CH}_4 + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{HCl}$, with the radicals cancelling out. This is a useful internal consistency check.

6. Exam Technique and Common Mistakes

AQA exam questions on this mechanism tend to follow predictable patterns. You may be asked to 'describe' the mechanism (state what happens in sequence), 'explain' a step (give the reason with reference to radicals and bond breaking), or 'suggest' a termination product given two named radicals. Each command word demands a different response structure.

When writing mechanism equations, always include the radical dot notation (•). Omitting the dot is a common error that loses marks, because it implies you are writing a normal covalent species rather than a radical. Similarly, always write UV light as a condition above the arrow in the initiation step — not as a reactant on the left-hand side.

Before attempting a full 6-mark response, pause and check that your radical equations are balanced and every radical carries a dot. That bridge from quick recognition to full written mechanism is exactly where AQA often differentiates answers.