Free-Radical Chlorination of Methane Revision Notes | A Level AQA Chemistry
Methane and chlorine react by free-radical substitution under UV light, forming chloromethane and HCl. Free AQA notes and 11 questions answered.
Updated 5 Sept 2026 · Revise Right Team
Free-Radical Chlorination of Methane is part of Alkanes in A Level AQA Chemistry.
6 min
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What this lesson covers
Methane and chlorine react by free-radical substitution under UV light, forming chloromethane and HCl. Free AQA notes and 11 questions answered.
Explain the free-radical substitution mechanism of methane with chlorine, including initiation, propagation, and termination steps
Describe the role of UV light in initiating the reaction and the nature of each radical species involved
Write equations for each stage of the mechanism using correct notation and half-arrows for radical movement
Evaluate why a mixture of products forms rather than a single substituted product
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Use this lesson to stay inside Alkanes while connecting the detail back to the wider Organic chemistry topic.
Free radical: A species with one unpaired electron, represented by a dot (e.g. Cl•). Highly reactive intermediate formed by homolytic fission of a covalent bond.
Homolytic fission: The breaking of a covalent bond in which one electron from the bonding pair goes to each atom, producing two radicals. Represented by half-headed (fishhook) curly arrows.
Free-radical substitution: A reaction mechanism in which a hydrogen atom in an alkane is replaced by a halogen atom via a chain reaction involving radical intermediates, requiring UV light to initiate.
Chain reaction: A self-sustaining sequence of reactions in which reactive intermediates (radicals) are continuously regenerated in the propagation steps, until termination occurs.
Revision notes
Read the core explanations from Free-Radical Chlorination of Methane before testing yourself from memory.
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:
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.
$\text{Cl}_2 \xrightarrow{\text{UV}} 2\text{Cl}\bullet$ — UV light causes homolytic fission of the Cl–Cl bond. Label this 'Initiation'. The UV condition goes above the arrow.
$\text{Cl}\bullet + \text{CH}_4 \rightarrow \text{CH}_3\bullet + \text{HCl}$ — Cl• abstracts a hydrogen atom from methane. A methyl radical and HCl are produced. Label this 'Propagation step 1'.
$\text{CH}_3\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}\bullet$ — The methyl radical reacts with Cl2. Chloromethane is produced and Cl• is regenerated, sustaining the chain. Label this 'Propagation step 2'.
$\text{Cl}\bullet + \text{Cl}\bullet \rightarrow \text{Cl}_2$
$\text{CH}_3\bullet + \text{Cl}\bullet \rightarrow \text{CH}_3\text{Cl}$
$\text{CH}_3\bullet + \text{CH}_3\bullet \rightarrow \text{C}_2\text{H}_6$ — 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.
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.
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.
Product
Formula
Number of substitutions
Chloromethane
$\text{CH}_3\text{Cl}$
1
Dichloromethane
$\text{CH}_2\text{Cl}_2$
2
Trichloromethane
$\text{CHCl}_3$
3
Tetrachloromethane
$\text{CCl}_4$
4
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.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-05.
Alkanes are largely unreactive under normal conditions because their C–H and C–C bonds are strong and non-polar, offering no obvious site for attack by ions or polar reagents. They do react with chlorine under UV light, which breaks the Cl–Cl bond into radicals.
Why is chlorinating methane a poor way to make a pure product?
Chlorinating methane is a poor way to make a single pure product because free-radical substitution lacks selectivity: chloromethane can be substituted further to CH₂Cl₂, CHCl₃ and CCl₄, and termination gives traces of ethane. Only careful control of the methane to chlorine ratio favours one product.
What is a chlorine radical?
A chlorine radical, Cl•, is a chlorine atom with an unpaired electron, formed when UV light breaks the Cl–Cl bond by homolytic fission so that each atom keeps one electron. It is highly reactive and abstracts a hydrogen atom from methane in the first propagation step.
What is the mechanism for the reaction of methane with chlorine?
Methane and chlorine react by free-radical substitution under UV light. Initiation: Cl₂ → 2Cl•. Propagation: Cl• + CH₄ → CH₃• + HCl, then CH₃• + Cl₂ → CH₃Cl + Cl•. Termination: any two radicals combine, such as CH₃• + Cl• → CH₃Cl.
Why is ethane formed when methane reacts with chlorine?
Trace amounts of ethane form because two methyl radicals can combine in a termination step: CH₃• + CH₃• → C₂H₆. Detecting ethane is useful evidence that the reaction genuinely proceeds via methyl radicals.
How do you get mainly chloromethane from methane and chlorine?
To favour chloromethane, use a large excess of methane. A Cl• radical is then statistically more likely to meet a CH₄ molecule than a CH₃Cl molecule, so monosubstitution is favoured, though a single pure product still cannot be obtained reliably.
What products form when methane reacts with excess chlorine?
With excess chlorine, chloromethane undergoes further substitution, giving dichloromethane (CH₂Cl₂), trichloromethane (CHCl₃) and tetrachloromethane (CCl₄). The result is a mixture of all four chlorinated products, along with unreacted methane and trace amounts of ethane.
What is the overall equation for methane and chlorine?
The overall equation for the monosubstitution of methane with chlorine is CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g), with UV light as the condition. It is the sum of the two propagation steps, with the radicals cancelling out.
Why does methane not react with chlorine in the dark?
Methane does not react with chlorine in the dark because the Cl–Cl bond cannot be broken thermally under normal conditions. UV light provides the energy for homolytic fission of Cl₂ into two chlorine radicals, which starts the chain reaction.
How is dichloromethane formed from methane?
Dichloromethane forms when chloromethane undergoes further substitution: a Cl• radical abstracts a hydrogen from CH₃Cl, and the radical produced reacts with Cl₂ to give CH₂Cl₂. Continued substitution gives CHCl₃ and then CCl₄, especially with excess chlorine.
What are common mistakes in free-radical mechanisms?
Common errors in free-radical mechanisms are omitting the radical dot, which implies a normal covalent species rather than a radical, and writing UV light as a reactant on the left-hand side instead of as a condition on the arrow in the initiation step. Every radical equation must also balance.