Chlorination of alkanes — AQA A-Level Chemistry
Test yourself on Chlorination of alkanes with AQA A-Level practice questions.
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Chlorination of alkanes explained
Methane and chlorine do not react in the dark, but in ultraviolet light they react readily: CH₄ + Cl₂ → CH₃Cl + HCl.
Read the full explanation
The chlorine replaces a hydrogen atom, so this is a substitution, not an addition; the hydrogen that leaves is carried away as hydrogen chloride. The reaction does not stop at chloromethane. Any CH₃Cl formed is itself attacked, giving CH₂Cl₂, then CHCl₃ and finally CCl₄, so the product is always a mixture. Using a large excess of methane keeps most molecules at the mono-substituted stage, while an excess of chlorine drives the mixture towards the more highly chlorinated products. That poor selectivity is why this route is a laboratory and industrial nuisance rather than a clean synthesis, and why questions often ask you to predict the full set of products.
Students should be able to: explain this reaction as a free-radical substitution mechanism involving initiation, propagation and termination steps.
The mechanism has three named stages. Initiation: ultraviolet light supplies enough energy to break Cl₂ homolytically, one electron going to each atom, giving two chlorine radicals, Cl₂ → 2Cl•. Propagation runs twice: Cl• + CH₄ → •CH₃ + HCl, then •CH₃ + Cl₂ → CH₃Cl + Cl•. A radical is used and a radical is regenerated in each step, so one initiation event can produce thousands of product molecules; that is why the two steps add to give the overall equation. Termination is any step that removes radicals by joining two of them, such as •CH₃ + Cl• → CH₃Cl, 2•CH₃ → C₂H₆ or 2Cl• → Cl₂. The ethane in the product mixture is direct evidence that termination happens.
Your focus
- Write balanced equations for each successive chlorination of methane, from CH₃Cl through to CCl₄.
- Predict which product dominates when methane is in large excess and when chlorine is in large excess, giving a reason.
- Explain why this reaction is unsuitable for preparing a pure sample of chloromethane.
Show all 6 objectives
- Write the initiation, both propagation and at least two termination steps for the chlorination of methane.
- Classify an unfamiliar step as initiation, propagation or termination by counting the radicals on each side.
- Explain why a trace of ultraviolet light produces a large amount of chloromethane, referring to radical regeneration.
Chlorination of alkanes exam tips
Marking Points
- one mark for the balanced equation CH₄ + Cl₂ → CH₃Cl + HCl, with hydrogen chloride shown as the second product
- one mark for ultraviolet light (sunlight) as the essential condition
- one mark for naming further substitution products such as CH₂Cl₂, CHCl₃ or CCl₄
- one mark for saying an excess of methane favours mono-substitution while an excess of chlorine favours further substitution
- one mark for the initiation step Cl₂ → 2Cl• together with ultraviolet light as the condition
- one mark for each correct propagation step, with a radical on both sides of the arrow
- one mark for a termination step in which two radicals combine and no radical is produced
- one mark for the overall equation CH₄ + Cl₂ → CH₃Cl + HCl obtained by adding the two propagation steps
Examiner Tips
- 💡Write the condition as ultraviolet light every time; "light" alone is often not enough for the mark.
- 💡If asked for all products, work upwards one chlorine at a time so you do not miss CHCl₃.
- 💡Check your equation balances for hydrogen before chlorine; a missing HCl is the usual cause of failure.
- 💡Sort any step by counting radicals: in gives one and out gives one for propagation, two in and none out for termination.
- 💡Add your two propagation steps together; if the radicals do not cancel, one of them is wrong.
- 💡If you use half-headed arrows, use them consistently; they are not required, but incorrect ones lose credit.
- 💡Missing radical dots are penalised once within a clip, never credited, so check every species carries its dot before you move on.
Common Mistakes
- writing the reaction as an addition, giving a product such as CH₄Cl₂, when an alkane is saturated and can only substitute
- leaving HCl out of the equation, so the hydrogen atoms do not balance
- saying a catalyst is needed rather than ultraviolet light
- claiming chloromethane is the only product, when a mixture of all four chloromethanes is formed
- omitting the dot on Cl• or •CH₃, so the species is read as an ion or a molecule
- writing initiation as Cl₂ → Cl⁺ + Cl⁻, which is heterolytic fission and gives ions, not radicals
- labelling a step as termination when a radical still appears on the right-hand side
- using double-headed curly arrows in a free-radical mechanism, which AQA penalises