WJEC A-Level Chemistry AS Unit 2, June 2025: Question 7
10 marks · Medium difficulty · Structured Questions
Describe the mechanism of free-radical substitution of methane with chlorine, explain the formation of side-products, and evaluate reaction conditions to optimize the yield of chloromethane.
Practise this questionQuestion
Question text
7. A reaction used to produce chloromethane is initiated by shining ultraviolet light into a mixture
of methane gas and chlorine gas.
(a) (i) Name the mechanism for this reaction. [1]
(ii) Give the reaction mechanism for the production of chloromethane from methane
and chlorine. Show the initiation and propagation stages. [3]
Initiation
Propagation
(b) (i) Explain why the mixture produced in this reaction could contain ethane. [1]
(ii) Explain why the mixture produced in this reaction could contain dichloromethane.
[2]
(c) The original reaction to produce chloromethane was carried out in an aluminium-lined
container. The gas mixture was exposed to ultraviolet light for a prolonged period of
time. This method produced large quantities of side-products.
05 © WJEC CBAC Ltd. (2410U20-1)
UV lamp
glass
aluminium foil
methane and chlorine mixture
A possible improvement to this method, designed to minimise the formation of
side-products, involves passing the mixture of methane and chlorine through a narrow
transparent tube wrapped around an ultraviolet light source.
methane and chlorine mixture
UV lamp
7 products
(i) Suggest why this improvement could lead to fewer side-products in the mixture. [1]
(ii) This method was tested using different flow rates of the methane and chlorine
mixture to find the best yield of chloromethane. The composition of the product
mixture was found and recorded below.
06 Flow rate©WJEC CBAC Ltd. (2410U20-1)Percentage of product mixture
/ cm3 s–1
CH4 Cl2 CH3Cl CH2Cl2 C2H6
12 1 12 48 37
26 3 43 28 20
3 10 5 72 8 5
4 34 17 42 5 2
5 58 29 10 2 1
Use this data to determine the optimum flow rate for the production of
chloromethane.
Explain the results recorded at this flow rate. [2]
Mark scheme
Show the mark scheme
Marks available
Question Marking details
AO1 AO2 AO3 Total Maths Prac
7 (a) (i) (free) radical substitution
(ii) initiation
Cl2 → 2Cl• (1)
propagation
Cl• + CH4 → HCl + •CH3 (1)
•CH3 + Cl2 → CH3Cl + Cl• (1)
(b) (i) award (1) for either of following
two methyl radicals combine (in a termination reaction) 1 1
•CH3 + •CH3 → C2H6
(ii) free radical substitution / chain reaction continues
/ CH3Cl + Cl• → •CH2Cl + HCl (1)
chloromethyl radical / •CH2Cl reacts with Cl2 / Cl• (forming
dichloromethane / CH2Cl2)
/ •CH2Cl + Cl2 → CH2Cl2 + Cl•
/ •CH2Cl + Cl• → CH2Cl2 (1)
Marks available
AO1 AO2 AO3 Total Maths Prac
(c) (i) award (1) for either of following
shorter exposure time to UV 4
limits reaction time
fewer chlorine radicals produced
accept suggestion that methane/chlorine may react with
aluminium in the original method
(ii) 3 cm3 s–1 (1)
do not credit second mark if incorrect flow rate given
award (1) for any indication that this flow rate gives enough time
for most of the reactants to be used up or not enough time for
chloromethane to react to form side-products
Question 7 total 5 2 3 10 0 0
How to answer it
Free Radical Substitution: Chlorination of Methane
This 10-mark question evaluates core organic reaction mechanisms and practical reaction optimisation:
- Mechanism identification: Naming free-radical substitution.
- Mechanism equations: Writing initiation and propagation steps with radical dots placed correctly.
- Explaining side-products: Deducing termination products (ethane) and multi-substitution propagation products (dichloromethane).
- Practical problem-solving: Evaluating experimental modifications (flow system vs. batch system) and interpreting flow-rate data to find optimal reaction conditions.
Mechanism Identification
Name the mechanism for this reaction [1 mark]
✅ Correct Answer
(Free) radical substitution
❌ Common Errors
- Writing just "Substitution" (too vague; loses the mark).
- Confusing with other mechanisms: e.g. "Electrophilic substitution" or "Nucleophilic substitution".
Initiation and Propagation Steps
Give the reaction mechanism showing initiation and propagation stages [3 marks]
✅ Correct Mechanism Equations
Initiation:
Propagation:
•CH₃ + Cl₂ → CH₃Cl + Cl•
[1 mark] for first propagation step (forming HCl and •CH₃).
[1 mark] for second propagation step (forming CH₃Cl and regenerating Cl•).
💡 Key Knowledge
- Initiation: Homolytic fission occurs when UV light breaks the covalent Cl–Cl bond, producing two chlorine radicals.
- Propagation Step 1: The Cl• radical removes a hydrogen atom from methane, forming stable HCl and leaving a methyl radical (•CH₃).
- Propagation Step 2: The •CH₃ radical attacks a Cl₂ molecule to form chloromethane and regenerates the Cl• radical, keeping the chain reaction going.
❌ Examiner Pitfalls to Avoid
- Omission of unpaired electrons: Forgetting the radical dot (•) loses the equation mark instantly.
- Incorrect hydrogen abstraction: Writing Cl• + CH₄ → CH₃Cl + H• is chemically incorrect and gets 0 marks. Always form HCl first!
- Dot placement: In •CH₃, the unpaired electron is hosted on the carbon atom; writing CH₃• or •CH₃ is acceptable, but ensure it's clearly distinct from punctuation.
Formation of Reaction By-products
✅ Part (b)(i): Formation of Ethane [1 mark]
Explanation: Two methyl radicals combine in a termination reaction.
✅ Part (b)(ii): Formation of Dichloromethane [2 marks]
Step 1: The reaction continues by further substitution (a chlorine radical attacks the chloromethane product):
Step 2: The chloromethyl radical reacts with chlorine (propagation) or another chlorine radical (termination):
or •CH₂Cl + Cl• → CH₂Cl₂
[1 mark] for chloromethyl radical forming CH₂Cl₂.
🧠 Exam Technique: Explaining By-products
- Ethane (larger alkane): Always explained by the termination step combining two carbon-containing radicals ( •CH₃ + •CH₃ → C₂H₆ ).
- Polysubstituted alkanes (e.g. CH₂Cl₂, CHCl₃): Occur because as chloromethane accumulates, it competes with methane to react with newly formed Cl• radicals (further radical substitution).
Reaction Engineering & Data Interpretation
✅ Part (c)(i): Why the Improvement Works [1 mark]
Any one of the following explanations:
- Shorter exposure time to UV light.
- Limits the reaction time / residence time.
- Fewer chlorine radicals are produced.
- Alternative: Prevents reactants interacting/reacting with aluminium lining as in the original method.
📐 Part (c)(ii): Data Analysis [2 marks]
1. Optimum Flow Rate: 3 cm³ s⁻¹ [1 mark]
2. Explanation of Results: [1 mark]
- At this flow rate, there is enough time for most reactants to react (only 10% CH₄ and 5% Cl₂ remain unreacted),
- BUT not enough time for the product (chloromethane) to react further to produce high levels of side-products (only 8% CH₂Cl₂ and 5% C₂H₆).
[1 mark] for explaining why (balancing consumption of reactants with prevention of side-product formation). Note: Second mark dependent on correct flow rate.
🧠 Examiner Commentary: How to Analyse Flow Rate Trends
The gas mixture spends too long under UV light. Nearly all reactants react, but high exposure causes excessive multiple substitution ( 48% CH₂Cl₂ ) and termination ( 37% C₂H₆ ).
The mixture passes through too quickly. Minimal side-products form, but reactants do not have time to react, leaving mostly unreacted gas ( 58% CH₄ and 29% Cl₂ ).
The "Sweet Spot": 3 cm³ s⁻¹ yields an outstanding 72% CH₃Cl, striking the ideal compromise between residence time and selectivity.
Topics
Organic Chemistry · 2.5 Hydrocarbons
Question and mark scheme from the WJEC A-Level Chemistry examination, AS Unit 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.