OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2019: Question 2

1 mark · Medium difficulty · Multiple Choice

Identify the propagation step in the free-radical substitution mechanism of butane reacting with chlorine.

Practise this question

Question

Multiple choice question 2 asking to identify a propagation step in the chlorination of butane. Four equations are given: A shows Cl2 forming two chlorine radicals, B shows a chlorine radical and a chlorobutane radical forming dichlorobutane, C shows chlorobutane and a chlorine radical forming dichlorobutane and a hydrogen radical, and D shows a chlorine radical and chlorobutane reacting to form a chlorobutane radical and HCl. An answer box and a mark allocation of [1] are at the bottom.
Question text

2 Butane reacts with chlorine in the presence of ultraviolet radiation to form a mixture of organic

products.

Which equation shows a propagation step in the mechanism for this reaction?

A Cl2 → •Cl + •Cl

B •Cl + •C4H8Cl → C4H8Cl2

C C4H9Cl + •Cl → C4H8Cl2 + •H

D •Cl + C4H9Cl → •C4H8Cl + HCl

Your answer

[1]

Mark scheme

Show the mark scheme Mark scheme for question 2 showing the correct answer is D.

2 D 1 AO2.1

How to answer it

Free Radical Substitution: Identifying Propagation Steps

What this question tests

This question assesses your understanding of the radical substitution mechanism of alkanes with halogens (specifically butane and chlorine). You must be able to recognise and distinguish between the three stages of a radical mechanism: initiation, propagation, and termination, focusing on the defining feature of propagation steps—that a free radical is used up and another one is generated.

Question 2 Analysis

Free Radical Mechanism Equations

✅ Correct Answer: Option D

•Cl + C₄H₉Cl → •C₄H₈Cl + HCl

Why it's correct: This equation features a free radical ( •Cl ) on the reactant side and generates a new free radical ( •C₄H₈Cl ) on the product side. This is the hallmark of a propagation step.

💡 Key Knowledge

  • Initiation: Homolytic fission of a halogen molecule to form two radicals (e.g., Cl₂ → 2•Cl). Starts with no radicals, produces radicals.
  • Propagation: A radical reacts with a non-radical to form a new molecule and a new radical. (Radicals on both sides).
  • Termination: Two radicals combine to form a single non-radical molecule. Starts with radicals, produces zero radicals.

🧠 Exam Technique

Count the radicals! Look at the types of particles on both sides of the equation:

  • Radical + Non-radical → Non-radical + Radical (Propagation)

❌ Common Errors

  • Option A shows initiation (molecule forming two radicals).
  • Option B shows termination (two radicals combining to make a stable molecule).
  • Option C invents an impossible hydrogen radical ( •H ) product instead of generating a carbon-based alkyl/chloroalkyl radical correctly balanced with HCl.
Mark Scheme Allocation: 1 mark awarded for selecting D (AO2.1 - Application of chemical knowledge).

Topics

Module 4: Core organic chemistry · 4.1 Basic concepts and hydrocarbons

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.