AQA A-Level Chemistry Paper 3, June 2022: Question 22

1 mark · Medium difficulty · Multiple Choice

Identify the reaction mechanism when 2-bromopropane reacts with bromine to form 2,2-dibromopropane.

Practise this question

Question

Multiple-choice question 22 asks: '2-Bromopropane reacts with bromine to form 2,2-dibromopropane. What is the name of the mechanism of this reaction?' Four options are provided: A Electrophilic addition, B Elimination, C Free-radical substitution, and D Nucleophilic substitution.
Question text

22 2-Bromopropane reacts with bromine to form 2,2-dibromopropane.

What is the name of the mechanism of this reaction?

[1 mark]

A Electrophilic addition

B Elimination

C Free-radical substitution

D Nucleophilic substitution

Mark scheme

Show the mark scheme Mark scheme table row for question 22 showing the correct answer is option C (AO2), described as 'Free-radical substitution', worth 1 mark.

22 C (AO2) 1 Free-radical substitution

How to answer it

Mechanism Identification: Halogenation of Haloalkanes

What this question tests

This question assesses your ability to deduce reaction mechanisms based on functional groups and reagent combinations:

  • Recognising that replacing a hydrogen atom with a halogen atom on a saturated carbon chain is a free-radical substitution.
  • Distinguishing substitution mechanisms (free-radical vs nucleophilic) based on attacking species.
  • Eliminating distractor mechanisms (electrophilic addition, elimination) by checking whether unsaturation is present or formed.
Question 22 (1 Mark)

Identifying the Mechanism for Forming 2,2-Dibromopropane

Reaction: CH₃CH(Br)CH₃ + Br₂ → CH₃C(Br)₂CH₃ + HBr

✅ Correct Answer

C: Free-radical substitution

Award 1 mark for selecting option C (Assessment Objective AO2).

💡 Key Knowledge

  • Alkane / Haloalkane + Halogen (X₂): Reacts via free-radical substitution under UV light.
  • A C–H bond is broken and replaced with a C–Br bond:
    CH₃CH(Br)CH₃ + Br₂ → CH₃C(Br)₂CH₃ + HBr
  • Nucleophilic substitution requires an electron-pair donor with a lone pair or negative charge (e.g. OH⁻, CN⁻, NH₃), not molecular bromine (Br₂).

🧠 Exam Technique & Step-by-Step Elimination

  1. Look at the starting material: 2-bromopropane is saturated (contains only single C–C and C–H / C–Br bonds). Therefore, it cannot undergo electrophilic addition (rules out A).
  2. Look at the product: 2,2-dibromopropane has gained a bromine atom while maintaining a saturated carbon skeleton; it has not formed a C=C double bond, so this is not an elimination (rules out B).
  3. Identify the reagent type: Bromine (Br₂) acts as a source of halogen free radicals (Br•) via homolytic fission, not as a nucleophile. Therefore, it is free-radical substitution (rules out D).

❌ Common Errors & Pitfalls

  • The "Haloalkane Reflex": Seeing a haloalkane and automatically picking Nucleophilic substitution. Remember: haloalkanes undergo nucleophilic substitution when attacked by nucleophiles (e.g. aqueous NaOH, KCN, NH₃). Here, it reacts with elemental bromine (Br₂), which introduces an additional halogen via radical chemistry.
  • Confusing Addition with Further Substitution: Students sometimes confuse the addition of Br₂ across an alkene (electrophilic addition) with the replacement of H on an alkane/haloalkane by Br₂ (free-radical substitution).

Topics

Organic Chemistry · 3.3.1 Introduction to Organic Chemistry · 3.3.2 Alkanes

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.