AQA A-Level Chemistry Paper 3, June 2025: Question 10

1 mark · Easy difficulty · Multiple Choice

Identify which diagram correctly shows the curly arrows in the mechanism for the elimination reaction of bromoethane.

Practise this question

Question

Question 10 asks: Which diagram shows a correct step in the mechanism for an elimination reaction of bromoethane? Four options are given, labeled A to D, displaying displayed formulas of bromoethane interacting with a hydroxide ion via curly arrows. Option A shows the hydroxide attacking the carbon with the bromine. Option B shows the hydroxide attacking a hydrogen on the adjacent carbon, the C-H bond electrons moving to form a C=C double bond, and the C-Br bond breaking onto the bromine. Option C and D show incorrect arrow directions from the hydrogen or carbon to the hydroxide ion.
Question text

10 Which diagram shows a correct step in the mechanism for an elimination reaction

of bromoethane?

[1 mark]

A

B

C

D

Mark scheme

Show the mark scheme Mark scheme for question 10 indicating that the correct answer is B, awarded 1 mark (AO1), showing the mechanism diagram where the lone pair on hydroxide attacks a beta-hydrogen, the C-H bond moves to form a double bond between the two carbons, and the C-Br bond breaks to bromine.

10 B 1 (AO1)

How to answer it

AQA A-Level Chemistry • Halogenoalkanes

Mechanism of Elimination in Bromoethane

What this question tests

This question assesses your ability to identify and represent curly-arrow mechanisms accurately for the base-catalysed elimination of a haloalkane (producing an alkene), distinguishing it from nucleophilic substitution.

  • The role of the hydroxide ion (:⁻OH) acting as a base (proton acceptor) rather than a nucleophile.
  • Correct curly arrow origin (lone pair or covalent bond) and destination (atom or bond).
  • Understanding which hydrogen atom is abstracted (adjacent / β-carbon) and how the C=C double bond forms simultaneously with the loss of the halide leaving group.

Question 10 Breakdown

Mark Allocation: 1 Mark (AO1)

✅ Correct Answer: B

Option B correctly displays all three concerted curly arrow movements required in the elimination mechanism:

  1. An arrow starting from the lone pair on :⁻OH pointing directly to a hydrogen atom on the carbon adjacent to the C–Br carbon.
  2. An arrow starting from the middle of that C–H bond pointing to the C–C bond, representing the formation of the C=C π-bond.
  3. An arrow starting from the C–Br bond pointing directly to the Br atom, showing heterolytic fission to form Br⁻.

💡 Key Knowledge

  • Elimination Conditions: Hot, ethanolic potassium/sodium hydroxide under reflux ( OH⁻ in ethanol ).
  • Role of OH⁻: Acts as a Brønsted-Lowry base, removing a proton (H⁺) from the β-carbon.
  • Overall Equation:
    CH₃CH₂Br + OH⁻ → CH₂=CH₂ + H₂O + Br⁻
  • Contrast with Substitution: In aqueous conditions at lower temperatures, OH⁻ acts as a nucleophile, attacking the δ+ carbon atom bonded to the halogen.

📐 Mechanism Walkthrough (Concerted E2 Step)

Examiners look for three precise curly arrows:

  1. Arrow 1 (Proton Abstraction): Originates at the lone pair of electrons on the oxygen of :⁻OH and terminates at a hydrogen atom on the methyl carbon (C2).
  2. Arrow 2 (Double Bond Formation): Originates from the electron pair in the C–H single bond and points directly between the two carbon atoms (C–C bond).
  3. Arrow 3 (Halide Leaving): Originates from the electron pair in the C–Br bond and points to the bromine atom.

❌ Why Options A, C, and D are Incorrect

  • Option A is wrong: It depicts nucleophilic substitution (OH⁻ attacking the C–Br carbon), not elimination.
  • Option C is wrong: The arrow is shown starting from a hydrogen on the carbon that also carries the bromine atom (α-carbon), and points backwards toward the OH⁻ instead of the base attacking the proton.
  • Option D is wrong: It also targets the hydrogen on the same carbon as the bromine, which cannot form an alkene double bond across the adjacent carbons.

🧠 Exam Technique & Examiner Tips

  • Identify the reaction type first: The question asks for elimination. That immediately rules out nucleophilic attack on the carbon skeleton (eliminating diagram A).
  • Check the carbon atom: For elimination to yield an alkene, the proton removed must come from a carbon adjacent to the C–halogen bond. If the arrow targets the C bonded to Br, it is wrong.
  • Arrow tails and heads matter: Ensure curly arrows always start at a pair of electrons (either a lone pair or the centre of a covalent bond) and end where a new bond or lone pair forms.

Topics

Organic Chemistry · 3.3.3 Halogenoalkanes · 3.3.1 Introduction to Organic Chemistry

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.