OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2025: Question 8

1 mark · Easy difficulty · Multiple Choice

Identify the correct statement regarding the role of hydrogen bromide in its reaction with ethene.

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Question

Multiple-choice question 8: 'Ethene can be reacted with hydrogen bromide. Which statement about this reaction is correct?' Option A states 'H⁻ acts as a nucleophile.' Option B states 'Br⁺ acts as an electrophile.' Option C states 'Hδ+—Brδ- acts as an electrophile.' Option D states 'Hδ+—Brδ- acts as a nucleophile.' An answer box is provided at the bottom with 1 mark allocated.
Question text

8 Ethene can be reacted with hydrogen bromide.

Which statement about this reaction is correct?

A H– acts as a nucleophile.

B Br+ acts as an electrophile.

C Hδ+—Brδ– acts as an electrophile.

D Hδ+—Brδ– acts as a nucleophile.

Your answer

[1]

Mark scheme

Show the mark scheme Mark scheme table row showing the entry for question 9 with answer D, allocated 1 mark, and guidance note 'ALLOW A'. The row for question 8 is partially cut off above it.

8 C 1

How to answer it

Electrophilic Addition of Hydrogen Bromide to Ethene

WHAT THIS QUESTION TESTS

Core mechanism concepts in organic chemistry (OCR Chemistry A):

  • Reaction classification: Identifying that alkenes undergo electrophilic addition due to the electron-rich C=C π-bond.
  • Role of the reagent: Recognising that polar diatomic molecules like H–Br act as electrophiles because the electron-deficient Hδ+ accepts an electron pair.
  • Definitions: Distinguishing an electrophile (an electron-pair acceptor) from a nucleophile (an electron-pair donor).
QUESTION 8 (MULTIPLE CHOICE)

Identifying Reagent Roles in Alkene Addition

Full Mark Breakdown & Mechanism Analysis

✅ Correct Answer: C

Hδ+—Brδ- acts as an electrophile.

In the addition of hydrogen bromide to an alkene:

  • Bromine is more electronegative than hydrogen, setting up a permanent dipole: Hδ+—Brδ-.
  • The C=C double bond has a region of high electron density in its exposed π-bond.
  • The Hδ+ atom is electron-deficient and accepts a pair of electrons from the C=C bond. Because HBr accepts an electron pair, the entire molecule acts as an electrophile.
Mark Scheme Reference: 1 mark awarded for selecting C.

💡 Key Knowledge & Mechanism

  • Electrophile: An electron-pair acceptor.
  • Nucleophile: An electron-pair donor.
  • Step 1: The curly arrow starts from the C=C double bond (π-bond) and points directly to the Hδ+ of the H–Br molecule.
  • Heterolytic Fission: Simultaneously, the H–Br bond breaks heterolytically, with the curly arrow moving from the bond onto the Brδ- atom, releasing a bromide ion (:Br-).
  • Intermediate: A positively charged carbocation intermediate (CH3—CH2+) is formed.
  • Step 2: The bromide ion (:Br-) acts as a nucleophile, donating a lone pair to the positive carbon to yield bromoethane.
Option Statement Status Examiner Explanation
A H- acts as a nucleophile Incorrect A hydride ion (H-) is never formed here. Hydrogen carries a partial positive charge (Hδ+) and accepts electrons; it does not donate a pair.
B Br+ acts as an electrophile Incorrect Bromine is more electronegative than hydrogen, so it carries a partial negative charge (Brδ-), leaving as a bromide ion (:Br-). There is no Br+ species in this mechanism.
C Hδ+—Brδ- acts as an electrophile Correct The polar HBr molecule approaches the electron-rich C=C bond and accepts an electron pair via its δ+ hydrogen atom. By definition, it is the electrophile.
D Hδ+—Brδ- acts as a nucleophile Incorrect A nucleophile donates an electron pair. The electron-rich species donating electrons here is the alkene (specifically the π-bond of ethene), not HBr.

🧠 Exam Technique: Mechanism Questions

  • Look at the reaction title: The reaction is classified as Electrophilic Addition. That immediately tells you the attacking reagent (HBr) is an electrophile, instantly ruling out options A and D.
  • Check polarities: Remember periodic electronegativity trends (F > O > Cl > Br > C > H). In H–Br, Br is significantly more electronegative than H, which confirms that H is δ+ and Br is δ-.
  • Deduce who accepts electrons: Electrons always flow from negative/electron-rich to positive/electron-poor. The π-electrons flow to the Hδ+, so HBr is accepting electrons (the hallmark of an electrophile).

❌ Common Misconceptions & Traps

  • Confusing the role of Bromine: In halogenation with Br2, an induced dipole creates a Brδ+ that attacks first. Students sometimes misapply this to HBr and wrongly guess that bromine is the electrophilic center (Option B).
  • Misidentifying ions: Assuming H- forms because "acids lose H". In HBr, heterolytic fission gives H to the alkene and leaves the electron pair on Br to make Br-, never hydride (H-).
  • Reversing Nucleophile and Electrophile: Confusing the donor (ethene) and the acceptor (HBr). Ethene donates π-electrons; HBr receives them.
📝 Note on Mark Scheme Clipping:

The mark scheme image provided shows row 9 ( 9 | D | 1 | ALLOW A ), which corresponds to the subsequent question on the paper. For Question 8, the unambiguous correct answer is C.

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 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.