Explain the mechanism by which ethene reacts with hydrogen bromide to form bromoethane, including the role of the π bond and the nature of each step.

IB DP Chemistry Standard Level (2023 syllabus) — R3.3 Electron sharing reactions · Explain · 4 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

Model answer (4 marks)

The π electrons of ethene act as a nucleophile and attack the δ+ hydrogen of H–Br. HBr behaves as an electrophile. This gives a carbocation intermediate when H bonds to one carbon, leaving the other carbon positively charged. The bromide ion then attacks the carbocation to give bromoethane. The overall process is electrophilic addition.

Examiner tips

Use the word ‘electrophilic addition’ to summarise the mechanism
Show the π electrons as the nucleophile and HBr as the electrophile
Mention the carbocation intermediate and the final attack by Br⁻
Keep the answer concise – 4 marks only

Common mistakes

Confusing the role of HBr (electrophile vs nucleophile)
Forgetting to state the carbocation intermediate
Using vague terms like ‘reaction’ instead of ‘electrophilic addition’

Mark scheme (4 marks)

The π electrons of ethene act as a nucleophile / electron donor and attack the δ+ hydrogen of the polar H–Br bond.
H–Br acts as the electrophile; HBr is described as (acting as) an electrophile in this reaction.
A carbocation intermediate forms when H bonds to one carbon, leaving the other carbon with a positive charge.
The bromide ion (Br⁻) then attacks the carbocation to form bromoethane; the overall mechanism is electrophilic addition.

Key terms in this question

π bond