2-bromopropane can react with potassium hydroxide dissolved in ethanol to produce propene. Explain why this reaction produces an alkene rather than an alcohol, and describe the type of reaction taking place.

Eduqas A-Level Chemistry — 2.6 Halogenoalkanes · Explain · 5 marks · View as Markdown

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

Halogenoalkanes can undergo different reactions depending on the conditions used. When potassium hydroxide is dissolved in ethanol rather than water, the product formed is different.

Model answer (5 marks)

The reaction is an elimination.
1. KOH in ethanol supplies OH⁻, which behaves as a strong base.
2. The base abstracts a proton from a β‑carbon (adjacent to the C–Br bond).
3. Simultaneously the C–Br bond breaks, releasing Br⁻ (or HBr).
4. The removal of H⁺ and Br⁻ generates a C=C double bond.
5. The product is propene, an alkene, rather than an alcohol.

Examiner tips

  • Identify the reaction as elimination (E2).
  • Show that OH⁻ acts as a base, not a nucleophile.
  • Explain β‑hydrogen abstraction and C–Br cleavage.
  • State that a C=C bond forms.

Common mistakes

  • Confusing the reaction with substitution (SN2).
  • Claiming the product is an alcohol or that OH⁻ acts as a nucleophile.
  • Failing to mention the β‑hydrogen or the formation of the double bond.

Mark scheme (5 marks)

  1. Identifies the reaction as elimination
  2. States that the hydroxide ion acts as a base (not a nucleophile) in this reaction
  3. States that a hydrogen atom is removed from a carbon adjacent to the carbon bonded to bromine
  4. States that the bromine (or bromide ion / HBr) is also removed / lost
  5. Explains that a C=C double bond forms between the two carbons, producing an alkene

Key terms in this question

alkene

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