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Elimination — hydroxide ions
Year 1
AQA 3.3.3
Nucleophilic substitution & elimination · KOH in ethanol · hot, ethanolic
Watch every step, with the electron flow and reasoning shown.
Reagents: KOH in ethanol · Conditions: hot, ethanolic
Step 1 of 1
- 1.OH⁻ acts as a BASE here: its lone pair attacks a hydrogen on the β-carbon (the carbon next to C–Br).
- 2.The C–H bonding pair moves in to become the new π bond between the two carbons.
- 3.Simultaneously the C–Br bond breaks heterolytically, releasing bromide.
Water, bromide and propene form in one concerted step — OH⁻ acted as a base, not a nucleophile.
Product: propene
Watched it? Now draw it yourself — that's what the exam asks.
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