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3.3.2Organic Chemistry

Elimination Reactions of Halogenoalkanes — Oxford AQA International A Level Chemistry Revision Notes & Topical Questions

5 min read 5 parts Lesson 13 of 55

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Part 1 of 5The Reaction

3.3.2 Elimination Reactions of Halogenoalkanes
1
A halogenoalkane is heated with hydroxide ions dissolved in ethanol, with no water present.
2
The hydroxide ion acts as a base rather than as a nucleophile, removing a hydrogen atom from a carbon next to the C–X bond.
3
The electrons of the C–H bond move to form a C=C double bond between the two carbons.
4
At the same time the C–X bond breaks and the halide ion leaves.
5
The overall equation is CH₃CH₂Br + KOH → CH₂=CH₂ + KBr + H₂O, forming ethene.

Part 2 of 5Substitution or Elimination

Substitution conditions
  • Aqueous NaOH at room temperature or warm favours nucleophilic substitution and gives an alcohol.
Elimination conditions
  • Ethanolic NaOH at a high temperature favours elimination and gives an alkene.
Type of halogenoalkane
  • Primary halogenoalkanes tend to give substitution, while tertiary ones tend to give elimination.
Role of OH⁻
  • In substitution it donates a lone pair to carbon; in elimination it accepts a proton from an adjacent carbon.
Both can happen
  • A mixture of products is often formed, and the conditions only shift the balance.
FeatureSubstitutionElimination
ReagentNaOH(aq)NaOH in ethanol
Temperaturewarmhot
Productalcoholalkene
Role of OH⁻nucleophilebase

Part 3 of 5Isomeric Products

Isomeric Products
2-bromobutane
  • Hydrogen can be removed from carbon 1 or from carbon 3, giving two different alkenes.
Products
  • But-1-ene is formed by loss of a hydrogen from carbon 1, and but-2-ene from carbon 3.
E/Z isomers
  • But-2-ene itself can exist as the E isomer or the Z isomer, so three alkenes may be formed in total.
Symmetrical halogenoalkane
  • If every adjacent carbon is equivalent, as in bromoethane, only one alkene can form.

Part 4 of 5Worked Example

Question Write the equation for the reaction of 2-bromopropane with ethanolic potassium hydroxide and state why only one alkene forms.

  1. Hydroxide removes H from a carbon adjacent to C–Br, and Br⁻ leaves.
  2. The equation is CH₃CHBrCH₃ + KOH → CH₃CH=CH₂ + KBr + H₂O.
  3. Both carbons adjacent to C–Br are CH₃ groups and are equivalent, so only propene can form.

Answer

  • CH₃CHBrCH₃ + KOH → CH₃CH=CH₂ + KBr + H₂O;
    • propene is the only alkene.

Part 5 of 5Exam Checklist & Tips

3.3.2 – Key Points for the Exam

  • Elimination Removal of a small molecule from a larger one to form a double bond.
  • Base A species that accepts a proton.
  • Ethanolic Dissolved in ethanol rather than water.
  • Dehydrohalogenation Removal of hydrogen halide from a halogenoalkane.

Common Mistakes

  • Writing aqueous conditions for elimination; ethanolic conditions are required.
  • Forgetting the water product in the elimination equation.
  • Describing OH⁻ as a nucleophile in elimination, when it acts as a base.
  • Missing that two different adjacent carbons can give different alkene isomers.

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