E2 mechanism: elimination from 2-bromopropane

With a strong base and heat, halogenoalkanes undergo elimination instead of substitution. In E2 the base removes a hydrogen from the carbon next to the C–Br carbon while bromide leaves, all in one step.

Try it in the builder →  Change the reactants or conditions and generate your own mechanism.

Elimination (E2)

Ethoxide ions act as a base: one removes a proton from a carbon next to the C–Br carbon, the C–H bonding electrons form a C=C double bond, and bromide leaves. The product is propene.

Overall reaction

1. 1Concerted eliminationconcerted (E2)

   ethoxide ion

   +

   2-bromopropane

   propene

   +

   ethanol

   +

   bromide ion

   Electron movement

   1. A lone pair on the ethoxide oxygen removes a hydrogen from the carbon next to the C–Br carbon.
   2. The C–H bonding electrons move in to form the C=C double bond.
   3. The C–Br bond breaks and bromine leaves as Br⁻.

   All three electron movements happen together through one transition state. For this to work, the H being removed and the Br leaving are best lined up on opposite sides of the C–C bond (anti-periplanar).