EXPLAINING THE ELIMINATION REACTIONS PRODUCING ALKENES FROM SIMPLE HALOGENOALKANES

This page guides you through the elimination mechanism for the reaction between simple halogenoalkanes like 2-bromopropane and hydroxide ions from, for example, sodium hydroxide.

Elimination involving more complicated halogenoalkanes and the competition between elimination and substitution in these reactions are dealt with on separate pages. These are essential parts of this topic, and you should follow the links at the bottom of this page if you haven't already done so.

The role of the OH^- ion in an elimination reaction

Hydroxide ions have a very strong tendency to combine with hydrogen ions to make water - in other words, the OH^- ion is a very strong base.

In an elimination reaction, the hydroxide ion hits one of the hydrogen atoms in the CH₃ group and pulls it off. This leads to a cascade of electron pair movements resulting in the formation of a carbon-carbon double bond, and the loss of the bromine as Br^-.

The complete elimination mechanism

• The OH^- ion takes one of the hydrogens from the CH₃ group, but it only needs the hydrogen nucleus (a hydrogen ion). That means that the two electrons which originally joined the hydrogen to the carbon aren't being used any more.

• Those two electrons move to form a double bond between the two carbon atoms.

• The approach of those electrons repels the electrons in the carbon-bromine bond right out onto the bromine, throwing the bromine off as a negative ion.

The attack could equally well have been on any of the other hydrogens on the left-hand carbon, or on any on the right-hand one - it simply depends on what the OH^- ion hit.