20.3 – Elimination Reactions
20.3.1 – Describe, using equations, the elimination of HBr from bromoalkanes
Halogenoalkanes can undergo both substitution and elimination reactions. The conditions determine the outcome of the reaction. When bromoethane is reacted in hot alcoholic sodium hydroxide (NaOH) solution, ethene is formed in an elimination reaction.
20.3.2 – Describe and explain the mechanism for the elimination of HBr from bromoalkanes
There are two types of elimination reactions:
E2 – Bimolecular Elimination
This occurs when the halogenoalkane reacts with a strong base, and involved a single step. The base removes a hydrogen atom whilst the leaving group (X) simultaneously begins to leave. As a result, the rate is dependent on the concentration of both the base and the halogenoalkane.
In this reaction, the OH– acts as a base and accepts a proton (H+) from the halogenoalkane to make water (H2O). A halide ion is released and a double bond forms. This is similar to SN2 mechanisms. However, in the presence of a strong base (RO–), E2 reactions are favoured over nucleophilic substitution.
E1 – Unimolecular Elimination
Tertiary halogenoalkanes typically undergo E1 reactions. In this type of reaction, there are two steps. The rate is dependent on the slowest step, where the halide ion is lost. The fast step is the formation of water and the double bond.
In the slow step, a carbocation is formed and the halide is released. In the fast step, the double bond forms, releasing a H+ ion.
Remember it is the conditions of the reaction – warm alcohol – that determine whether an elimination reaction will occur.