20.3 - Elimination reactions
20.3.1: Describe, using equations, the elimination of HBr from bromoalkanes
Bromoalkanes give alkenes by dehydrohalogenation using an ethanolic solution of sodium hydroxide:
The solvent ethanol makes the hydroxide ion an even stronger base by converting some of it into the ethoxide ion:
The ethoxide ion is more basic due to the +I (positive induction) effect of the alkyl group increasing the electron density of the lone pair.
20.3.2: Describe and explain the mechanism for the elimination of HBr from bromoalkanes
The mechanism first involves removal of a proton (abstraction) by the base. This happens to the hydrogen atom which is on the carbon adjacent to the carbon holding the bromine atom. This hydrogen atom is said to be an 'alpha-hydrogen':
This produces a negatively charged species (a carbanion), which then rearranges by moving the negative charge lone pair into a pi orbital between the two carbon atoms with the loss of the bromine atom as a bromide ion:
The product is an alkene, in this case ethene.
Asymmetric bromoalkanes can give two different products if they have two non-equivalent alpha-hydrogen atoms.
In the above example the postion of the double bond in the product is either on carbon #1 or on carbon #2.