IB syllabus > organic (hl) > 12.2 

These notes were written for the old IB syllabus (2009). The new IB syllabus for first examinations 2016 can be accessed by clicking the link below.

IB syllabus for first examinations 2016

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:

NaOH + CH3CH2Br CH2CH2 + NaBr + H2O

The solvent ethanol makes the hydroxide ion an even stronger base by converting some of it into the ethoxide ion:

OH- + CH3CH2OH CH3CH2O- + H2O

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':

OH- + CH3CH2Br -CH2CH2Br + H2O

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:

-CH2CH2Br CH2CH2 + Br-

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.

OH- + CH3CHBrCH2CH3 CH3CH=CHCH3 or CH2=CHCH2CH3

In the above example the postion of the double bond in the product is either on carbon #1 or on carbon #2.


Resources

Elimination reactions of haloalkanes