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Chlorination of alkanes




AQA Content

Explain this reaction as a free-radical substitution mechanism involving initiation, propagation and termination steps.

Specification Notes

The reaction of methane with chlorine.


Free radical substitution
Alkanes are notoriously unreactive – they are non-polar, don’t have a functional group, lack a reactive double bond and have strong covalent carbon-carbon and carbon-hydrogen bonds. They generally need an input of energy to get them to react. Halogens, however, will readily react with alkanes in the presence of ultraviolet light.

Ultraviolet light will break the covalent bond in chlorine, for example, with an electron from the bond associating with each chlorine atom. This is known as homolytic fission (splitting) and leads to the creation of free radicals. The radical species is shown with a dot to the side...

(Notice that the arrows are 'harpoons' showing the movement of one electron)

Radicals are very reactive and will 'attack' any available molecule

There are 3 distinct parts...
The halogen bond is broken, through energy supplied by UV light, to form two radicals via homolytic fission of chlorine.
The progression of the substitution reaction through a chain reaction
Reactive free radicals attack unreactive alkanes.
Homolytic fission of the C-H bond occurs producing an alkyl free radical.
This can attack another chlorine molecule to form the halogenoalkane and regenerate the chlorine free radical (this then repeats the cycle).
Mixture of products formed.
With enough chlorine, eventually get full substitution (C2Cl6)

Occurs when two free radicals form an unreactive molecule – you get multiple products:

Free Radical Substitution of Methane



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