AQA A Level Chemistry

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3.2.4 Chlorination of Alkanes

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Free Radical Substitution

 Free-radical substitution of alkanes

  • Alkanes can undergo free-radical substitution in which a hydrogen atom gets substituted by a halogen (chlorine/bromine)
  • Since alkanes are very unreactive, ultraviolet light (sunlight) is needed for this substitution reaction to occur
  • The free-radical substitution reaction consists of three steps:
    • In the initiation step, the halogen bond (Cl-Cl or Br-Br) is broken by UV energy to form two radicals
    • These radicals create further radicals in a chain reaction called the propagation step
    • The reaction is terminated when two radicals collide with each other in a termination step

  • Alkanes can undergo free-radical substitution in which a hydrogen atom gets substituted by a halogen (chlorine/bromine)
    • Ultraviolet light (sunlight) is needed for this substitution reaction to occur

  • The free-radical substitution reaction consists of three steps

Hydrocarbons Colour Change of Solution, downloadable AS & A Level Chemistry revision notes

The fact that the bromine colour has disappeared only when mixed with an alkane and placed in sunlight suggests that the ultraviolet light is essential for the free radical substitution reaction to take place

Initiation step

  • The first step of the free-radical substitution reaction is the initiation step
  • In the initiation step the Cl-Cl or Br-Br is broken by energy from the UV light
  • Each atom takes one electron from the covalent bond
  • This produces two radicals in a homolytic fission reaction

One reactant rightwards arrow with UV on top two radicals

Cl–Cl rightwards arrow with UV on top 2Cl 

Br–Br rightwards arrow with UV on top 2Br 

Propagation step

  • The second step of the free-radical substitution reaction is the propagation step 
  • This refers to the progression (growing) of the substitution reaction in a chain reaction
  • Free radicals are very reactive and will attack the unreactive alkanes
  • A C-H bond breaks homolytically 
    • Remember: Homolytic fissions is where each atom gets one electron from the covalent bond
  • An alkyl free radical is produced

CH3CH3 + ClCH2CH3 + HCl

OR

CH3CH3 + BrCH2CH3 + HBr

  • This can attack another chlorine / bromine molecule to form a halogenoalkane and regenerate the chlorine / bromine free radical

CH2CH3 + Cl2 → CH3CH2Cl + Cl 

OR

CH2CH3 + Br2 → CH3CH2Br + Br 

  • The regenerated chlorine / bromine free radical can then repeat the cycle

  • This reaction is not very suitable for preparing specific halogenoalkanes as a mixture of substitution products are formed
  • If there is enough chlorine / bromine present, all the hydrogens in the alkane will eventually get substituted
  • For example, ethane could be substituted to become chloroethane and then further substituted:
    • First substitution:

CH3CH3 + Cl → CH2CH3 + HCl

CH2CH3 + Cl2 → CH3CH2Cl + Cl 

    • Second substitution:

 CH3CH2Cl + Cl → CH2CH2Cl + HCl

CH2CH2Cl + Cl2 → CH2ClCH2Cl + Cl

    • Third substitution:

CH2ClCH2Cl + ClCHClCH2Cl + HCl

CHClCH2Cl + Cl2 → CHCl2CH2Cl + Cl

  • This process can continue until full substitution has occured
  • So, ethane would become:
    • Hexachloroethane, C2Cl6, with chlorine
    • Hexabromoethane, C2Br6, with bromine

Termination step

  • The final step in the substitution reaction is the termination step
  • This is when the chain reaction terminates (stops) due to two free radicals reacting together and forming a single unreactive molecule

Two radicals → one product

  • Multiple products are possible, dependent on the radicals involved
  • For example, in the single substitution of ethane with chlorine:

CH2CH3 + Cl → ClCH2CH3 

CH2CH3 + CH2CH3 → CH3CH2CH2CH3 

Cl + Cl → Cl2

Exam Tip

If you are asked to give an equation for the termination step of a free radical reaction / mechanism, you should not give the equation reforming the original halogen as this is often marked as "ignore" on mark schemes 

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