Chlorination of Alkanes (AQA A Level Chemistry): Revision Note

Free Radical Substitution

 Free-radical substitution of alkanes

• Alkanes undergo free-radical substitution reactions in which a hydrogen atom is replaced by a halogen atom, such as chlorine or bromine

• Because alkanes are relatively unreactive, ultraviolet (UV) light is required to initiate the reaction

  • The UV light causes homolytic fission of the halogen molecule, forming free radicals that begin the substitution process

• 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

1. 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 halogen radicals in a homolytic fission reaction

• The dot (•) represents an unpaired electron, indicating that a radical has been formed:

One reactant two radicals

Cl–Cl  2Cl^• 

Br–Br  2Br^• 

2. 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 fission is where each atom gets one electron from the covalent bond

• An alkyl free radical is produced

CH₃CH₃ + Cl^• → ^•CH₂CH₃ + HCl

OR

CH₃CH₃ + Br^• → ^•CH₂CH₃ + 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 is 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 + Cl^• → ^•CHClCH₂Cl + HCl

^•CHClCH₂Cl + Cl₂ → CHCl₂CH₂Cl + Cl^•

• This process can continue until full substitution has occurred

• So, ethane would become:

  • Hexachloroethane, C₂Cl₆, with chlorine

  • Hexabromoethane, C₂Br₆, with bromine

3. 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