The Free Radical Substitution Mechanism (Edexcel International A Level Chemistry)

• The free-radical substitution reaction consists of three steps:

Initiation step

• In the initiation step, the Cl-Cl or Br-Br is broken by energy from the UV light
• This produces two radicals in a homolytic fission reaction

The first step of the free-radical substitution reaction is the initiation step, in which two free radicals are formed by sunlight

Propagation step

• The propagation step 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 (each atom gets an electron from the covalent bond)
  • An alkyl free radical is produced
  • This can attack another chlorine/bromine molecule to form the halogenoalkane and regenerate the chlorine/bromine free radical
  • This free radical can then repeat the cycle

The second step of the free-radical substitution reaction is the propagation step in which the reaction grows in a chain reaction

• • 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 (eg. ethane will become C₂Cl₆/C₂Br₆)

The free-radical substitution reaction gives a variety of products and not a pure halogenoalkane

Termination step

• The termination step is when the chain reaction terminates (stops) due to two free radicals reacting together and forming a single unreactive molecule
  • Multiple products are possible

The final step in the substitution reaction to form a single unreactive molecule

Limitations of the free radical mechanism

• In the termination step, there are a number of possibilities 
• Remember that termination involves any free radical bonding with another free radical
• If we have two ^•CH₃ radicals, they can bond to form ethane, CH₃CH₃
  • ^•CH₃ + ^•CH₃ → CH₃CH₃
  • If we are trying to form a chloroalkane, then ethane is an impurity

Further substitution

• Excess chlorine present when reacted with methane in the presence of UV light will promote further substitution and could produce CH₂Cl_2, CHCl_3, CCl₄
• Further substitution can occur as follows
  • CH₃Cl + ^•Cl → HCl + ^•CH₂Cl
  • ^•CH₂Cl + Cl₂ → CH₂Cl₂ + ^•Cl
• These reactions could occur
  • CH₃Cl + Cl₂ → CH₂Cl₂ + HCl 
  • CH₂Cl₂ + Cl₂ → CHCl₃ + HCl
  • CHCl₃ + Cl₂ → CCl₄ + HCl

Substitution of different carbon atoms 

• If we have an alkane with a middle carbon such as propane, substitution can occur here
• Propagation steps for the substitution of propane with excess bromine in the presence of UV light on the middle carbon are as follows:
  • CH₃CH₂CH₃ + ^•Br → CH₃^•CHCH₃ + HBr
  • CH₃^•CHCH₃ + Br₂ → CH₃CH(Br)CH₃ + ^•Br
• If the question asks for the halogen to be substituted onto a middle carbon, you must show the radical dot in the correct place, so on the electron-deficient carbon