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
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
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 type 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 type 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