AS-Level Alkanes

  • Covalent bonds can break in two ways

    • Heterolytic fission: bond breaks unevenly and both bonded electrons go to one atom, creating positive and negative ions.

    • Homolytic fission: bond breaks evenly and each bonded atom get one electron, forming free-radicals.

  • Halogens can react with alkanes in free-radical substitution, the mechanism occurs as a ‘chain’ reaction with three stages

    • Initiation - U.V. light is needed to start the reaction and cause homolytic fission of the halogen molecule, creating two halogen radicals.

    • Propagation – radical species react with the alkane and get substituted into the molecule, creating further radicals.

    • Termination – two radical species combine to create a covalent bond and terminate the chain as the product is not a free radical.


Free Radical Substitution (of alkanes)


In a covalent bond, a pair of electrons is shared between two atoms. When a covalent bond breaks, it can break in two ways.  The breaking of a bond is called fission.


In one scenario, both bonding electrons go to one of the atoms. This creates a negative and a positive ion. The species produced are different types. This kind of bond breaking is called heterolytic fission and is the most common type in organic chemistry.










In the second scenario, the bonding electrons can be shared between the two atoms when the bond breaks. This produces two products that have one unpaired electron each.


A species with one unpaired electron is called a radical.  Both species formed are the same type. This kind of bond breaking is called homolytic fission. Although not as common as hetereolytic fission, homolytic fission produces radical species that are highly reactive so is of great interest to organic chemists.


A radical is written with a • symbol to show one unpaired electron (i.e. Br•).




Free-radical substitution of alkanes with halogens


In a substitution reaction, a reacting species is substituted (swapped) for a bonded species in a compound.


Alkanes can react with halogens in the presence of UV light. A halogen is substituted for a hydrogen atom in the alkane to form a halogenoalkane.



The process happens in stages. These stages link together to gives us a ‘mechanism’ that describes how products are formed from the reactants.



Methane with chlorine substitution


Initiation step


A halogen and an alkane will not react without UV light. The UV light is required to form a halogen radical.


In this case the UV light gives energy to the covalent bond in Cl , so that it undergoes homolytic fission (see above). Producing two Cl• radicals.




Propagation step


Once radicals have been produced in the initiation step, they are free to react with the alkane present (in this case methane).

As radicals contain an unpaired electron, unless a radical is reacting with another radical, the products produced must also contain an unpaired electron. We therefore produce new radicals. These new radicals can then react again, producing the desired product and the same species as our starting radical.


Termination step


When two radicals react together, two unpaired electrons come together and a covalent bond forms. This removes the radicals and ends the chain reaction. Any two radicals can react to end the chain process.




It is important to note that further substitution reactions can happen between the halogenoalkane formed and chlorine radicals to give us di-chloromethane, tri-chloromethane and tetra-chloromethane (carbon tetrachloride).