Halogenation of Alkanes (DP IB Chemistry): Revision Note
Halogenation of alkanes
Stability of alkanes
Alkanes are relatively stable / unreactive due to the strengths of the C–C and C–H bonds and their non-polar nature
Strength of bonds
Alkanes consist of carbon and hydrogen atoms which are bonded together by single bonds
Unless a lot of heat is supplied, it is difficult to break these strong C-C and C-H covalent bonds
This decreases the alkanes’ reactivities in chemical reactions
Lack of polarity
The electronegativity of the carbon and hydrogen atoms in alkanes are almost the same
This means that both atoms share the electrons in the covalent bond almost equally
Pauling electronegativity values for the elements

As a result of this, alkanes are nonpolar molecules and have no partial positive or negative charges (δ+ and δ– respectively)
Structural formula of ethane showing bond polarities

Alkanes, therefore, do not react with polar reagents
They have no electron-deficient areas to attract nucleophiles
And also lack electron-rich areas to attract electrophiles
Alkanes only react in combustion reactions and undergo substitution by radicals
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
Proving that energy from UV light is required for radical reactions with halogens

The free-radical substitution reaction consists of three steps
Initiation step
In the initiation step, the halogen bond (Cl-Cl or Br-Br) is broken by UV energy to form two radicals
The covalent Cl-Cl bond 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
Cl–Cl 2Cl•
For more information about the initiation step, see our revision note about homolytic fission
Propagation step
The halogen free radicals are very reactive and will attack the unreactive alkanes
One of the methane C-H bond breaks homolytically to produce an alkyl radical
CH4 + Cl• → •CH3 + HCl
The alkyl radical can attack another chlorine molecule to form a halogenoalkane
This also regenerates the chlorine free radical
•CH3 + Cl2 → CH3Cl + Cl•
The regenerated chlorine free radical can then repeat the cycle
For example, the chlorination of ethane is:
ethane + chlorine radical → ethyl radical + hydrogen chloride
CH3CH3 + Cl• → •CH2CH3 + HCl
ethyl radical + chlorine molecule → chloroethane + regenerated chlorine radical
•CH2CH3 + Cl2 → CH3CH2Cl + Cl•
This reaction is not very suitable for preparing specific halogenoalkanes as a mixture of substitution products is formed
If there is enough halogen present, all the hydrogens in the alkane will eventually get substituted
For example, the chlorination of ethane could continue:
chloroethane + chlorine radical → radical + hydrogen chloride
CH3CH2Cl + Cl• → •CH2CH2Cl + HCl
radical + chlorine molecule → 1,2-dichloroethane + regenerated chlorine radical
•CH2CH2Cl + Cl2 → CH2Cl2 + Cl•
This process can repeat until hexachloroethane, C2Cl6, is formed
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
For example, the single substitution of ethane by chlorine can form:
ethyl radical + chlorine radical → chloroethane
•CH2CH3 + Cl• → CH3CH2Cl
ethyl radical + ethyl radical → butane
•CH2CH3 + •CH2CH3 → CH3CH2CH2CH3
chlorine radical + chlorine radical → chlorine molecule
Cl• + Cl• → Cl2
Examiner Tips and Tricks
Make sure you practice and are able to write out these equations, especially the propagation steps
Students frequently get the propagation steps wrong, by showing the formation of a hydrogen radical produced in propagation
This step (CH3CH3 + Cl• → CH3CH2 Cl + H•) does not happen:
Do not fall into this trap!
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