Free radical substitution header image.


Free radical substitution reactions

Ultraviolet light (UV) has more than enough energy to break the covalent bonds holding some molecules together; the covalent bonds will break in a homolytic manner resulting in the formation of atoms with an unpaired electron. These atoms with unpaired electrons are called free radicals and they are very reactive. The image below shows how the covalent bond between the two atoms in a chlorine molecule (Cl2) can be broken homolytically by UV light to form two chlorine free radicals. The covalent bond in a chlorine molecule splits homolytically to form two chlorine free radicals when UV is shone onto it

The bond enthalpy of the chlorine covalent bond in a chlorine molecule is only 242kJmol-1. This means that a photon of UV light has more than enough energy to break or cleave the Cl-Cl bond and form two chlorine free radicals; as shown above. The two equations shown below also show this homolytic bond cleavage taking place. Since each chlorine atom has an identical electronegativity value when the bond breaks each chlorine atom is able to attract its own electron and form the two chlorine free radical containing an unpaired electron.

Equations to show the homolytic bond cleavage of a chlorine chlorine bond in a chlorine molecule.

Halogenalkane formation

Image to show that halogenalkanes can be made by reacting a halogen with an alkane in the presence of UV light

Alkanes are chemically unreactive substances largely due to the fact that the C-H bond in an alkane molecule is non-polar and so is not susceptible to attack by electrophiles or nucleophiles. However alkanes will react violently with chlorine in the presence of sunlight or a camera flash, which is used to start or initiate the reaction. The sunlight contains UV radiation which is the same type of radiation responsible for sunburn and over exposure to UV radiation can also prematurely age the skin, however for our purposes the ultraviolet radiation will supply enough energy to break the relatively weak bond in a chlorine molecule and form two chlorine free radicals. These free radicals are high reactive since they contain an unpaired electron and will immediately react with the first "thing" they come into contact with which in our case is likely to be an alkane molecule.

Halogenalkanes

For example chlorine (Cl2) will react with methane (CH4) to form a mixture of chloroalkanes as shown below. The chlorine free radicals produced in this reaction will replace or substitute for the hydrogen atoms on the methane molecule to form a halogenalkane or haloalkane molecule. A mixture of four halogenalkane molecules is likely produced in this reaction; these are shown in the image below; but it is possible to adjust the initial reaction mixture to obtain mainly one of these halogenalkane molecules as the main component in the mixture produced.

3d models of the products from the reaction of chlorine with methane.  Free radiacl substitution reaction. A mixture of chlorine and methane is palced in a plastic bottle and a flash gun is used to set off the reaction.

The mechanism of free radical substitution reactions

This reaction of chlorine with methane is an example of a free radical substitution reaction, these reactions are free radical chain reactions that proceed via 3 steps; these 3 separate steps are called:

Step1 - initiation

The initiation step generates or produces the free radicals that are needed to start the reaction between the chlorine molecules and the methane molecules. It is the presence of a small but constant number of these reactive free radicals that allows the reaction to proceed. The free radicals which are needed to maintain this chain reaction are constantly being used up and regenerated while the reaction takes place, the reaction will stop once the concentration of free radicals present drops.

For example if a mixture of methane and chlorine are placed in a plastic bottle and kept in the dark then no reaction occurs, however if a bright camera flash is set off then a very violent reaction starts. The camera flash will produce photons of light with enough energy to break the Cl-Cl bond homolytically to form 2 chlorine free radicals, we can show this as:


Equation to show the homolytic bond fission or cleavage in a chlorine molecule.
The photons of UV light do not contain enough energy to break the stronger C-H bonds in methane

Step2-propagation

The key to understanding these free radical substitution reactions are the free radicals. It is the presence of these reactive free radicals that enables the chain reaction to start and indeed to continue and produce the product molecules. The two steps that happen in the propagation steps are shown in the equations below. These reactions are examples of a chain reaction. This basically means that a chlorine free radical produced in the initiation step will react with a methane molecule and this leads to the formation the product as well as another chlorine free radical; which can further react with another reactant molecule, methane in this case and so the reaction is essentially one big loop.

The two equations below represent the propagation steps, both equations start and end with the formation of a chlorine free radical - the perfect set-up for a reoccuring loop or chain reaction!

In the propagation stage of this reaction a chlorine free radical reacts with a reactant molecule, the methane (CH4) in this case to form a methyl free radical and hydrogen chloride gas. The methyl free radical then goes on to react with the other reactant, chlorine gas to form the product, chloromethane and a chlorine free radical is regenerated to carry on the chain reaction. So we started the propagation step with a chlorine free radical and we finish the propagation step with the same chlorine free radical being produced. The high reactivity of free radicals and the fact that both the equation above are exothermic results in this chain reaction being very violent and very fast.

Equations to show the propagation steps in the free radical reaction between chlorine and methane.

These two equations can be combined into an overall equation for the propagation step by simply cancelling out similar species on both sides of the equations as shown below:

Combining the two equations in the propagation steps of a free radical substitution reaction will give the overall equation for the reaction.

Step3 -termination

The number of free radical present in the reacting mixture at any one time is quite small; however as we have seen in the propagation steps as one free radical is used up another is always produced so the number of free radicals stays fairly constant throughout the reaction. Since the number of free radicals is low the probability of them reacting together is quite low. However when they do combine together two free radicals will form a stable molecule and the chain reaction will come to an end at this point.

To write equations for the termination steps simply react together any two free radicals that are present in the reaction mixture from the propagation steps and the initiation step. In the examples above there are two free radicals present; a chlorine free radical and a methyl free radical so the possible termination steps will simply be:

Equations for the termination steps in the free radical reaction of chlorine and methane.

Problems with free radical substitution reactions

In the propagation step the overall result is that one of the hydrogen atoms on a methane molecule is replaced by a chlorine atom to form chloromethane. In the first propagation step a chlorine free radical reacts with methane molecule to form a methyl free radical and hydrogen chloride gas. However as the overall reaction proceeds the methane gas being a reactant will be getting used up and the amount present will start to fall; at the same time the amount of chloromethane present from the propagation step will start to increase. This means that ultimately in the first step in the propagation reaction the chloromethane will take the place of methane and this will lead to the formation of dichloromethane:

Equations to show the formation of dichloromethane in the free radical substitution reaction of methane and chlorine gases.

However you can probably see where this is going! The amount of dichloromethane present will start to rise and the amount of chloromethane present will start to fall as the reaction proceeds; so the dichloromethane will take the place of the chloromethane in the first propagation step and trichloromethane will be formed, remember that every time an alkane molecule goes through the propagation step all that happens is a hydrogen atom on the alkane molecule is replaced by a chlorine atom. So the trichloromethane molecule will then go through the same process as the dichloromethane and tetrachloromethane will be formed; at this point all the hydrogen atoms on the methane molecule have been replaced by chlorine atoms. This means that the final products of this reaction will be a mixture of chloromethane, dichloromethane, trichloromethane and tetrachloromethane - not ideal!!

We can however change the reaction conditions to make chloromethane the main product of the reaction simply by using a large excess of methane. In the first step of the propagation reaction the chlorine free radical will react with the first "thing" it comes into contact with. So if we add a large excess of methane then it is likely to come into contact with a molecule of methane and less likely to meet say a molecule of chloromethane. By the same argument if we want to produce tetrachloromethane as the main product then simply use an excess of chlorine. This will ensure that all the hydrogen atoms on the methane are replaced by a chlorine atom.

Key Points

Practice questions

Check your understanding - Questions on free radical substitution reactions

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