Free radical substitution heading

Free radical substitution reactions

Homolytic bond fission or cleavage will form free radicals. Free radicals contain an unpaired electron and are extremely reactive species which are normally produced as intermediates in reactions.

In the example shown below the bond enthalpy of the chlorine 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.



Alkanes are chemically unreactive substances, due largely to the fact the 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 or camera flash will supply enough energy to break the relatively weak bond in a chlorine molecule and form chlorine free radicals , these will immediately react with the first "thing" they come into contact with, which is likely to be an alkane molecule.

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 produced but it is possible to adjust the reaction mixture to obtain mainly one of these halogenalkane molecules as the main component in the mixture produced.

This reaction of chlorine with methane is an example of a free radical substitution reaction and it proceeds via 3 steps, these 3 separate steps are called:


Step1 - initiation

The reaction of methane with chlorine proceeds via free radicals. The initiation step generates or produces these free radicals. It is the presence of a small but constant number of these reactive free radicals that allows the reaction to proceed.

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 the 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:

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 products. The two steps that happen in the propagation process 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 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.

In the two equations below represent the propagation steps. They both 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. 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 and release energy results in this chain reaction being very violent and very fast.

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:

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 progation 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 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. 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:

Problems with free radical substitution

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 to form a methyl free radical and hydrogen chloride gas. However as the overall reaction proceeds the methane being a reactant will be getting used up and the amount present will start to fall, at the same time the amount of chloromethane, the product 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:

However you can probably see where this is going! The amount of dichloromethane present will start to rise and the amount of chloromethane 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. This molecule will then go through the same process 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 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 arguement if we wanted 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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