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 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.
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.
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.
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:
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:
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.
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:
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:
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:
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.