So the alkanes are not supposed to be that reactive.
Actually that's not so really, just another one of those chemistry myths you meet in school courses, across the world not just in the UK or the US.
Given the quantity of hydrocarbon alkanes separated from crude oil per annum they had better be reactive in some way or other or we are going to waste so much bonded carbon atom and turn them into greenhouse causing CO2 through using them as fuels - a not very creative approach to say the least.
One very imaginative reaction of the alkanes puts a halogen atom in the molecule and that halogen can be pretty labile, replaceable with other groups and deliver on a variety of petrochemical precursors.
The halogen atoms usually used to illustrate this reaction are chlorine atoms in the form of molecular chlorine.
And the simplest example alkane usually used in the illustration is methane CH4.
But what we say below about chlorine and methane applies equally to say bromine and hexane.
Now if you are thinking like a chemist you will be asking yourself how are we going to have two gases react together and in what proportions?
Even more so what will the result be if the proportions of gases reacting are different?
If methane is in excess what will the product(s) be?
If chlorine is in excess how will the products be different if at all?
So let's think first about how this reaction can occur.
I'm going to assume you have some familiarity with organic reaction mechanisms.
Reaction mechanisms reveal the possible stages a reaction goes through to produce products from reactants.
Here is the overall reaction of excess methane with chlorine:
This is a substitution reaction because a chlorine atom takes the place of a hydrogen atom in methane.
The conditions used are symbolised with hν which refers to high energy uv light.
A Rescue Box explains the hν symbol here:
Now all the compounds involved in the reaction are gases.
When the reaction takes place four key observations can be made:
Let's see how these observations help us to work out or elucidate the reaction mechanism.
The light seems very important especially its energy value per photon why is this?
Let's look at some bond energies for the relevant molecules:
Cl2 = Cl• + Cl• +242kJ/mol homolytic fission of chlorine molecule
Cl2 = Cl+ + Cl- +1147kJ/mol heterolytic fission of chlorine molecule
CH4 = CH3• + H• +435kJ/mol homolytic fission of C-H bond in methane
CH4 = CH3- + H+ +1306kJ/mol heterolytic fission of C-H bond in methane
[You should realise that I have selected a relevant set of reactions here from a list of over 10 possible reactions that the uv light could cause to happen to chlorine or methane molecules in the mixture.]
If the gas mix is exposed to photons with energies around 300kJ/mol then absorption of light of wavelength around 400nm would provide sufficient energy for the homolytic fission of chlorine molecules to occur.
As you can see, the other bond fissions require much more energy per mole of bonds.
[What's more if the reaction is carried out in an inert non-polar solvent then the effect of the solvent will be to suppress the formation of ions.]
So we can suggest that the initial change in this mixture of gases exposed to uv light will be the production of chlorine free radicals like this:
Note the use of one half arrow to show the movement of each bond electron to each chlorine atom.
This is called the Initiation Step in the reaction mechanism.
What could happen next?
Well chlorine free radicals have great energy of movement and will collide with molecules nearby almost immediately.
Some will collide with other nearby chlorine radicals and reverse the initiation step.
But others will collide with methane molecules.
So which of the two possible reactions below are likely to occur?
1. CH4 + Cl• = CH3Cl + H• ΔH = +85kJ/mol
2. CH4 + Cl• = CH3• + HCl ΔH = +4kJ/mol
Now one of our initial observations was that no molecular hydrogen is detected in the final products.
But if reaction 1. occurred then some molecular hydrogen would be produced by the combination of hydrogen radicals.
Furthermore, it is more likely that a less endothermic reaction would occur more readily.
So reaction 2. occurs after the initiation step.
But this reaction still leaves a very reactive methyl free radical.
What could happen to this free radical?
There is still enough chlorine available for the reactive methyl free radical to collide with a chlorine molecule like this:
3. Cl2 + CH3• = CH3Cl + Cl•
If it did, the reaction would regenerate the initial chlorine free radical that came from the initiation step.
And yes that is what happens many times over.
Reactions 2. and 3. form what is called the Propagation Step.
This is how we said "In the right light, for each photon of light absorbed, many thousands of chloromethane molecules are produced."
What's more combining the energy profiles of reactions 2 and 3 shows that the overall propagation process is exothermic even if the reaction 2 is endothermic as you can see below:
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We call the combination of reaction 2 and reaction 3 a chain reaction.
But this isn't the end of the process.
You should see by now that reactive free radicals can react quite easily with each other and when they do that terminates the chain reaction.
So here are three possible termination reactions:
Cl• + Cl• = Cl2
Cl• + CH3• = CH3Cl
and also not surprisingly:
CH3• + CH3• = C2H6
These are all called Termination Steps
The overall mechanism looks like this:
This mechanism fits the conditions in which there is excess methane but what if there were excess chlorine instead what would change?
Well here let me suggest that other chlorination products form like CH2Cl2, CHCl3 and CCl4.
Can you satisfy yourself that you could write equations for the formation of these other products and show a mechanism for the reactions too?
And lastly, why is it that lead tetramethyl Pb(CH3)4 introduced into the reaction mixture increases the rate of formation of chloromethane?
Pages on the "Mole" and "Using the Mole" in chemical calculations are here
Actually that's not so really, just another one of those chemistry myths you meet in school courses, across the world not just in the UK or the US.
Given the quantity of hydrocarbon alkanes separated from crude oil per annum they had better be reactive in some way or other or we are going to waste so much bonded carbon atom and turn them into greenhouse causing CO2 through using them as fuels - a not very creative approach to say the least.
One very imaginative reaction of the alkanes puts a halogen atom in the molecule and that halogen can be pretty labile, replaceable with other groups and deliver on a variety of petrochemical precursors.
The halogen atoms usually used to illustrate this reaction are chlorine atoms in the form of molecular chlorine.
And the simplest example alkane usually used in the illustration is methane CH4.
But what we say below about chlorine and methane applies equally to say bromine and hexane.
Now if you are thinking like a chemist you will be asking yourself how are we going to have two gases react together and in what proportions?
Even more so what will the result be if the proportions of gases reacting are different?
If methane is in excess what will the product(s) be?
If chlorine is in excess how will the products be different if at all?
So let's think first about how this reaction can occur.
I'm going to assume you have some familiarity with organic reaction mechanisms.
Reaction mechanisms reveal the possible stages a reaction goes through to produce products from reactants.
Here is the overall reaction of excess methane with chlorine:
 |
| methane + chlorine = chloromethane + hydrogen chloride |
This is a substitution reaction because a chlorine atom takes the place of a hydrogen atom in methane.
The conditions used are symbolised with hν which refers to high energy uv light.
A Rescue Box explains the hν symbol here:
Now all the compounds involved in the reaction are gases.
When the reaction takes place four key observations can be made:
- Molecular hydrogen (H2) is absent during the course of the reaction, it is never detected in this reaction.
- The reaction can go at about 300oC but it proceeds at much higher rate in sunlight at room temperature.
- Light of a wavelength of around 400nm or less, with an energy of around 300kJ/mol of photons or more, is every effective in increasing the reaction rate.
- In the right light, for each photon of light absorbed, many thousands of chloromethane molecules are produced.
Let's see how these observations help us to work out or elucidate the reaction mechanism.
The light seems very important especially its energy value per photon why is this?
Let's look at some bond energies for the relevant molecules:
Cl2 = Cl• + Cl• +242kJ/mol homolytic fission of chlorine molecule
Cl2 = Cl+ + Cl- +1147kJ/mol heterolytic fission of chlorine molecule
CH4 = CH3• + H• +435kJ/mol homolytic fission of C-H bond in methane
CH4 = CH3- + H+ +1306kJ/mol heterolytic fission of C-H bond in methane
[You should realise that I have selected a relevant set of reactions here from a list of over 10 possible reactions that the uv light could cause to happen to chlorine or methane molecules in the mixture.]
If the gas mix is exposed to photons with energies around 300kJ/mol then absorption of light of wavelength around 400nm would provide sufficient energy for the homolytic fission of chlorine molecules to occur.
As you can see, the other bond fissions require much more energy per mole of bonds.
[What's more if the reaction is carried out in an inert non-polar solvent then the effect of the solvent will be to suppress the formation of ions.]
So we can suggest that the initial change in this mixture of gases exposed to uv light will be the production of chlorine free radicals like this:
This is called the Initiation Step in the reaction mechanism.
What could happen next?
Well chlorine free radicals have great energy of movement and will collide with molecules nearby almost immediately.
Some will collide with other nearby chlorine radicals and reverse the initiation step.
But others will collide with methane molecules.
So which of the two possible reactions below are likely to occur?
1. CH4 + Cl• = CH3Cl + H• ΔH = +85kJ/mol
2. CH4 + Cl• = CH3• + HCl ΔH = +4kJ/mol
Now one of our initial observations was that no molecular hydrogen is detected in the final products.
But if reaction 1. occurred then some molecular hydrogen would be produced by the combination of hydrogen radicals.
Furthermore, it is more likely that a less endothermic reaction would occur more readily.
So reaction 2. occurs after the initiation step.
But this reaction still leaves a very reactive methyl free radical.
What could happen to this free radical?
There is still enough chlorine available for the reactive methyl free radical to collide with a chlorine molecule like this:
3. Cl2 + CH3• = CH3Cl + Cl•
If it did, the reaction would regenerate the initial chlorine free radical that came from the initiation step.
And yes that is what happens many times over.
Reactions 2. and 3. form what is called the Propagation Step.
This is how we said "In the right light, for each photon of light absorbed, many thousands of chloromethane molecules are produced."
What's more combining the energy profiles of reactions 2 and 3 shows that the overall propagation process is exothermic even if the reaction 2 is endothermic as you can see below:
We call the combination of reaction 2 and reaction 3 a chain reaction.
But this isn't the end of the process.
You should see by now that reactive free radicals can react quite easily with each other and when they do that terminates the chain reaction.
So here are three possible termination reactions:
Cl• + Cl• = Cl2
Cl• + CH3• = CH3Cl
and also not surprisingly:
CH3• + CH3• = C2H6
These are all called Termination Steps
The overall mechanism looks like this:
This mechanism fits the conditions in which there is excess methane but what if there were excess chlorine instead what would change?
Well here let me suggest that other chlorination products form like CH2Cl2, CHCl3 and CCl4.
Can you satisfy yourself that you could write equations for the formation of these other products and show a mechanism for the reactions too?
And lastly, why is it that lead tetramethyl Pb(CH3)4 introduced into the reaction mixture increases the rate of formation of chloromethane?
Pages on the "Mole" and "Using the Mole" in chemical calculations are here
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