Alkanes with an excess of chlorine in the presence of UV light

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brettsullivan
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I have a question on free radical reactions.

-why the propagation step happens. With propane and longer chained hydrocarbons anyways.
With propane I have been told that the species produced in the propagation step is C3H6Cl2.
^Why does this happen and why is it this specific species?

A bit off topic but why does this reaction not happen with other halogens, and why is it specific to chlorine?


I am an A level student so any help would be appreciated.
Thanks,
Brett
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ThatGuy107
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The CL-CL bond is weak, and therefore the UV has sufficient energy to cause homolytic fission of this bond, forming 2 Cl free radicals.
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Jfueofkrnb
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Free radical substitution reactions can occur with any chlorine or bromine . Eg. Chlorination of methane, or bromination of ethane. This is when a hydrogen from a hydrocarbon chain is replaced with Cl or Br. Multiple substitutions can occur producing many products. If enough halogen is present, all the hydrogens in the hydrocarbon chain can be replaced.

For example: chlorination of methane
Chlorine is a diatomic molecules so exists as Cl2
UV light causes homolytic fission of a halogen molecule. This is when the covalent bond breaks and each atom receives one of the electrons from the covalent bond - this forms radicals.

Initiation Cl2 —> Cl• + Cl•
Propagation CH4 + Cl• —> CH3• + HCl
CH3• + Cl2 —> CH3Cl + Cl•
Termination 2Cl• —> Cl2
2CH3• —> C2H6
CH3• + Cl• —> CH3Cl
Here a mixture of products is produced by random collisions between radicals (radicals are very reactive)
The desirable product is CH3Cl

I hope this makes sense!
Last edited by Jfueofkrnb; 3 years ago
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ChemicalGoat
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Hi there,
I'm in my final year of college so hopefully I can help

For the first part, are you taking this out of an exam question, or just generally confused? Fine for either, it'd just help me with explaining if you have an exam question for me to read through.

As to why this only happens with chlorine - It doesn't! (From what I know) Other halogens can form radicals, they just require more energy to break bonds (In other words, a higher frequency of UV light) then chlorine. I believe fluorine radicals can form too, but due to the alkane in your question only being in a solution of excess chlorine and not fluorine, only chlorine radicals form.

I believe this is correct, but please correct me if I'm wrong, radicals aren't my strongest topic in Chemistry and I don't remember encountering a chemistry question like this before
(Original post by brettsullivan)
I have a question on free radical reactions.

-why the propagation step happens. With propane and longer chained hydrocarbons anyways.
With propane I have been told that the species produced in the propagation step is C3H6Cl2.
^Why does this happen and why is it this specific species?

A bit off topic but why does this reaction not happen with other halogens, and why is it specific to chlorine?


I am an A level student so any help would be appreciated.
Thanks,
Brett
1
reply
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