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Chemistry Free Rad Substitution URGENT question

Hi guys. I'm struggling to write the mechanism (initation, propagation and termination stages) for the reaction of

CH3CH2Cl + 5 Cl2 --> CCl3CCl3 + 5 HCl

Do you take into account balanced numbers? How would you approach this?

Reply 1

bl0bf1sh

does this page (https://www.chemguide.co.uk/mechanisms/freerad/multisubcl.html) on multiple substitutions help?

Reply 2

Hercule-Poirot

Original post by bl0bf1sh

does this page (https://www.chemguide.co.uk/mechanisms/freerad/multisubcl.html) on multiple substitutions help?

Not a lot, since the equation we have is not one we've seen before and also has balanced numbers at the front

Reply 3

TriplexA

Original post by Hercule-Poirot

Some key tips:

The first propagation step the acid is always formed from the reaction of the radical + molecule (HX where X is the halogen)
The free radical is always regenerated in the last step (termination)

(edited 2 years ago)

Reply 4

scimus63

There are 5 hydrogens in the chloroethane molecule and 10 chlorine atoms in the 2 x Cl2 molecules. The chlorine is in excess and will replace all the hydrogen atoms in the chloroethane.

The reaction will simply go through the propagation steps until all the hydrogens are ultimately replaced.

Tho I think you will end up with a mixture of products from dichoroethane all the way tohexachloroethane simply because the 5 fold excess of chlorine I think will probably not be enough to chlorinate all the molecules fully.

https://science-revision.co.uk/A-level_organic_free_radical_substitution.html

might help but does not have the mechanism for multiple substitution as bl0bfsh has posted above

Reply 5

charco

Study Forum Helper

Original post by Hercule-Poirot

First of all there many steps in the mechanism as you are substituting 5 chorine atoms, one at a time.
But each step is fundamentally the same.

Initiation requires the formation of a chlorine free radical from the chlorine molecule using UV light:

Cl₂ ==> 2Cl·

Then you start the propagation steps:

Cl· + CH₃CH₂Cl ==> ·CH₂CH₂Cl + HCl
Cl₂ + ·CH₂CH₂Cl ==> ClCH₂CH₂Cl + Cl·

That's the first hydrogen substituted. Now repeat these previous two equations four more times, before the termination step:

Cl₃CCCl₂· + Cl· ==> Cl₃CCCl₃