Chlorination mechanism

Chlorination mechanism....

In chlorination, with chlroine gas and methane, why does the UV light cause the chlorine atom to split into free radicals? I know that it "supply the energy", but what actually happens?

Thanks.

movies
Yes, you form two Cl radicals. One Cl radical can then abstract a hydrogen atom from methane to make HCl and a methyl radical. The methyl radical can then combine with another Cl radical to make the product, chloromethane.

chem_tr
Gold Member
Well, Movies answered the question fairly well, but I think there is something with $$h\nu$$; the energy in photon causes the sigma bond to be homolytically (just in the middle) cleaved, giving away two radicals, as Movies also wrote. The rest is very clear from his post.

movies
Yes, chem_tr is right. I guess I glossed over how that actually happens. I have always thought of it as exciting the vibration of the Cl2 molecule until the Cl-Cl bond is essentially "ripped" apart.

Yeah I think it's to do with Eistien's E=hf and homolytic fission occurs in Cl2. The free radical is very reactive, and can therefore substitute a hydrogen from methane.

GCT
Homework Helper
You should see this in the perspective of free energy diagram, if you wish to "know what happens." It should be given in your text.

Thats really quite an advanced question. If you want to know the real answer, you learn it in either third semester physics (modern physics) or second semester physical chemistry.

It has to do with the energy state of the electrons in Cl-Cl bond. In this case, it happens that the electrons in the sigma bond absorb light in the frequency of the UV region, because their energy states are matched by the energy of the photon, where E = hf. The excited electrons jump up to a higher energy state, which causes the bond to "break" and disassociate homolytically.

Thanks so-crates - could you please elaborate further on how the bond actually breaks?

chem_tr