How Does UV Light Initiate Chlorine Atom Splitting in Chlorination?

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The discussion focuses on the mechanism of chlorine atom splitting in chlorination, specifically when chlorine gas (Cl2) interacts with methane (CH4) under UV light exposure. UV light provides the necessary energy to excite the electrons in the Cl-Cl sigma bond, leading to homolytic fission and the formation of two chlorine radicals (Cl·). These radicals can then abstract hydrogen from methane, resulting in the formation of hydrochloric acid (HCl) and a methyl radical (CH3·), which subsequently reacts with another Cl radical to produce chloromethane (CH3Cl).

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  • Understanding of chlorination reactions and free radical mechanisms
  • Knowledge of UV light's role in photochemistry
  • Familiarity with concepts of homolytic bond cleavage
  • Basic principles of quantum mechanics, specifically E=hf
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  • Learn about the principles of photochemistry and the effects of UV light on molecular bonds
  • Explore the concept of free energy diagrams in chemical reactions
  • Investigate the role of sigma and sigma* orbitals in bond breaking
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Cheman
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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. :wink:
 
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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.
 
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.
 
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.
 
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?
 
As he said, a single sigma bond with two electrons can be cleaved homolytically, if each electron is a sigma bonding and sigma* antibonding orbital; in this case, the bond is homolytically broken. The photonic energy is very high to supply this excitation.
 

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