Free Radical Reactions- Topic 10 Organic Chemistry

[WesPotts2k16]

So I'm getting ready for a difficult organic test and one of the few things I still don't understand is how one can predict the products of a free radical equation, such as:

 

 

 

                                       Cl* + CH₄ = CH₃* +HCl

 

More specifically: How did the Methyl group become radical, and how did Cl drop its radical??

What method do i use to predict products in a free radical equation??

Please help

[kw0573]

The Cl• is not the most stable form of chloride; it's just an individual chloride neutral atom. The atom's natural form is Cl₂, a diatomic gas, which is already not stable, but Cl• is even more unstable. One way to reach stability is by conforming to the octet rule and get an electron, such as from hydrogen in a methane molecule. The free radical just means a single electron is transferred, instead of an electron pair.
The mechanism is typically described in three stages: initiation, propagation, termination.
Initiation: the diatomic halogen undergoes homogeneous homolytic (thanks to Msj chem for correction) fission (Cl₂ ---> 2Cl•)
Propagation: the Cl• attacks a methane: Cl• + CH₄ ---> HCl + CH₃•, THEN CH₃• + Cl₂ ---> CH₃Cl + Cl• 
Termination: any two radicals bond: 2Cl• ----> Cl₂; Cl• + CH₃• ----> CH₃Cl; 2CH₃• ---> CH₃CH3

So the reaction you provided is part of a series of the steps in halogenation of an alkane. In either part of the propagation step a radical reacts with a non-radical. Since we only have Cl•, CH₃•, Cl₂, CH₄, it's not hard to remember what species are reacted if we keep in mind the goal to make CH₃Cl. Also in this case it's not a methyl group, it's just a methane molecule. 

For a longer hydrocarbon chain, it doesn't matter which carbon you put the Cl because IB does not require you to know which Cl position is most stable.  

Let me know if this helps or if I made it more confusing.

Edited January 21, 2016 by kw0573

[Msj Chem]

The bond in the chlorine molecule (Cl₂) breaks in homolytic bond fission (in the presence of UV light) with each chlorine atom taking one electron from the single bond.

This produces two chlorine radicals (Cl•), each with an unpaired electron, which makes them very reactive.

One of the chlorine radicals reacts with a methane molecule, removing a hydrogen atom which produces a methyl radical (CH₃•)

Cl• + CH₄ ---> HCl + CH₃•

The reaction between the chlorine radical and the H atom from the methane molecule produces HCl.

 

In the next propagation step, the methyl radical reacts with Cl₂ to produce chloromethane (CH₃Cl) and another chlorine radical.

CH₃• + Cl₂ ---> CH₃Cl + Cl•

 

Here's a video that might help:

https://youtu.be/ZpQZXo0xZF0