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Free radical halogenation question

  1. Aug 19, 2009 #1
    1. The problem statement, all variables and given/known data

    The initiation step in chlorination of methane is as follows...

    http://upload.wikimedia.org/wikipedia/en/thumb/3/32/MethaneChlorinationMechanismInitiation.svg/260px-MethaneChlorinationMechanismInitiation.svg.png [Broken]

    We get two chlorine radicals as a result.

    Next we have two propagation steps

    http://upload.wikimedia.org/wikipedia/en/4/48/MethaneChlorinationPropagationStep.svg [Broken] (won't let me embed the pic, sorry guys!)


    Ok, my understanding of this is... in the initiation step, the Cl2 was broken into two Cl radicals. ONE Cl radical moved on to the next step and reacted with the methyl group forming a methyl radical and HCl. THEN, the methyl radical reacted with Cl2 to form chloromethane and a chlorine radical.

    My question is, Where does the SECOND Cl radical that was produced in the initiation step go? Because only ONE Cl radical reacted with the methyl group and then the methyl radical reacted with Cl2. In my mind, there is a Cl radical unaccounted for, but this is not explained in any example I can find (in a book or online)?

    Thanks!!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 19, 2009 #2

    Ygggdrasil

    User Avatar
    Science Advisor

    It performs the same reaction with other molecules of methane. Eventually, you can get termination reactions that get rid of the radicals (often, this involves two radicals bonding together, e.g. Cl• + Cl• --> Cl2).
     
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