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Organic Chemistry Mechanism Question

  • Thread starter DoctorB2B
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  • #1
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Homework Statement


I am given isobutane as my starting material, Br2 and benzoyl peroxide as my radical initiators. My final product is the isobutane with Br attached instead of H.

I have to design the mechanism. I have tried what seems like an endless combination of mechanism attempts, and can't seem to get it. I do not have to include the termination steps, only the initiation and propagation steps.

The Attempt at a Solution


-I homolyzed the Br atom and the benzoyl peroxide at the same time in step one.
-In step two, I took one of the benzoyl peroxide radicals and attempted to remove the H from isobutane creating a radical off the alkyl group.
-In step three, I homolyzed another Br2 molecule added it onto the isobutane radical.
-In step four, I didn't know what to do.

Evidently I have my steps all screwed up. I don't know when I'm supposed to homolyze the peroxide and then what I'm supposed to do with it. Any help would be greatly appreciated.
 

Answers and Replies

  • #2
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i never came across benzoyl peroxide. I would suggest that it would break Br2 into Br+ and Br-. The Br+ is accepted by the benzene ring (electrophilic substitution rxn) and the Br- would attack the hydrocarbon as a radical.

but i really don't know.
 
  • #3
chemisttree
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In isobutane you have three methyl groups and one methine. That is a 9:1 ratio of primary substitution vs tertiary. One would think that you would get a 9:1 ratio of the 1-bromo product relative to the 2-bromo product but remember that bromine is less reactive than chlorine in free radical radical reactions and is much more selective.

Benzoyl peroxide can be thought of as a free radical initiatiator (like you would think of heat or light in that regard). Which bond (C-H vs. Br-Br) is most likely to homolytically cleave in the first step of your mechanism?

Hint: Look at the bond energies of these two types of bonds.
 

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