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Anti-Mark Question

Discussion in 'DAT Discussions' started by BeaverLover, Aug 2, 2015.

  1. BeaverLover

    BeaverLover SDN Gold Donor
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    [​IMG]

    Why did second anti-Mark rxn add H to #2 when both 1&2 equally substituted?
     
    #1 BeaverLover, Aug 2, 2015
    Last edited: Aug 3, 2015
  2. Carbohydrates

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    You numbered the chain wrong. You have to number the chain from right to left, which mean your C3 suppose to be C1.
     
  3. orgoman22

    orgoman22 DAT DESTROYER
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    This is poorly written.....confusing indeed. Assuming the author of the problem meant excess HBr/ ROOR.....it was simply 2 consecutive anti-Markovnikov additions. Anti-Markovnikov addition only works when HBr/ROOR is used. This reaction is very very important to know, and is used in organic synthetic procedures on a wide scale.

    Hope this helps.

    Dr. Romano
     
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  4. OP
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    BeaverLover

    BeaverLover SDN Gold Donor
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    Sorry Dr. Romano, I thought I had it figured out but I'm still confused. I didn't understand the reasoning as to why the second anti-Markovnikov addition had added an H to #2 carbon when both 2+3 are equally substituted. I googled around and it seemed that they would produce a equal mixture of product if the carbons were equally substitued, but the carbons in the examples that I saw were both secondary carbons.

    Based on the picture above I would have guessed the second anti-Mark reaction would have added the hydrogen to #3 to form a secondary carbocation and produce 1,2 dibromopropane since both carbons are equally substituted, whereas adding the hydrogen to #2 would be less favorable because of the formation of a primary carbocation (which I believe leads to the formation of the final product depicted above). How am I viewing this wrong? Wouldn't the primary carbocation undergo a hydride shift anyways to become 1,2 dibromopropane and not the answer above? I just don't seem how the final answer is correct :(
     
  5. Futuredentista62

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    This goes through a radical intermediate, not a carbocation intermediate.
    The intermediate formed also has the radical on the most substituted carbon because, in the mechanism, the bromine adds first, not the hydrogen as we're used to seeing.
     
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  6. JTLresces

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    Is the radical more stable on the C with Br because Br is highly ekectronegative to stabilize it?
     
  7. Futuredentista62

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    The radical does not for on the carbon with the bromine. It's more stable there because it is on the secondary carbon, not the primary carbon.
    Take a look at the mechanism. Hope it helps.
    28.jpg
     
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    BeaverLover

    BeaverLover SDN Gold Donor
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    Interesting...didn't see the formation of a radical, thx.
     
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