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Carbocation Stability on Fused Rings

  1. Sep 16, 2010 #1
    So I just had a question on a quiz (did not go well) about carbocation stability on the fused rings bicyclo[2.2.1]heptane, with the positive charge on a bridgehead carbon and 2-methylbicyclo[2.2.1]heptane with the positive charge on C2. The question was which is more stable and why?

    The question stated that the second option is far more stable, but I have little to no idea why. Both are tertiary carbons, so that doesn't factor in. I answered that the first option was more symmetrical, so will be less likely to deprotonate at the bridgehead, but in retrospect this doesn't make sense as deprotonation wouldn't leave a positive charge in the first place.

    A friend of mine said that the bridgehead carbon is under a ton of stress geometrically, and to think of it as an intermediate in a hydrohalogenation across what used to be a double bond. So with respect to this, why would the charge not be placed on the bridgehead? I'm thinking of this in terms of tertiary vs. secondary carbons (maybe this is where I'm going wrong?), and a tertiary carbon is more likely to hold a charge.

    Can anyone help? I won't get a chance to ask about this in detail until next Tuesday and I'm really bothered by this question so I don't want to wait. Let me know if I need to clarify on anything.
     
  2. jcsd
  3. Sep 16, 2010 #2
    I think it is because the tensile stress, a carbocation usually go to reorder in a more stable specimen, and c+ in specimen 1 is really unestable.
    Specimen two is a quasi normal sp2, but specimen 1 its really stressed because of the bridge.
     
  4. Sep 17, 2010 #3

    chemisttree

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    That's right. In the first case the carbocation is for, essentially, an sp3-hybridized bridgehead carbon and in the second case it is for your standard, run-of-the-mill sp2-hybridized carbon. Tying back those other carbons surrounding the bridgehead carbon prevents it from rehybridizing completely to sp2 in the first case.
     
  5. Sep 18, 2010 #4
    Ah, back to hybridization; I should have known. Thanks for the help.
     
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