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Photon336

  • one year ago

Using concentrated sulfuric acid, why is the mechanism leading to a five membered ring favored over a six membered ring? How and why does this happen? Will attach problem in a bit

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  1. Photon336
    • one year ago
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    Here is the problem

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  2. Photon336
    • one year ago
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    I know the problem already gives the product that you would get from this reaction but why would this mechanism below not apply here? Hydroxyl group gets protonated followed by loss of water to generate a carbonation followed by a methyl shift and then formation of a double bond.

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  3. anonymous
    • one year ago
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    Here 2,2 dimethylcyclohexanol undergoes ring contraction in order to increase stability. It does form 1,2 dimethylcyclohexene but the stability of isopropylidene cyclopentane is greater. The formation of the former is a result of methyl migration and the latter the result of C-2-C-3 bond migration.

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  4. anonymous
    • one year ago
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    I hope this helps ................

  5. abb0t
    • one year ago
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    Well, you get two products. Remember that in organic chemistry, it is quite difficult to control a reaction and the way it proceeds. You can get a high yield of a product you desire, but will most likely not yield 100%, as ideal as you make the conditions. But for the reaction you have here, you can get both. Let me show you.

  6. abb0t
    • one year ago
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    However, the problem is specifically asking you to show mechanism for the isopropylidenecyclopentane product. But, you're not incorrect in your first proposed product of 1,2-dimethylcyclohexene.

  7. abb0t
    • one year ago
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    However, the problem is specifically asking you to show mechanism for the isopropylidenecyclopentane product. But, you're not incorrect in your first proposed product of 1,2-dimethylcyclohexene.

  8. Photon336
    • one year ago
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    @abb0t @Ritika6 Thanks for both of your explanations!

  9. Photon336
    • one year ago
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    if you guys are wondering, if found an explanation online as to why the 5 membered ring is more stable it probably doesn't apply here as much "When you have an open chain molecule in solution, that molecule has lots of potential rotations and positions that it can assume. this makes the molecule very happy because it is more disordered/less constrained. when you want to forma ring you are effectively freezing out some of the motions available to that chain. the bigger the chain, the more it was moving freely before the ring formation and the more difficult it will be to for the target ring. this is why it is much slower to forma six membered ring, compared to a five membered ring, even though the stability of six membered rings (once formed) is roughly equivalent other of a five membered ring" chm.uri.edu.

  10. Photon336
    • one year ago
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    |dw:1437588284810:dw| basically the way I understood that was, the shorter chained molecule has fewer rotation available to it than the molecule that leads to the 6 membered ring and it would be faster to form the 5 membered ring. but yeah @abb0t I saw what you did for the second one, that's interesting though how you get 2 products. thanks again.

  11. abb0t
    • one year ago
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    assuming you have a FLAT cyclohexane, then yes, you would have immense strain, but the cyclohexane stability relies on chair conformation, which has 0 kcal/mol of energy ring strain.

  12. abb0t
    • one year ago
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    however, this isn't true for biologically important rings, such as glucose.

  13. Photon336
    • one year ago
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    I hate those chair conformations.. had to draw out so many of those.

  14. Photon336
    • one year ago
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    I want to try to find some problems on sugars just to see what you guys think

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