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will this undergo E1,E2,SN1, or SN2?

Discussion in 'DAT Discussions' started by joonkimdds, Dec 25, 2008.

  1. joonkimdds

    joonkimdds Senior Member
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    [​IMG]

    I was wondering which reaction will this undergo.

    It doesn't have aprotic but I think that doesn't mean it's impossible to undergo SN2.

    it has small unhindered base/nucleophile so I think E2 is not an answer.

    It has strong base/nucleophile that prefers SN2 and E2.

    it has protic that prefers SN1 and E1.

    I don't know which one it is.
     
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  3. sciencegod

    sciencegod Super Member
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    e2. it has a very strong nucleophile so on a secondary halide e2 will be prefered.
     
  4. chessxwizard

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    E2 since it's a strong base which creates a pi bond that is conjugated with the ring system.
     
  5. joonkimdds

    joonkimdds Senior Member
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    But shouldn't it be hindered base to make it E2?
    Ethoxide is an unhindered base/nucleophile.

    I thought that
    strong base/nucleophile = both SN2 and E2
    but if it's hindered = E2
    if not hindered = SN2
     
  6. chessxwizard

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    Read my response and think about it.
     
  7. Panther85

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    yup agreed.... that ring is gonna trump all of them
     
  8. doc3232

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    I think E1/E2 because it is in the benzylic position. The carbocation is VERY stable because of the aromatic ring. I don't think you can say E2 and not E1 despite the strong base.
     
  9. sciencegod

    sciencegod Super Member
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    good point:thumbup:
     
  10. sciencegod

    sciencegod Super Member
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    as it happens, the reaction would be e2 on a cyclohexane also.
     
  11. 5HTburb

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  12. sciencegod

    sciencegod Super Member
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    its still e2 if the protic solvent is the conjugagte acid of the strong nucleopihile.
     
  13. Sublimation

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    If you enter any chemistry laboratory, and ask the TA or researchers to define which mechanism a reaction will undergo or describe the different mechanisms. You will get an array of answers. yes this is an predominantly E2 reaction, however because there is a lack of heat, you can just as easy argue an sn1 due to stabilization of the carbocation by the benzene ring (resonance stabilized). However, in a lab you will end up with both mixtures and obviously due to the fact that we have a secondary carbon the major product will be a zeitsev << I think i speeled that wrong, Alkene. Here this chart helped me out alot in orgoI. Hope this helps. :)
     

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