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KOH alcohol elimination

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

  1. joonkimdds

    joonkimdds Senior Member
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    In DAT destroyer, KOH, alcohol brings elimination and it forms C=C with more substituted carbon thus more stable and i think that's called Zaitchev.

    When I googled and found this page
    www.funeducation.co.uk/a2resources/resources/organic_a2.ppt
    (plz go down to page #13)

    it shows that it forms C=C with less substituted carbon and I think that's called hoffman.

    which one is correct?
     
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  2. Sublimation

    5+ Year Member

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    Yes both are correct there is a slight difference here. When a chemist want to make a compound that has a alkene that does not follow zeitchevs rule, they place a very bulky base to eliminate, KOH will not in any case produce a hoffman product its just not bulky enough, however, a t-butoxide on the other hand will give u a hoffman product (also depending on the variables in the reaction).
     
  3. joonkimdds

    joonkimdds Senior Member
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    Thanks for ur answer.

    How should I memorize this though? When I see KOH(alc), should I assume that answer can be either one(hoff and zaitchev).
     
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  4. sciencegod

    sciencegod Super Member
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    for your purpoeses any elimination of alcohols will yield zaitchev unless its t-butoxide or LDA. theose 2 will yield hoffman.
     
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  5. joonkimdds

    joonkimdds Senior Member
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    Thank you .

    But what about CH3-CHCH3-CHBr-CH3? (from the link posted)
    this is neither t-butoxide nor LDA (or is it?) but it yields Hoffman.
     
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  6. sciencegod

    sciencegod Super Member
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    its bec of the high steric hindrance. don't think you have to worry bout this for the dta. just remember what I told you.
     
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  7. joonkimdds

    joonkimdds Senior Member
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    I think I just found out why it can't form C=C with tertiary carbon.
    Because that will make it go beyond octet.
    (CH3)3 C=C is impossible :)
     
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