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Ochem help- Robinson Annulation

Discussion in 'Pre-Pharmacy' started by jmcfa002, Jun 2, 2008.

  1. jmcfa002

    jmcfa002 UCSF SoP c/o 2014
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    I'm in my second quarter of ochem, and I'm studying the "robinson annulation"

    I'm having a hard time with it, does anyone have a good way of explaining it?

    Thanks
     
  2. OrllY

    OrllY Awesome
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    Michael + Aldol.
    just look for those acidic hydrogens and see what you can do with them.
     
  3. AbsoluteEthanol

    AbsoluteEthanol PHARMACY STUD
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    couldn't have said it better myself! occam's razor at its finest here!

    p.s. i could never understand why people ask chemistry/biology questions on SDN. Is this seriously your main resource when you have a problem you cannot solve? i should hope not!
     
  4. jmcfa002

    jmcfa002 UCSF SoP c/o 2014
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    No, its not my main source. I studied it for about 20 minutes today and couldn't figure it out, and I'm going to school tomorrow to get help. I've been studying it and trying to figure it out better before I go to ask for help. I was thinking that someone may have an easy way to do it.

    I don't know Michael's addition, so maybe that is the problem.
     
  5. ExpressMaiL

    ExpressMaiL alea iacta est
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    The annulation is a great reaction.
     
  6. OrllY

    OrllY Awesome
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    i may have written a paper on this...
     
    #6 OrllY, Jun 2, 2008
    Last edited: Jun 8, 2008
  7. Sparda29

    Sparda29 En Taro Adun
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    Are you serious? Just download BitComet, and then go to PirateBay and download any cracked version of office.
     
  8. jmcfa002

    jmcfa002 UCSF SoP c/o 2014
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    or use open office, it can do .doc
     
  9. rycetrix

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    Robinson Annulation...I just taught my students how to do this so yea..(I am a tutor)

    So here we go:

    It is as some people have stated above, it is a Michael followed by an aldol. A Michael addition is basically an enolate + an alpha, beta-unsaturated carbonyl. The enolate adds to the double bond of the alpha, beta-unsaturated carbonyl. The product of Michael reaction is a 1,5-dicarbonyl compound. The aldol part is basically the Michael product undergoes an intramolecular attack. This is performed by making an enolate on part of the Michael product, and then it will attack the other carbonyl since this is a dicarbonyl compound (2 carbonyls). And then the molecule will undergo aldol condensation to get the desired product, which should have an alpha, beta-unsaturated carbonyl in it. And that would be the Robinson Annulation product.

    It is quite hard to describe the reactions without actually showing you what is going on.

    Let me know if this helps.
     

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