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orgo question

Discussion in 'DAT Discussions' started by joonkimdds, May 7, 2008.

  1. joonkimdds

    joonkimdds Senior Member

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    complete the following reaction

    CH3CH2CO2H + CH3CH2MgBr => ?


    a) propanoate ion + ethane
    b) 3-pentanone
    c) ethyl propanoate
    d) 3-pentanol
    e) 1-propanol + ethane

    what do you guys think :)
    I want some explanation, too.
    I will post the answer unless someone comes up with the answer.
     
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  3. AdaAda87

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    It should be C) because it's a grindard reaction and since its not an ester, it wont attack twice, so it should be C.
     
  4. doc3232

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    So thinking of the MgBr molecule as having a neg charge
    First thing that would happen is that it will rip the H+ off the carboxylic which is quite acidic (leaving the proponoate in ion form)
    that will form ch3ch3

    Extra ideas: Then adding more grignard (I can't spell or its spellcheck) will attack the carbonyl and the Oxygen will hold the neg charge which is more stable
    OH- is not a good leaving group
    I hope I explained some of it well.

    Answer is A.

    I just studied this in my class :)
     
  5. Orgodox

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    The answer is A. Grignard does usually attack but it also a base and it will remove the H from the acid.

    I dont think adding more grignard will attack the carbonyl carbon once it has removed the H from the acid. Other wise you would end up with two oxygens both with negative charges.
     
  6. doc3232

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    Yes it would...because the neg charges are on two different molecules and the MgBr that is floating around will stabilize those neg charges.
    Grignards are very very strong bases, they will definately attack the carbonyl.
    Think about like this, do you want the neg charge on the Oxygen or the Carbon?
     
  7. joonkimdds

    joonkimdds Senior Member

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    I thought grignard reagent(R-MgBr) reacts with Carbonyl group(C=O) makes MgBr to be separated from R and only R gets attached to carbonyl carbon. This is the reason I put B as the answer.

    But the correct answer is 'A' like most of you said.

    My question is....
    How do I know that the grignard reaction(what I just explained) doesn't happen and
    acid base reaction occurs instead?
     
  8. doc3232

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    You have a reaction mentality right now. Just remember how basic ch3 is. It is like pKa 30 (I am guessing) and the alcohol is no more than 10. So that reaction will happen sooo fast. However, it will take too long for the ch3ch2 to get to the carbonyl compared to taking a H.
     
  9. userah

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    question. if the reaction was instead

    CH3CH2COCl + CH3CH2MgBr =>

    would you then get 3-pentanone because Cl is such a good leaving group?
     
  10. doc3232

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    So first attack the carbonyl then Cl can leave easily as you said (actually best leaving group that is commonly used).
     
  11. Orgodox

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    With the grignard you would attack the carbonyl the Cl would leave as you said, but the grignard attacks the carbonyl again and with a little acid protonates the O- to forms 3-Ethyl-3-pentanol
     
  12. userah

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    do grignards attack twice because the carbonyl carbon is still positively charged? and if that's the case, as long as you have a carbonyl group (not a carboxylic acid), does the grignard always attack twice?
     
  13. doc3232

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    Ya, your right, but that is only with excess, however....

    It will attack twice if there is excess, but what you should be asking is which is more reactive: CH3-COCl or CH3COCH3 assuming you added CH3MgBr to CH3-COCl
    I would think the latter is more reactive because the Cl can offer electrons to the carbonyl through resonance which would prohibit the grignard from getting in.
    So even without excess addition of grignard, the major product will be CH3CO-(CH3)2 with neg charge on the Oxygen.
     
  14. Orgodox

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    Well there has to initially be a good leaving group on so that after the grignard attacks the carbonyl, the c=o double bond can reform and kick something off. If that is the case then yes it will attack twice.
     
  15. Orgodox

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    Check DAT Destroyer Number 97. It clearly says in the answer key that when you have an ester you must assume 2 moles of grignard are present! Also there are other examples like number 6 where 2 moles are implied.
     
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  17. doc3232

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    I don't have Destroyer...
    Can you explain that again or give an example?
     
  18. userah

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    the solution just states that if you have an ester and you're reacting it with Grignard, you should always assume that there are 2 moles of Grignard.

    my question is to orgodox. In the example I gave, the reactant is not an ester, but it's still attacked twice by grignard. Is this because after the first attack by grignard, the carbonyl reforms thereby giving it a positive charge and thus leaving it for attack by grignard again?
     
  19. Orgodox

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    You got it! Exactly- the carbonyl reforms by kicking off the Cl-. Then like any other carbonyl there is positive character on the carbon as you mentioned and this is where the grignard attacks again.

    Also, I believe you can apply that 2 mole assumption written in DAT Destroyer #97 to any carbonyl attached to a good leaving group not just an ester. Since an acyl halide specifically in the example you asked about an Acetyl chloride is very reactive and has a good leaving group attached to the carbonyl carbon, I dont see why this would be any different than the ester where we assume 2 moles!
     
  20. userah

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    so 2 moles of grignard are assumed whenever you have a good leaving group. When can you not assume 2 gignards; like if you have -OH, NH2 etc? what are other good examples of poor leaving groups? sorry i'm just trying to fully understand it.
     
  21. doc3232

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    You can't make these assumptions on the real DAT despite what dat destroyer says. It is actually complicated, it depends on whether the first product (Grignard attacks carbonyl and Cl leaves, carbonyl reforms) or if the initial reactant is more reactive. Resonance on the carbonyl makes the molecule nonreactive.
     
  22. Orgodox

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    Dont you think DAT Destroyer was written FOR THE DAT. It is obviously written that you take what it says INTO the dat!
     
  23. doc3232

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    Not exactly, the real DAT can say anything it wants. DAT Destroyer is nothing more than prep. The real DAT questions will be 100% clear, no reason to argue with me...
     

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