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Reaction of acetone with oxalic ester

  1. Jan 20, 2005 #1
    Hello again

    What is/are the product(s) when acetone reacts with oxalic ester in presence of sodium ethoxide? I would be grateful if someone could explain mechanistically.

    Thanks and cheers
    vivek
     
  2. jcsd
  3. Jan 20, 2005 #2

    chem_tr

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    Acetone will give the reaction via its enol form, thus, [tex]H_3C-C(OH)=CH_2[/tex] form will be responsible for the reaction. If an ester is reacted with an alcohol in the presence of a base, transesterification (ester exchange) will occur, so the product would be oxalic acid dienolate, I think. The mechanism probably involves the deprotonized acetone (enolate form) to attack the ester's carboxylic carbon, making the previous esteric alcohol be removed as its alcoholate. So the brief reaction should be like this:

    [tex]H_3C-C(O^-)=CH_2 + ROOC-COOR \longrightarrow H_3C-C(=CH_2)-(O-OC-CO-O)-C(=CH_2)CH_3 + 2RO^-[/tex]
     
  4. Jan 20, 2005 #3

    GCT

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    Perhaps its a addition elimination reaction, with the acetone as the nucleophilic component attacking the carbonyl carbon and under the right conditions.......a cyclic compound.
     
  5. Jan 21, 2005 #4
    Thanks for your help. According to the answer I have, there are two producs: one of which is cyclic.
     
  6. Jan 21, 2005 #5
    I'll try and post the products here but I couldn't figure out how the mechanism works...(to be precise, what the mechanism is).
     
  7. Jan 21, 2005 #6

    chem_tr

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    If you write the products, we will try to devise a mechanism for them.
     
  8. Jan 21, 2005 #7

    GCT

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    The addition elimination mechanism can be found through the index of your text.
     
  9. Jan 26, 2005 #8
    I know but I couldn't figure out how a cyclic product was formed. Okay well I'll try and look at this again.

    Thanks
    vivek
     
  10. Jan 26, 2005 #9

    GCT

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    It's probably an intramolecular reaction involving the same mechanism.
     
  11. Jan 29, 2005 #10
    I had an idea...dunno if its generally okay...

    Sodium ethoxide being a relatively strong base can abstract the acidic alpha hydrogen atom from a ketone (since abstraction from an ester would lead to relatively unstable intermediates--in particular one involving crossconjugation) and the carbanion thus formed could attack as a nucleophile on the 'carbonyl carbon of the ester'. The intramolecular attack of the oxygen atom (with a negative charge now) would then force the (otherwise weak leaving group) oxy-substitutent out of the compound thus forming a so called Claisen Schmidt condensation product.

    Having said this, I get one answer correct but not the second one. There's still a mystery (I am still looking up all the books I can find) as to how the cyclic product can be formed...and I know that once I figure it out, I am gonna have to bang my head into the nearest inelastic wall... :grumpy:

    Cheers
    vivek
     
  12. Jan 29, 2005 #11
    Yeah I've figured out both the products...thanks.
     
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