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Author: Subject: Potassium Salt of Acetonedicarboxylic Acid Monomethyl Ester
ronstark
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[*] posted on 27-10-2016 at 06:06
Potassium Salt of Acetonedicarboxylic Acid Monomethyl Ester


There are many known routes to Acetonedicarboxylic Acid from Citric Acid (anhydrous or monohydrate). One would have to oxidize Citric Acid with H2O2 + a catalyst or fuming H2SO4 or Oxone.

From Acetonedicarboxylic Acid (which is unstable) with MeOH + KOH would yield the desired compound.

Am I missing something here or what cause I think that if you start from Citric Acid with Oxone in MeOH would yield the same compound with the advantage that you skip one reaction.

Would be great that this works since a spectra is hard to get.

[Edited on 27-10-2016 by ronstark]
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Tsjerk
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[*] posted on 28-10-2016 at 05:18


MeOH under oxidizing conditions isn't going to give you a methyl ester.
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bmays
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[*] posted on 29-10-2016 at 13:12


H2SO4 could be used to oxidize then also form the methyl ester, but i tried and it doesn't work so well. Won't work with a base. If you get good yeild from Oxone that might be best. I'm not sure the best way to form the ester but i know in hot H2SO4 it will all decompose at a significant rate.
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29-10-2016 at 15:31
zed
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[*] posted on 30-10-2016 at 15:04


Hmmmm. Somewhere, sometime, I gained the impression this material was obtained via a cyclic anhydride intermediate.
http://www.chemicalbook.com/ChemicalProductProperty_EN_CB650...

[Edited on 30-10-2016 by zed]
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ronstark
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[*] posted on 1-11-2016 at 08:57


Acetonedicarboxylic Acid from H2SO4:

3C6H8O7 + H2SO4 = 5C5H6O5 + 7HCOOH + SO3

wanted product from Acetonedicarboxylic Acid:

2C5H6O5 + 8CH3OH + 3KOH = 3C6H7O5K + 6H2O

so... for 10g Acetonedicarboxylic Acid (0.06 molar) you need 5.76g KOH (0.10 molar)

if you start from Citric Acid in Oxone to form Acetonedicarboxylic Acid, first, you prepare the medium:

K5H3S4O18 + 5H2O = H8S4O18 + 5KOH

and then you add Citric Acid just like in H2SO4... you get exactly the same molar amount of KOH, that I think it can connects to the formed Acetonedicarboxylic Acid to a side chain forming COOK and on the other side, MeOH added after a period of time, when the reaction is mostly over, on reflux, to form COOMe.

Monitoring the rx will be great on TLC plates.

[Edited on 1-11-2016 by ronstark]
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bmays
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[*] posted on 3-11-2016 at 17:31


Have you tried? Extract the methyl ester with ether? The problem is if water is present the ester will be hydrolyzed by KOH, you need acid. Or you are saying acetonedicarboic acid forms a potassium salt and is esterfied. Maybe.

[Edited on 4-11-2016 by bmays]
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ronstark
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[*] posted on 4-11-2016 at 16:11


That is exactly the point. The final product has to stay in water. It's very similar to the later link but a "CH3-" (from "CH3-O-") is replaced by "K-" ("K-O-C"). A Potassium Salt but only on the left. On the right it has to stay the same ("C-O-CH3").
http://m.chemicalbook.com/ChemicalProductProperty_EN_CB92205...

[Edited on 5-11-2016 by ronstark]
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