I have a little bit of problem trying to figure out the mechanism for the synthesis of Meldrum Acid. Meldrum acid is synthesis by a condensation reaction of acetone with malonic acid in acetic anhydride.

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First, the carboxyl OH will be acetylated by acetic anhydride. But I can't figure out how the two ester bonds are being formed subsequently. I'm thinking that there's an acid-catalysed aldol condensation somewhere that results in the removing of water. Perhaps by attack of the enol (from acetone) to the acetylated carbon. But I'm not really sure about it either.

Any thoughts?


1 Answer 1


The first step (reaction with acetic anhydride) creates a mixed anhydride, this is a good electrophile with acetate as a leaving group. This is nucleophilically attacked by the oxygen of acetone (drawn as its enol in the scheme, but it is easier to visualise it as acetone) to give a cationic intermediate which is immediately attacked by the -OH of the neighbouring carboxyl group to give the product.

  • $\begingroup$ Thank you, I can understand how this eventually gets me to the product. But is there a reason why the carbonyl oxygen is the nucleophile here? Seems quite unusual for a carbonyl oxygen to attack a carbonyl carbon. $\endgroup$
    – derrick
    Commented Mar 11, 2020 at 15:30
  • $\begingroup$ The mixed anhydride is very reactive, comparable to an acyl halide. The carbonyl has sufficient electron density to attack (or consider the enol the attacking species). I would emphasise that the intermediate that results reacts very quickly indeed with the carboxyl OH and that reaction is irreversible, so you only need a tiny percentage reaction for it to move forward. $\endgroup$
    – Waylander
    Commented Mar 11, 2020 at 15:46

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