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I found this equation to turn this compound into an ketoester but I am unsure of the reaction mechanism.

I postulate that the alcohol attacks the carboxyl carbon, the O becomes formally negative. It reforms the double bond and the leaving is the O in the ring which takes the electrons in the bond.

Here my thought differs from the image (gotten from one of the United States Patent Office Patents). I thought the end product would be an ester OHCCCCOOR. Alcohol then alkene and at the end an ester but the picture shows a ketone and no alcohol.


1 Answer 1


You are correct that the alcohol atacks the carbonyl then the ring oxygen leaves in a manner analagous to ester transesterification. The intermediate generated is an enol which picks up a proton to for the ketone.

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image from here: 1

  • $\begingroup$ Oh keto-enol Tautomer. I sometimes have trouble seeing where this can happen. Is there a way to know where this would occur or do you simply have to see many examples and at a certain point you can just see it? $\endgroup$
    – bobsburger
    Nov 6, 2020 at 16:34
  • 1
    $\begingroup$ Most of the time the keto form prevails unless there's extended conjugation. It is worth noting that in this molecule enolisation towards the ester group is the more favoured. $\endgroup$
    – Waylander
    Nov 6, 2020 at 16:59

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