1
$\begingroup$

enter image description here

Image source

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.

$\endgroup$
4
$\begingroup$

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.

enter image description here

image from here: 1

$\endgroup$
2
  • $\begingroup$ 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 '20 at 16:34
  • $\begingroup$ 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 '20 at 16:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.