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I recently had an assessment which required me to draw the mechanism of decarboxylation of a $\beta$-ketoacid. However, it is slightly peculiar because there is the presence of $\alpha$, $\beta$-unsaturation. If the normal mechanism was drawn, then there would be an $\ce {sp}$-hybridised carbon in the enol intermediate after the $\ce {CO2}$ leaves. The hydrolysis and decarboxylation steps are similar to these shown in a publication by Takeda, Tsuboi & Sakai (1973). The only difference being that the question I attempted specified that the acid used to effect decarboxylation is $\ce {H2SO4}$.

enter image description here

To make it seem more logical, after the ester hydrolysis, I included an additional protonation step at the start of the decarboxylation, which involves protonating the double bond and forming an oxonium ion. Then, the normal decarboxylation mechanism proceeded. After which, there is tautomerisation back to the keto form, with shifting of the two double bonds to generate the bullatenone. Would that be the likely mechanism that is operating?

enter image description here

Out of curiosity, I would also like to ask if the decarboxylation would still take place without the $\alpha$ oxygen atom or the $\ce {Ph}$ group. I believe it could still take place but it is much less likely since it will now involve the formation of a secondary carbocation with no resonance stabilisation.

Reference

Takeda, S.; Tsuboi, K.; Sakai, T. A new synthesis of bullatenone. Chem. Lett. 1973 , 2, 425-426.

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  • $\begingroup$ Can you draw you suggested mechanism $\endgroup$ – Waylander Apr 29 at 12:57
  • $\begingroup$ @Waylander I understand that my description is unclear. I can only draw it on paper and not on any drawing softwares. $\endgroup$ – Tan Yong Boon Apr 29 at 12:58
  • $\begingroup$ @Waylander I have uploaded an image of what I drew. Please have a look. $\endgroup$ – Tan Yong Boon Apr 29 at 13:05
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    $\begingroup$ That looks pretty reasonable. I think it would still work without the Ph group though perhaps at a higher temperature. $\endgroup$ – Waylander Apr 29 at 13:16
  • $\begingroup$ @Waylander: You are absolutely correct. Still with Ph, it needed heating at $\pu{70 ^{\circ}C}$ for $\pu{9 h}$ to yield 54% (when X = H). $\endgroup$ – Mathew Mahindaratne Apr 29 at 22:07

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