1
$\begingroup$

In Part 1 of the synthesis of Maoecrystal V (found on synarchive), the last step subjects a $\ce {\alpha, \beta}$-unsaturated ketone to $\ce {I2}$ and pyridine, as shown below:

enter image description here

I am wondering what is the mechanism that takes place for the elimination process. It is unlikely to be an E2 process since that requires anti stereochemistry for the $\ce {H}$ and $\ce {I}$ that are being eliminated. After anti-addition of the $\ce {I2}$ molecule to the double bond, the $\ce {H}$ and $\ce {I}$ atoms would both be syn to each other. Since this is a ketone with acidic $\ce {\alpha}$ hydrogens, a E1cB mechanism is likely operating and thus, we observe the elimination of the $\ce {\alpha}$ hydrogen and the iodine atom on the $\ce {\beta}$ position. Is my hypothesis correct?

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ This is a reasonable hypothesis. It could also be proceeding by a Baylis-Hillman type halogenation via the pyridinium enolate. $\endgroup$
    – Waylander
    Jan 4, 2022 at 9:51
  • $\begingroup$ Yeah, it's rather pyridine being eliminated. $\endgroup$
    – Mithoron
    Jan 5, 2022 at 1:21
  • $\begingroup$ @Mithoron Well... E1cB or Baylis-Hillman, both are just equally likely hypotheses $\endgroup$
    – Tan Yong Boon
    Jan 5, 2022 at 1:53

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.