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:

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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?

  • 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 $\endgroup$ Jan 5, 2022 at 1:53


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