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I've been stuck with this for some time. I just don't see it. As the reactant is Pd(0), I think oxidative addition will happen to the bromine. Then I would say an insertion reaction happens. However this would form a 7-membered ring which isn't as stable as a 5-or 6-membered ring and isn't even in the final product.

Then I tried complexating it with the double bond next to the ether. However I don't see what use the Pd could have from there. Maybe deprotonate the alpha-carbon which would give a hapto-3 complexation? But what next, I wouldn't know. Any help would be appreciated!

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Nicco: Of course you can't "see it". If you were given the correct structure, the solution to your post would be straightforward. There should be a double bond in the lactam ring as shown in compound 1. The reaction is conducted in acetonitrile (bp. 82oC) at reflux. These conditions allow for a Claisen rearrangement to occur to form ketolactam 2. Subsequent oxidative addition of palladium(0) [ligands ignored] to the C-Br bond creates the Pd(II) species 3 which adds to the double bond to afford cyclized species 4. Reductive elimination of Pd(0) regenerates the catalyst as alkene 5 is formed. The role of $\ce{Et3N}$ is to neutralize HBr that is formed. Addendum: Just found the original paper: https://doi.org/10.1016/0040-4039(94)88380-7

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