Sciencemadness Discussion Board

Heck reaction - why the formation of 5 membered ring?

francis - 6-11-2011 at 02:02

Hi,

Our lecturer has given us a sheet on the Heck reaction. One of the reactions I have drawn below:


The reagent Pd(OAc) should be Pd(OAc)2, I drew it incorrectly.


The product is 3-methyl-1H-indole according to the sheet.

I would have thought it would be 1,2-dihydroquinoline? (which I drew under the product).

Why does the five-membered ring form, rather than the six membered ring?

[Edited on 6-11-2011 by francis]

bbartlog - 6-11-2011 at 05:26

Which is sort of the same as asking why the reaction takes place at the second carbon rather than the terminal one... looks like an intramolecular application of Markovnikov's rule, almost, but as to the actual mechanism I have no clue.

francis - 6-11-2011 at 06:23

I tried drawing a mechanism...forgive my dodgy handwriting.

I found a segment in my book that says, "Electron donating groups lead to attack at the end of the alkene (closest to the substituent)...these rxns must be domination by the interaction of the filled p orbital of the alkene with an empty d orbital on Pd."



Could the -NH2-CH2- end of the alkene be considered electron donating? I wouldn't have thought it would be strong enough.

The book says also, "when there is a choice as to which hydride can be lost to form the alkene, the stability of the possible product alkenes often governs the outcome."

But it appears there's only one hydrogen possible to remove by beta-hydride elimination?