[Rakkoonn]

What is the mechanism of the Doebner modification for Knoevenagel condensation?

The final step looks like this:



but what leads to that? I think I can't use piridine the same way piperidine is used in the Knoevenagel condensation so should it react with the RCHO? If so then what does the reaction look like? A drawing of the first steps would be appreciated

thanks in advance

[kremar]

well, the first step is is the formation of the pseudo-enolate, with the pyridine acting as a base on H-CH(COOH)2.
this pseudo-enolate reforms the diacid and attacks the aldehyde as one could expect. since the alcoxide formed is more basic than a carboxilic acid, one could expect a fast proton exchange to occur, yielding the pseudo-aldol you have depicted in your step.

your step shows the rest. in this case, the decarboxilation doesnt occur via the usual mechanism, since we dont have an acid catalyst.

i'd expect the proton exchange and the decarboxilative elimination to be concertated, since that would imply the 6-center transition stante that's accepted for the usual beta-ketoacid decarboxilations, but that's just me...