Which proton would yield the most stable conjugate base?

[MechRocket]

[PLAIN]http://img638.imageshack.us/img638/5747/asdfnt.jpg

My workbook says that protons A & B are both more acidic than C, with the answer being A (I do get that A should be more acidic than B).

[PLAIN]http://img52.imageshack.us/img52/9800/asdfmi.jpg

I chose C though because de-protonation of C would access the more conjugated system. I would think there is more electron delocalization if you de-protonate C compared to A or B.


I posted this question on another forum and got the response that the reason is because the alpha hydrogens are closer to the Oxygen (electron-withdrawing), but I am not entirely convinced. In my undergrad Orgo course, I was told to always go with the proton that yields that more stable conjugate base, and I would think the base that results from de-protonation at C is clearly much more stable than the one that results from either A or B.

Is this a matter of kinetic control vs. thermo control or something? That's the only thing I can think of, where at low temperatures the proximity to the Oxygen makes A & B more acidic while at higher temperatures the yield would increasingly favor de-protonation of C. Just something I'm putting out there, but it's really just a shot in the dark.

 

[SpectraCat]

[PLAIN]http://img638.imageshack.us/img638/5747/asdfnt.jpg

.. I would think the base that results from de-protonation at C is clearly much more stable than the one that results from either A or B.

I think you have come to that conclusion a little hastily. Have you tried to draw resonance structures for deprotonation at A or B? The key stabilizing feature of the resonance system you drew for deprotonation at C is that it includes a structure where the negative charge is localized on the oxygen in the anion.

All three deprotonation sites result in enolate anions ... you might want to look up keto-enol tautomerism to help understand this a little better.

Is this a matter of kinetic control vs. thermo control or something? That's the only thing I can think of, where at low temperatures the proximity to the Oxygen makes A & B more acidic while at higher temperatures the yield would increasingly favor de-protonation of C. Just something I'm putting out there, but it's really just a shot in the dark.

 

[MechRocket]

I think you have come to that conclusion a little hastily. Have you tried to draw resonance structures for deprotonation at A or B? The key stabilizing feature of the resonance system you drew for deprotonation at C is that it includes a structure where the negative charge is localized on the oxygen in the anion.

All three deprotonation sites result in enolate anions ... you might want to look up keto-enol tautomerism to help understand this a little better.

I do think I have a good handle on keto-enol tautomerism. If you draw those resonance structures out (de-protonation of A & B), you will see that the conjugated system upon de-protonation of C is still much more extensive because the aldehyde is a Beta-unsaturated aldehyde.

I drew these side by side for you to compare (de-protonation of A #1 vs. de-protonation of C, #2):

[PLAIN]http://img692.imageshack.us/img692/1476/asdfhug.jpg