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