[biocamson]

And please give some explanation, thanks a lot!

[discodermolide]

Do you have any ideas?

[biocamson]

Do you have any ideas?

I initially thought it should be A coz it next to the carbonyl group, but seems it is wrong.
I know it should be C and sth related to the delocalization of electrons(not so sure), but can you please tell the reason? thx

[discodermolide]

If you remove a proton from the various positions you will get a carbanion. Try then drawing out resonance structures for them and see which are most likely. This will help get you to the answer.

[biocamson]

If you remove a proton from the various positions you will get a carbanion. Try then drawing out resonance structures for them and see which are most likely. This will help get you to the answer.

indeed i have done that, but still quite confused with the reason behind... may you please give the explanation? thx

[orgopete]

For a 'non-standard' suggestion, please look at my explanation here: http://www.chemicalforums.com/index.php?topic=83298.0

So let me give the result of a different problem, the enolate of cyclohex-2-enone. If done with a strong base (LDA) at low temperatures, one gets the kinetic product. If an excess of the cyclohexanone is present (as a proton source), the thermodynamic enolate results. The kinetic and thermodynamic products are the enolates of 2- and 1-hydroxycyclohexa-1,3-diene, respectively. So the kinetic product is the enolate from the most acidic hydrogen while the thermodynamic product is that of the most stable enolate.

I would argue that the proximity to the C=O leads to the kinetic product. For the thermodynamic product, the inductive properties or effect are being transferred through and added to the inductive effect of the C=C. The dominate effect of delocalization of the non-bonded electrons of the oxygen will be greater in the enolate of 1-hydroxycyclohexadiene (or a greater resonance stabilization). The net result will be an increase in the acidity of the CH2-hydrogens. I would argue the present problem is similar. A double bond will increase the acidity of vicinal hydrogens and the inductive effect of the carbonyl group will added to that increase (by resonance through the double bond), but because it is now at a greater distance, the inductive effect will be reduced.

I caution anyone reading this in that how I understand and use resonance effects is different than the common usage and has often led (if not most?) to the negative responses you see attached to my answers. None the less, I am of the opinion this explanation succeeds in answering the why question better than looking at the resonance structures of the conjugate bases, especially as it explains why different pKa values can be present for common conjugate bases, e.g., acetone v 2-hydroxypropene.

[Babcock_Hall]

If you remove a proton from the various positions you will get a carbanion. Try then drawing out resonance structures for them and see which are most likely. This will help get you to the answer.

indeed i have done that, but still quite confused with the reason behind... may you please give the explanation? thx

All else held equal, the more resonance structures one can draw, the more stable a carbanion (or other high-energy intermediate) is.  Can you show us what you have drawn?

[biocamson]

can anyone give a simpler explanation.... thx

[Babcock_Hall]

Can you show us the resonance structures you have drawn?  That would help greatly.