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I've been taught that aromaticity is often a very good stabilization factor. Then I came across this question in which another compound was said to have more stability.

Upon deprotonation arrange in the order of stability.

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Why is S more stable than P?

(the answer given is (a); from Himanshu Pandey Organic Chemistry chapter General Organic Chemistry)

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  • $\begingroup$ Indeed the comparison is between totally different molecules and/or their different carbocations. I can even push further saying that in principle one doesn't know without observation or calculations. Of course an answer is possible and that given is correct. Consider the effect of the two C=O in S. It is stronger than the moderate effect in Cyclopentadyenil cation. $\endgroup$ – Alchimista Dec 22 '17 at 9:12
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    $\begingroup$ I agree that this is not answerable without recourse to either (1) memorising pKa ranges; or (2) experiment (either in the lab, or in silico). I would recommend the former. If you look up the pKas, a typical 1,3-dicarbonyl compound has pKa 10. Cyclopentadiene has pKa 16 ish, so it is pretty close, actually. 1,3-Dithiane has pKa... I’m not sure, but I’d guess around 30 or 35. Of course, a lower pKa means stronger acid means more stable conjugate base. $\endgroup$ – orthocresol Dec 22 '17 at 9:33
  • $\begingroup$ I am not sure lab experimentation would work either. How do you work with (P) when it's prone to self-dimerize? It looks like the student is supposed to know pKa values by rote. Sucks to be given that kind of question. $\endgroup$ – Oscar Lanzi Dec 22 '17 at 17:21
  • $\begingroup$ Where can i get the pka ranges? Not all of them i just want to get a rough idea. $\endgroup$ – Avnish Kabaj Dec 23 '17 at 6:40
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    $\begingroup$ @AvatarShiny A very extensive listing of pKa values is here; obviously it makes no sense to memorise all of them, but they are conveniently categorised based on functional groups, so you should be able to get a general feeling of relative pKas as well as how much they can be affected by substituents. You will also probably see more of these as you do more chemistry. $\endgroup$ – orthocresol Dec 27 '17 at 22:25
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The situation in S and P is very similar. Essentially you have a negative charge $\alpha$ to two double bonds which will delocalize these charges.
In the case of cyclopentadiene, this will be stabilized because the anion is aromatic. But still, there will be a charge which is supported by carbon atoms.

In the case of the 1,3-diketone, the charge will also be delocalized by two double bonds, but instead of an aromatic anion of five members, the charge will be supported by only three atoms. However, two of them are oxygens which can hold a negative charge much better than a carbon.

The electronegativity of carbon is 2.55 while for oxygen it is 3.44. Note that the electronegativity scale ranges from 0.7 for $\ce{Fr}$ to 4.0 for $\ce{F}$. This ability of supporting negative charge is much bigger than the small contribution of aromaticity to the stabilization of a carbanion.

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