I am confused about how one can determine the stability of carbanions using pKa-values. I know that the stability of carbanions can be determined by the inductive effect, hybridization of the charged-bearing atom and resonance, but when searching about how pKa can do so I don't find anything. The lower the pKa-value, the stronger the acid and hence, the weaker the conjugate base, does this have anything to do with the stability since weak bases are unreactive and thus more stable?
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$\begingroup$ Except for the mistake SteffX has pointed out, I think your argument is valid but you need to pay attention to the context of "stability". $\endgroup$– Weijun ZhouCommented Jan 28, 2018 at 15:39
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2 Answers
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As commented by SteffX, less the $\mathrm{p}K_\mathrm{a}$, more stronger the acid. More stronger acid means that the conjugate base or the anion formed after donating the proton is more stable compared to the carbanion of an acid of a higher $\mathrm{p}K_\mathrm{a}$ value. So, if the anion formed is a carbanion, then lower the $\mathrm{p}K_\mathrm{a}$, the stable the carbanion is.
answered Jan 28, 2018 at 15:50
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$\begingroup$ Why the downvote? Have I written anything wrong? $\endgroup$ Commented Jan 29, 2018 at 3:46
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The stability of every molecule is related to the spreading of charges: the more you can share a charge among atoms, the more stable the molecule is.
For a carbanion, every group WITHDRAWING electrons to the carbanion center will stabilize it (i.e. the negative charge will be shared by several atoms)
For a carbocation, every group GIVING electrons to the carbocation will stabilize it (i.e. the localized positive charge will be compensated for... at least partially)
