I came across a question which asks to work out the stability order for the following species-

b. benzene carbanion

c. $\ce{R2C=CH-}$

Can the lone pair on phenyl carbanion participate in resonance? If yes, then how? Also, how is that more stable than the vinylic carbanion?

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    $\begingroup$ No it doesn't participate in resonance. $\endgroup$
    – Mithoron
    Commented Feb 25, 2015 at 16:46
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    $\begingroup$ Then how come it is more stable than the primary vinylic carbanion ? $\endgroup$ Commented Feb 25, 2015 at 16:48
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    $\begingroup$ Again... stable towards what exactly? I find these (I suppose exam) question rather weird asking about stability of compounds without a proper context. $\endgroup$
    – Jori
    Commented Feb 25, 2015 at 16:52
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    $\begingroup$ @Jori, unfortunately these questions are a lot to come by, at least here. Hmm, I put them like this: The compound that is more stable produces lesser heat upon decomposition. $\endgroup$
    – M.A.R.
    Commented Feb 25, 2015 at 20:55
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    $\begingroup$ Also, why close this question? It is not homework. I think OP could have tried to add some more of his own thought, but I consider it on-topic. If we do not want to scare too many newcomers away I think we should not have a too harsh policy. $\endgroup$
    – Jori
    Commented Feb 25, 2015 at 21:39

2 Answers 2


Can the lone pair on benzene carbanion participate in resonance?

As Jori has pointed out, the carbanion lone pair is orthogonal to the pi system of the aromatic ring. Therefore it cannot participate in resonance. Here is a drawing that illustrates this point.

Structure of benzene anion

how is it more stable than vinylic carbanion?

The $\mathrm{p}K_\mathrm{a}$ approach discussed in @Jori's answer becomes problematic when we are trying to look at a small difference between two large $\mathrm{p}K_\mathrm{a}$s. A $\mathrm{p}K_\mathrm{a}$ of $43$ means $1$ ionized molecule exists in every $10^{43}$ molecules. Certainly difficult to measure experimentally and so a significant error is attached to these estimates. In fact, using the data from Evan's $\mathrm{p}K_\mathrm{a}$ table we find a $\mathrm{p}K_\mathrm{a}$ of $43$ for the benzene $\ce{C-H}$ proton and a $\mathrm{p}K_\mathrm{a}$ of $50$ for an ethylenic proton, suggesting that the phenyl carbanion might be more stable than the vinyl carbanion.

An alternate way to examine this question involves comparing the hybridizations of the phenyl and ethylenic $\ce{C-H}$ bonds. Since bonds with more $\mathrm{s}$-character will stabilize electrons better than bonds with less $\mathrm{s}$-character (because the $\mathrm{s}$-orbital is lower energy than a $\mathrm{p}$-orbital), whichever bond has more $\mathrm{s}$-character should produce a more stable carbanion.

The $\ce{C-H}$ bond in benzene is exactly $\mathrm{sp^2}$ hybridized as required by the symmetry of the benzene molecule. Experimentally, we know that the $\ce{H-C-H}$ angle in ethylene is $117^\circ$. Using this fact along with Coulson's theorem (reference1, reference2, reference3) we can determine that the $\ce{C-H}$ bond in ethylene is $\mathrm{sp}^{2.2}$ hybridized.

The $\mathrm{sp^2}$ hybridized orbital in benzene has more s-character than the $\mathrm{sp}^{2.2}$ hybridized orbital in ethylene. This would lead us to suspect that the phenyl carbanion is more stable than the vinyl carbanion.

Of course, once we remove these protons we expect the resulting carbanions to undergo further rehybridization in order to further stabilize their lone pair (carbanion) electrons. It seems like a reasonable first approximation that they would both rehybridize similar amounts and the "relaxed" phenyl carbanion would remain more stable than the "relaxed" vinyl carbanion. But in any case, the transition state leading to the phenyl carbanion will be more stable than the transition state leading to the vinyl carbanion because of the hybridization \ $\mathrm{s}$-character reasons presented above.

Edit: OP's comment on hybridization

Molecules are not just $\mathrm{sp, sp^{2}}$ or $\mathrm{sp^3}$ hybridized. While the $\ce{C-H}$ bonds in methane are exactly $\mathrm{sp^3}$ hybridized (as required by symmetry), the $\ce{C-H}$ bonds in $\ce{CH3F}$ are not (see here for a discussion). Bonds can have any hybridization index (the superscript to the right of "$\mathrm{p}$") between zero and infinity. Hybridization indices are not limited to integer values, nature can mix $\mathrm{s}$ and $\mathrm{p}$ orbitals in whatever ratio is required to create the most stable bonds and molecules.

  • $\begingroup$ Cant we say that the vinylic one is more unstable due to electron releasing effect of R groups? $\endgroup$
    – Archer
    Commented Nov 7, 2018 at 4:31
  • $\begingroup$ @Abcd Yes, if there are alkyl groups located near the carbanionic center then they could destabilize either carbanion inductively as you suggest. However, I think this effect would be secondary to the hybridization effect discussed above, $\endgroup$
    – ron
    Commented Nov 7, 2018 at 14:19

The lone pair on the benzene is in the plane of the ring and hence orthogonal to the conjugated $\pi$-system so that it cannot interact with it. It is $\ce{sp^2}$ hybridized just like the vinylic carbanion, which also cannot resonate. Furthermore, both compounds lack an electron withdrawing group, so there is no inductive effect either. Looking at the pKa values for both compounds, which usually gives a good indication about the stability of the anion, we see that there is indeed hardly any difference between them (pKa of benzene ~43 vs pKa of propene ~44 vs pKa of ethylene ~44).

In your question you only show us option (b) and (c). It would be nice if you could show the whole question including all possibilities (also as much as possible in text please, so the search engines can find it!)

  • $\begingroup$ I got the order partially right. So I did not the whole question. $\endgroup$ Commented Feb 25, 2015 at 21:56
  • $\begingroup$ Jori, there are basically two different types of protons in propene; those attached to an $\ce{sp^2}$ carbon and those attached to an $\ce{sp^3}$ carbon. Which type of proton in propene is responsible for the pKa you reported? $\endgroup$
    – ron
    Commented Feb 25, 2015 at 22:15
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    $\begingroup$ @ron I assumed it would be for the $\ce{sp^2}$ since I suspected that the $\ce{sp^2}$ hybridization was of more importance than the allylic conjugation. Comparison with the pKa of 3,3-Dimethyl-1-butene would give a more definite answer for sure, but I cannot find it anywhere. However comparing with the pKa of ethylene, which is also ~44, gives a good hint in this direction. I simply listed that table because it listed both types of compounds at the same place. $\endgroup$
    – Jori
    Commented Feb 25, 2015 at 23:06
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    $\begingroup$ @Martin You commented that the vinyl carbanion will adopt a linear configuration. That may be true for a vinyl cation, but is it also true for the vinyl carbanion? I would have thought that the carbanion would be bent with a barrier to inversion (just like an imine). The lone pair would be more stable in an sp2 orbital rather than a p orbital. Hybridization is a way to answer the OP's question. The C-H bond in benzene is $\ce{sp^2}$ while the C-H bond in ethylene is $\ce{sp^{2.2}}$, consequently the phenyl carbanion is more stable. If this question is reopened I will write more. $\endgroup$
    – ron
    Commented Mar 3, 2015 at 3:27
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    $\begingroup$ I was wrong about the geometry, see J. Am. Chem. Soc., 1978, 100 (19), pp 6007–6012. The lone pair is still not sp2 though. I think hybridisation is a poor concept to explain stability in any case (as long as we are not talking about highly accurate VB theory calculations and even then it is a stretch). I am tired of talking about hybridisation though. (If you want this question to be reopened, you have to rework it yourself I fear.) $\endgroup$ Commented Mar 3, 2015 at 4:22

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