Chemistry - Which is more stable? A phenyl carbanion or a disubstituted vinylic carbanion?

Solution 1:

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.

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how is it more stable than vinylic carbanion?

The pKa approach discussed in @Jori's answer becomes problematic when we are trying to look at a small difference between two large pKas. A pKa 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 pKa table we find a pKa of 43 for the benzene $\ce{C-H}$ proton and a pKa 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 s-character will stabilize electrons better than bonds with less s-character (because the s-orbital is lower energy than a p-orbital), whichever bond has more s-character should produce a more stable carbanion.

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

The $\ce{sp^2}$ hybridized orbital in benzene has more s-character than the $\ce{sp}^{2.2}$ 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 \ s-character reasons presented above.

Edit: OP's comment on hybridization

Molecules are not just $\ce{sp, sp^{2}}$ or $\ce{sp^3}$ hybridized. While the $\ce{C-H}$ bonds in methane are exactly $\ce{sp^3}$ 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 "p") between zero and infinity. Hybridization indices are not limited to integer values, nature can mix s and p orbitals in whatever ratio is required to create the most stable bonds and molecules.

Solution 2:

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!)