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Which is more stable? A phenyl carbanion or a disubstituted vinylic carbanion?
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+ Biochemistry
+ Stability
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Naziru Imam
Which is more stable? A phenyl carbanion or a disubstituted vinylic carbanion?
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.
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.
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.
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.
@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,More
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.
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.
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.
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.
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