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Finding order of stability of 4 carbanions using resonance and inductive effects

I am confused about the stability of carbanion III and IV in the image, I think III is more unstable because of electron donating effect (+R) of $\ce{-OCH3}$ group but my book says III is more stable than IV. In compound III, $\ce{-OCH3}$ would donate its lone pair and would make the carbanion unstable but no such effect happens in compound IV. $\ce{-OCH3}$ in compound III would also have -I effect to stabilize the carbanion but I have read that for $\ce{-OR}$ group, +R effect is more than its -I effect i.e., its electron donating nature is more than its electron withdrawing nature through inductive effect, +R means electron donating by resonance (In $\ce{-OR}$, $\ce{R}$ is any alkyl group)

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    $\begingroup$ To answer this question, I have been trying to find the pKa values of the four parent acids, but I have only found acetone. It is the only compound of the four listed in the well known Bordwell and Evans tables. Ethyl acetate (not methyl) is listed, so I supposed that could be used, but I cannot find an authoritative source for acetaldehyde and methyl vinyl ketone. I am frustrated by my inability to easily search for papers that report pKa values. $\endgroup$ – Ben Norris Feb 11 '16 at 14:05
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    $\begingroup$ The picture clearly says that it is abput resonance stability. So drawing all possible resonance structures for every substance should help probably. $\endgroup$ – pH13 - Yet another Philipp Feb 11 '16 at 21:58
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I think in 3, carbanion is resonance stabilised. While in 4, C=C-C=O are in conjugation, thus will not stabilise carbanion.

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    $\begingroup$ Why does conjugation prevent stabilisation? I see no reason for this. $\endgroup$ – bon Feb 12 '16 at 10:06
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I think in both (III) and (IV), Cross Conjugation is present which generally prevents resonance from taking place.

  • In (III), Both CH3O and CH2minus can give electrons to the C in middle.

  • In (IV), After resonance, CHminus(on left) and CH2minus(on right), can give electrons to the C in middle.

Supposedly, in (III), O will not prefer to give it's LP electrons because of high Electronegativity. In (IV), CHminus(on left) can comparatively give electrons more easily than O. Also +M effect of CH2minus is greater than O(with LP).

Therefore, CROSS CONJUGATION EXTENT in (III) is less than in (IV).

Hence, (III) is more stable than (IV).

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