4
$\begingroup$

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)

$\endgroup$
2
  • 1
    $\begingroup$ 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
    Commented Feb 11, 2016 at 14:05
  • 1
    $\begingroup$ The picture clearly says that it is abput resonance stability. So drawing all possible resonance structures for every substance should help probably. $\endgroup$ Commented Feb 11, 2016 at 21:58

2 Answers 2

1
$\begingroup$

I think in 3, carbanion is resonance stabilised. While in 4, C=C-C=O are in conjugation, thus will not stabilise carbanion.

$\endgroup$
1
  • 2
    $\begingroup$ Why does conjugation prevent stabilisation? I see no reason for this. $\endgroup$
    – bon
    Commented Feb 12, 2016 at 10:06
1
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.