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Of 4-nitrobenzaldehyde and 4-aminobenzaldehyde, which has the higher $\ce{K_{eq}}$ of hydration?

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Steric factors can likely be ignored in this problem since we are dealing with rather structurally similar compounds.

I suspect the problem comes down to recognizing that a more polar compound is more favorably solvated by water.

I know that the nitro group is always electron-withdrawing, and I drew a few resonance structures as well. The resonance structures for both, however, seem to suggest that both are rather polarized (given the distribution of formal charges).

Resonance forms for 4-nitrobenzaldehyde

  • I expect the ones with the oxygen possessing the incomplete octet are very minor resonance contributors.
  • Therefore it seems that the main thing about this molecule is that the nitro group is withdrawing electron density from the ring and the aldehyde group.

enter image description here

Resonance forms for 4-aminobenzaldehyde

  • The resonance structures here are all somewhat reasonable.

enter image description here

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  • $\begingroup$ I think you need to consider the relative electrophilicities of the two compounds... $\endgroup$ – jerepierre Sep 2 '14 at 22:40
  • $\begingroup$ @jerepierre i would expect the p-nitrobenzylaldehye to be more electrophilic, but what does that have to do with hydration? $\endgroup$ – Dissenter Sep 2 '14 at 23:08
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    $\begingroup$ What is the reaction that you're trying to determine the Keq of? You might also want to write that out. $\endgroup$ – jerepierre Sep 2 '14 at 23:25
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I think you're asking about the following equilibrium

enter image description here

This equilibrium will shift to the left as the aldehyde becomes more stabilized. So the question becomes, which R group, $\ce{NO2}$ or $\ce{NH2}$, will tend to stabilize the aldehyde the most? A substituent that further delocalizes the positive charge shown in the dipolar resonance form would tend to be stabilizing. Just like in electrophilic aromatic substitution, additional stabilizing resonance structures can be drawn involving the amino substituent, but similar structures involving the nitro group are not as significant due to electrostatics (2 adjacent positive charges). Therefore, since the amino will stabilize the aldehyde better than the nitro substituent, the equilibrium involving the amino substituent should lie further to the left, and the aldehyde with the nitro substituent should be more hydrated.

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  • $\begingroup$ How come in your hydrated picture you have two OH groups? Is hydration not the same as solvation? $\endgroup$ – Dissenter Sep 4 '14 at 0:12
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    $\begingroup$ No, hydration is not the same as solvation. In the case of carbonyl compounds, when one speaks of hydration, and especially the equilibrium constant for this process, one is speaking about the addition of water across the carbonyl double bond to produce a gem-diol. I thought perhaps you were really asking about solvation. That's why I wrote, "I think you're asking about the following equilibrium." $\endgroup$ – ron Sep 4 '14 at 0:24

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