Although the phenolate ion has more resonance structures (4) compared to acetate ion (2), acetate is more stable because it has two equivalent resonance structures of same energy.

Why does having equivalent resonance structures give more stability? Shouldn't a compound with more resonance structures be more stable?



One important thing to know is that what we call "resonance structure" is a byproduct of our chemical notation which can't describe the structure of some compounds effectively using only one chemical structure. Personally I find the old term mesomeric structure more appropriate (meso- Greek mésos in the middle; -merism from Gk. merismos "dividing, partition") because resonance is already used for a phenomena, a real phenomena, in physics that has nothing to do with resonance in chemistry. Furthermore the compound is not continuously changing it's structure between the possible resonance structures, like the term resonance seems to suggest, but is a "normal steady" compound that simply has an electron distribution that cannot be described with a single chemical structure.

Why does having equivalent resonating structure give more stability?

The resonance energy is due to electron delocalization, so more the electrons are delocalized in the molecule higher is the resonance energy ( this means a decrease in the overall molecule energy). This is clear in the acetate ion - we can draw a resonance hybrid like this: enter image description here

Here the electrons are equally distributed (delocalized) between the two oxygen atoms.

However for some compounds the real structure is not always a structure right in the middle between the different resonance structures; the first assumption is true only if the two structures are two equivalent structures. So it can occur that the electrons are not well delocalized between all the atoms hence the resonance energy is lower and so the overall stability.

In the case of phenolate ion I think the first structure in the figure 2 gives a greater contribution to the final structure of the phenolate ion so the electrons are not well delocalized, but if I'm not wrong are mainly in the oxygen atom. Unfortunately I can't find the energy of the different resonance structures. enter image description here

Finally, however, I think that you should study case by case if the number of the resonance structures or their contributions are determinant to the overall resonance energy taking into account valency and electronegativity of the atoms involved.

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    $\begingroup$ By delocalizing the charge into the phenyl ring, you would put electrons in an antibonding (aromatic) orbital. This is of course possible, but since the electronegativity of Oxygen is greater than in carbon, not very likely. If however you put an external field around the molecule ([point] charges or non-protic solvent) you could see more negative charges at the indicated atoms. Resonance energy is a mere mathematical concept which is not measureable and strongly dependent on the definition. {I'd give additional +1 for your preamble.} $\endgroup$ – Martin - マーチン Apr 5 '14 at 10:09
  • $\begingroup$ @Martin thanks a lot for the comment always very useful, I think that resonance energy can be consider as the difference between the true energy of the molecule and the energy calculated without taking in account the delocalization. $\endgroup$ – G M Apr 6 '14 at 22:12

You can estimate stability by the number of resonance structures you can draw. For the acetate ion, there are 2 resonance structures, while for the protonated form (acetic acid) only one structure is possible, which favors deprotonation. For the phenolate ion, you can formulate 4 structures, with the negative charge delocalized into the benzene ring. For phenol, you can draw 4 structures as well, although 3 of them have a positive formal charge at the oxygen. This is why phenol is weakly acidic, and quite acidic for an alcohol.

Edit: The content of this post can also help understanding the different acidities of phenol and acetic acid: Why is ethanoic acid more acidic than phenol?

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  • $\begingroup$ If acetic acid is favored in the deprotonated form, why is the Ka so low? $\endgroup$ – LDC3 Apr 2 '14 at 13:32
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    $\begingroup$ The Ka of an acidic/basic functional group varies depending on its molecular environment, i.e. the substituents it is attached to. The methyl group does not stabilize the negative charge of the carboxylate group in the acetate ion, which lowers the Ka. However, electron-withdrawing groups, like in chloroacetic acid, for example, stabilize the negative charge by inductive effects, and the resulting Ka is higher. $\endgroup$ – Jannis Andreska Apr 2 '14 at 14:08

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