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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?

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  • $\begingroup$ Hückel theory might already give a nice picture without dwelling on QM. $\endgroup$ – Kexanone Feb 17 at 17:12
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Preamble

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 structures 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:

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.

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
  • $\begingroup$ @GM So isn't neccesary that more resonance structures will to greater stability right? is also about the stability (energy) of each resonance structure? $\endgroup$ – ado sar Aug 20 '20 at 14:08
  • $\begingroup$ @adosar correct $\endgroup$ – G M Aug 20 '20 at 14:34
  • $\begingroup$ @GM But isn't this contradictory to the fact that the greater the delocalization of electrons (number of resonance structures) the greater the stability? $\endgroup$ – ado sar Aug 21 '20 at 15:48
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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.

I would argue the phenolate ion has five mesomeric structures, two with the charge assigned to oxygen (analogous to the two resonance structures of phenol), and three with the charge assigned to the ring carbon in ortho and para positions (see G M's answer).

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

I don't think the statement is true in this general form. It is always possible to make mesomeric structures a little bit less equivalent (for instance, through isotope-labeling of one of the oxygen atoms in acetate). This does not change the acidity of acetic acid much.

To give another example, we can compare the acidity of propene, acetaldehyde and formic acid. When you remove a proton from each of these, you get three anions that have multiple mesomeric structures.

enter image description here

The ones for propene and formic acid are symmetric, the one for acetaldehyde is not. Formic acid is most acidic, followed by acetaldehyde with propene least acidic. I would observe that for formate, the negative charge is assigned to oxygen for both mesomeric structures whereas for the deprotonated propene, it is assigned to a carbon atom. The enolate anion has the charge on oxygen for the more significant mesomeric structure (shown) and on carbon for the less significant one (not shown).

The argument for phenol vs acetic acid is similar. So it matters what the significant mesomeric structures are and which atom type carries the negative charge. When mesomeric structures are symmetric, it ensures that they have the same significance. If you compare apples to apples (i.e. anions where the negative charge is assigned to carbon for all mesomeric structures), it helps to distribute the charge (having the carbons with assigned charge at large distances), and having many possible locations for the charge.

For phenolate, the large number of mesomeric structures is less relevant because the ones with the charges assigned to oxygen are the most significant (the only ones that don't force us to write a structure lacking alternate double bonds in the benzene ring).

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    $\begingroup$ Yes, exactly! Having multiple equivalent resonance structures isn't itself a reason for stability; it's because it leads to greater delocalisation or distribution of charge, and that's merely one factor in the grand scheme of things. $\endgroup$ – orthocresol Feb 23 at 0:30
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    $\begingroup$ No quantum, no bounty... @orthocresol $\endgroup$ – Karsten Theis Feb 23 at 1:45
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    $\begingroup$ True — but an answer nonetheless, and a useful one at that... $\endgroup$ – orthocresol Feb 23 at 1:52
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    $\begingroup$ @MilanPaul, that's kind of you to say but it is misplaced: you are well within your rights to award the bounty to any answer you wish (I'm not sure if the system lets it give it to yourself), but not to my tiny comment, which hardly stands on its own and is a mere restatement of Karsten's (and others') points. Just as well the system definitely does not allow that. :-) $\endgroup$ – orthocresol Feb 23 at 16:28
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    $\begingroup$ @orthocresol To be fair, I did base my answer on chemistry.stackexchange.com/a/6160 and chemistry.stackexchange.com/a/41810, which is the cool thing about a community such as the one contributing to this site. $\endgroup$ – Karsten Theis Feb 23 at 16:44
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I would like to give a very rough answer as to why equivalent resonance structures have a stabilizing effect.

To begin, we assume that resonance can be represented as a discrete n-state system. Feynman has used this simple approximation to explain many of the essential features of chemical bonding and resonance in the Feynman Lectures on Physics Vol III, Chapters 8 - 10.

To simplify we begin with a molecule which can have 2 different canonical structure based on the usual rules. Let the energies of the two states be $E_1$ and $E_2$ .

The Hamiltonian of this two state system will be: $$H = \pmatrix{E_1& 0\\0&E_2}$$

The canonical structure are themselves eigen-states here. However in the real world we find more symmetry to the molecule/different bond lengths/angles than that is predicted by our canonical structures. This is because the two canonical structures are not the actual eigen states.

We assume that the new eigen states however can be represented as a superposition of the 'old' states.

Therefore in the most general case the Hamiltonian expressed in the 'old' base states become: $$H = \pmatrix{E_1 & H_{12}\\H_{12}^*&E_2}$$

We find the energy eigenvalues from this matrix.

The energy eigenvalue with the lower energy in this case is:

$$ E^{'}_1 = \frac{E_1 + E_2}{2} - \frac{\sqrt{(E_1 - E_2)^2 + 4H_{12}H_{12}^*}}{2} $$

Here without loss of generality let us assume that $E_2 \ge E_1$

So taking the difference $D = E_1 - E^{'}_1$ which is the energy difference between the old base state and the new eigenstate with lower energy: $$ D = E_1 - E^{'}_1 = \frac{E_1-E_2}{2} + \frac{\sqrt{(E_1 - E_2)^2 + 4H_{12}H_{12}^*}}{2} $$

This $D$ is a measure of the stabilization due to resonance.

Setting $E_1 - E_2 = x \le 0$, we get $$D = \frac{x}{2} + \frac{\sqrt{x^2 + 4H_{12}H_{12}^*}}{2}.$$

The graph of this function is:
enter image description here

So we see that for $x \le 0$ , $D$ is maximum when $x = 0$, i.e. $E_1 = E_2$. One way this can happen is when our base states are symmetrical or in other words when the molecule has equivalent resonance structures.

However one should be careful of the comparison being made here. We are not comparing the molecule to some other molecule. All that is said here is that the stabilization of the molecule due to resonance is maximum when the canonical structures are equivalent (for a 2-state system).

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    $\begingroup$ It should be $D = E_1' - E_1$. It's always final minus initial state. It will give you a negative value, which makes sense, since a stabilization will formally release energy. $\endgroup$ – Kexanone Feb 20 at 10:16
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    $\begingroup$ "The answer given here is not based on the Molecular orbital theory which should actually be used." There is no resonance in MOT, so you cannot answer this question in that way. What you are describing here reminds me a bit about quantitative Valence Bond Theory, but I have a feeling that the reasoning here is circular. If $E_1 \approx E_2$, then neither can. structure leading to $E_1$, nor $E_1$ is a good approximation to the el. structure, hence you need resonance. The truth still remains any 'resonance energy' is a purely hypothetical value which only seems to carry any physical meaning. $\endgroup$ – Martin - マーチン Feb 21 at 17:53
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    $\begingroup$ @Martin-マーチン Resonance energy need not be "real" (ie an observable) but rather explains why some structures are observed instead of simple eg Lewis/basic VB theory structures that do not invoke resonance. Isn't your statement akin to saying that perturbation theory gives "imaginary" results, or that linear combinations of orbitals are not "real" solutions? $\endgroup$ – Buck Thorn Feb 22 at 6:49
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    $\begingroup$ I'm not really sure why we're spending all of this effort on the situations where resonance structures are more or less stable. Resonance structures are just a model for how things work and not a great one at that. Insofar as it doesn't represent reality, any attempt to explain a fake thing is also fake. There's no reason to explain that equivalent resonance structures are "good" because that's the setup for the fake model that most agrees with reality. $\endgroup$ – Zhe Feb 23 at 1:46
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    $\begingroup$ I'm amazed how ppl are struggling with appreciating this approach, when it's not all to different from LCAO. Hell, I could even say MOs are fake, since they are just an artifact of the Slater determinant approximation. In that sense the only thing that really matters are the multielectron wave functions, which are the solution to the electronic Schrodinger equation. $\endgroup$ – Kexanone Feb 23 at 15:34
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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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