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3-carboxycyclobutane-1-carboxylate

Select correct order of bond length of the above bond P, Q, R & S:

(A) P > R > S > Q
(B) Q > R = S > P
(C) S > Q > R > P
(D) None of these

I think the answer should be (D) because in this molecule, both carboxylic acid groups are equally acidic and the $\ce{H+}$ may migrate from one acid moiety to the other. This would be somewhat similar to the idea of resonance, although it isn't actually the same idea.

However, the given answer is (B) which we will get if this structure remains static. Because of the proton transfer, I don't think this can happen. Am I right or not?

Moreover, if we were to use some spectroscopic method to study the molecule, would we get an image in which the molecule has four equivalent bond lengths (due to proton transfer), or not?

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    $\begingroup$ Over time, yes, the bond lengths will all average out to be equal. But you said it yourself: this isn't resonance. At any one instant, this molecule has three different equilibrium bond lengths. $\endgroup$ – orthocresol May 21 at 4:16
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    $\begingroup$ That's a really good question to ponder about. The general answer is beyond the scope of this question, but it comes down to how "fast" your spectroscopic method is. To put it very simply, if it takes $x$ seconds (or microseconds, nanoseconds, whatever) to obtain the information that you are trying to get, then what you will observe is an average over those $x$ seconds. Bear in mind that everything has a finite time, including the interaction of photons with the system. There is no such thing as an instant method. So it comes down to how fast the system is moving, vs how fast your method is. $\endgroup$ – orthocresol May 21 at 8:14
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    $\begingroup$ Of course, for photons you need to phrase that in a more quantum-y way, using e.g. energy-time uncertainty principle. But that's the general idea, at least. If you use a method that is "slow"er than proton transfer between the acids (e.g. NMR), then what you may find is that all four oxygens (and hence C–O bonds) appear to be equivalent. If you use a method that is "fast" enough (maybe IR??), then you may well find that they are inequivalent. But the question is probably asking in the context of a (hypothetical) infinitely fast method. $\endgroup$ – orthocresol May 21 at 8:21
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    $\begingroup$ @orthocresol thank you for your such a detalied explanation. I appreciate your help $\endgroup$ – Harsh jain May 21 at 8:26
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    $\begingroup$ No problem. You've touched on a very important idea for all analytical methods, so kudos. I will try to gather my thoughts properly into an answer at some point in time. $\endgroup$ – orthocresol May 21 at 8:29
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It's most certainly true that proton transfer between the two acid groups can occur and is facile (particularly in an intermolecular manner). However, that only means that over a sufficiently long period of time, the four C–O equilibrium bond lengths will all average out to be the same. It does not mean that at any one instant in time, they are all the same.*

As you said, this is not the same thing as resonance. Resonance in a carboxylate group is a phenomenon whereby at all points in time, the two bonds are equal. There is no flip-flopping between two different structures.


Before we get to the answer for the actual question, I'll talk a little bit about your last question.

You bring up the idea of using spectroscopy to measure bond lengths, which is a very good point. The general answer is beyond the scope of this question, but it basically comes down to how "fast" your spectroscopic method is. To put it very simply, if it takes $x$ seconds (or microseconds, nanoseconds, whatever) to obtain the information that you are trying to get, then what you will observe is a time average over those $x$ seconds. Bear in mind that every method of spectroscopy takes a finite time, including the interaction of photons with the system. There is no such thing as an instant method; there are merely faster and slower ones. So it comes down to how fast the system is moving versus how fast your method is.

It's exactly analogous to taking a picture of a moving object, like a fast car.† If you try that with an phone camera with a long exposure time you will probably just get a blur. But a high-speed camera is good enough to take clear snapshots of Formula 1 cars during the race.

Of course, for photons you need to phrase that in a more quantum mechanical manner, using e.g. the energy–time uncertainty principle. But that's the general idea, at least. If you use a method that is "slower" than proton transfer between the acids (perhaps NMR), then what you may find is that all four oxygens (and hence C–O bonds) appear to be equivalent. If you use a method that is "fast" enough (maybe IR?), then you may well find that they are inequivalent.


Back to your question.

Generally, it is the instantaneous (static) picture which chemists deal with. That means that, at least for the purposes of this question, the dynamic motion of the system is not relevant, even though you are very correct to bring up the proton transfer. Equivalently, we could say that these questions are usually meant in the context of a (hypothetical) infinitely fast measurement method.

An infinitely fast method would be able to tell the difference between the $\ce{CO2H}$ and the $\ce{CO2-}$, because at any one point in time, one of them is protonated and one isn't. They can't be in a state where they're both half-protonated.

But an instantaneous method can't differentiate between the two oxygens in the carboxylate group. Those are entirely equivalent by resonance. And therefore you get three different bond lengths: Q > (R = S) > P.


* Of course, we are neglecting vibrational motion of the bonds, which technically mean that at any point in time it's far more likely that all the bonds are different. For example, even in the carboxylate group in which both oxygen atoms are equivalent (due to resonance), if an asymmetric stretch occurs then the instantaneous bond lengths will be different. Here, when we talk about "bond lengths", we are only referring to equilibrium bond lengths. We are also neglecting single bond rotations...

† I am not an expert on photography, so you may want to check out Photography Stack Exchange.

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