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In kryptomerism assignment, there was a question to compare more stable tautomeric form between 4-pyridone (1) and 4-pyridinol (2):

1: 4-pyridone; 2: 4-pyridinol

Answer: 4-pyridone.

My friend gave two plausible reasons:

  1. $\ce{C=O}$ is more stable than $\ce{C=N}.$

  2. $\ce{-N=}$ will have greater repulsion between the lone-pair and the double bond than the same in $\ce{-O=}.$

Though, not being convinced by his explanation, I have two questions. First, is the answer correct? And second, what's the reason if it's correct?

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  • $\begingroup$ In my opinion, the aromaticity gained by 4-pyridinol would offset the energy considerations mentioned by @Zenix's friend, making it more stable. $\endgroup$ – Aniruddha Deb Dec 30 '19 at 13:00
  • $\begingroup$ @AniruddhaDeb both are aromatic $\endgroup$ – Zenix Dec 30 '19 at 13:14
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    $\begingroup$ @Zenix researched and answered accordingly. You're pretty sharp for someone in eighth grade :) $\endgroup$ – Aniruddha Deb Dec 30 '19 at 13:52
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I don't think this is an either this or that situation. As with most of the equilibria we know it depends very much on the circumstances we look at it. Many examples of keto-enol-tautomerism are heavily influenced by the solvent. Due to time and resource restrictions I won't be able to go in as much detail as I did in explaining whether enolates get protonated at the carbon or oxygen atom, but I ran a few calculations showing some of the controversy.

The calculations were carried out using Gaussian 16 Rev. B.01 at the DF-B97D3(BJ)/def2-TZVPP level of theory, and thermal corrections were obtained from normal coordinate analysis (in the harmonic approximation) at the optimised structures at the same level of theory using $T=\pu{273.15 K}$ and $p=\pu{1 atm}$ (because that is the G16 standard, sorry for the non-SI). The polarisable continuum model was employed except for the gas phase. All energies are in $\pu{kJ mol^-1}$ for reaction \eqref{tauto}, which means that negative values indicate that 4-pyridinol is more stable than 4-pyridone and vice versa for positive values:* $$\ce{4-pyridinol <=> 4-pyridone}\tag1\label{tauto}$$

\begin{array}{lrrr} \text{solvent} & \Delta E_\mathrm{el} & \Delta E_\mathrm{o} & \Delta G \\ \hline \text{gas} & -3.4 & -4.2 & -3.7 \\ \text{water} & 21.0 & 18.7 & 18.9 \\ \text{benzene} & 6.9 & 5.2 & 5.5 \\ \hline \end{array}

From this we can imply (although there are too few points) that in condensed phase 4-pyridone is the predominant species, while in vacuum it would be 4-pyridinol. Unfortunately, at this time I cannot offer a satisfactory (and easy) explanation for this. The values above have not been calibrated, they are a mere snapshot. I don't want to wander deep into hand-wavy territory and propose a rule of thumb based on bond energies and/or similarities to other molecules.

It is probably true that water is able to stabilise a possible negative charge at the oxygen and at the same time stabilising the positive charge of the proton bonded to nitrogen, but that's about as much of an educated guess I would like to give up.

I want to caution everybody with expressions like the aromatic form when referring to a resonance contributor. Aromaticity is a concept which applies to a (sub-) structure and as such to all resonance contributors. This is why both tautomers are aromatic regardless of the depiction chosen.
In a similar way I want to caution against the argumentation with bond energies. Whenever resonance comes into play, we will have fractional bond orders, which deviate significantly from the conditions under which the tabulated values have been measured (or calculated).

The above notes apply directly to the two reasons your friend offers:

My friend gave two plausible reasons:

  1. $\ce{C=O}$ is more stable than $\ce{C=N}.$

  2. $\ce{-N=}$ will have greater repulsion between the lone-pair and the double bond than the same in $\ce{-O=}.$

There is neither an isolated $\ce{C=O}$, nor a $\ce{C=N}$ bond, as they are both part of the same π-system. The lone pairs cannot (directly) interfere with the π-system as they are orthogonal.

I wish I could offer you a better explanation, but with most chemistry there is nothing easy about simple systems.


Notes:

  • There is a more extensive theoretical study available, which compares THF, water, and methanol and focuses more on the solvent model. However, obtained values are in a similar region:

    Nagy, P. I.; Alagona, G.; Ghio, C. Theoretical Investigation of Tautomeric Equilibria for Isonicotinic Acid, 4-Pyridone, and Acetylacetone in Vacuo and in Solution. J. Chem. Theory Comput. 2007, 3 (4), 1249–1266. DOI: 10.1021/ct6002252.

  • * For those unfamiliar with the notations: $\Delta E_\mathrm{el}$ refers to the electronic energy in the Born-Oppenheimer approximation; $\Delta E_\mathrm{o}$ refers to the observed energy, i.e. it is the former including the zero-point energy correction; $\Delta G$ refers to the Gibbs energy.

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    $\begingroup$ When I have more time, I might come back to this and add some more calculations. And if I find a nice way to look at the solutions and analyse them I will add that, too. $\endgroup$ – Martin - マーチン Jan 5 at 18:07
  • $\begingroup$ Thank you for correcting me and cautioning me not to jump into hasty conclusion. Do you have access to ACS papers mentioned by HahaHaha in his answer. $\endgroup$ – Zenix Jan 5 at 18:51
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    $\begingroup$ @Zenix I wasn't really correcting you, merely pointing out a different view. And after all you have had some doubts about your approach that needed clearing up. It's a good instinct to question things as they don't always are as easy as it seems, or less so: obvious. I currently do not have access to the publication mentioned by Haha; I have my doubts about how comparable this molecule is with the ones we're looking at. $\endgroup$ – Martin - マーチン Jan 5 at 19:34
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    $\begingroup$ & the references it cites: doi.org/10.1021/ja00417a027, doi.org/10.1016/S0040-4020(01)83237-3, doi.org/10.1021/jo01296a002, doi.org/10.1016/S0040-4039(01)86850-7. Feel free to use all, any, or none; I'm too lazy to read them properly. $\endgroup$ – orthocresol Jan 5 at 21:05
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    $\begingroup$ @Zenix The answers you are looking for probably does not exist; at least there won't be a simple explanation, which you can summarise in two keywords. I agree with orthocresol that any explanation of the solvation-stability relation won't be trivial; I share the fear that it might be impossible to explain the gas phase at all. I don't know whether there is enough experimental data to correlate the findings to (I believe I read somewhere that they decompose before being able to measure them in the gas phase). I hope you won't be too disappointed $\endgroup$ – Martin - マーチン Jan 6 at 11:27
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This Article states the following about 4-pyridone in its abstract:

Aromatic resonance energies for 4-pyridone and its analogues are similar to those for the 2-pyridone series.

This means that in order to find out how 4-pyridone is so stable, all we have to do is take a look at it's sister, 2-pyridone. 2-pyridone and it's tautomeric form, 2-pyridinol have the following resonance structures respectively, as shown in this question.

Resonance of 2-pyridinol Resonance of 2-pyridinone

The major reason of the stability of 2-pyridone is that Its aromatic form has the negative charge on the more electronegative element and the positive charge on the more electropositive element. More about the tautomerism of 2-pyridone can be found on Wikipedia, however for the purposes of this explanation I'll restrict myself to the above concept.

Applying this concept to the resonance of 4-pyridone and 4-pyridinol respectively, we obtain the following structures:

Resonance of 4-pyridinol

Resonance of 4-pyridinone

This explains why 4-pyridone is more stable than 4-pyridinol

References:

  • Cook, Michael J.; Katritzky, Alan R.; Linda, Paolo; Tack, Robert D.: Aromaticity and tautomerism. Part II. The 4-pyridone, 2-quinolone, and 1-isoquinolone series. J. Chem. Soc., Perkin Trans. 2, 1973, 1080-1086
  • Forlani L.; Cristoni G.; Boga C.; Todesco P. E.; Del Vecchio E.; Selva S.; Monari M. (2002). Reinvestigation of tautomerism of some substituted 2-hydroxypyridines. Arkivoc. XI: 198–215.
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  • $\begingroup$ I have a doubt, isn't in the question asked here,OP make fundamental mistakes in counting no. of π electrons. Also your answer is contrary to what answer says, and acc. to you 2-pyridinol is more stable, which is not the case $\endgroup$ – Zenix Dec 30 '19 at 14:11
  • $\begingroup$ Sorry I fixed a couple typos above. As for the answer, the $\ce{\pi}$ electrons seem in order. Nitrogen’s orbitals in 4-pyridinone are sp3 hybridised and the lone pair electrons in the sp2 orbital do not contribute to aromaticity $\endgroup$ – Aniruddha Deb Dec 30 '19 at 14:20
  • $\begingroup$ "On the 2-pyridone resonance forms (right) we have on the left side of the equilibrium a non aromatic (8pi e−) uncharged molecule and on the right side an aromatic (6pi e−) charged molecule", this is quoted from above link. How are there 8π e- and aren't both aromatic? $\endgroup$ – Zenix Dec 30 '19 at 15:21
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    $\begingroup$ There are no sp3 nitrogens in any of these structures. Mind you, it's not possible for hybridisation to change between two different resonance forms of the same thing. $\endgroup$ – orthocresol Dec 30 '19 at 20:19
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    $\begingroup$ The bottom image uses the wrong arrows, these symbolise tautomerism, but the correct form would be $\ce{<->}$ for resonance (or mesomerism [outdated]). $\endgroup$ – Martin - マーチン Dec 31 '19 at 1:44
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This answer does seem right. In the reaction I have drawn here, a similar logic prevails, you would be shocked to know that the equilibrium shifts to the right here in an ethanolic solution. The reason simply is the very high $\ce{C=O}$ bond energy. $\ce{C=O}$ has a bond energy of more than $\pu{100 kJ/mol}$ than the $\ce{C=N}$ and $\ce{C=C}$ bond energies (data from here). So, both these compound being aromatic, the debate of resonance energy dies, so it is a perfect explanation.

Here is a solid ACS paper[1] which says the same, the source for my claim even if you cannot access the paid version if you are not an ACS member, the first page is free, you can read clearly, that the mixture exists largely in the quinoid form.

On page 72 in the paper by Suzanne Slayden and Joel Liebmann titled The organic thermochemistry of hydroxylamines oximes hydroxamic acids and their derivatives, there is proof as well as justification of the same.

tautomerisation of quinone oximes and nitrosoarenols[*J. Am. Chem. Soc.* **1934,** *56* (3), 732–735](https://doi.org/10.1021/ja01318a062)

References:

  1. Anderson, L. C.; Yanke, R. L. The Tautomerism of Quinoneoxime-para-nitrosophenol Systems1. J. Am. Chem. Soc. 1934, 56 (3), 732–735. DOI: 10.1021/ja01318a062.
  2. Rappoport, Z.; Liebman, J. F. (eds.): The Chemistry of Hydroxylamines, Oximes and Hydroxamic Acids, Volume 1. John Wiley & Sons, 2008. (See a preview on books.google.co.in.)
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    $\begingroup$ Thanks for the help andselik, but what should I cite when the reaction was given by my teacher in-class notes? Is citation avoidable in such situations, because i do not have many books at my disposal to search from where the reaction was given. Thanks for the software I will try. $\endgroup$ – Haha Hahaha Dec 30 '19 at 16:10
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    $\begingroup$ Regarding citing the reaction by your teacher: it's a gray zone. Most likely, the teacher adapted it from somewhere else, so ideally that original source should be cited. If finding this out is deemed impossible, then post it as is (e.g. citing your teacher's private lecture notes), but be prepared for the critique for the lack of credibility of such source. $\endgroup$ – andselisk Dec 30 '19 at 16:20
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    $\begingroup$ found it @Zenix I guess ACS is trustworthy enough $\endgroup$ – Haha Hahaha Dec 31 '19 at 15:19
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    $\begingroup$ Please do not use edit statements, rewrite your post as necessary to form a continuous text. New readers will not care about the timely progression of writing; for those interested there is the history available. I have cleaned up your post a bit also using the mhchem package (look here and here. Also note that not all content on google books is licensed everywhere; it would be better to quote the relevant portions and leave a link as reference. $\endgroup$ – Martin - マーチン Jan 3 at 12:22
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    $\begingroup$ For the chemistry at hand: I do think that the molecules you cite a fundamentally different from what was asked in the question, and I would not be too sure to make this leap in argument. $\endgroup$ – Martin - マーチン Jan 3 at 12:25

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