# Do symmetric hydrogen bonds in neutral molecules exist?

As far as I know, there have only few truly symmetric hydrogen bonds been observed. Unquestionable is the existence of it in the bifluoride ion, $$\ce{[F-H-F]-}$$, see also here. There are a couple of more, like in a water hydroxyl complex $$\ce{[HO\bond{~}H2O]- @Cu(110)}$$,[1] or in the hydrogen disuperoxide anion $$\ce{[O2-H-O2]-}$$.[2]

Long time it was believed that malonaldehyde and acetylacetone[3] form keto-enol tautomers with symmetric hydrogen bonds. The reason for this is the lack of resolution of the experiments.

I recently came across the article about substituted malonaldehydes.[4] According to this, if you chose R1 to be an amine and R2 a nitro group, you could end up with a single well on the potential energy surface.

However, there is one response, that left me a bit puzzled. Henry Rzepa wrote:

Searches of the Cambridge crystal structure database will reveal quite a number of instances of symmetrical hydrogen bonds. These are mostly homonuclear, ie N…H…N or O…H…O. Years ago, we came across a more unusual example, of an (almost) symmetrical N….H….O heteronuclear example (http://dx.doi.org/10.1039/C39890001722 ).

So apparently there exist many. Unfortunately he doesn't extend us the courtesy of even a single example. If you look on Google Scholar, you do indeed find a few examples of symmetric hydrogen bonds. However, most of them are either "crystallograpically" symmetric, for charged species, or both.

Also Evangelista et al. conclude their paper about malonaldehydes with the following statement, which leads me to believe, that there is no such thing as a symmetrical hydrogen bond in a neutral molecule:

A remaining goal is the identification of a substituted malonaldehyde with no barrier at all, i.e., a C2v equilibrium geometry.

Since then a few more years have passed, but my search did not turn up any more results.
Do symmetric hydrogen bonds in neutral molecules exist?

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• What about diborane? Does that even count as hydrogen bonding? – SendersReagent Mar 3 '16 at 8:43
• @DGS This is actually quite an interesting stand. I did not think about that. I really need to wrap my head around that and maybe make a few searches if anyone has ever called it a hydrogen bond. – Martin - マーチン Mar 3 '16 at 8:50
• The big difference there is that normal hydrogen bonds are 4-electron, 3-atoms. In the case of diborane, they are 2-electron, 3-atom bonds. Could almost call them hydride bonds instead of hydrogen bonds. – SendersReagent Mar 3 '16 at 9:01
• @DGS Yes that's correct. They are also the prime example for 3c2e-bonding, covalent through and through. For the sake of the question, it is probably better to exclude them. – Martin - マーチン Mar 3 '16 at 9:06
• @SendersReagent IUPAC defined hydrogen bond as "an attractive interaction between a hydrogen atom from a molecule or a molecular fragment X–H in which X is more electronegative than H..." currentscience.ac.in/Volumes/110/04/0495.pdf So because boron is less electronegative than hydrogen, diborane does not involve hydrogen bonds. – DavePhD Apr 11 '16 at 18:14

It appears that symmetric hydrogen bonds can exist between neutral molecules, even between water molecules! But it takes 60 GPa of pressure to make water to bond like that.

Article named very descriptively "Compression of Ice to 210 gigapascals: Infrared Evidence for a Symmetric Hydrogen-Bonded Phase" proves it:

Protonated and deuterated ices ($\ce{H2O}$ and $\ce{D2O}$) compressed to a maximum pressure of 210 gigapascals at 85 to 300 kelvin exhibit a phase transition at 60 gigapascals in $\ce{H2O}$ ice (70 gigapascals in $\ce{D2O}$ ice) on the basis of their infrared reflectance spectra determined with synchrotron radiation. The transition is characterized by soft-mode behavior of the $\nu_3$ $\ce{O-H}$ or $\ce{O-D}$ stretch below the transition, followed by a hardening (positive pressure shift) above it. This behavior is interpreted as the transformation of ice phase VII to a structure with symmetric hydrogen bonds.

On the other hand, article "The Hydrogen Bond in the Solid State" mentions:

First example of an $\ce{O-H-N}$ hydrogen bond with a centered position of the proton: pentachlorophenol/4-methylpyridine at 100 K characterized by neutron diffraction studies (...)

Unfortunately:

Even very small chemical changes in the system lead to loss of the symmetry (...)

so symmetric hydrogen bonds can exist also in very low temperatures.

References

Goncharov, A F; Struzhkin, V V; Somayazulu, M S; Hemley, R J; Mao, H K. Science 273.5272

Steiner, Thomas Angew. Chem. Int. Ed. 41 1 71

• This is certainly an interesting and unexpected take on the matter. I admit I was not hoping for something like this. Thanks for digging it up. – Martin - マーチン Apr 22 '16 at 6:20
• I think I read sth similar about carboxylic acid dimers (also pressure symmetrization). I think it's quite possible also in normal pressure in certain situations, maybe I'll find more... – Mithoron Apr 23 '16 at 1:38

This question crossed my mind during a lab today because we were making $\ce{Ni(dmg)2}$, which is shown below.

Well first, this is charged in some sense, but it is net-neutral. Also, the configuration shown above is only one way to draw it, and it seems as if everything should be symmetrical and thus that hydrogen should be equally shared between the two oxygens which I believe ought to be classified as a symmetric hydrogen bond on the basis of the fact there must be 4 electrons present here where that hydrogen could be equally shared. As opposed to something like diborane which was being discussed in the comments and is better classified as a 3 centre 2 electron bond.

Indeed, I dug around and found this paper which is quite old but also very interesting! They measure the infrared spectrum of this compound and a related copper compound and find a peak at about $2322\ \mathrm{cm^{-1}}$ which is an absurd shift in frequency from a normal $\ce{O-H}$ vibration. They verify this by forming the deuterated version and finding a large frequency shift down.

They say themselves, "the infrared data interpreted in terms of the abnormal isotope effect indicate that solid $\ce{Ni(dmg)2}$ and $\ce{Ni(emg)2}$ have symmetrical $\ce{OHO}$ and $\ce{ODO}$ bonds."

So then what is this "abnormal isotope effect"? Well, they also studied $\ce{Cu(dmg)2}$ and measured the $\ce{OH}$ and $\ce{OD}$ frequencies. They measured $2382\ \mathrm{cm^{-1}}$ and $2370\ \mathrm{cm^{-1}}$ respectively. Yes, a shift of $12\ \mathrm{cm^{-1}}$ upon deuteration. This effect had apparently been observed before, and the interpretation is that when the hydrogen is present, the hydrogen bond there is symmetrical, but when the oxygen is deuterated, the deuterium bonds more strongly to the oxygen (for some unknown reason to me) and thus is shared unequally which raises the frequency (despite the lowering due to a larger reduced mass).

So not sure if that's exactly what you're looking for, but I thought it close enough to your description to post here.

Obviously what remains to be seen is whether or not this actually corresponds to a single well on the potential energy surface. I would guess that it does simply because of the symmetry present in the system along with the evidence presented above.

EDIT:

Thanks to Martin for running a calculation to see whether or not there is really a single well, and despite some good experimental evidence, he found the hydrogen bond to be asymmetric.

This paper also indicates the same. So alas we haven't found an answer yet.

References:

Caton Jr, J. E., & Banks, C. V. (1967). Inorganic Chemistry, 6(9), 1670-1675.

Bruce-Smith, I. F., Zakharov, B. A., Stare, J., Boldyreva, E. V., & Pulham, C. R. (2014). The Journal of Physical Chemistry C, 118(42), 24705-24713.

• This is very interesting indeed. I will run a calculation on it to check for the single well. It could be a similar case to malonaldehyde/acetylacetone, where it appears to be symmetric at first glance only. For now, have my upvote. – Martin - マーチン Oct 20 '16 at 6:38
• I've done a really quick calculation (DF-BP86/def2SVP) and found that the complex in the equilibrium geometry has only $C_\mathrm{2h}$ symmetry. The bond lengths are $\mathbf{d}(\ce{O-H})=110.2~\mathrm{pm}$ and $\mathbf{d}(\ce{O\bond{~}H})=137.0~\mathrm{pm}$. After that I found a paper, that confirms my result: Ian F. Bruce-Smith et. al, J. Phys. Chem. C 2014, 118 (42), 24705–24713. In conclusion: unfortunately not a single well, not a symmetric hydrogen bond. – Martin - マーチン Oct 20 '16 at 10:06
• Ah bummer. I guess it's never completely safe to trust a paper from 1967 cause they obviously couldn't run the calculations we can. Nonetheless, it's a pretty interesting interaction! – jheindel Oct 20 '16 at 18:25
• Not trust is maybe too strong. I think the papers published at the time are trustworthy. You just have to factor in development of analysis tools, increased resolution of experiments, etc. Nevertheless, prior to your post I did not think about organometallics to achieve symmetric hydrogen bonds; while not the answer I was hoping for, it provides an interesting new direction. I would like to encourage you to include the reference I found in your last paragraph, just to make it clear. Not all people read the comments. In any case, thank your very much for your effort. – Martin - マーチン Oct 21 '16 at 5:42
• Couldn’t the discrepancy between $\ce{O-H\bond{...}O}$ and $\ce{O-D\bond{...}O}$ be explained by tunneling? Deuterium, being twice as heavy as protium has a reduced probability of tunneling, which could macroscopicly be interpreted as ‘sticking’ to one side more. (CC@Mart) – Jan Oct 22 '16 at 0:55

In Schaefer's paper,[1] which you linked, the authors describe nitromalonamide as being the compound with the lowest calculated barrier to intramolecular proton transfer, or tautomerisation.

Earlier this month (Feb 2019), Perrins and Wu used NMR isotope shifts to show that nitromalonamide possesses a symmetric hydrogen bond.[2] This is "the first case, to be compared with the FHF anion, of a neutral species with a single symmetric structure in solution and with a centred hydrogen".

The authors showed this by synthesising a mixture of oxygen-18 isotopologues. It's easier to explain this starting with an example of a hydrogen bond that is not symmetrical, such as 4-cyano-2,2,6,6-tetramethylheptane-3,5-dione, which the authors used. Consider the three possible isotopologues:

• For the 16O2 isotopologue, because the equilibrium constant $$K$$ is equal to $$1$$, both C–3 (indicated by a solid arrow) and C–5 (dashed arrow) have the same chemical shift (they interconvert between a "keto" carbon and an "enol" carbon, but because they spend equal amounts of time in both tautomeric forms, the averaged shift for both carbons is the same).
• The same can be said about the 18O2 isotopologue. However, although both carbons here (brown arrows) have the same chemical shift as each other, they do not have the same chemical shift as the carbons in the 16O2 isotopologue (blue arrows). This is because the adjacent 18O leads to a slight isotope shift.
• For the 16O18O isotopologue, one might naively think that the purple carbon (C–3) should have the same chemical shift as the brown carbons in the 18O2 isotopologue. This would be true if this equilibrium constant were equal to $$1$$. However, by virtue of introducing unsymmetrical isotope labelling, the equilibrium constant is no longer $$1$$; more specifically, it favours the C–18OH tautomer, which has a lower vibrational zero-point energy. Therefore, in this isotopologue, the purple and green carbons differ from each other in chemical shift, and also differ from the shifts in the previous two isotopologues, for a total of four peaks.

On the other hand, in nitromalonamide, which has a symmetrical double bond and therefore does not exist in an equilibrium of two tautomers, we have a different situation:

• For the 16O2 isotopologue, both blue carbons are equivalent by symmetry and hence exhibit the same shift.
• For the 18O2 isotopologue, both brown carbons are equivalent and hence have the same shift as each other, but differ from the shift in the 16O2 isotopologue because of the aforementioned isotope shift.
• In the 16O18O isotopologue, now, because there is no equilibrium to speak of, C–3 (brown arrow) has the same chemical shift as the two carbons in the 18O2 isotopologue.[3] So, we end up with only two peaks.

This is, of course, what is seen in the NMR spectra (of all three isotopologues together):

Notes and references

1. Hargis, J. C.; Evangelista, F. A.; Ingels, J. B.; Schaefer, H. F. Short Intramolecular Hydrogen Bonds: Derivatives of Malonaldehyde with Symmetrical Substituents. J. Am. Chem. Soc. 2008, 130 (51), 17471–17478 DOI: 10.1021/ja8060672.
2. Perrin, C. L.; Wu, Y. Symmetry of Hydrogen Bonds in Two Enols in Solution. J. Am. Chem. Soc. ASAP DOI: 10.1021/jacs.8b13785.
3. Theoretically, there should be a three-bond isotope shift arising from the remote oxygen being of a different isotope. However, I suspect that this effect is too tiny to be measured. Three-bond isotope shifts are known, especially with H/D, but a cursory search of the Internet didn't reveal anything about 16O/18O.