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.
Steven Bachrach 03 Feb 2009 http://comporgchem.com/blog/?p=113

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?

  1. T. Kumagai, M. Kaizu, H. Okuyama, S. Hatta, T. Aruga, I. Hamada, and Y. Morikawa, Phys. Rev. B 2010, 81, 045402
  2. S.J Knak Jensen , I.G Csizmadia, Chem. Phys. Lett. 2000, 319 (3-4), 220-222.
  3. Daryl L. Howard, Henrik G. Kjaergaard, Jing Huang, and Markus Meuwly, J. Phys. Chem. A 2015, 119 (29), 7980–7990.
  4. Steven Bachrach, 03 Feb 2009, Malonaldehydes: searching for short hydrogen bonds Also: Jacqueline C. Hargis, Francesco A. Evangelista, Justin B. Ingels and Henry F. Schaefer III, J. Am. Chem. Soc. 2008, 130 (51), 17471–17478.
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    $\begingroup$ What about diborane? Does that even count as hydrogen bonding? $\endgroup$ Commented Mar 3, 2016 at 8:43
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    $\begingroup$ @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. $\endgroup$ Commented Mar 3, 2016 at 8:50
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    $\begingroup$ 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. $\endgroup$ Commented Mar 3, 2016 at 9:01
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    $\begingroup$ @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. $\endgroup$ Commented Mar 3, 2016 at 9:06
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    $\begingroup$ @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. $\endgroup$
    – DavePhD
    Commented Apr 11, 2016 at 18:14

4 Answers 4


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.


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.


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.

  • $\begingroup$ 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. $\endgroup$ Commented Oct 20, 2016 at 6:38
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    $\begingroup$ 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. $\endgroup$ Commented Oct 20, 2016 at 10:06
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    $\begingroup$ 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. $\endgroup$ Commented Oct 21, 2016 at 5:42
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    $\begingroup$ 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) $\endgroup$
    – Jan
    Commented Oct 22, 2016 at 0:55
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    $\begingroup$ @Jan So then are you saying that the protium tunnels back and forth between oxygens which macroscopically makes the hydrogen bond appear to be symmetric? $\endgroup$
    – jheindel
    Commented Oct 22, 2016 at 3:52

It appears that, perhaps, water may have symmetric hydrogen bonds between oxygen atoms, but it would take at least 60 GPa of pressure, to make water to bond like that. I guess water stops being molecular, because of that, though.

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

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 (...)


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

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


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

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    $\begingroup$ 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. $\endgroup$ Commented Apr 22, 2016 at 6:20
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    $\begingroup$ 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... $\endgroup$
    – Mithoron
    Commented Apr 23, 2016 at 1:38
  • $\begingroup$ @Mithoron I have been reading a bit about ice and I came across this paper (Benoit, M., Marx, D., & Parrinello, M. (1998). Nature, 392(6673), 258.) which uses DFT/PIMD trying to simulate the transition to ice X (the phase described in the paper you link). They note that this phase actually does not have a single well, but still has a double well with the proton symmetrically delocalized across both wells via zero point energy. They do find a transition to a single well at even higher pressure, but from what I can tell this has not been found experimentally. $\endgroup$
    – jheindel
    Commented Jul 2, 2019 at 19:04

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".

Symmetric hydrogen bond in nitromalonamide

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:

Isotopologues of enol form of 4-cyano-2,2,6,6-tetramethylheptane-3,5-dione

  • 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:

Isotopologues of enol form of nitromalonamide

  • 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):

NMR spectra of both enols

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.
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    $\begingroup$ It seems like we should clarify if by a symmetric hydrogen bond, we mean a hydrogen bond where the proton exists in a single symmetric well with the minimum halfway between two groups, or if we mean any symmetrically distributed proton. In reading this paper, it seems to me that they cannot distinguish whether this is a single or double well because if the barrier is as small as some of the theory suggests, the proton could be evenly shared by having a relatively large zero point energy, which I think would result in the same NMR results. I wonder if this makes sense and what you think? $\endgroup$
    – jheindel
    Commented May 22, 2019 at 0:10
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    $\begingroup$ I find this very interesting and enlightening, but I still have doubts. On the NMR (what did they actually measure btw) timescale it appears as symmetrical, but most calculations show a double well with a tiny barrier (small enough that you may pick a DFA and receive varying results). So while this is probably as close as we get so far, it is not really what I was looking for. $\endgroup$ Commented May 22, 2019 at 9:23
  • $\begingroup$ @jheindel My interpretation of the paper is as follows: if the potential was a double well, then for the 16O/18O mixed isotopomer the two minima (or the two lowest vibrational states) would have different energies, which would result in different shifts for the 16O-bound carbon and the 18O-bound carbon. This doesn’t matter if the interconversion on the NMR time scale is fast or slow (it is most definitely fast), because the Boltzmann-averaged shifts for those two carbons (which is the relevant quantity for fast equilibria) will still be different (as illustrated by the cyano compound). $\endgroup$ Commented May 22, 2019 at 9:59
  • $\begingroup$ On the other hand, if it’s a single well, then there cannot be any such asymmetry, as there is just one minimum. In principle, a perfectly symmetrical double well cannot be differentiated from a single well, which is (I think) your line of questioning (forgive me if I misinterpret). That is true for the 16O/16O and 18O/18O isotopomers, where you only see one peak regardless of double/single well, but the authors rely on the fact that the asymmetry in a double well for the 16O/18O isotopomer is detectable via chemical shifts. @Martin-マーチン (sorry, comment ping limitations; and it was 13C NMR). $\endgroup$ Commented May 22, 2019 at 10:03
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    $\begingroup$ I’m not totally sure about the case where the barrier is so shallow that the zero-point energy of the molecule is larger than the barrier. That’s somewhat pathological and I’m not sure if you can measure such a thing experimentally at all, since you can’t actually detect molecules without zero-point energy to probe the shape of the PES underneath that (maybe you can, but certainly not within what I know). Then I suppose the only recourse is to computational methods. $\endgroup$ Commented May 22, 2019 at 10:13

I believe the anhydrous Hydrogen Fluoride ring (HF)6 would be an example. I'm not certain, and can't find proof in literature, but it definitely looks the part! enter image description here

(image was drawn myself)

As we would expect it releases heat when forming from HF, and the reaction 6HF -> (HF)6 has an enthalpy of $\pu{-167kJ/mol}$ according to an experiment by Franck and Meyer as referenced in Fluorine Chemistry, Volume 5 . This is on the order of what we might expect from 6 Hydrogen bonds. The moiety is stable at STP while in an HF atmosphere with a very neutral Gibbs of formation from HF of $\pu{-0.8kJ/mol}$, according to thermodynamics data taken from JANAF for HF and (HF)6. This said, these tables may have some issues, as we're exploring in another question.

In the gas phase, Hydrogen Fluoride also forms dimers, trimers, tetramers, and chains, but I think the example of the ring is easiest to see as an actual neutral bond, and not something akin a Zwitterion.

I'm still hunting for better references on this, but I did find a bulletin quoting a 1921 paper by Langmuir, of all people, saying:

"In double molecules such as H4O2 (in ice), H2F2, and in compounds such as KHF2, etc., it seems that the hydrogen nuclei instead of forming duplets with electrons in the same atom, form duplets in which the two electrons are in different atoms. The hydrogen nucleus itself thus acts like a bond in such a case. Latimer and Rodebush have made a somewhat similar suggestion in regard to hydrogen nuclei acting as bonds. They consider that the hydrogen nucleus acts on two pairs of electrons: one pair in each of the two atoms... Since the first layer of electrons in all atoms contains only two electrons, it seems probable that the hydrogen in this case also holds only two electrons and that these form the definite stable group which we have termed the duplet."

This sounds very akin to what we're hunting for, but interpreting the spectroscopic data would be unfortunately beyond my skill level. The latest and greatest the 1970s had to offer on the topic are listed in this reference on Fluorine compounds by JANAF with (HF)6 on page 152 (1202 as listed), where they provide several references for the spectral lines. Thanks to anyone willing to give it a peak, or who may happen to know better on the topic.

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    $\begingroup$ This is an interesting approach to the question, and certainly has the potential to contain symmetric hydrogen bonds, as of the nature of the biflouride ion, but that brings us back to being ionic, or having large ionic contributions. I am concerned with the auto-protolysis equilibrium, $\ce{2HF <=> H+ + HF2-}$. I also ran a quick calculation on the system, it appears that a twelve-membered ring is a saddle point. While the idea remains interesting, I'm afraid it appears to not head in the right direction. $\endgroup$ Commented May 23, 2019 at 13:37
  • $\begingroup$ Thank you very much for giving it a look! If I'm understanding your point about the Bifluoride ion, it's that the system may be behaving partially as a salt of (HF2)- bound to H+? Electrostatically, I'd be surprised by the stable ring shape as compared to a more compact form. There's also a stable trimer (HF)3, which I think couldn't be rationalized as easily in this way, as there's not an even number of Fluorines to form Bifluorides. Could be a trick of resonance, of course. Happy to go with your instinct regardless, and thanks again for looking more deeply into the idea! $\endgroup$ Commented May 23, 2019 at 17:01
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    $\begingroup$ Names of elements and compounds aren't capitalised. As far as your comment goes, F-H-F in pure HF is countered by solvated cation H2F+. Oligomers of HF have "normal" h-bonds, if one would have symmetric one, it would dominate all others. $\endgroup$
    – Mithoron
    Commented May 25, 2019 at 22:55
  • $\begingroup$ I'd like to learn more about these gas phase ions, especially their thermodynamics data like Gibbs energies. Do you know of anywhere I can read more? $\endgroup$ Commented May 26, 2019 at 1:53

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