Data Point 1 - quantum chemical calculations

Data Point 1


Key points in paper:

  • There is some sort of interaction between the hydrogen and the terminal oxygen in the cis isomer; the author leave it up to the reader to decide whether one should consider that a hydrogen bonding interaction or a dipole-dipole interaction. Question - the interaction seems to fit the definition of H-bonding; why raise the possibility of it being a dipole-dipole interaction?
  • The authors discuss resonance contributors and use bond lengths along with other factors to determine the main contributors for the cis and trans isomers. Their analysis reflects ron's analysis perfectly.
  • This is perhaps the most significant point and it is also discussed on ron's excellent analysis:

enter image description here

  • Bent's rule is not explicitly invoked in the paper's analysis but it is implied.

Data Point 2 - Professor's claim - unsourced

enter image description here

Data Point 3 - Another Book's Claim

enter image description here

My questions

1) I have three sets of bond angles for the HONO molecule. The first and the last agree with each other in a general sense: cis-HONO exhibits a larger $\ce{O-N-O}$ bond angle than does the trans conformer. The middle picture - my professor's claim - contradicts the other two. Cis-HONO exhibits a smaller $\ce{O-N-O}$ bond angle than trans-HONO.

Who's right? From the replies I have gleaned that hydrogen bonding is likely to have a small to non-existent role in HONO given that hydrogen-bonding might actually increase (pseudo) ring strain.

I also think that my professor's argument regarding free-rotation along a single bond axis is a bit simplistic at best; in isolation, yes, single bonds can rotate, but so can double bonds (you just have to expend enough energy to break the pi-bond). Plus, there are barriers to free rotation of single bonds. You can't rotate single bonds willy nilly all the time. And according to the quantum mechanical calculation, the cis isomer exhibits a slightly higher energy than the trans isomer.

1a) Over what maximum distance can hydrogen bonding still exert a notable force? I'm curious. With this information I might try to do some trig using the QM data and see if the H and the O in the cis-conformer even come close enough for there to be hydrogen bonding.

2) Can cis and trans terminology be used to describe HONO? Usually these two terms are used with respect to identical atoms or functional groups. How about anti and syn? Or eclipsed and non-eclipsed?

3) Tautomerization. Why can't the proton be attached to the N? What's the barrier to tautomerization?

4) Ionic resonance structures - are these valid in the aqueous phase? Seems that if we have a lone proton hanging around in water it would be immediately solvated and taken away from the rest of the molecule ... this was the issue the prof raised. An obvious counter to his issue is that resonance structures don't exist discreetly; the ionic resonance structure is just a contributor - a partial representation of the true bonding nature in nitrous acid. It doesn't actually suggest that the proton is separated from the rest of the molecule. Dunno why he raised that issue ...

5) Authors of one paper also mention that nitric acid exhibits a similar widening of the cis isomer's ONO bond angle ... what might the two different bond angles for nitric acid be? Hell, does nitric acid even have cis/trans isomers? I don't think so ...

Regarding intramolecular hydrogen bonding in HONO

The below paper is based on the premise that hydrogen bonding exists within nitrous acid.

The influence of intramolecular hydrogen bonding on the photodissociation of cis-HONO is probed via measurements on the OH fragment ejected by specific [...] stretching (ν2) vibrational levels of the à state. Due to hydrogen bonding between the H atom and the terminal oxygen, the ν2 motion is coupled to the in-plane HON bend ν3. Since the latter evolves into fragment rotation, the rotational energy and anisotropy of the OH product increases with the number of ν2 quanta. By contrast, ν2 in trans HONO is a relatively isolated [...] vibration and thus does not influence the OH photofragment's properties.

  • 1
    $\begingroup$ Follow up? chemistry.stackexchange.com/q/10091/4945 I cannot see that there is an answer there, but it is the same paper linked. Maybe it would be good removing one of these questions. $\endgroup$ – Martin - マーチン Aug 25 '14 at 6:09
  • 1
    $\begingroup$ I was thinking about the question title and especially the use of "unexpectedly" in light of the following. From previous questions I know that you know that, if two things cannot be interconverted by a symmetry operation, then they must be different. So right off the bat we know that the two conformers must have different bond angles and lengths. What's left to discuss is "which direction" and "how much." For me, the question was, "why does cis HNO2 have a larger ONO angle than its trans counterpart?" $\endgroup$ – ron Aug 25 '14 at 13:36
  • 1
    $\begingroup$ The reference in my answer provides experimental data. Maybe you could ask a TA in the class to get a copy of the paper, so we can see what the author's say about this "cis" effect. Maybe your professor would like a copy too. $\endgroup$ – ron Aug 25 '14 at 14:28
  • 1
    $\begingroup$ @Dissenter "5) Authors of one paper also mention that nitric acid exhibits a similar widening of the cis isomer's ONO bond angle ... what might the two different bond angles for nitric acid be?" The ONO angle where the terminal O is cis or trans to the terminal H. "Cis ONO ~ 115.88 degrees, trans ONO ~ 113.85 degrees, see en.wikipedia.org/wiki/Nitric_acid $\endgroup$ – ron Aug 26 '14 at 21:40
  • 1
    $\begingroup$ @Dissenter, I'm definitely not certain, but it may be worth considering that negative hyperconjugation is occurring between the nitrogen lone pair and the $\sigma^{*}_{OH}$ antibonding orbital (analogous to what's purported to occur in molecules such as (Z)-1,2-difluoroethene, halogenated diazenes, etc.) in the cis rotamer. I think that would be consistent with the widening bond angles, shortening of the $\ce{N-O}$ bond, and lengthening of the $\ce{O-H}$ bond given in the QM calculations you cite. $\endgroup$ – Greg E. Aug 26 '14 at 22:44

These are very small effects that we are discussing, so who really knows what's behind them. The experimental measurement data was reported in, "Microwave spectrum, structure, dipole moment and quadrupole coupling constants of cis and trans nitrous acids, A. Peter Cox, Alan H. Brittain and David J. Finnigan. Trans. Faraday Soc., 1971,67, 2179-2194, DOI: 10.1039/TF9716702179." The abstract states, "The structural parameters are markedly different in the two isomers and have been discussed, together with the torsional barrier and relative stability of the isomers, in terms of a cis interaction." If anyone has access to the paper, perhaps they can let us know what the authors have to say about the nature of this "cis interaction."

Below are some of the resonance structures that can be drawn for the two isomers. First note that the cis-

enter image description here

isomer has the lone pairs on the same side of the molecule (lone pair - lone pair interactions), presumably responsible for the cis isomer being about 1/2 kcal/mol less stable.

If we examine the (hyperconjugative) resonance structures III and IV in the cis isomer, we might expect some stabilizing, through-space electrostatic attraction of the opposite charges between the positively-charged proton and the negatively-charged, terminal oxygen.

For arguments sake, let's assume that the trans isomer is adequately described by resonance structure I, while the cis isomer is best described by resonance structures III-IV. In this case we would expect the following structural differences between the cis and trans isomers:

  • A longer O-H bond in the cis isomer due to the larger O-H separation in cis resonance structures III and IV, than in trans I (this is analogous to longer C-H bonds present on the carbon next to a carbocation due to hyperconjugation)
  • A shorter internal N-O bond in the cis isomer due to resonance structure III which has a double bond in this position
  • A longer terminal N-O bond in the cis isomer due resonance structure III which has a single bond in this position
  • An increased H-O-N bond angle in the cis isomer because the structures cis III and IV involve roughly $\ce{sp^2}$ hybridized O-H oxygen, whereas in resonance structure trans I the O-H oxygen is roughly $\ce{sp^3}$ hybridized (actually shifted towards an even more constrained angle due to the effect of the lone pair); as we shift hybridization of the O-H oxygen from $\ce{sp^3}$ to $\ce{sp^2}$, we expect the H-O-N angle to widen.
  • An increased O-N-O bond angle in the cis isomer due to Bent's rule (a tendency to direct hybrids of increased s-character towards more electropositive substituents); a partial negative charge resides on both oxygen atoms in cis III and IV, whereas trans-1 has no negative charge on oxygen; therefor, the oxygens in cis III and IV are more electropositive (less electronegative) and the nitrogen will direct hybrid orbitals higher in s-character towards them as compared to trans I; higher s character in the bond means a larger O-N-O bond angle in cis III and IV compared to trans I

So perhaps very different contributions from the possible resonance structures to the real descriptions of cis- and trans-nitrous acid can account for the observed structural differences between these two isomers.

Side Notes: 1) hydrogen bonding is unlikely due to the attendant ring strain as noted by Ben Norris, 2) the oxygen-oxygen interaction should be electrostatically attractive.

  • $\begingroup$ Why would the O-O interaction be electrostatically attractive @ron? $\endgroup$ – Dissenter Aug 25 '14 at 4:01
  • 1
    $\begingroup$ Your comment #1) in resonance structure 2 the oxygens have opposite charges, this should lead to some oxygen-oxygen attraction in the real molecule; #2) yes, it's all about resonance structures and which ones best describe the real molecule; #3) my first thought, before I looked at the experimental bond angles, was that cis should have a smaller ONO angle due to N-O lone pair repulsions, your prof's angles are in accord with that rationale but don't seem to match the experimental data; #4) yes, the formal charges should add to zero, I didn't show all lone pairs. $\endgroup$ – ron Aug 25 '14 at 13:10
  • 1
    $\begingroup$ They have to, everything has to be different between cis and trans isomers, rotamers, or anything else where the two entities in question cannot be superimposed. If they can't be superimposed by a symmetry operation, then they are different. That means different in all aspects, including electron distribution. As I mentioned earlier, they must be different, all that's left for us to discuss is, by how much and in what direction. $\endgroup$ – ron Aug 25 '14 at 16:12
  • 1
    $\begingroup$ After studying molecules for a while you develop a feeling for which isomer might be more stable, the same applies here. There are general rules (see the sections entitled "Major and Minor Resonance Contributors" and "Resonance Contributor Preference Rule" at chem.ucla.edu/harding/notes/notes_14C_resonance.pdf $\endgroup$ – ron Aug 25 '14 at 20:43
  • 1
    $\begingroup$ I found a print copy in my university library; I will either go tonight or tomorrow and get to the bottom of this issue :D $\endgroup$ – Dissenter Aug 26 '14 at 1:29

What potential stabilization might be derived from an intramolecular hydrogen bond in cis-HONO, is counteracted by strain generated by the bad geometry. In order for the H and the distal O to get close enough to have a bonding interaction, cis-HONO starts to suffer from van der Waals repulsion between the two oxygen atoms. Note that cis-HONO is higher in energy than trans-HONO on your table.

If the hydrogen bond existed, cis-HONO would be a pseudo-four-member ring. For cycloalkanes, four-member rings have almost as much ring strain (110 kJ/mol) as three-member rings (120 kJ/mol). The origin of this ring strain is much more likely van der Waals repulsion than any sort of "angle strain" due to "improper" bond angles. Meanwhile, $\ce{O-H-O}$ hydrogen bonds average at 21 kJ/mol. The ring strain that would result from hydrogen bonding in cis-HONO would destabilize the molecule more than any stabilization from a hydrogen bond.

  • $\begingroup$ The four membered ring would not explain the widening of the ONO bond angle, rather a narrowing. $\endgroup$ – Martin - マーチン Aug 25 '14 at 6:27
  • 1
    $\begingroup$ The fact that the four-member ring destabilizes the molecule when the hydrogen bond is present explains why the hydrogen bond does not overcome the van der Waals repulsion (which is the reason for the bond angle widening). $\endgroup$ – Ben Norris Aug 25 '14 at 11:28
  • $\begingroup$ Where would there be van der Waals repulsion? The intramolecular hydrogen bond can never explain the widening. I also believe that there would be an atractive force acting between the oxygens, destabilising the nitrogen oxygen bonds. The concept of ring strain is a derived concept from hybridisation, which is also only a mathematical concept. The derived numbers are qualitatively arbitrary. The true reason for the widening must lie somewhere else. We are nonetheless talking half a kcal difference, which is nothing at room temperature, also 3 degree is nothing in terms of vibration. $\endgroup$ – Martin - マーチン Aug 25 '14 at 16:05
  • $\begingroup$ @Martin Heats of combustion and heats of hydrogenation also support the concept of ring strain. $\endgroup$ – ron Aug 25 '14 at 16:35
  • $\begingroup$ @ron They only support the relative instability towards hydrogenation or combustion products. There is no way to absolutely measure ring strain, as there is no way to prove hybridisation. Numbers assigned to ring strain are always derived from reactions and I would rather look a six membered rings to be extraordinarily stable, than four membered rings being especially strained. But that is just my opinion and that does rarely matter here. I believe the concept of ring strain will be an ongoing debate till the world ends... $\endgroup$ – Martin - マーチン Aug 25 '14 at 16:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.