I have a small molecule (23 atoms) for which I want to verify parameters in a molecular mechanics force field. small molecule

My first step is to look at the potential energy as a function of the dihedral angle, defined by these four atoms: define dihedral

The results for the dihedral scan, rotating the guanidino tail from 0 to 360 degrees in increments of 5, looks like this: scan1

Alright, something is funky around the 0 and 180 degree structures. Honing in on 180, and performing a QM scan with increments of every 0.1 degrees, I see this:scan2

There is a gap between 177.1-178.2 and between 181.5-182.6, where the QM optimization failed. Checking the plateau and either side of the plateau, what I find is that the plateau structures are all planar, whereas on either side of the plateau, some hydrogens come out of the plane (consistent on both sides of the plateau). For example, this is an overlay of two structures: blue = 178.2° = planar = higher energy. Red = 177.1° = nonplanar Hs = lower in energy. (front view and bottom side view) view1 view2

Let's assume that the QM calculations are to be trusted (this was run using a few different levels of theory). A mere 1° causes this shift between planarity and nonplanarity. Intuitively, if we think about the ring system being the same in both cases, this makes no sense, as a planar molecule should be better conjugated and thus lower in energy, correct? What other factors might account for this phenomenon?

  • 1
    $\begingroup$ I'm not sure I understand what all these calculations mean, but I can offer something that may be of assistance: by Hückel's rule, cyclodecapentaene should be aromatic, but due to the various types of ring strain, including the steric clash of the hydrogens of carbon-1 and carbon-6, the ring is bent into being nonplanar, and has no aromatic character. If these two carbons are bridged, howerver, aromaticity is achieved. Perhaps something like this occurring here. $\endgroup$ – ringo Dec 2 '16 at 2:37
  • $\begingroup$ There is no need to go that far into the details. The answer to your titular question is: because of steric problems. Think of biphenyl, or hexahelicene for that matter. $\endgroup$ – Ivan Neretin Dec 2 '16 at 5:41
  • $\begingroup$ > There is a gap between 177.1-178.2 and between 181.5-182.6, where the QM optimization failed. What do you mean by this? Fail as in the run was not completed successfully? From experience, Gaussian 09 gets a bit funky with flat molecules and 180 degree angles, which could cause the calculation to fail - have you tried the optimization with the "nosymm" keyword? $\endgroup$ – tetrahydrofuran Dec 23 '16 at 10:40
  • $\begingroup$ @tetrahydrofuran - yes, failed as in the run was not completed successfully. This was run using the Psi4 software package. $\endgroup$ – halcyon Dec 25 '16 at 7:14

I reproduced your calculations with PBE/def2-SVP (using Orca). The energy "jump" is even smaller than with MP2, however the causes are probably the same. So here is my answer:

  1. The planar structure is a Transition State, with one negative frequency.
  2. The dihedral you are scanning is probably not the best to describe this twisting.

Now in more detail. Here you can see the Minimum Energy Path (calculated with pDynamo using the NEB method) between the two symmetrically related minima. enter image description here You can see that, as expected, the path is smooth.

The main change between the structures is the pyramidalisation of the terminal NH$_2$. This can be described by the N-C-N-H dihedral, which also changes smoothly as you can see here: enter image description here

I don't think the reason why the terminal NH$_2$ is not planar is steric problems as @Ivan Neretin suggests. I think its more a question of nitrogen hybridisation, which is between sp$^2$ and sp$^3$. It is the same reason why aniline is not planar. This is discussed in these two questions:

Hope this helps!

  • 1
    $\begingroup$ Ramon, thank you for the in-depth analysis! Questions I had regarding your answer: (1) The planar structure is a Transition State, with one negative frequency. Does that mean you believe that you think the planar structure is actually higher in energy and not just an artifact of the calculation? (2) The dihedral you are scanning is probably not the best to describe this twisting. Can you explain further? (3) I don't think the reason why the terminal NH2 is not planar is steric problems Could it be that steric problems lead to mixed N hybridization? $\endgroup$ – halcyon Jan 11 '17 at 18:08
  • 2
    $\begingroup$ (1) Yes. I don't think it is an artefact, it is higher in energy as a planer ammonia is higher than the pyramidal structure (in fact, a planer NH$_3$ is also a TS) (2) I mean that this dihedral is not the one that best describes the reaction coordinate from the TS to the minima. And that is why you get a jump in the energy. The other dihedrals I suggest may be a better scan (though I haven't tried) (3) Could be, but I don't thing so. I think it is mainly electronic. It's a competition between sp$^3$ hybridization typical of N and sp$^2$ hybridization from some resonant forms. $\endgroup$ – Ramon Crehuet Jan 18 '17 at 15:36

Depending on the geometry of the hydrogen on the "imine" moiety of the guanadine, you could be creating a very strong syn-pentane interaction. In either case, either the lone pair or the hydrogen is being quite sterically crowded with with N-H's on the 5-membered ring.

The planar form of this species actually looks pretty bad. I would expect a bit of out plane distortion to relieve some of the strain.

  • $\begingroup$ The syn-pentane looks like it could be relieved by the hydrogen pointing to the right, though. $\endgroup$ – Jan Dec 3 '16 at 0:56
  • $\begingroup$ Well, you'll still have a lone pair there... Unlike a methyl group, there's no free rotation to point that out of plane. $\endgroup$ – Zhe Dec 3 '16 at 1:33
  • $\begingroup$ But the hydrogen can form a cyclic hydrogen bond between the two nitrogens. $\endgroup$ – Jan Dec 3 '16 at 1:44
  • $\begingroup$ Good point... :/ $\endgroup$ – Zhe Dec 3 '16 at 2:05

An artificial jump like this usually points to convergence issues. You could try reading the wavefunction from the previous step to converge to nearly the same SCF solution. If it is not that then it could have to do with how the geometries are obtained. You might want to read the previous wavefunction and tweak the geometry converger to take smaller steps to stay in the minima and not kick out to a higher energy local minimum.


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.