Posts Tagged ‘Resonance’

Imaging normal vibrational modes of a single molecule of CoTPP: a mystery about the nature of the imaged species.

Thursday, April 25th, 2019

Previously, I explored (computationally) the normal vibrational modes of Co(II)-tetraphenylporphyrin (CoTPP) as a “flattened” species on copper or gold surfaces for comparison with those recently imaged[1]. The initial intent was to estimate the “flattening” energy. There are six electronic possibilities for this molecule on a metal surface. Respectively positively, or negatively charged and a neutral species, each in either a low or a high-spin electronic state. I reported five of these earlier, finding each had quite high barriers for “flattening” the molecule. For the final 6th possibility, the triplet anion, the SCF (self-consistent-field) had failed to converge, but for which I can now report converged results.†

charge

Spin

Multiplicity

 ΔG, Twisted Ph,
Hartree		 ΔG, “flattened”,
Hartree

ΔΔG,

kcal/mol
-1 Triplet -3294.68134 (C2) -3294.64735 (C2v) 21.3
-3294.60006 (Cs) 51.0
-3294.37012 (D2h) 195.3
Singlet -3294.67713 (S4) -3294.39418 (D4h) 175.6
-3294.39321 (D2h) 178.2
-3294.56652 (D2) 69.4
FAIR data at DOI: 10.14469/hpc/5486

I am exploring the so-called “flattened” mode, induced by the voltage applied at the tip of the STM (scanning-tunnelling microscope) probe and which causes the phenyl rings to rotate as per above. This rotation in turn causes the hydrogen atom-pair encircled above to approach each other very closely.‡ To avoid these repulsions, the molecule buckles into one of two modes. The first causes the phenyl rings to stack up/down/up/down. The second involves an all-up stacking, as shown below. Although these are in fact 4th-order saddle points as isolated molecules, the STM voltage can inject sufficient energy to convert these into apparently stable minima on the metal surface.

All syn mode, Triplet anion

The up/down/up/down “flattened” form (below) shows a much more modest planarisation energy than all the other charged/neutral states reported in the previous post, whereas the all-up isomer (which on the face of it looks a far easier proposition to come into close contact with a metal surface) is far higher in free energy.

The caption to Figure 3 in the original article[1] does not explicitly mention the nature of the metal surface on which the vibrations were recorded, but we do get “The intensity in the upper right corner of the 320-cm−1 map is from a neighbouring Cu–CO stretch” which suggests it is in fact a copper surface. Coupled with the other observation that in “contrast to gold, the Kondo resonance of cobalt disappears on Cu(100), suggesting that it acquires nearly a full electron from the metal (see Extended Data Fig. 2),” the model below of a triplet-state anion on the Cu surface seems the most appropriate.

Syn/anti mode, Triplet anion with C2v symmetry

There is one final remark made in the article worth repeating here: “This suggests that the vibronic functions are complex-valued in this state, as expected for Jahn–Teller active degenerate orbitals of the planar porphyrin.26” Orbital degeneracy can only occur if the molecule has e.g. D4h point group symmetry, whereas the triplet anion stationary-point shown in the figure above has only C2v symmetry for which no orbital degeneracies (E) are expected. Enforcing D4h symmetry on Co(II) tetraphenylporphyrin results in eight pairs of H…H contacts of 1.34Å,‡ which is an impossibly short distance (the shortest known is ~1.5Å). Moreover this geometry has an equally impossible free energy 176 kcal/mol above the relaxed free molecule. Visually from Figure 3, the H…H contact distance looks even shorter (below, circled in red)! A D2h form (with no E-type orbitals) can also be located.

Singlet, Calculated with D4h symmetry. Click for vibrations.

Singlet, Calculated with D2h symmetry. Click for vibrations.

Taken from Figure 3 (Ref 1).

These totally flat species are calculated to be at 13 or 12th-order saddle points, with the eight most negative force constants having vectors which correspond to up/down avoidance motions of the proximate hydrogen pairs encircled above and the remaining being buckling modes of the entire ring.

So to the mystery, being the nature of the “flattened” CoTPP on the copper metal surface, as represented in Figure 3 of the article.[1] Is it truly flat, as implied by the article? If so, the energy of such a species would be beyond the limits of what is normally considered feasible. Moreover, it would represent a species with truly mind-blowing short H…H contacts. Or could it be a saddle-shaped geometry, where the phenyl rings are not lying flat in contact with the metal but interacting via the phenyl para-hydrogens? That geometry has not only a much more reasonable energy above the unflattened free molecule, but also acceptable H…H contacts (~2.0.Å) However, would such a shape correspond to the visualised vibrational modes also shown in Figure 3? I have a feeling that there must be more to this story.

†These convergence problems were solved by improving the basis set via adding “diffuse” functions, as in (u)ωB97XD/6-311+G(d,p). ‡If the crystal structure for these species is flattened without geometry optimisation, the H-H distance is around 0.8Å

References

  1. J. Lee, K.T. Crampton, N. Tallarida, and V.A. Apkarian, "Visualizing vibrational normal modes of a single molecule with atomically confined light", Nature, vol. 568, pp. 78-82, 2019. https://doi.org/10.1038/s41586-019-1059-9

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Imaging vibrational normal modes of a single molecule.

Thursday, April 18th, 2019

The topic of this post originates from a recent article which is attracting much attention.[1] The technique uses confined light to both increase the spatial resolution by around three orders of magnitude and also to amplify the signal from individual molecules to the point it can be recorded. To me, Figure 3 in this article summarises it nicely (caption: visualization of vibrational normal modes). Here I intend to show selected modes as animated and rotatable 3D models with the help of their calculation using density functional theory (a mode of presentation that the confinement of Figure 3 to the pages of a conventional journal article does not enable).

I should start by quoting some pertinent aspects obtained from the article itself. The caption to Figure 3 includes assignments, which I presume were done with the help of Gaussian calculations. Thus in the Methods section, we find … The geometry of a free CoTPP molecule is optimized under tight convergence criteria using Gaussian 09 (ref. 33). The orientationally averaged Raman spectrum and vibrational normal modes are calculated with the geometry of a free molecule … All the calculations mentioned above are performed at the B3LYP/6-31G* level with the effective core potential at the cobalt centre. Armed with this information, I looked at the data included with the article (the data supporting the findings of this study are available within the paper. Experimental source data for Figs. 1–4 are provided with the paper) but did not spot any data specifically relating to those Gaussian 09 calculations; in particular any data that would allow me to animate some vibrational normal modes for display here. No matter, it is easy to re-calculate, although I had to obtain the basic 3D coordinates from the Cambridge crystal data base (e.g. entry IKUDOH, DOI: 10.5517/cc6hj4b) since they were unavailable from the article itself. At this point some decisions about molecular symmetry needed to be made (the symmetry is not mentioned in the article), since it is useful to attach the irreducible representations (IR) of each mode as a label♥ (lacking in Figure 3). The crystal structure I picked has idealised S4 symmetry, but it could be higher at D2d or lower at C2.

The next issue to be solved is how many electrons to associate with the molecule. Tetraphenylporphyrin has 347 electrons and the free molecule would be expected to be a doublet spin state (with the quartet as an excited state). Were the vibrational modes calculated for this state? Perhaps not since I then found this statement: The physisorbed CoTPP is positively charged on gold, as demonstrated through TERS measurements using CO-terminated tips24 and through the Smoluchowski effect29…. In contrast to gold, the Kondo resonance of cobalt disappears on Cu(100), suggesting that it acquires nearly a full electron from the metal (see Extended Data Fig. 2). So it seems worth calculating both the cation and the anion singlets as well as the neutral doublet. But at this stage we do not know for certain what spin state the Gaussian 09 assignments in Figure 3 were done for, since there is no data associated with the article to tell us, only that they were done for the free molecule (nominally a doublet).

There is one more remark made in the article we need to take into account: After lowering the sample bias to approach the molecule and scanning at close range, the molecule flattens. Its phenyl rings, which in the free molecule assume a dihedral angle of 72°, rotate to become coplanar (see Extended Data Fig. 1b). Evidently, the binding energy of the phenyl groups to copper overcomes the steric hindrance in the planar geometry. So it might be useful to calculate this “flattened” form to see how much steric repulsion energy needs to be overcome by that binding of the phenyl groups to the surface of the metal. 

Finally, I decided to not try to replicate exactly the reported calculations (B3LYP/6-31G(d)) since this type of DFT mode does not include any dispersion attraction terms; moreover by today’s standards the basis set is also rather small. So here you have an ωB97Xd/6-311G(d,p) calculation, with tight convergence criteria (integral accuracy 10-14 and SCF 10-9; again we do not know what values were used for the article). To ensure that my data is as FAIR as possible, here is its DOI: 10.14469/hpc/5461

charge Multiplicity ΔG, Twisted Ph
Hartree		 ΔG, Co-planar Ph
Hartree		 ΔΔG, kcal/mol
0 Doublet -3294.58693 -3294.48867 61.7
0 Quartet -3294.58777 -3294.51985 42.6
+1 Singlet -3294.35473 -3294.24973 65.9
+1 Triplet -3294.40821 -3294.33092 48.5
-1 Singlet -3294.67713 -3294.56652 69.4

Starting with a singlet cation as a model, the intent is to compare the “free molecule” energy with that of a flattened version where the dihedral angles of the phenyl rings relative to the porphyrin ring are constrained to ~0° rather than ~72°. This emerges as a 4th order saddle point (a stationary point with four negative roots for the force constant matrix). Such a property means that each co-planar phenyl group is independently a transition state for rotation. The calculated geometry overall is far from planar, having S4 symmetry. The image below in (a) shows how non-planar the molecule still is; (b) an attempt to orient it into the same position as is displayed in Figure 3 of the article.[1]

Singlet cation. Click on the image to get a rotatable model.

The free energy ΔG is 65.9 kcal/mol higher than the twisted form, which means that according to the model proposed, the binding energy of the phenyl groups to copper must recover at least this much energy. If we consider a cationic porphyrin interacting with an anionic metal surface as an ion-pair, then this is perhaps feasible. It is difficult however to see how more than two of the phenyl rings can simultaneously interact with a flat metal surface.

Next, the triplet state of the cation, again a 4th-order saddle point with a rotational barrier of ΔG‡48.5 kcal/mol; the triplet being 33.6 kcal/mol lower than the singlet using this functional (singlet-triplet separations can be quite sensitive to the DFT functional used).

Triplet cation. Click on the image to get a rotatable model.

Next, the neutral doublet, another 4th-order saddle point and below it the quartet state, which this time is just a 2nd-order saddle point (an interesting observation in itself).

Neutral Doublet

Neutral Quartet

Finally, the “flattened” singlet anion, which also emerges as a 4th-order saddle point (the triplet state has SCF convergence issues which I am still grappling with).

Singlet anion

To inspect the vibrational modes of any of these species, click on the appropriate image to open a JSmol display. Then right-click in the molecule window, navigate to the 3rd menu down from the top (Model – 48/226), where the frames/vibrations are ordered in sets of 25. Open the appropriate set and select the vibration you want from the list of wavenumbers shown. The preselected normal mode is the one identified in Figure 3 as 388 cm-1, the symmetric N-Co stretch (I note the figure 3 caption refers to them as vibrational frequencies; they are of course vibrational wavenumbers!). You can also inspect the four modes shown as negative numbers (correctly as imaginary numbers) to see how the phenyl groups rotate. If you want to analyze the vibrational modes using other tools (the free Avogadro program is a good one), then download the appropriate log or checkpoint file from the FAIR data archives at 10.14469/hpc/5461.

I conclude by noting that the aspect of this article which I presume reports the Gaussian normal vibrational mode calculations (Figure 3, caption Bottom, assigned vibrational normal modes), has been a challenging one to analyse.† Neither the charge state nor the spin state of these calculations is clearly indicated in the article (unless I missed it somewhere). The barriers to flattening out the molecule by twisting all four phenyl groups are unreported in the article, but emerge as substantial from the calculations here. The various species I calculated (summarised in the table and figures above) are all predicted to be non-planar. In the absence of provided coordinates with the article, the visual appearances (bottom row, Figure 3) are the only information available. These certainly appear flat and rather different from my projections shown above or below.

All of which amounts to a plea for more data and especially FAIR data to be submitted, providing information such as the charge and spin states used for the calculations, along with a full listing of all the normal mode vectors and wavenumbers. The article is only a letter at this stage; perhaps this information will appear in due course!

†As noted above I have not attempted a direct replication, not least because there is no reported data to which any replication could be compared. ♥The IRs of each vibrational mode are displayed along with the wavenumber when the 3D JSmol display is shown with a right-mouse-click.

References

  1. J. Lee, K.T. Crampton, N. Tallarida, and V.A. Apkarian, "Visualizing vibrational normal modes of a single molecule with atomically confined light", Nature, vol. 568, pp. 78-82, 2019. https://doi.org/10.1038/s41586-019-1059-9

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The Graham reaction: Deciding upon a reasonable mechanism and curly arrow representation.

Monday, February 18th, 2019

Students learning organic chemistry are often asked in examinations and tutorials to devise the mechanisms (as represented by curly arrows) for the core corpus of important reactions, with the purpose of learning skills that allow them to go on to improvise mechanisms for new reactions. A common question asked by students is how should such mechanisms be presented in an exam in order to gain full credit? Alternatively, is there a single correct mechanism for any given reaction? To which the lecturer or tutor will often respond that any reasonable mechanism will receive such credit. The implication is that a mechanism is “reasonable” if it “follows the rules”. The rules are rarely declared fully, but seem to be part of the absorbed but often mysterious skill acquired in learning the subject. These rules also include those governing how the curly arrows should be drawn.† Here I explore this topic using the Graham reaction.[1]‡

I start by noting the year in which the Graham procedure was published, 1965. Although the routine representation of mechanism using curly arrows had been established for about 5-10 years by then, the quality of such representations in many articles was patchy. Thus, this one (the publisher will need payment for me to reproduce the diagram here, so I leave you to get it yourself) needs some modern tidying up. In the scheme below, I have also made a small change, using water itself as a base to remove a NH proton, rather than hydroxide anion as used in the article (I will return to the anion later). The immediate reason is that water is a much simpler molecule to use at the start of our investigation than solvated sodium hydroxide. You might want to start with comparing the mechanism above with the literature version[1] to discover any differences. 

The next stage is to compute all of this using quantum mechanics, which will tell us about the energy of the system as it evolves and also identify the free energy of the transition states for the reaction. I am not going to go into any detail of how these energies are obtained, suffice to say that all the calculations can be found at the following DOI: 10.14469/hpc/5045 The results of this exercise are represented by the following alternative mechanism.♥

How was this new scheme obtained? The key step is locating a transition state in the energy surface, a point where the first derivatives of the energy with respect to all the 3N-6 coordinates defining the geometry (the derivative vector) are zero and where the second derivative matrix has just one negative eigenvalue (check up on your Maths for what these terms mean). Each located transition state (which is an energy maximum in just one of the 3N-6 coordinates) can be followed downhill in energy to two energy minima, one of which is declared the reactant of the reaction and the other the product, using a process known as an IRC (intrinsic reaction coordinate). The coordinates of these minima are then inspected so they can be mapped to the conventional representations shown above. New bonds in the formalism above are shown with dashed lines and have an arrow-head ending at their mid-point; breaking bonds (more generally, bonds reducing their bond order) have an arrow starting from their mid-point. The change in geometry along the IRC for TS1 can then be shown as an animation of the reaction coordinate, which you can see below.

Don’t worry too much about when bonds appear to connect or disconnect, the animation program simply uses a simple bond length rule to do this. The major difference with the original mechanism is that it is the chlorine on the nitrogen also bearing a proton that gets removed. Also, the N-N bond is formed as part of the same concerted process, rather than as a separate step.

Shown above is the computed energy along the reaction path. Here a “reality check” can be carried out. The activation free energy (the difference between the transition state and the reactant) emerges as a rather unsavoury ΔG‡=40.8 kcal/mol. Why is this unsavoury? Well, according to transition state theory, the rate of a (unimolecular) reaction is given by the expression: Ln(k/T) = 23.76 – ΔG‡/RT where T is temperature (~323K in this example), R = is the gas constant and k is the unimolecular rate constant. When you solve it for ΔG‡=40.8, it turns out to be a very slow reaction indeed. More typically, a reaction that occurs in a few minutes at this sort of temperature has ΔG‡= ~15 kcal/mol. So this turns out to be an “unreasonable” mechanism, but based on the quantum mechanically predicted rate and not on the nature of the “curly arrows”. And no, one cannot do this sort of thing in an examination (not even on a mobile phone; there is no app for it, yet!) I must also mention that the “curly arrows” used in the above representation are, like the bonds, based on simple rules of connecting a breaking with a forming bond with such an arrow. There IS a method of computing both their number and their coordinates “realistically”, but I will defer this to a future post. So be patient!

The next thing to note is that the energy plot shows this stage of the reaction as being endothermic. Time to locate TS2, which it turns out corresponds to the N to C migration of the chlorine to complete the Graham reaction. As it happens, TS2 is computed to be 10.6 kcal/mol lower than TS1 in free energy, so it is not “rate limiting”.

To provide insight into the properties of this reaction path, a plot of the calculated dipole moment along the reaction path is shown. At the transition state (IRC value = 0), the dipole moment is a maximum, which suggests it is trying to form an ion-pair, part of which is the diazacylopropenium cation shown in the first scheme above. The ion-pair is however not fully formed, probably because it is not solvated properly.

We can add the two reaction paths together to get the overall reaction energy, which is no longer endothermic but approximately thermoneutral. Things are still not quite “reasonable” because the actual reaction is exothermic.

Time then to move on to hydroxide anion as the catalytic base, in the form of sodium hydroxide. To do this, we need to include lots of water molecules (here six), primarily to solvate the Na+ (shown in purple below) but also any liberated Cl–. You can see the water molecules moving around a lot as the reaction proceeds, via again TS1 to end at a similar point as before.

The energy plot is now rather different. The activation energy is now lower than the 15 kcal/mol requirement for a fast reaction; in fact ΔG‡= 9.5 kcal/mol and overall it is already showing exothermicity. What a difference replacing a proton (from water) by a sodium cation makes!

Take a look also at this dipole moment plot as the reaction proceeds! TS1 is almost entirely non-ionic!

To complete the reaction, the chlorines have to rearrange. This time a rather different mode is adopted, as shown below, termed an Sn2′ reaction. The energy of TS2′ is again lower than TS1, by 9.2 kcal/mol. Again no explicit diazacylopropenium cation-anion pair (an aromatic 4n+2, n=0 Hückel system) is formed.



Combing both stages of the reaction as before. The discontinuity in the centre is due to further solvent reorganisation not picked up at the ends of the two individual IRCs which were joined to make this plot. Note also that the reaction is now appropriately exothermic overall.

So what have we learnt?

  1. That a “reasonable” mechanism as shown in a journal article, and perhaps reproduced in a text-book, lecture or tutorial notes or even an examination, can be subjected in a non-arbitrary manner to a reality check using modern quantum mechanical calculations.
  2. For the Graham reaction, this results in a somewhat different pathway for the reaction compared to the original suggestion.
    1. In particular, the removal of chlorine occurs from the same nitrogen as the initial deprotonation
    2. This process does not result in an intermediate nitrene being formed, rather the chlorine removal is concerted with N-N bond formation.
    3. The resulting 1-chloro-1H-diazirine does not directly ionize to form a diazacyclopropenium cation-chloride anion ion pair, but instead can undertake an Sn2′ reaction to form the final 3-chloro-3-methyl-3H-diazirine.
  3. A simple change in the conditions, such as replacing water as a catalytic agent with Na+OH–(5H2O) can have a large impact on the energetics and indeed pathways involved. In this case, the reaction is conducted in NaOCl or NaOBr solutions, for which the pH is ~13.5,♣ indicating [OH–] is ~0.3M.
  4. The curly arrows here are “reasonable” for the computed pathway, but are determined by some simple formalisms which I have adopted (such as terminating an arrow-head at the mid-point of a newly forming bond). As I hinted above, these curly arrows can also be subjected to quantum mechanical scrutiny and I hope to illustrate this process in a future post.

But do not think I am suggesting here that this is the “correct” mechanism, it is merely one mechanism for which the relative energies of the various postulated species involved have been calculated relatively accurately. It does not preclude that other, perhaps different, routes could be identified in the future where the energetics of the process are even lower. 

†This blog is inspired by the two students who recently asked such questions. ‡In fact, you also have to acquire this completely unrelated article[2] for reasons I leave you to discover yourself. ♥You might want to consider the merits or demerits of an alternative way of showing the curly arrows. Is this representation “more reasonable”? ♣I thank Ed Smith for measuring this value for NaOBr and for suggesting the Graham reaction in the first place as an interesting one to model.

References

  1. W.H. Graham, "The Halogenation of Amidines. I. Synthesis of 3-Halo- and Other Negatively Substituted Diazirines<sup>1</sup>", Journal of the American Chemical Society, vol. 87, pp. 4396-4397, 1965. https://doi.org/10.1021/ja00947a040
  2. E.W. Abel, B.C. Crosse, and D.B. Brady, "Trimeric Alkylthiotricarbonyls of Manganese and Rhenium", Journal of the American Chemical Society, vol. 87, pp. 4397-4398, 1965. https://doi.org/10.1021/ja00947a041

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What are the highest bond indices for main group and transition group elements?

Sunday, March 4th, 2018

A bond index (BI) approximately measures the totals of the bond orders at any given atom in a molecule. Here I ponder what the maximum values might be for elements with filled valence shells.

Following Lewis in 1916[1] who proposed that the full valence shell for main group elements should be 2 (for the first two elements) and 8 (the “octet“), Bohr (1922[2]), Langmuir (1919-1921[3]) and Bury (1921[4]) extended this rule to include 18 (the transition series) and 32 (the lanthanides and actinides). If we assume no contributions from higher Rydberg shells (thus 3s, 3p, 3d for carbon etc) and an electron pair model for orbital population (which amounts to the single-determinantal model), then the maximum bond index for hydrogen (and helium) would be 1, it would be 4 for main group elements, and then what?

For the special case of hydrogen, I have previously identified (for a hypothetical species) a bond index of 1.33, due mostly to a high Rydberg occupancy of 1.19e. The more normal BI is <1.0, as noted for this hexacoordinated hydride system. My current estimate for the maximum bond index for main group elements is <4.5. Thus for SF6, it has the value of ~4.33 and that includes a modest occupancy of Rydberg shells of 0.36e = 0.18 BI. Exclude these and it is close to 4.

Move on from group 16 to group 6 and you get compounds such as Me4CrCrMe44- or ReMe82- where the metal bond indices are ~6.5.‡ Compounds such as Cr(Me)6 (BI = 5.6)  and W(Me)6 (BI = 6.1) are rather lower. This is a long way from 18/2 = 9. The lanthanides and actinides[5] are unlikely to reveal many large BIs (32/2= 16 maximum value) since they are often ionic and the wavefunctions may be too complex to allow a simple index such as a BI to be safely computed.

So if we are hunting for record BIs, the transition elements are the place to hunt. Can a BI of 6.5 be beaten? Can it even approach 9, its maximum value? Does anyone know of candidate molecules? 

‡FAIR Data doi: 10.14469/hpc/3352.

References

  1. G.N. Lewis, "THE ATOM AND THE MOLECULE.", Journal of the American Chemical Society, vol. 38, pp. 762-785, 1916. https://doi.org/10.1021/ja02261a002
  2. N. Bohr, "Der Bau der Atome und die physikalischen und chemischen Eigenschaften der Elemente", Zeitschrift f�r Physik, vol. 9, pp. 1-67, 1922. https://doi.org/10.1007/bf01326955
  3. I. Langmuir, "Types of Valence", Science, vol. 54, pp. 59-67, 1921. https://doi.org/10.1126/science.54.1386.59
  4. C.R. Bury, "LANGMUIR'S THEORY OF THE ARRANGEMENT OF ELECTRONS IN ATOMS AND MOLECULES.", Journal of the American Chemical Society, vol. 43, pp. 1602-1609, 1921. https://doi.org/10.1021/ja01440a023
  5. P. Pyykkö, C. Clavaguéra, and J. Dognon, "The 32‐Electron Principle", Computational Methods in Lanthanide and Actinide Chemistry, pp. 401-424, 2015. https://doi.org/10.1002/9781118688304.ch15

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π-Resonance in amides: a crystallographic reality check.

Saturday, September 5th, 2015

The π-resonance in amides famously helped Pauling to his proposal of a helical structure for proteins. Here I explore some geometric properties of amides related to the C-N bond and the torsions about it.

Scheme

The key aspect of amides is that a lone pair of electrons on the nitrogen can conjugate with the C=O carbonyl only if the lone pair orbital is parallel to the C-O π-system. We can define this with the O=C-N-R torsion angle (and equate 0 or 180° with the p-orbitals being parallel). In the above definition, each R can be either 4-coordinate C (to avoid alternative conjugations) or H and the C-N bond is specified as being cyclic. As usual the R-factor is < 5%, no errors, no disorder.

First, the C-N torsion, which adopts values of either 0 or 180°. Notice that whilst the anti R-group shows no more than about 20° deviation from 180°, it does have a small tail tending towards longer C-N distances of >1.4Å. The hotspot is for the syn R-group.  Here there is a strong trend that as the dihedral deviates from 0° the C-N bond very clearly elongates. As the π-π overlap decreases, the bond elongates from the hot spot value of ~1.34Å to 1.41Å at 50°. The greater propensity of the syn-R to twist may be because it incurs more steric hindrance or perhaps because we have defined the C-N bond to be part of a cycle.

Scheme

Next, we plot the C-N distance against the torsion R-N-C-R’, which defines how planar the nitrogen is. A value of 180° is planar and the hot-spot is here. But as the planarity decreases down to almost tetrahedral (110°) the C-N bond elongates to  1.41Å. Notice one rather intriguing aspect;  from 180° to 160° or so, there is little response from the  C-N bond, but the elongation really accelerates from 140° to 110°. A little twisting hardly affects the π-π overlap, but it really starts to matter for twists of >50°.

Scheme

Finally a plot of the C-N vs the C-O distances. As the C-N increases, the C-O contracts, this being a nice summary of the π resonance in amides. 

Scheme

We have not seen any surprises, but this statistical exploration of crystal structures at least puts some numbers on the changes in bond lengths as a result of conjugative resonance.

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