# Diastereomers and H-NMR

What effect does stereochemistry have on the number of signals in an H-NMR? For example, if you consider the two structures below, would the number of signals and splitting differ due to the restricted rotation? If so, which hydrogens are being affected?

EDIT: This is what I think the splitting would look like.

• Enantiomers have identical spectra. What you show are two different diastereomers. Different boiling point, different chemical behaviour, different spectrum.
– Karl
Aug 27 '17 at 10:25
• How about you predict what the spectrum of each looks like, and I correct you if necessary? Not absolute chemical shifts, just the number of peaks and their splitting?
– Karl
Aug 27 '17 at 11:33
• The methylene protons are equivalent in the first structure but not the second. Aug 27 '17 at 12:10
• Note that cyclobutane rings are typically not planar but puckered. If I am not missing anything, the left-hand compound has $C_\mathrm{s}$ point group at most while the right-hand one has at least $C_\mathrm{2v}$.
– Jan
Aug 27 '17 at 14:55

The first thing you need to take into account is that cyclobutane rings are not planar, as they are drawn on paper, but puckered. Thus, a three-dimensional structure of the two isomers would look something like this:

I expect the structure on the left to be rather stable as it is, in the way I drew it, minimising steric interactions. Any other puckering would can either be traced back to that one by rotation or mirroring or put the two methyl groups into axial position which is sterically undesirable. Thus, we can expect only one significant species in solution.

This species should have $C_\mathrm{2v}$ symmetry if I didn’t miss anything. Using the principal axis — the one through the centre of the cyclobutane ring — we can identify four sets of homotopic hydrogens. We therefore expect four signals in the proton NMR spectrum.

Coupling constants, especially those along the ring, should be clearly visible and identifyable (if not confounded due to signal overlap).

The species on the right is less rigid. It can interconvert easily between different puckered states and in each one there will be one methyl group in axial position and one in equatorial position. Still, all structures can be traced back to the one I’ve drawn through mirroring and rotation unless I missed something. So here again, we have only one set of signals at all temperatures.

This species has a lower symmetry: $C_\mathrm s$. There is no rotation symmetry axis, so none of the ring protons are actually homotopic. Transforming one onto another is only possible by mirroring, hence they are enantiotopic. Since the methyl groups and the tertiary hydrogens are also non-identical, I expect a total of six signals at low temperature. However, repuckering would transform the axial into the equatorial methyl group and vice-versa. Therefore, I expect only a set of three signals to be visible at room temperature due to rapid interconversion.

These three signals would be:

• the methyl group
• the tertiary proton
• the set of secondary protons

Since the protons rapidly exchange positions, I would expect the coupling constants at room temperature to be averaged. At very low temperatures, after freezing the rearrangements, these should become distinct.

So yes, rigidness of a ring due to stereochemistry does affect the number and type of signals. cis-1,3-dimethylcyclobutane is very rigid and unlikely to transform from one arrangement to another. This leads to a clean, full set of signals. trans-1,3-dimethylcyclobutane, on the other hand, interconverts rapidly between different orientations, mapping the otherwise non-identical methyl groups and protons onto each other. It therefore shows less signals than it should at low temperature.

A similar effect can be observed in cyclohexane and derivatives, where cooling to low temperatures also freezes the ring flip and causes the signals of axial and equatorial protons to be distinct.