# How can Hydrogen NMR be used to distinguish between these two isomers?

I am having trouble explaining how and why the spectra of these two compounds would differ:

I am sure it has something to do with how the protons will couple. In the first isomer (left), the protons are anti-periplanar whereas in the second (right), I think they are staggered? So the first may be able to couple stronger than the second?

• I think this will be very difficult because both groups are quite small, so chair inversion may be facile. If the fraction of both conformers is significant, the peaks will just be a smear and you likely will find it difficult to measure any coupling constants... – Zhe Sep 25 '17 at 13:22
• Also these are anomers so be careful with pH... – Mithoron Sep 25 '17 at 14:39
• The difference is through space. – Alchimista Sep 25 '17 at 15:31
• what does this mean? – gamma1 Sep 25 '17 at 20:14
• See the nice answer – Alchimista Oct 26 '17 at 13:01

This issue aside, it is very easy to distinguish the two once you know you have a stable conformation. The key is the Karplus relationship which allows you to link a vicinal coupling constant to the dihedral angle between the two hydrogens. In general, a $^3J_\ce{HH}$ coupling constant will be very high ($^3J_\ce{HH} > \pu{9Hz}$) if the dihedral angle is either $0^\circ$ or $180^\circ$ — typically, if the hydrogens are anti-configured. On the other hand, the coupling constant becomes minimal for values around $90^\circ$ ($^3J_\ce{HH} < \pu{2Hz}$). For a $60^\circ$ gauche arrangement, a coupling constant of $^3J_\ce{HH} \approx \pu{3 Hz}$ is expected.
So the coupling between $\ce{H{1}}$ and $\ce{H{2}}$ would be around $\pu{9Hz}$ or $\pu{10Hz}$ in the first case but only around $\pu{3Hz}$ in the second — if the configuration is long-living on the NMR timescale.