I know that aromatic rings exhibit diamagnetic ring currents which causes the protons outside the ring to go downfield in H-NMR. Antiaromatic compounds exhibit paramagnetic ring currents which have the opposite effect. How does the difference in two electrons between aromatic and antiaromatic systems cause such a dramatic shift in the properties of the ring current? Why does diamagnetic current induce a downfield while the paramagnetic current induces an upfield shift?

I've found this question: Ring currents of anti-aromatic and aromatic systems but the answer doesn't address how the different currents give rise to different effects in H-NMR.

  • $\begingroup$ The second answer chemistry.stackexchange.com/a/57413/17952 to the linked question addresses exactly that. $\endgroup$ – Karl Aug 25 '16 at 19:35
  • $\begingroup$ What is unrealistic about it? This is what molecular orbital theory predicts and the resulting diamagnetic/paramagnetic ring currents match the prediction. If you study other magnetic systems, you always find the situation that diamagnetic substances have all electrons paired up (indeed, superconductors are perfectly diamagnetic, magn. susc. = -1, due to the cooper pairs) while all paramagnetic substances have unpaired electrons. $\endgroup$ – S. Burt Aug 29 '16 at 20:48
  • $\begingroup$ Because of Jan-Teller effect the electrons are paired in cyclobutadiene. $\endgroup$ – EJC Aug 29 '16 at 21:53
  • $\begingroup$ Good point! (I clearly answered before thinking clearly about the question). I would write off any discussion of anti-aromatic ring currents as fictitious/theoretical since you can't create a stable anti-aromatic system. It's expected to behave a certain way, but kind of a pointless discussion since it can't be observed. Although the ring-current model is a simple/approximate model, it does work quite well: many beautiful examples of very negative chemical shifts for protons dangling above a benzene ring or on the inside of annulenes, etc. vs. those around the perimeter of the ring. $\endgroup$ – S. Burt Oct 26 '16 at 18:35

Your Answer

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

Browse other questions tagged or ask your own question.