According to this website,

For two protons to be magnetically equivalent they not only have to have the same chemical shift, but they must also each have the same J coupling to other magnetic nuclei in the molecule.

(Emphasis in orginal text)

One example of this is 1,1-difluoroethylene, where the two protons couple to the fluorines differently (each one is cis- to one fluorine and trans- to another, so JHF is different than JH'F and JHF' is different than JH'F'). Why does this same logic not apply to ethene? I would think that each proton couples differently to the protons on the opposite side of the double bond, leading to magnetic inequivalence. This answer says that the protons are equivalent through symmetry, but if the criterion for magnetic equivalence is having the same coupling constant with each magnetic nucleus, I think symmetric molecules could still have magnetically inequivalent protons. Are there cases where symmetry leads to magnetic equivalence, despite differences in coupling constants?

Thank you.

  • $\begingroup$ The same logic does apply. The protons in 1,1-difluoroethylene are equivalent, only they couple to the different fluorines differently. Same thing here, only without fluorines. $\endgroup$ Nov 15, 2022 at 8:06

1 Answer 1


Magnetic equivalence is more rigorously defined in Levitt's Spin Dynamics (2nd ed., secction 17.5):

Magnetic equivalence is a strong form of chemical equivalence. A set of spins is magnetically equivalent IF

  • Condition 1: the spins have the same chemical shifts


  • (Condition 2a: the spins have identical couplings to all other spins in the molecule


  • Condition 2b: there are no other spins in the molecule.)

Condition 1 is the so-called 'chemical equivalence', and condition 2 is what differs between the cases of ethene and 1,1-difluoroethene.

In 1,1-difluoroethene, condition 2a is not fulfilled (cis and trans H–F couplings are different), and condition 2b is not fulfilled (when considering the equivalence of the two protons, there are obviously other spins, namely the fluorines).

In ethene, condition 2b is fulfilled: we can consider the four protons as a single group because they are all chemically equivalent, and there are no other spins present (ignoring the low-natural abundance 13C, etc.)

A more mathematical explanation of this is shown in Appendix A.9 of Levitt. Long story short, the H–H J-coupling interactions in ethene (of the form $\sum_{ij} 2\pi J_{ij} (\mathbf{I}_i \cdot \mathbf{I}_j)$) commute fully with the chemical shift interactions ($\sum_i \omega I_{iz}$) as well as the detection operator ($\sum_i I_{ix}$), so can be dropped from the Hamiltonian when calculating expectation values. nb: in the copy I have, eqn. (A.37) has a typo. It should read

$$\hat{B} = 2\pi J'(\hat{\mathbf{I}}_1\cdot\hat{\mathbf{I}}_2 + \hat{\mathbf{I}}_2\cdot\hat{\mathbf{I}}_3 + \hat{\mathbf{I}}_1\cdot\hat{\mathbf{I}}_3).$$

  • $\begingroup$ When we are determining whether protons are magnetically equivalent, do we only compare the couplings to chemically non-equivalent protons? Is this why we consider ethene’s four chemically equivalent protons to be one group? As another example, would the four protons in p-dinitrobenzene be magnetically equivalent because there are no chemically non-equivalent magnetic nuclei to compare couplings to, whereas in p-xylene, H2 and H6 would be magnetically equivalent and H3 and H5 would be magnetically equivalent, since we only compare couplings to the chemically non-equivalent methyl hydrogens? $\endgroup$
    – Akash
    Nov 15, 2022 at 18:36
  • $\begingroup$ @Akash First statement: Yes, that's right. Dinitrobenzene: Yes, that's right. The spectrum would consist of only one peak. Xylene: I'm not sure where you're going with this. The methyl hydrogens (all six of them) are chemically equivalent. In any case, it's probably not the best example, because the J-coupling is zero or so close to zero that it doesn't impact the spectrum. p-Difluorobenzene might be a better example; in that case I'd agree with your analysis. The spectrum has second-order effects. You can check all these out at sdbs.db.aist.go.jp $\endgroup$ Nov 15, 2022 at 22:44
  • $\begingroup$ I meant that the methyl hydrogens and the hydrogens on the aromatic ring are chemically inequivalent, so you would need to examine the coupling between H3 and the methyl hydrogens and H2 and the methyl hydrogens, and since these couplings are different, H2 and H3 would be magnetically inequivalent. Was p-xylene not a good example because J coupling across 4 bonds is so small? $\endgroup$
    – Akash
    Nov 16, 2022 at 2:45
  • $\begingroup$ @Akash Oh, I see. Yes, you would be correct in principle. In practice, the coupling is probably too small to really see any second-order effects. (Perhaps that might be different if you got a high-resolution spectrum and looked really closely, but I don't have the ability to get a sample myself and record a spectrum.) It's worth pointing out that the quantity of interest is not 4J itself, but rather the difference in the two couplings of interest (one could say the 'extent of magnetic inequivalence'). 5J is basically zero, and 4J is so close to zero, though, so 5J - 4J is pretty negligible. $\endgroup$ Nov 16, 2022 at 9:02

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