I recently had a discussion where I was told that in an $\ce{^1H}/\ce{^13C}$-HMBC experiment coupling through an $\ce{C-O-C}$ ether bond would not be observed with the reasoning that O is not NMR active. That somehow surprised me as I thought only the "end nodes" in the coupling needed to be NMR active. So I have the following two questions:

enter image description here

  1. Do the nuclei in-between need to be NMR-active?

  2. Does the end node X need to be NMR-active?


2 Answers 2


It is incorrect to say that oxygen block the HMBC coupling because it be magnetically inactive.

The theoretical explanation is that the coupling of spins $^nJ_{\ce{XY}}$ is mediated by the electrons i.e. transferred completely independently of the nuclei on the way. And I can provide you with multiple examples that show how coupling crosses over magnetically inactive nuclei:

  1. I myself in my bachelor’s thesis have distinguished the two different ortho-nitrobenzyl ethers of 1-(4-bromophenyl)-1,2-ethandiol by HMBC coupling. It was always possible to locate a coupling from one side’s hydrogen to the other side’s carbon in HMBC in spite of the magnetically inactive oxygen in-between.

  2. During my Ph.D. thesis, I performed a number of $\ce{^29Si}$-HMBC experiments to determine which hydroxy groups were protected by the silyl group used. Again, this transferred across a magnetically inactive oxygen.

  3. Remember that $\ce{^12C}$ itself is magnetically inactive. In your standard $^3J_{\ce{CH}}$ 3-bond-HMBC-experiment, the coupling and thus the cross-peak is transferred along $\ce{H-C-C-^13C}$. I didn’t spell it out explicitly but considering each carbon has only a $1~\%$ chance of being $\ce{^13C}$, it’s much more likely for the chain to be $\ce{H-^12C-^12C-^13C}$, i.e. involving only two magnetically active nuclei.

  4. Finally, your standard hydrogen spectrum exhibits a lot of $^3J_{\ce{HH}}$ coupling. Again, this crosses over two carbon nuclei; the probability of both carbons being $\ce{^13C}$ is $0.01~\%$. If only magnetically active nuclei were able to transfer coupling, then the uncoupled hydrogen signals should amount to $99.99~\%$ of a signal with the coupling signal having only $0.01~\%$ of the intensity. Clearly not what is being observed. In fact, $98~\%$ of the coupled hydrogen signal is due to molecules without magnetically active carbons between the vicinal hydrogens.

The final node has to be NMR-active, otherwise you wouldn’t be able to observe magnetisation.

  • $\begingroup$ I see, nice examples. I also observed HMBC coupling over inactive nuclei. So this basically means that J coupling does not need NMR-active nuclei. Right? $\endgroup$ Dec 4, 2016 at 12:52
  • 1
    $\begingroup$ @ketbra Beginning and end must be active, or else you wouldn’t observe anything. Nuclei in between do not need to be active. The explanation is that the magnetic spin information is transferred by electrons. $\endgroup$
    – Jan
    Dec 4, 2016 at 12:58

As Jan as demonstrated, observation of long range HMBC couplings across oxygen is commonplace. The key factor that dictates the appearance of coupling across multiple bonds the correct alignment of intervening bonds between the two coupled nuclei. The two end nuclei obviously need to be NMR active to be coupled to each other. The intervening nuclei do not.

Coupling arises from the favourable alignment of the σ and π orbitals of the bond pathway between the two nuclei. If this overlap of bonds is favourable, a couplng will be seen. For saturated systems, this long range coupling is usually confined to a well constrained W-coupling. It is also why long range coupling is most commonly observed in highly constrained cyclic systems.

And, although just 0.04% naturally abundant, oxygen-17 is actually NMR active.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.