What nuclei need to be NMR active in an HMBC experiment? 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:




Do the nuclei in-between need to be NMR-active?
Does the end node X need to be NMR-active?
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:


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.
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.
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.
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.