Heteronuclear J-Couplings: Part 1

All my previous discussions of coupling have been focused on homonuclear coupling – you know between ‘like’ nuclei – proton-proton, of course, being the most common type.  The communication of nuclear spins, however, is not limited to nuclei of the same ‘type’.  Atoms that have a nuclear spin (i.e., I ≠ 0) are affected by the same phenomena that cause protons to communicate and ultimately split each other’s NMR signal (the mechanism of which I discussed here).

If you’re concerned that it’s about to get much more complicated you can relax – well, you can mostly relax!  Once you’ve worked out solving and assigning homonuclear splitting patterns – you have a pretty good handle on the situation!

The first step is to think about the types of coupling that are possible given a molecular structure (i.e., what nuclei are present, if they are NMR active, and what is the natural abundance of the NMR active isotope).  There are three cases:

1)   I =  ½ and natural abundance of the NMR active isotope is approximately 100% (e.g., 1H, 19F, 31P, 89Y, 103Rh).
2)   I = ½ but the natural abundance of NMR active isotope is not ~100% (e.g., 13C, 117/119Sn)
3)   I ≠ 0 or ½.

For the sake of this discussion I’m going to omit the last case and focus on I = ½ nuclei.

Case 1. You already have homonuclear splitting down?  EASY!

These spectra can be solved in the same manner as you would with homonuclear couplings.  The ‘n + 1’ and Pascal’s triangle rules still apply.

For example, in the case of fluoroacetonitrile (FH2CCN) in the 1H NMR spectrum you will see one signal because the two protons are equivalent (due to free rotation around the C-C bond) and they ‘see’ one adjacent 19F (I = ½) neighbour.  This means that the signal will be a 1 : 1 doublet.  Right?  Yeah of course! 1 + 1 = 2!


Instead of seeing vicinal couplings (i.e., 3 bond or 3JH-X ), like is most commonly observed in proton NMR with only homonuclear splititng patterns, we’re observinggeminal couplings (i.e., 2 bond or 2JH-X).

Similarly in the 19F NMR spectrum we will observe one signal (there’s only one fluorine atom) but it will be split by 2 equivalent protons.  2 + 1 = 3, so we should observe a triplet with the trusty 1 : 2 : 1 intensity pattern!


Coupling constants are determined by taking the peak separation and multiplying it by the number of Hz per ppm of the spectral window.  So:

1H NMR: 2JH-F =    (5.32 – 4.56 ppm) x (60.16 Hz/ppm) = 46 Hz
19F NMR: 2JH-F =    (231.95 – 232.76 ppm) x (56.41 Hz/ppm) = 46 Hz
OR   (232.76 – 233.58 ppm) x (56.41 Hz/ppm) = 46 Hz

Not so difficult right?  Admittedly it can get a little trickier if the molecule is more complex and you’re observing both homo- and heteronuclear coupling.  In this case, it can be helpful to differentiate between these by predicting the expected pattern and looking at the coupling constants.  Typically, we see different ranges depending on what the ‘type’ of coupling is and what atoms are involved.  The magnitude of the coupling constant depends on things like bond strength, s-character of the bond and bond angle.

From: Silverstein, R. M.; Webster, F. X.; Kiemle, D. J.; “Spectrometer Identification of Organic Compounds” 7th Ed. John Wiley & Sons Inc.: USA

Oh, okay, that seems easy enough…right?

What happens if the neighbouring nuclei have isotopes that aren’t 100% NMR active?  Organic molecules, of course, are chock-full of carbon which is 98.93% 12C (NMR silent!) and only 1.07% 13C!  Stay tuned….