# The type of H-NMR spectra observed for S=PF2H and other inorganic molecules similar to this

I was hoping to learn more about the types of splitting observed in inorganic molecules such as S=PF2H. I am familiar with the typical splitting patterns associated with the study of organic molecules studied by C and H NMR. We know that H and F have a spin of 1/2, in reading about the effects of fluorine in on H NMR, it appears that fluorine relaxes in a similar way to protons when pulsed by a radio frequency. Since the two nuclei have the same spin but are different types of nuclei altogether and impact each other according to the Nuclear Overhauser Effect, what kind of signals would be given by an H NMR for the molecule? Would we see signals corresponding to fluorine given that it has the same spin as the proton?

• To be honest, I'd say you're overthinking it; for spin-1/2 nuclei relaxation behaviour (and NOEs) don't really matter, at least to a first approximation. Yes you would see H–F coupling in the 1H spectrum. No you wouldn't see 19F peaks unless you specifically looked for them, because 19F peaks appear at a different frequency altogether. It's like how you don't see 13C peaks in 1H spectra, but you could if you went hunting for them and acquired a 13C spectrum. Mar 19, 2021 at 16:30
• @orthocresol I think you are right, I probably am overthinking it quite a bit. When I think about coupling, I am typically associating it with the constant that might be found between the distance between, for example, the center-most peak in a triplet and the two smaller peaks to either side. So if I am understanding you correctly, yes coupling will occur, but no, we would not observe signals for the fluorine atoms, so you would probably see a singlet for something like S=PF2H. Mar 19, 2021 at 16:35
• No, the proton is a triplet, and all three peaks in that triplet belong to the proton. You will see all of them. No part of the triplet belongs to the fluorine. There's also 31P which is NMR active, but ignore that for now, I suppose.... Mar 19, 2021 at 16:44
• @orthocresol so suppose we ran an H-NMR experiment with S=PF2H, we would observe a triplet for the single proton due to the two neighboring fluorines, but we would not observe a different triplet due to one fluorine having a hydrogen and a fluorine neighbor? Mar 19, 2021 at 16:49
• "we would not observe a different triplet" Yes and no. You indeed wouldn't observe a different signal for the fluorine (unless you were to record a 19F spectrum). It won't be a triplet, though: the two fluorines are equivalent and so you should not see coupling between the two fluorines. It would just be a doublet: fluorine coupled to proton. Note that everything we've talked about so far will in practice be further split by 31P (spin-1/2 too). So the 1H peak is in fact a doublet of triplets. Mar 19, 2021 at 17:22

This spectrum is qualified as other inorganic molecules similar to $$\ce{S=PF2H}$$ molecule. What I want to show is alike different coupling constants in $$\mathrm{^1H \ NMR}$$ due to the molecular structure, bond distance, etc., the coupling constants in $$\mathrm{^{31}P }$$ and $$\mathrm{^{19}F \ NMR}$$ are also changed as well. For instance, $$^1J_\ce{HH}$$ is $$17$$-$$\pu{20 Hz}$$ in $$\mathrm{^1H \ NMR}$$ organic compound. However, $$^1J_\ce{PH}$$ is $$\pu{836 Hz}$$ for this particular molecule ($$\ce{PF2H(^{15}NH2)2}$$) while $$^2J_\ce{PH}$$ is $$\pu{11.8 Hz}$$ for same molecule (c.f., avarage $$^2J_\ce{HH}$$ is $$\approx \pu{7 Hz}$$ for a hydrocarbon with more than 2 carbons).
This particular spectrum consists of 90 peaks due to additional coupling of $$\ce{P-^{15}N}$$ coupling. If the compound is not labelled with $$\ce{^{15}N}$$, the number of peaks would reduced to 30 peaks, a relatively simplified spectrum, but still complicated coupling pattern of doublet of triplet of quintet.