# NMR for an octahedral complexes sources of peaks

I was given the following information to assign to a Compound C and justify why the splitting patterns are as given:

I was confused about both the peaks.

I recall giving a previous question regarding NMR here.

Using the first bit of information

singlet in the proton-decoupled $$\ce{^{31}P}$$

I can't seem to understand why, because there is no plane of symmetry going across the plane of the octahedral 4 ligands (from left to right) so they wouldn't see "the same electronic environment."

Also, regarding the $$\ce{^1H}$$ I tried to justify this by saying that as there were 2 different $$\ce{P}$$ atoms in axial but they both faced different environments, the splitting would be triplet too, with the same splitting constant, hence a triplet, but I was unsure about this.

Why is the proton-decoupled $$\ce{^{31}P}$$ only displaying a singlet?

• The phosphorus atoms are enantiotopic. Thus they have the same chemical shift. – Zhe Jun 14 at 18:02
• @Zhe Hi there, I have just googled the definition of "enantiotopic" as being a proton in an enantiomer since I didn't come across this specific term previously. So, does this mean that as we are not in a chiral solvent, they will display the same shift? And one basic question, why is it enantiotopic? I just want to clarify this fully before I can move on! – vik1245 Jun 14 at 18:14
• One P looks up and sees four ligands in the plane, the other one looks down and sees the same arrangement, only mirrored. The two are chemically identical. – Karl Jun 14 at 18:24
• Feel free to write the answer yourself. But you do see the mirror plane through the metal atom and the four ligands CO,CO,MeCN and H, right? – Karl Jun 14 at 19:45
• As a follow up to my first comment, the logic is that enantiotopic groups are in identical but mirrored environments, so their shifts are identical because the NMR environment is achiral. Groups with identical shifts cannot split each other. – Zhe Jun 14 at 22:00