Thanks very much, I will try to get that right.

I agree with Michael that the 13C NMR spectrum of Weinert's sample (HO)2P(O)CH2CH2P(O)(OH)2 is not a first order spectrum, and cannot be analyzed by simple inspection. It is, as Michael pointed out, an AA'XX' spin system with A and A' the carbon and X and X' the phosphorous nuclei. AA'XX' spin systems can result in as many as 10 lines, but often some of these lines are missing. The resulting spectrum will depend on the various J values (and in same cases in their signs).  Assuming that Weinert's spectrum has high enough S/N and resolution to show all the lines, and  if JAA' is zero, should only 6 lines be left? Weinert's posted spectrum shows 5 lines, and, therefore, may be more complex. In my opinion, complete analysis would require computer simulation/iteration.

@Michael, what do you think?

Would the 13C natural abundance of ~1% simply just break symmetry and produce an AXX'?

The asymmetry caused be the 13C may produce a small chemical shift difference between the 31P nuclei (a few Hz) and therefore the 31P nuclei show a coupling to each other.

I obtained a "similar" spectrum using some VERY rough numbers, using the excellent program WINDNMR.

It is difficult to estimate the resolution from the posted spectrum, but I doubt that the w1/2 is any better than 3-5 Hz. If so, additional transition(s) might not have been detected.

I also would like to see a 13C NMR spectrum with better signal/noise, and higher resolution. If the S/N is good enough, omitting line broadening  might be enough to reveal any additional fine structure. However, if there were greater S/N, a negative LB could also be helpful. Zero filling is also an option to improve the (apparent) resolution. Better S/N and higher resolution might  show extra line(s) in the 13C spectrum and could help in the interpretation of the pattern, especially with the help of computer simulation.

What does the 31P spectrum look like? Weinert only said "very good". 

P.S. @Michael - would you please share with me the methyl ester paper you referred to in your question. 

Thanks for the great discussion. I Attached the 31P spectrum. It appeard to me all the time to be a singulett at 26.15 ppm. Now having a closer look on it, i can recognize some satellites.

I can try to get a spectrum at a higher field. That would take some time, though. Unfortunately I dont have fast access to an NMR with more than 300 MHz.

The mentioned paper dealing with methyl esters is by Li-Biao Han and is shown in the SI: http://dx.doi.org/10.1016/j.tetlet.2016.06.079

The 13C satellites in the calculated spectrum (X2 at ~99% and  AXX' at ~1%) are a reasonable match for the signals in the experimental 31P spectrum.

A 31P spectrum with a longer acquisition time and more scans would be useful to reduce the "ringing" artifacts and improve the S/N.

I concur this is very decent agreement between the observed and calculated results.

The compound has for exchangeable protons.  What would be rather interesting to discover is what happens to the C13 spectrum at the different pKa values of the compound.  Seems to me that when all protonated or all de-protonated,  C13 NMR structure could be different when either  one or three  proton sites match the pKa values.

I agree - no need to obtain a 13C spectrum at higher field strength. You could build up the S/N by accumulating  more scans (e.g., overnight ), then do zero filling, and use a small negative LB; you might be able to get a better resolved spectrum while not sacrificing too much S/N. Then simulate/iterate the spectrum. 

Best wishes in your research. AF

Have a look at DOI:10.1002/hc.20188

There is only one 13C in the molecule. Viewing from this, the two 31P are inequivalent (although they have the same chemical shift, neglecting a very, very small isotopic shift from 12C/13C) because one has a one bond JCP and the other one has a two bond JCP. Thus, this yields a higher order spectrum, which can be analysed via a computer program. It yields the very interesting parameter JPP.

Even if it is a bit late, an additional remark on this interesting topic: in carbon NMR you observe a molecule that has three NMR active nuclei (one 13C, two inequivalent 31P because of lacking symmetry in the observed species, as has been pointed out by ETK Haupt). So it is not an AA'XX' but an ABX spin system (X carbon, A and B phosphorus) which may show in its X part up to six lines. Five were observed, two may coincide, so there is nothing wrong. If you consult some old-fashioned text books on NMR, you will find these phenomena discussed, but, unfortunately, most modern texts leave them out. So I frequently get asked about such "oddities" in the carbon NMR spectra of "symmetric" phosphorus compounds, but there is nothing special at all...