[check12]

Hello,
I have to find the structure of a compound, and I've only got the 1H, 13C and 31P NMR Spectra. The molecular weight is about 388.x.
At the 31P (1H decoupled) there is only one peak at 0.7 ppm (Ester? Phosphane?)
I have 13C peaks at:
2x 203 (quaternary carbon), 148(quaternary carbon or CH2), 2x 135 (quaternary carbon or CH2), 130 (CH oder CH3), 2x 121 (CH oder CH3), 2x 65 (CH oder CH3), 53 (most likely OCH3), 2x 50 (CH oder CH3), 2x 38 (quaternary carbon or CH2), 3x 21 and 20 (CH oder CH3)

the double peaks are coming very close (the difference is about 0.04 to 0.06ppm)

apart from the 1H, where I have got 4 symmetric aromatic H's at ~7ppm, 4x 3H's and 5x multipletts (5x 1H; even one sextett and pentett which are split in doublets)

Due to the phosphorous I think I get more signals in the 13c and 1h because of the spin spin coupling.

Can somebody at least help me with the 13C peaks?
I think the two peaks at 203 must be ketones.
If I assume that there is no benzene present, then I have only this aromatic peaks: 148, 135, 130, 121 and two of them couples with phosphorous (or some other Spin 1/2 particles like fluor?) to a dublett. So I have probably a aromatic heterocycle with 4 carbons?

what can be the peaks at 65 and 38 ? Is it a Isopropyl-group?

If somebody is interested in solving this challenge, I can provide the spectra as well as the 2d spectra.
thanks!

[frenchy]

Hi, do you know what atoms are present in your molecule or do you have an elemental analysis?