I recorded VT NMR of my sample, highly hydroscopic, using D2O as a reference inside capillary. My sample contains only water impurity. In the spectra I have got two peaks extra, one is broad and other one is sharp and the broad one is not observed at high temperature (i.e., 80oC) but with changing temperature its chemical shift is changing while that of the sharp one undergoes no change. So which one is for D2O and which one for water and is the change of chemical shift for rupture of H-bond?
In the file blue one is high temperature and the red one is low temperature.
Hi, yes the water resonance does shift with temperature. I assume that you are locking the spectrometer on the D2O signal. The 2H resonance shifts with almost the same coefficient as H2O. For this reason the lock solvent signal will not move. But everything else will and exchangeable protons of your sample will change even more. The shift of water is 0.01 ppm/degree.
In addition exchange processes might go from slow exchange to fast exchange when going from ambient to 80C. It would me more helpful if one could see more of the spectrum and also spectra at intermeidate temperatures to more easily follow the shifts relative to temperature
Clemens
Thank you Clemens for your answer but I expect the peak being shifted is HOD peak of D2O as the triplet of D2O can be inside the broad peak. Further will peak shift of HOD of D2O and HOH of water (water is with sample and D2O inside the capilary) not be same? I have recorded spectra at 10oC interval. What do you think about satellite peak for the sharp peak at 4.7 ppm because I got few such peaks at intermediate temperature.
The satellite peaks at 4.7 are either vibrations or spinning side bands. Were you spinning the sample?
It is easily possible, that the HDO signal inside the capillary and the H2O outside have different shifts. Can you send me a stacked plot of the entire spectrum at several temperatures. You can send me a direct message.
The situation as you describe and as I understand is as follows:
a) you observe two peaks, one sharp, one broad
b) the shift of the sharp one does not depend on temperature nor does its linewidth
c) the broad one has a temperature-dependent shift and linewidth
My conclusion is:
a) the sharp signal is due to the water in the lock capillary: Of course, the chemical shift of water is dependent on temperature, but you will not be able to see this since you use this resonance for the deuterium lock system and this system will compensate (automatically) for any change - that's its purpose - and the signal will stay "fixed"
b) the broad signal is due to water traces in your solution, the interaction with other components is also temperature-dependent but in a different manner than in pure water, so it will move (relative to that in the capillary)
c) there are some solvents (e.g. acetone) that allow to discriminate the signals of H2O and HOD, and with some luck it is possible to detect the triplet structure of the HOD signal (reference: Fulmer et al., Organometallics vol 29 (2010) p 2176, Supp. Info). However, this difference is much smaller than the one you observe!
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