Hi Dal,

The peak u say at 3.33 is due to the amount water present in DMSO-d6 (not due to HOD peak, it will come even downfield). As dmso is highly miscible with water, during handling DMSO-d6 absorbs moisture and the peak at 3.33 is due to the moisture present. If the DMSO-d6 is being used for long time, usually the water peak comes bigger than the residual solvent peak in NMR.

regards

Deb

There are good papers listing the NMR peaks of common solvents in different NMR solvents.  OPRD:  "NMR Chemical Shifts of Trace Impurities: Industrially Preferred

Solvents Used in Process and Green Chemistry"  is one (and might be free) and J. Org. Chem.  vol. 62 no. 21 1997 is another.

Thanks for your time. So then for the peak appear at 3.33, is it right to label it as the solvent peak? What is the right way to label the peak?

Dr. Paech's response is correct.

Here is the link to the updated publication on NMR chemical shifts of trace impurities / residual solvents. Check the entry against DMSO on Page 2177 and the footnote.

http://pubs.acs.org/doi/abs/10.1021/om100106e

To avoid spectra dominated by the solvent signal, most 1H NMR spectra are recorded in a deuterated solvent. However, deuteration is never "100%", so signals for the residual protons are observed

[1] Ideally speaking  pure deuterated DMSO should  show no peaks in 1H NMR spectroscopy so it is commonly used as an NMR solvent. However, commercially available samples are not 100% pure and a residual DMSO-d5 1H NMR signal is observed at 2.50ppm (quintet, JHD=1.9Hz).[ From Wikipedia]

[2]Peak at 3.33ppm is  due to  D- in fact due to HDO while a peak at  39.5 ppm dueto 13^C

[3] For explaining  the reason for  D peak appearing at 3.33 ppm  while 1^H at 2.5ppm  lies in the basic principle of NMR as:

(a)Needless to say that NMR technique  involves nuclear spins(I) and have  mI  components which may  be parallel or anti parallel to  applied external magnetic field (B0).

(b)The transition of the nucleus from one spin state to the other state require energy

               E=2μN B0. --------------------------------------------------------------(1)                   

© This energy can be an electromagnetic radiation applied in such a way that the magnetic vector of radiation is oscillating in a plane perpendicular to that of applied field. (d)When the frequency of magnetic vector component of the electromagnetic radiation is equal to the precessing frequency of the magnetic moment of the nucleus, the two are said to be in resonance and the energy is transferred from the electromagnetic radiation to the nucleus under observation. When resonance occurs, the energy is transferred from one spin state to the other.

 [III]Again without involving rigorous mathematics, in practice, the absolute value of Bo cannot be determined very accurately. Therefore, it is convenient to compare the Bo required for resonance of a certain chemical system to that required of a standard. The difference is referred as Chemical Shift (δ ) which is given as:

                 δ = Bo- Bo (reference) / Bo (reference)------------------(2)

[IV] Since Deuterium's [D or 2^H] nuclear spin properties which differ from the light hydrogen usually present in organic molecules, NMR spectra of hydrogen/protium are highly differentiable from that of Deuterium. A few distinguishing properties are

(a)     I value of D =1 while that of 1^H is 1/2

(b)     D  is quadrupole while  1^H is not a quadrupole

        ( c )  Magnetic dipole moment of D is 0.857 µN   while for ^1H it is 2.7928456

( All these parameters are reported in Wikipedia) 

[V] With different   µN  values  for D and !^H,  the resonances for  D and 1^H will occur at  different Bo (eqn:1) and thus δ for D and 1^H will also differ (eqn:2)

This is HDO signal. Its chemical shift is changes at heating or cooling. If you left the vessel open, then DMSO it absorbs water from the atmosphere and its amount will increase. This water can be removed only with the use of molecular sieves. 

I think the auto calibration of did not worked properly in such cases the scale drifts. 

Yes, you may assign the peak to HDO or even just water.  Some people just report that the peak is due to water.  As DMSO is also very hygroscopic, great care must be taken to prevent that peak from appearing in an NMR taken with d6-DMSO.

Thank you for your time. I will look into the papers suggested. 

If its broad singlet, i hope it is water peak. DMSO-D6 must show quintate and with auto calibration it appears near to 2.5 ppm. If you use any internal standard for ex. TMS, then the peak position will shift to 3.04 ppm.

Hi Dal,

The peak u say at 3.33 is due to the amount water present in DMSO-d6 (not due to HOD peak, it will come even downfield). As dmso is highly miscible with water, during handling DMSO-d6 absorbs moisture and the peak at 3.33 is due to the moisture present. If the DMSO-d6 is being used for long time, usually the water peak comes bigger than the residual solvent peak in NMR.

regards

Deb

A peak at 3.33 is the water that exist in DMSO-d6. You can dry DMSO using molecular sieves activated. To regenerate 4A Molecular Sieve heat to a temperature of 250 to 450ºC.

Regards

Cristina

Hi, the peak at 3.33 is likely to be HOD/water. Your deuterated DMSO will absorb atmospheric moisture over time. The DMSO peak will be a sharper multiplet peak at 2.50.

To avoid excessive moisture, either use ampoules of DMSO-d6, or apply a septum to your bottle and flush with dry nitrogen gas before and after each use. Also, store nor tubes and caps in a dessicator. 

Best of luck.

Shailesh. 

Peak at at delta is 2.5, of DMDO-d6 undeuterated solvent ,and peak appeared at 3.33 is because of moisture in it.Please check.

I support what Madhukar B.D said peak at 2.5 is DMSO-d6 and at 3.33 is the DMSO Moisture. not any other thing

Madhukar said the exact answer