Friday, April 18, 2008

The 13C and 13C DEPT Spectrum of "Acetone-d6"

A student recently asked me why her solvent resonance for acetone-d6 was showing up in her 13C DEPT spectrum. Since the methyls of acetone-d6 have no protons they will not show up in a DEPT spectrum however, the solvent is typically bought at 99.9 atom % deuterium which means that there is a very small amount of acetone-d5 (CD3 - CO - CD2H). The - CD2H group will show up as a positive signal. Both 13C and 13C DEPT spectra are shown below for "acetone-d6". In the 13C spectrum in the bottom trace, one can see the expected 1:3:6:7:6:3:1 septet for a spin I = 1/2 nucleus coupled to three equivalent spin I = 1 nuclei. In the DEPT spectrum in the upper trace one sees only the small fraction of protonated carbons and the spectrum is a 1:2:3:2:1 quintet resulting from a spin I = 1/2 nucleus coupled to two equivalent spin I = 1 nuclei. One can also see the isotope effect of 0.254 ppm between the two isotopomers. Although the quintet is also present in the 13C spectrum, it is not seen simply because it is too small in comparison to the septet.

8 comments:

Anonymous said...

Pentet?

Quintet.

:)

Dave

Glenn Facey said...

Dave,

Hehehe, thanks ... right. Will change....

Glenn

Anonymous said...

Glenn, nice post. In the DEPT spectrum, it looks like you can still see a small amount of the d6 species barely above the baseline. I realize it's pretty insignificant, but what do you suspect this is due to? Slight mis-settings in pulse widths?

Glenn Facey said...

nmrfreak,

Yes, imperfect pulses or perhaps signals carried through from the 13C pulses alone.

Glenn

Edwin van der Eide said...

Hi Glenn,

I agree that an I = 1/2 nucleus coupled to three equivalent spin I = 1 nuclei will give a 1:3:6:7:6:3:1 septet. However, I think that an I = 1/2 nucleus coupled to two equivalent spin I = 1 nuclei will give a 1:2:3:2:1 quintet, not a 1:3:4:3:1 quintet.

Regards,
Edwin

Glenn Facey said...

Dear Edwin,

Thank you for your observation. You are of course correct. An I = 1/2 nucleus coupled to two equivalent spin I = 1 nuclei will give a 1:2:3:2:1 quintet, not a 1:3:4:3:1 quintet. I have corrected the post.

Glenn

Unknown said...

In the 13 C spectrum in the bottom trace, one can see the expected 1:3:6:7:6:3:1 septet for a spin I = 1/2 nucleus coupled to three equivalent spin I = 1 nucle... can u explain briefly pls?!!?

Glenn Facey said...
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