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Topic: Deuterium  (Read 1650 times)

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Offline kapital

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Deuterium
« on: December 01, 2015, 06:06:10 PM »
Why are signal of deuterated solvents not seen on H-NMR, while they are(?) NMR active(based on the fact that they do couple with protons)?

Offline Corribus

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Re: Deuterium
« Reply #1 on: December 02, 2015, 09:37:15 AM »
Deuterium peaks are not observed in an 1H NMR spectrum because 2H resonates at a very different frequency than 1H. 1H does couple to 2H but the coupling constant is incredibly small (I looked around and it seems the JHH and JHD values are around a ratio of 1 to 6; another value I found puts it at about 0.01-0.02 ppm). So, I guess if your resolution is good enough you could see it. As I understand, if you deliberately deuterate specific positions in a complex molecule, the resonances of the deuterated positions in the proton NMR spectrum will disappear, and the resonances of protons adjacent to the deuterated position will slightly broaden. The broadening is due to JHD. (See, for example, Figure 7.2.7 here.)
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Babcock_Hall

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Re: Deuterium
« Reply #2 on: December 02, 2015, 09:45:10 AM »
One source I consulted some time ago suggested that the ratio of the two kinds of coupling constants was the same as the ratio of gyromagnetic values of the two nuclei, which is about 1/6.5.  When I have measured 2JHD values, they fall around 2-3 Hz (roughly speaking), which is in the expected range for geminal coupling.  The same holds true for one-bond carbon/deuterium coupling constants that I have looked at (not systematically).
EDT
We do see the deuterium in NMR, every time we look at the lock channel.
« Last Edit: December 02, 2015, 11:03:44 AM by Babcock_Hall »

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