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This is a question related to the question I posted earlier titled "H NMR integration". I tried adding one drop of methanol-d (99.0% atom %D) using a regular pipet to the chloroform-d solution of polymer, hoping that the proton in water can exchange with methanol-d so that the water peak can vanish. Although only a very small amount of methanol-d was added, the NMR spectrum is dominated by the proton signal from CH3 group of undeuterated methanol at ~3.5 ppm. I'm wondering how to reduce the signal of this peak? Just reduce the amount of methanol-d? Thank you.

asked Jun 06 '11 at 12:51

Davis%20Chen's gravatar image

Davis Chen

Not directly related to your question, but it might help you measure liquids with a microliter syringe instead of the glass pipette. A little drop of a very small molecule compound can easily overwhelm signal from your sample. - Evgeny Fadeev (Jun 07 '11 at 13:53)

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Prepare a mixture of Methanol-D(10%)in Chloroform-D(90%), and add one drop of this mixture to the Polymer solution. Then the CH3 signal should reduce to 10% of what the signal was for one drop of pure Methanol-D.



answered Jun 11 '11 at 03:24

SankarampadiAravamudhan's gravatar image


I used isopropanol-d8 instead. But I have trouble having the sample spin well. The SPIN ON/OFF light on BSMS control panel (Bruker 400) always flashes. What causes this light flashes? The addition of isopropanol? How do I solve this problem? Thanks in advance for your answer. - Davis Chen (Jun 13 '11 at 20:39)

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I think you have to add methanol-d4 instead of methanol-d to reduce the signal at 3.5 ppm.


answered Jun 07 '11 at 04:31

Thnagel's gravatar image


Yes.. you're right, I wasn't aware it's CH3OD. I thought it's CD3OD and peak at 3.5 ppm is the residual undeuterated CH3. - Davis Chen (Jun 07 '11 at 14:21)

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Asked: Jun 06 '11 at 12:51

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Last updated: Jun 13 '11 at 20:39

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