Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Organic Spectroscopy => Topic started by: Miffymycat on April 16, 2016, 08:21:26 AM
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In CW or FT pmr, as chemical shift increases, why do we say "downfield" implying a lower magnetic field strength, when deshielded protons need a higher flip energy requiring a higher magnetic field strength and / or higher radio frequency radiation? Am I having a senior moment - apologies if so ...
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The term doesn't make sense for FT nmr, so it doesn't make sense in the vast majority of modern nmr experiments, because the static B0 field is the same for all. In CW NMR (or EPR), the rf-frequency is constant and the B0 field is swept. In this case the "deshielded" nuclei need a slightly lower magnetic field to be on resonance.
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Thank you Irlanur. I am still unclear as to why a deshielded nucleus needs a weaker magnetic field in CW. Surely a deshielded proton experiences a greater effective field and a larger flip energy is therefore created and hence a stronger field needed for resonance?!!
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as I said. In CW, the rf frequency is held constant. let's say we irradiate with about 45 MHz an we'll sweep the B-field. Let's say the proton resonates when we set the B-field to exactly 1 Tesla. For a shielded nucleus, the B-field is slightly lower (because it's shielded), so it will not resonate at 1 Tesla, but at a slightly higher external field. -> the shielded nucleus resonates upfield.
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Thanks again. Your comment: "For a shielded nucleus, the B-field is slightly lower (because it's shielded) ..." still implies to me that this lower B-field should createa smaller energy gap for resonance!
PS by B0-field and B-field, do you mean applied and effective field strengths?
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still implies to me that this lower B-field should createa smaller energy gap for resonance!
yes!!! that's why you need a higher field to get it on resonance with a CONSTANT frequency!