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Reason for NMR results not being proportional with C-13

  1. Mar 20, 2013 #1
    I am trying to figure out why unlike H-1, NMR output on C-13 is not proportional to types of carbons. Which is to say in a molecule with two CH3 and two CH2 The peaks for the related hydrogens would be one unit tall and 2/3 unit tall. Meanwhile the mentioned carbons would not by nessity have the same peaks at all (even though they are even in number).

    Some poking around yielded this:
    From wikipedia. I think that is referring to the Nuclear Overhauser effect. Is that correct, if so does anyone have any articles explaining what that is, and why it has this effect on NMR output?

    Thanks,
    Michael
     
  2. jcsd
  3. Mar 21, 2013 #2
  4. Mar 21, 2013 #3
    Ok, so if I've got this right, "relaxation time" referrs to the time taken for an excited nuclear particle to return to it's normal spin. And charged nuclear (like hydrogen) can change the time for a spin to return to normal, generally speaking this change is an decrease of time taken to return to normal spin. And the longer the relaxation time, the more clear the data. As such if you have a methyl group data for the attached carbon will have a higher error (due to the protons attached to it, decreasing its relaxation time) than a second or third degree carbon (which has fewer protons near it to change its relaxation time). And it is just that increased degree of error that leaves the spikes in the data non-relational to the number of particular carbon configurations?
     
  5. Mar 22, 2013 #4

    chemisttree

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    NOE is the transfer of spin energy through space from one spin system to another. In a C-13 experiment, the protons have a strong through bond coupling that complicates the analysis tremendously so the protons are generally irradiated continuously to mask that particular coupling. This puts some spin energy into the proton systems. It is the transfer of this spin energy into the C-13 manifold (through space) that affects the intensity of the signals. It also removes the through bond coupling information, simplifying the interpretation of the spectra. As you might imagine, since the effectiveness of the spin transfer is through space, things like tertiary structure (tightly coiled vs open structure, for example), concentration of nearby spin systems (methyl vs methinyl), concentration, aggregation and so forth can affect the spin transfer so it is not as simple as counting nearby protons and correlating intensity but it does follow those trends. Just not enough to be able to integrate peaks.
     
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