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Offline blaisem

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NMR Spectrum explanation
« on: January 01, 2015, 09:27:05 AM »
While reading up on NMR, I encountered the following spectrum, titled Figure 4.



Two questions:

1. What are CH2A and CH2B?

My guess is due to the chirality of the neighboring amine-substituted carbon, for a racemic mixture two different sets of peaks would appear for the CH2 group?  Assuming this is correct, is this kind of proton-NMR splitting in racemic mixtures a standard consequence of chirality?

2. Why does the CH peak appear as a singlet?

I was expecting a 1:4:6:4:1 quintet from 3J-coupling to the neighboring CH2 and NH2 groups, mixed with a 1:3:6:7:6:3:1 septet from the deuterated methyl group, which I believe would result in a 5x7 = 35 peak splitting?

Thank you for any advice.
« Last Edit: January 01, 2015, 09:46:22 AM by blaisem »

Offline orgopete

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Re: NMR Spectrum explanation
« Reply #1 on: January 01, 2015, 11:33:27 AM »
The hydrogens adjacent to the chirality center are not chemically equivalent. A simple test is to replace them, for example with fluorine. The result would be an RR, RS, and SS, SR mixture.

In this case the chemical shifts are sufficiently different that you can see each hydrogen clearly. That may not be the case. See the 2-chloro butane shown here: http://faculty.tlu.edu/nmr/fac.html.

For the splitting, deuterium does not split and should simplify a spectrum. Because the CH2-hydrogens are different, they can, and in this case do have different J-values. The spectrum is run in TFA, so the NH2 splitting shows a time averaging effect resulting in the NH2 peak being broadened. It couples with the CH resulting in its being very broad as well.
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Offline Babcock_Hall

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Re: NMR Spectrum explanation
« Reply #2 on: January 01, 2015, 02:43:16 PM »
orgopete, I am not sure what you mean about deuterium not splitting.  It is occasionally difficult to observe splitting between deuterium and protium because the coupling constants are small, but in some cases it is quite apparent.

Offline blaisem

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Re: NMR Spectrum explanation
« Reply #3 on: January 01, 2015, 05:53:50 PM »
The hydrogens adjacent to the chirality center are not chemically equivalent. A simple test is to replace them, for example with fluorine. The result would be an RR, RS, and SS, SR mixture.

In this case the chemical shifts are sufficiently different that you can see each hydrogen clearly. That may not be the case. See the 2-chloro butane shown here: http://faculty.tlu.edu/nmr/fac.html.

Thanks for the reply.  The link was helpful.  I also found this page to be helpful for explaining diastereotopic hydrogens in NMR, too.

Quote
The spectrum is run in TFA, so the NH2 splitting shows a time averaging effect resulting in the NH2 peak being broadened.

Are you sure this is from the TFA? Shouldn't the peak be further downfield? An article states it's usually >10 ppm.

Quote
For the splitting, deuterium does not split and should simplify a spectrum. Because the CH2-hydrogens are different, they can, and in this case do have different J-values.

Why doesn't coupling with the the CH2-hydrogens split the CH peak?

orgopete, I am not sure what you mean about deuterium not splitting.  It is occasionally difficult to observe splitting between deuterium and protium because the coupling constants are small, but in some cases it is quite apparent.

I also thought Deuterated samples elicit JD-H coupling.  I think maybe the coupling is too small to resolved here.  One source says JD-H coupling is about 1/6th the magnitude of JH-H coupling, which on the scale of this spectrum would probably mean it isn't resolved , given how closely spaced the JH-H splitting appears in the other peaks.  That also might explain the broad singlet for CH.  At least, that's the best I can reason.  What do you think?
« Last Edit: January 01, 2015, 07:56:53 PM by blaisem »

Offline Babcock_Hall

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Re: NMR Spectrum explanation
« Reply #4 on: January 02, 2015, 11:28:26 AM »
My knowledge of NMR theory is imperfect, but I am under the impression that the coupling constants should scale with the magnetogyric ratios.  Therefore H-D coupling should be 6.5 times smaller than H-H coupling.  This is consistent with my own observations.

You might want to look a little bit into ABX pattern analysis.  Ignoring the deuterium coupling and any coupling that might be observed to the amino hydrogen atoms, this system should produce an ABX pattern.

Offline Irlanur

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Re: NMR Spectrum explanation
« Reply #5 on: January 02, 2015, 11:41:06 AM »
There are probably several reasons resulting in the broadening of the CH peak.

1) small, unresolved couplings
2) the fast exchange of the NH  protons
3) the nuclear quadrupole of Nitrogen, enhancing the relaxation of nearby protons and thus broadening the signal.

(4)as always, bad spectrometer handling)

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