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Hello, After reading the Q and A that was dated Oct2011, as an NMR student I ran into a question:

in the answer you read: "The 13C-19F isotope shift can be fairly large, so the satellites wouldn't be symmetric."

13C-19F isotope shift: they mean satellites due to 13C-19F coupling, right? So if it is so large, they are not symmetric in Hz scale but symmetric in ppm scale? I can't understand this!

I have seen asymmetric small peaks around 19F-spectrum of perfluorohexane. But I guessed they might be due to some other artifacts!

Thanks for the time, Amin

asked Mar 29 '12 at 07:23

Amin%20os's gravatar image

Amin os

updated Jun 14 '12 at 01:47

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"Isotope shift" refers to the shift of the resonance frequency of the fluorine nucleus, depending on whether it is bonded to 12C or 13C. The different carbon isotopes to which the fluorine can be bonded cause a difference in chemical shift.

As mentioned by Kirk Marat, these shifts can, in the case of measuring fluorine, be quite large. In routine cases they are often very small, so the difference between the frequency of proton signals bonded to 12C or 13C is negligible, so that the appearing doublet for the 13C case has its central frequency at the same point as the singlet for the 12C case. This results in symmetric satellites. For 13C-19F this difference becomes significant, meaning that the central frequency of the 13C-19F doublet does no longer coincide with the 12C-19F singlet frequency. This can then be seen as asymmetric satellite peaks in the spectrum.


answered Mar 29 '12 at 09:39

Pascal%20Fricke's gravatar image

Pascal Fricke

updated Mar 29 '12 at 09:40

Just to clarify, the spacing between the satellites is a scaler coupling and is thus field independent (constant value in Hz), while the centre of the satellites is the shift of the 13C isotopomer, and the difference between the 13C and 12C isotopomers is field dependent (constant in ppm). - Kirk Marat (Apr 03 '12 at 12:50)

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