Why signals of carbon and hydrogen atoms (in NMR spectra) are split by coupling to $\ce{^{19}F}$? Some spectra books only states that

$\ce{^{19}F}$ (natural abundance 100%) has a spin quantum number I of ½. The signals of carbon and H atoms up to a distance of about four bonds are split by coupling to $\ce{^{19}F}$.

but there is no explanation. Could someone please explain why $\ce{^{19}F}$ can couple with H or C and split their signals.


1 Answer 1


You always have a coupling if you have to NMR-active nuclei connected, there is nothing unusual about fluor here. But in the most common and simplest NMR experiments on small organic molecules you usually don't encounter such heteronuclear couplings.

The first important aspect is that many common nuclei in organic molecules are present mostly in an NMR-inactive isotope. Carbon and nitrogen for example are only present in the NMR-active isotopes 13C and 15N in tiny amounts. So you'll be able to see small satellite peaks sometimes, but no complete splitting. If only 1% of the nuclei is NMR-active, only 1% of the signal will be affected by the coupling.

The second aspect is that some experiments like 13C 1D experiments are typically recorded with decoupling. There is a coupling between 1H and 13C, but it is intentionally suppressed as it would complicate the spectra.

  • $\begingroup$ Thank you for the answer. It has solved my problem not only of splitting signals by 19F but also of some small peaks shouldering the main peaks in 1H-NMR. Now I know they're "satellite peaks". Excited to know that. Thank you so much. $\endgroup$ Commented May 22, 2015 at 1:06

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