Why are fluorine-19 NMR spectra less affected by paramagnetic centers than other common nuclei?

I know that 19F NMR spectra are less sensitive to paramagnetic centers as an empirical fact: it’s my experience and that’s what people in my lab have always been saying.


For instance, some of my samples are NMR silent in 1H, 13C, and 11B, but not 19F NMR. Why does 19F NMR appear to be an exception? Any helpful literature will also be appreciated.




$\ce{^{19}F}$ has similar NMR properties to $\ce{^1H}$ except in particular for the much larger chemical shift tensor terms and resulting larger chemical shift anisotropy (CSA). As a result CSA relaxation is a significant contributor to $T\_2$ relaxation at all field strengths/tumbling rates. For instance, considering a [simple model](https://doi.org/10.1007/s10858-019-00268-y)
$$\frac{1}{T\_{2,\text{CSA}}}=\frac{2}{15}(\sigma\_| - \sigma\_\perp)^2 \omega\_F ^2 \tau\_c \left( \frac23 + \frac{1}{2(1+\omega\_F^2\tau\_c^2)}\right),$$
the $2/3$ term in parentheses means that even under slow tumbling (large $\tau \_c $) the contribution of CSA to $T\_2$ relaxation can be significant in $\ce{^{19}F}$ , since the CSA term $(\sigma\_| - \sigma\_\perp)$ can be 50 times larger than for protons. CSA for protons is typically a minor source of relaxation at intermediate to low fields and in liquids (fast tumbling).


Since the CSA term already makes line broadening an issue for $\ce{^{19}F}$ at all fields/tumbling rates, it follows that adding another relaxation mechanism (due to paramagnetic interactions, for instance) results in a *relatively* small change in the total linewidth.


*References*


(1) Lu, M.; Ishima, R.; Polenova, T.; Gronenborn, A. M. $\ce{^{19}F}$ NMR Relaxation Studies of Fluorosubstituted Tryptophans. *J. Biomol. NMR* **2019,** *73* (8–9), 401–409. <https://doi.org/10.1007/s10858-019-00268-y>.


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