Friday, August 11, 2017
Boron Isotope Effects in Fluorine NMR Spectra
In previous posts on this BLOG, examples of 1H/2H and 12C/13C isotope effects were discussed. The figure below shows an example of a 10B/11B isotope effect observed in the 19F NMR spectrum of tetrabutyl ammonium tetrafluorobarate.
The spectrum clearly shows two resonances separated by 0.05 ppm with an intensity ratio of approximately 20:80 corresponding to the natural abundances of 10B and 11B, respectively. The low frequency resonance is due to 11BF4-. Since 11B is a spin I = 3/2 nuclide we observe a 1:1:1:1 quartet with J = 1.25 Hz corresponding to the one bond 19F - 11B coupling. The high frequency resonance is due to 10BF4-. Since 10B is a spin I = 3 nuclide we observe a very poorly resolved 1:1:1:1:1:1:1 septet with J ~ 0.4 Hz corresponding to the one bond 19F - 10B coupling.
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2 comments:
I am a bit confused, why does the natural abundance of boron have such a profound impact on the chemical shifts in the 19 F spectrum? Also, I have recently obtained the NMR of an organic trifluoroborate salt. I see the very similar 1:1:1:1 quartet corresponding to the 3 fluorine atoms and their coupling to the B11. However, the analogous peak corresponding to B10 is not present.
Anonymous,
The natural abundance of each boron isotope does not give rise to the isotope effect, it is the fact that each isotope is different slightly different in mass and the chemical bonds have very slightly different properties which give rise to differences in chemical shift. The natural difference in natural abundance between the isotopes will determine the integral ratio between the resonances. Other examples of isotope effects are here:
http://u-of-o-nmr-facility.blogspot.ca/2007/09/isotope-shifts-for-chloroform.html
http://u-of-o-nmr-facility.blogspot.ca/2008/10/dilute-d-2-o-in-benzene-d-6.html
http://u-of-o-nmr-facility.blogspot.ca/2012/10/isotope-effects-and-19-f-13-c-hmqc.html
http://u-of-o-nmr-facility.blogspot.ca/2008/06/spin-spin-coupling-between-equivalent.html
If you do not see the 11B/10B isotope effect in the 19F spectrum of your compound, it could be that the line width of the 19F resonance exceeds the isotope effect. Is there a small "shoulder" on the high frequency side of your 1:1:1:1 quartet? Is the signal-to-noise ratio sufficient to see the smaller signal?
Glenn
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