Eat Your Heart Out Mass Spec: Measuring 10B/11B Isotopic Ratio by NMR Spectroscopy

As I’m sure the readers of this blog know, NMR spectroscopy is used widely across all branches of chemistry due to its powerful structure elucidation capabilities and the inherently quantitative nature of the technique. Organic chemistry relies primarily on ¹H/¹³C experiments whereas inorganic chemistry can expand to other nuclei, like ³¹P and ¹¹B. However, there are many other applications for NMR other than just structural elucidation. Perhaps a lesser known application of NMR spectroscopy, is its ability to determine the isotopic ratio of elements! In this blog post I would like to demonstrate a method to determine the ¹⁰B/¹¹B isotopic ratio^[1] using our 60e instrument and ¹H NMR spectra! Why not use boron NMR? Well, the problem is that boron has two NMR active nuclei, as we discussed in this blog entry, and one cannot see ¹⁰B in ¹¹B spectra and vice versa. However, by using ¹H NMR spectroscopy we can see different splitting patterns from the proton atoms depending on which isotope they are bonded to.

For this experiment, the ¹H NMR spectrum of NaBH₄ in D₂O is recorded on the 60e instrument. The resulting spectrum is shown in Figure 1.

Figure 1. ¹H NMR spectrum of NaBH₄ in D₂O.

As you can see, the spectrum consists of a large 1:1:1:1 quartet with ¹J_B-H = 80.6 Hz and a 1:1:1:1:1:1:1 heptet with a ¹J_B-H = 27.0 Hz, both centred around −0.01 ppm. The splitting pattern observed in the spectrum is due to coupling of the proton signals in BH₄⁻ with the ¹¹B and ¹⁰B nuclei, which have nuclear spins of 3/2 and 3, respectively. This is a common phenomenon for these quadrupolar nuclei. Using the “2In + 1” rule, where I is the nuclear spin and n is the number of nuclei, it is easily determined that the ¹⁰B nucleus will split the proton signal into a heptet while the ¹¹B nucleus will give the quartet. Due to the inherently quantitative nature of the NMR experiment, comparing the sum of the areas of the heptet (¹⁰B) with the combined peak areas of the quartet (¹¹B) will give the isotopic distribution of the two boron isotopes. This results in an isotopic distribution of 18.35% for ¹⁰B and 81.65% for ¹¹B which compares very well with the known ratio of 19.90% : 80.10%.^[2]

While it is unlikely that this method could be applied to any other nuclei, this experiment provides an excellent introduction to isotopic ratios, coupling to quadrupolar nuclei, and a novel application of the NMR experiment beyond structural elucidation.

References

^[1] a) Walker, J. M.; Starks, R. J.; Gray, G. A.; Schoolery, J. N. Appl. Spectrosc. 1981, 35, 607;
b) Zanger, M.; Moyna, G. J. Chem. Ed. 2005, 82, 1390.
^[2] Coursey, J. S.; Schwab, D. J.; Dragoset, R. A. NIST Atomic Weights and Isotopic Compositions Home Page http://physics.nist.gov/Comp (accessed November 2021).