Why does NMR have an inherently low sensitivity?

It is well known that NMR analysis requires a higher concentration of analyte than any other spectroscopic method. For example, UV-Vis requires an analyte concentration range of only nM to µM, while NMR typically requires the analyte to be in the mM range (>1000 times more concentrated!). In this blog, we will demonstrate why NMR is considerably less sensitive than UV-Vis. We have chosen UV-Vis for this comparison as it is widely recognized as one of the most sensitive spectroscopic techniques.

The amount of signal observed in both spectroscopic methods is related to the difference in population between the two energy states. The energy difference (ΔE) is equal to Planck’s constant (h) multiplied by the frequency (ν), Equation 1. In NMR, a spin ½ in the presence of a magnetic field will generate two spin states (α and β), where the difference in energy between the two spin states will vary based on the magnetic field strength (B₀) and the nucleus gyromagnetic ratio (γ), Equation 2. On the other hand, in UV-Vis, the energy variation between the ground state and the excited state will depend on the wavelength of absorption (λ), Equation 3.

We can calculate the ΔE by solving Equations 2 and 3 using the values displayed in Table 1. The calculated energy difference between the α and β states for ¹H at 400 MHz is 2.65*10^-25 J, whereas the ΔE between the ground state and the excited state for the electronic transition, considering a wavelength of 500 nm, is 3.98*10^-19 J. To put this in perspective, if the energy difference for the UV-Vis system was equivalent to the length of a soccer field (~125 m), the corresponding energy difference for the NMR system would be the thickness of a human hair (~80 µm)! (Apologies to our math and physics geek readers, this was the best way I found to translate the energy differences). Keep in mind that the thermal energy of our system at 25 ºC (k_B*T) is equal to 4.11*10^-21 J, Equation 4. So, while in NMR the ΔE is around 15,000 times lower than the thermal energy, in UV-vis the ΔE is 100 times bigger than the thermal energy.

The next step is to calculate the population difference between the two states, but even before we do this calculation, it is clear that in NMR, the spins will be distributed between the two levels more evenly, while in UV-Vis, essentially all molecules will be in the ground state. Let’s do the calculation to get a concrete number…

The Boltzmann distribution defines the difference in population between the two energy levels calculated previously, Equation 4.

For UV-Vis, the population ratio between the ground and excited state is 1*10⁴², which means that essentially 100% of molecules will be in the ground state. However, for NMR, the population ratio between the levels is 1.000064, which means that the lower level (α) has only a tiny population excess (50.0016%). To visualize this, let’s assume we have one million ‘species’ in NMR (spins); there would be 500,016 species in the α state and 499,984 in the β state. However, in UV-Vis, we would have essentially 1,000,000 species in the ground state and 0 in the excited state. The population excess for NMR is 32, while in UV-Vis is 1,000,000! This difference in population is the source of the sensitivity, which explains why NMR has inherently low sensitivity compared to other analytical techniques such as UV-Vis.

Table 1: Values used to calculate energy variation and population ratio.

Despite the low sensitivity displayed by NMR compared to other analytical techniques, NMR is one of the most powerful characterization techniques available. Not only is NMR inherently quantitative, but it is also non-destructive, and unlike other techniques, it can provide detailed information of the molecular structure. For more information about our 100, or 60 MHz benchtop NMR spectrometer, or options for advanced experiments or pulse programming capabilities through SPINit, please contact sales@nanalysis.com or fill out this form.

References

^[1] Larive, C. K. and Korir, A. K. Topic 1.3: How does the population difference in NMR compare to the difference between electronic ground and excited states? in Quantitative NMR. It is an open book and can be accessed here.
^[2] Claridge, Timothy D. W., High-Resolution NMR Techniques in Organic Chemistry, 2nd edition, chapter 2, pp. 11-13.
^[3] https://physics.nist.gov/cuu/Constants/index.html (accessed December 2022).