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I've just been getting into organic and been reading about H-H related to $\mathrm{NMR}$. I've been seeing articles state how recasting of H-H is necessary and want to know how equations like below are derived.

$$σ_\mathrm{peak} = σ_\ce{HA} + \frac{Δσ}{1 + 10^{(\mathrm{p}K_\mathrm{a} - \mathrm{pH})}}\tag{1}$$

$$y = y_\mathrm{min} + \frac{y_\mathrm{max} - y_\mathrm{min}}{1 + 10^{(\mathrm{pH} - \mathrm{p}K_\mathrm{a})}}\tag{2}$$

Eq. 1: https://www.nature.com/articles/srep43748

"In the equation, $\sigma_\mathrm{peak}$ is the measured $\mathrm{NMR}$ chemical shift of the peak of interest, $\sigma_\ce{HA}$ is the chemical shift of the protonated form and $\Delta\sigma$ is the difference between $\sigma_\ce{HA}$ and the chemical shift of the deprotonated form of $\ce{HA}$ ($\sigma_\ce{A}$)."

Eq. 2: https://pubs.acs.org/doi/full/10.1021/ed3000028?mobileUi=0

"$y$ is the experimentally measured parameter (chemical shift, in this case), and $(y_{max} - y)$ and $(y - y_{min})$ are proportional to the concentrations of the deprotonated and protonated forms of the acid, respectively."

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  • $\begingroup$ Am I right in assuming that organic stands for organic chemistry and not organic food or agriculture; can I further assume that H-H stands for Henderson-Hasselbalch equation. Don't the articles you link to explain (or cite) how to derive these equations? $\endgroup$ – Martin - マーチン Oct 29 at 14:03
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The derivation of these expressions begins from the premise that the observed value of the NMR parameter (chemical shift) is an average over values for two states weighted by their mole fractions: $$\sigma = \sigma_1 \chi_1 + \sigma_2 \chi_2= \sigma_1 \chi_1 + \sigma_2 (1-\chi_1)$$ This is a reasonable assumption in NMR experiments if there are two states (protonated and deprotonated) with distinct shifts and exchange between the two states is fast compared to the chemical shift difference such that the observable signal is a single time average. This allows us to write that $$\frac{(\sigma-\sigma_2)}{(\sigma_1-\sigma_2)} = \chi_1 $$ Similarly we can write that $$\frac{(\sigma-\sigma_1)}{(\sigma_2-\sigma_1)} = \chi_2 $$ such that $$\frac{(\sigma-\sigma_2)}{(\sigma_1-\sigma)} = \frac{\chi_1}{\chi_2} $$ or, in terms of concentrations, that $$\frac{(\sigma-\sigma_2)}{(\sigma_1-\sigma)} = \frac{c_1}{c_2} $$ Introducing this last expression into the Henderson-Hasselbalch equation $$pK_a=pH+\log_{10}\left(\frac{c_{HA}}{c_{A}}\right)$$ gives upon rearrangement $$10^{(pK_a-pH)}=\left(\frac{c_{HA}}{c_A}\right)=\frac{(\sigma-\sigma_A)}{(\sigma_{HA}-\sigma)}$$ which can be further rearranged into $$1+10^{(pK_a-pH)}=\frac{(\sigma_{HA}-\sigma_A)}{(\sigma_{HA}-\sigma)}=\frac{\Delta \sigma}{(\sigma-\sigma_{HA})}$$ and then further into the expressions in the articles (note the sign convention for $\Delta \sigma$).

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  • $\begingroup$ Could you further elaborate on the last equation, specifically how it was rearranged to get 1? Is it just a matter of ratios? $\endgroup$ – cam Nov 2 at 19:15
  • $\begingroup$ @cam I just added 1 to both sides (farmost left and farmost right) of the preceding equation and used the identities $1=(\sigma_{HA}-\sigma)/(\sigma_{HA}-\sigma)$ and $\Delta \sigma = (\sigma_{A}-\sigma_{HA})$ $\endgroup$ – Buck Thorn Nov 2 at 19:19
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Fraction vs Ratio

If you have two species with the concentrations $c_1$ and $c_2$ and you want to know the fraction of species 2, $\chi_2$, you can to the following:

$$c_\mathrm{total} = c_1 + c_2$$

$$\chi_2 = c_2 / c_\mathrm{total} = \frac{c_2}{c_1 + c_2} = \frac{1}{1 + \frac{c_1}{c_2}} = \frac{1}{1 + \mathrm{ratio}} $$

In the last part, we divided both numerator and denominator by $c_2$ so it doesn't appear twice. This is how the mole fraction (or the percentage of one species, where percentages of species add up to 100%) is related to the ratio of the two species. This expression is commonly found in many contexts. (Try plugging in ratios such as 0:1 = 0, 1:1 = 1, 3:1 = 3 and 1:0 = infinity to verify it results in 100%, 50%, 25% and 0% species 2.)

NMR shift vs fraction of species 2

The NMR chemical shift (under conditions of fast exchange) will vary linearly between the shift of pure species 1 and that of pure species 2. So the lower the mole fraction of species 2, the closer the signal to that of pure species 1, and the higher the mole fraction of species 2, the closer the signal to that of pure species 2.

Putting it all together

We can chunk the equation given by the OP into the following:

  1. The observed shift is equal to the shift of species 1 plus the difference in shift between the species times the mole fraction of species 2. (Try some examples like 0%, 50% and 100% species 2 to see whether that makes sense to you.) $$σ_\mathrm{peak} = σ_\ce{HA} + Δσ \cdot \chi_2$$
  2. The mole fraction of species 2 is given by adding 1 to the ratio and taking the reciprocal, as shown above. $$\chi_2 = \frac{1}{1 + \mathrm{ratio}} $$
  3. The ratio of species is given by the Henderson-Hasselbalch expression, leading to the power of 10 expression found in the denominator (try pH = pKa or much more acidic or basic than the pKa to see how this expression evaluates in the range from zero to infinity). $$ \text{ratio} = 10^{(\mathrm{p}K_\mathrm{a} - \mathrm{pH})}$$

Now look at the expression again and see if you can identify those three chunks.

$$σ_\mathrm{peak} = σ_\ce{HA} + Δσ \frac{1}{1 + 10^{(\mathrm{p}K_\mathrm{a} - \mathrm{pH})}}$$

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