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I would like to ask a question about the ordering of the $\ce{X}$ spins on the right hand side of an $\ce{AX2}$ energy diagram.

I was given the diagram here below:

enter image description here

and asked to evaluate why this diagram results in a triplet. This is simple, since there are 4 possible energy transitions, but one of them is equivalent to the singlet transition, hence only 3 transitions would appear in the NMR spectrum.

Prior to this though, I was confused because I thought the ordering for the bottom 3 spins was incorrect. Taking $\ce{\alpha}$ as the lower energy spin, I thought the bottom 3 spins should read as:

$$\begin{align}&\beta \beta \\ \beta \alpha &~~~~ \alpha \beta \\ &\alpha \alpha \end{align}$$

and the energy levels would be drawn more appropriately. I attempted but failed to mention this to the lecturer during the presentation.

Am I correct in this instance?

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  • $\begingroup$ Neither of those diagrams make sense afai can tell. How can you have a transition that doesnt flip any spins? $\endgroup$ – Buck Thorn Nov 13 at 18:51
  • $\begingroup$ @BuckThorn I questioned whether the order of the bottom 3 spins was incorrect but I did think about the selection rule and why that was not obeyed in this slide $\endgroup$ – vik1245 Nov 13 at 19:09
  • $\begingroup$ @BuckThorn it is flipping the A spin. The A spin state is on the left. The X spin states (which don’t flip) are on the right. $\endgroup$ – orthocresol Nov 14 at 8:19
  • $\begingroup$ @orthocresol Gotcha, I misread the diagram. $\endgroup$ – Buck Thorn Nov 14 at 8:21
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    $\begingroup$ @BuckThorn that was a minor point, though. You are quite correct that the diagram is wrong in many ways. It is not a physically accurate description of coupling. The diagram implies that the splitting between the peaks in the triplet has a frequency on the order of an X spin flip, which is completely... well... off. $\endgroup$ – orthocresol Nov 14 at 8:25
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There are a few things that seem to be wrong with the diagram you present, but since your focus is on the relative energies of the $\alpha$ and $\beta$ spin states I'll address only this. If you check a reliable source such as Cavanagh et al.s NMR textbook [ref 1] it is explained that

the state with $m=+\frac{1}{2}$ is referred to as the $\alpha$ state, and the state with $m=-\frac{1}{2}$ is referred to as the $\beta$ state. If $\gamma$ [the gyromagnetic ratio] is positive, then the $\alpha$ state has lower energy than the $\beta$ state.

Therefore for protons $\alpha$ is lower energy (since $\gamma>0$). If the nucleus is $\ce{^{15}N}$ ($\gamma<0$), $\beta$ becomes the lower energy state.

So while your concern is valid, it is a relatively minor flaw (in my opinion) with the diagram. Good for you for noticing it.

Reference

  1. John Cavanagh, Nicholas Skelton, Wayne Fairbrother, Mark Rance, Arthur Palmer, III. Protein NMR Spectroscopy, Principles and Practice, 2nd Edition. Academic Press, 2006, ISBN: 9780121644918.
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