# Product operator analysis for CHn groups in HSQC and HMQC

I'm taking and advanced NMR course this semester and we are learning about product operator formalism. I have a homework where I'm supposed to apply this formalism to an HSQC and HMQC experiment. The problem I have is that I should use a 3 and 4 spin systems $$(\ce{CH2})$$ and $$(\ce{CH3})$$ groups instead of just simply $$(\ce{CH})$$ and we only did two spin systems in class. I understand how to use the formalism and how to apply it to the experiment, but I need let's say a concept clarification. We use basic pulse sequences for HSQC and HMQC.

In the exercise, we assume that the hydrogens do not couple to each other, only to the carbon. I did the example a found out that the result (in the product operator formalism it was the same observable term which with I was left at the end) was the same for $$\ce{CH3}$$ group as for $$\ce{CH}$$ after optimization of delay time. After I thought about it, I feel it makes sense. Since the three hydrogens will do the same as one hydrogen, only I will have a different intensity afterward in the spectrum and of course different shifts. I even calculated that the optimal tau delay is the same for both $$\ce{CH}$$ and $$\ce{CH3}$$ groups.

And since HMQC basically gives the same info, I assume it will work the same, and results for $$\ce{CH}$$ and $$\ce{CH2}$$ group will also give me the same.

Is this correct? Is there any other difference in what happens during the experiment to the $$\ce{CH}$$ and $$\ce{CH3}$$ group (or $$\ce{CH2})?$$

Just to clarify, it is not a multiplicity editing experiment, just a simple HSQC/HMQC applied to $$\ce{AX2}$$ and $$\ce{AX3}$$ spin systems.

Your analysis is generally correct. In the case where all the protons are equivalent, you don't need to worry about proton–proton coupling. The main thing you need to worry about is that when you have a term such as $$2I_{1z}S_x$$ you will get evolution of different proton–carbon couplings, represented by (e.g.) the Hamiltonian $$2\pi J_{IS} I_{2z}S_z$$. This could, in theory, generate more complicated product operators such as $$4I_{1z}I_{2z}S_y$$. You probably already know that

$$S_x \xrightarrow{2\pi J_{IS} I_{2z}S_z \tau} (\cos \theta) \, S_x + (\sin \theta)\, 2I_{2z}S_y$$

where $$\theta = \pi J_{IS}\tau$$. A term on a different nucleus, such as $$I_{1z}$$, does not evolve under this coupling, so you can "multiply" by $$2I_{1z}$$ on both sides* to get

$$2I_{1z}S_x \xrightarrow{2\pi J_{IS} I_{2z}S_z \tau} (\cos \theta) \, 2I_{1z}S_y + (\sin \theta)\, 4I_{1z}I_{2z}S_y$$

Thankfully you do not need to worry about this, because the only times that these can evolve are during the $$t_1$$ periods of both experiments, when there is magnetisation present on carbon. Conveniently, both experiments have a proton 180° pulse in the middle of $$t_1$$ which refocuses all C–H couplings anyway. Outside of the $$t_1$$ periods the magnetisation of interest is single-quantum on proton, so the only coupling that evolves is that one proton–carbon coupling, which behaves in exactly the same way as for a CH system.

For a more detailed treatment I suggest consulting Chapter 12 of James Keeler's Understanding NMR Spectroscopy, 2nd ed. (2010).

However, if the protons are not equivalent, then you will start to get evolution of H–H homonuclear couplings, which can't simply be refocused by a 180° pulse. For the HSQC this is less important as it just leads to a reduction in intensity from magnetisation that goes down non-useful coherence pathways. Because the magnetisation during $$t_1$$ is single-quantum carbon, it does not evolve under homonuclear couplings. However, for the HMQC, the multiple-quantum coherences present during $$t_1$$ will evolve under homonuclear couplings, which means that you will get multiplet shapes in the indirect dimension after Fourier transformation. In practice it is difficult to resolve these splittings and so it is just manifested as line broadening.

There are also considerations to be made when using a sensitivity-enhanced HSQC sequence, but I assume those are not relevant for your current situation. The interested reader is directed to J. Biomol. NMR 1994, 4, 301–306.

* To be more precise, this is because $$I_{1z}$$ commutes with the Hamiltonian. So, it also commutes with the unitary propagator $$U = \exp(-\mathrm i H\tau)$$, and we can write

$$2I_{1z}S_x \xrightarrow{H\tau} U(2I_{1z}S_x)U^\dagger = 2I_{1z} \cdot US_xU^\dagger$$