# 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.

For equivalent protons, the main thing you might need to worry about is that when you have coherences on carbon, such as $$S_x$$, you will get evolution of multiple proton–carbon couplings, represented by (e.g.) the Hamiltonian $$2\pi J_{IS} I_{1z}S_z + 2\pi J_{IS}I_{2z}S_z$$ for a $$\ce{CH2}$$ pair. This could, in theory, generate more complicated product operators such as $$4I_{1z}I_{2z}S_x$$. To analyse this, you can separately evaluate the effects of the two J-couplings. The first one goes as:

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

where $$\theta = \pi J_{IS}\tau$$. To evaluate the second coupling, the $$S_x$$ term becomes

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

and for the $$I_{1z}S_y$$ term, we can just "carry through" the $$I_{1z}$$ terms because they don't evolve under a Hamiltonian on different nuclei*:

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

So all in all, single-quantum coherence on carbon evolves under two different couplings to give four terms:

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

Thankfully, you do not need to worry about this, because the only times that single-quantum coherence on carbon exists is during the $$t_1$$ periods of the HSQC. Conveniently, the HSQC also has a proton 180° pulse in the middle of $$t_1$$, which refocuses all C–H couplings, so this whole evolution of $$J_{IS}$$ can be neglected. 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).

If the protons are equivalent, then the above also applies to the HMQC: the only time when there are coherences on carbon is during $$t_1$$, and there is also a 180° pulse in the middle of $$t_1$$ which removes the effects of carbon–proton couplings.

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 the HSQC $$t_1$$ is single-quantum carbon, it does not evolve under homonuclear couplings. However, for the HMQC which has multiple-quantum coherence present during $$t_1$$, the proton component will evolve under homonuclear couplings, which means that you will get multiplets 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.

As a final point, in the HSQC above, notice that when we allow single-quantum carbon coherence to evolve under $$J_{IS}$$, the $$\ce{CH2}$$ group obtains a phase factor of $$cos^2\theta$$, whereas a $$\ce{CH}$$ group would only have a factor of $$\cos \theta$$.

This is the basis of multiplicity editing in the HSQC: immediately after $$t_1$$, a spin echo of total duration $$\tau = 1/J_{IS}$$ is added in order to allow $$J_{IS}$$ to evolve. This choice means that $$\theta = \pi$$, so $$\cos\theta = -1$$: thus, $$\ce{CH}$$ groups are inverted and $$\ce{CH2}$$ groups are not. The resulting spectrum will therefore have different signs for peaks belonging to $$\ce{CH}$$ and $$\ce{CH2}$$ groups.

($$\ce{CH3}$$ groups get a factor of $$\cos^3\theta$$, so have the same sign as $$\ce{CH}$$ groups.)

* 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

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