0
$\begingroup$

I need to determine as many J values as I can for 3-methoxybenzaldehyde. The peaks overlap a lot, and I don't see the expected ddd or dt pattern that I would expect for the hydrogen on carbon 6. The HSQC I was given has the clearest peaks for the H NMR, so I attached that. Maybe someone with more NMR experience can help me tease apart the signals?

enter image description here

$\endgroup$
2
$\begingroup$

There's almost nothing you can do in this situation, unfortunately. Because of strong coupling and other issues, aromatic peaks often do not conform to the so-called $n+1$ rule, which makes it highly non-trivial to extract coupling constants. It is probably possible to "fit" coupling constants to the multiplet shape by simulating the spectrum for a given set of $J$'s and seeing which set best matches the observed spectrum.

The HSQC trace is unfortunately not useful in this regard. At first glance, it may look simpler than the 1D proton spectrum you undoubtedly have. But that is because HSQC spectra are invariably recorded with low resolution in the proton dimension. The main reason for that is because of the high-power 13C decoupling that is performed during the acquisition. Acquisition times therefore need to be kept to approximately 100 ms, or there is a serious risk of damaging the spectrometer. On the other hand, a 1D proton spectrum doesn't utilise 13C decoupling, and can have acquisition times of several seconds.

Assuming the spectral width is constant, a shorter acquisition time means poorer resolution (this is a property of the discrete Fourier transform).

$\endgroup$
1
  • 1
    $\begingroup$ The last point "Assuming the spectral width is constant, a shorter acquisition time means poorer resolution (this is a property of the discrete Fourier transform)." is very critical. The readers should keep in mind that the resolution with shorter acquisition time is poorer because of the frequency spacing with lesser number of data points. Larger the length of the data, smaller the frequency spacing on the x-axis. $\endgroup$
    – M. Farooq
    Mar 17 '20 at 3:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.