Why does spin-spin coupling cause splitting in NMR spectroscopy?

[BobP]

Hi.
I was wondering if you could please help me understand something about NMR spectroscopy (MRS).

I know that the chemical shift (Hz) of a particular chemical will vary with B0. As I understand (although this may be wrong) ppm is frequency independent so the peaks should always be in the same position

There are two things I do not understand though

1) Why does increasing the frequency increase the separation of peaks on a ppm curve if ppm is frequency indepndenant? (my lecturer said it did)

2) When trying to localise the position of a spectral signal in 3D space why can we NOT use a frequency encoded gradient but we can use a slice or phase encoded gradient (I assume this links with the previous question).

Thank you very much

 

[DrClaude]

1) Why does increasing the frequency increase the separation of peaks on a ppm curve if ppm is frequency indepndenant? (my lecturer said it did)

Increasing the frequency (or, more appropriately, the magnetic field) doesn't change the separation of the peaks in ppm, but it does allow for a better resolution (the peaks are narrower).

2) When trying to localise the position of a spectral signal in 3D space why can we NOT use a frequency encoded gradient but we can use a slice or phase encoded gradient (I assume this links with the previous question).

I don't know about this. Are talking about solid state NMR?

 

[BobP]

Increasing the frequency (or, more appropriately, the magnetic field) doesn't change the separation of the peaks in ppm, but it does allow for a better resolution (the peaks are narrower).

Thanks. Why does resolution increase?
Has it anything to do with a longer FID decay time? If so, how does this happen?

I don't know about this. Are talking about solid state NMR?

Don't think so. I am talking about acquiring a signal from a specific voxel in the brain, for example, using spectroscopy in a clinical setting. My lecturer said we cannot use frequency gradients but we can use other gradients. His reasoning was that ppm is affected by frequency but clearly it isn't so I am wondering why we cannot use them.thanks

 

[Dale]

1) Why does increasing the frequency increase the separation of peaks on a ppm curve if ppm is frequency indepndenant? (my lecturer said it did)

There are two reasons for that. One reason is simply SNR. Since you have higher SNR at higher field strengths two peaks that are blurred out by noise at los field strengths may be separable at higher field.

The second is due to shimming. A poor shim causes blurring in the spectrum. Say you can shim down to 10 Hz at both fields, then that is a smaller ppm blurring at the higher field.

2) When trying to localise the position of a spectral signal in 3D space why can we NOT use a frequency encoded gradient but we can use a slice or phase encoded gradient (I assume this links with the previous question)

You can, but the math gets substantially more complicated. Andrew Maudsley at University of Miami and Stefan Posse at New Mexico have each developed their own approaches for doing that.

Are you familiar with the k space approach for understanding encoding?

 

[BobP]

There are two reasons for that. One reason is simply SNR. Since you have higher SNR at higher field strengths two peaks that are blurred out by noise at los field strengths may be separable at higher field.

The second is due to shimming. A poor shim causes blurring in the spectrum. Say you can shim down to 10 Hz at both fields, then that is a smaller ppm blurring at the higher field.

Why does SNR increase with B0? thanks

You can, but the math gets substantially more complicated. Andrew Maudsley at University of Miami and Stefan Posse at New Mexico have each developed their own approaches for doing that.

Are you familiar with the k space approach for understanding encoding?

Sadly not :( Is it possible to explain why it is more complicated without going into it?

 

[Dale]

Why does SNR increase with B0? thanks

As the field strength increases you get more net longitudinal magnetization in the fully relaxed state. I don't remember the numbers exactly, but it is something like 6 excess protons per million at 1.5 T vs 12 excess protons per million at 3.0 T.

 

[BobP]

As the field strength increases you get more net longitudinal magnetization in the fully relaxed state. I don't remember the numbers exactly, but it is something like 6 excess protons per million at 1.5 T vs 12 excess protons per million at 3.0 T.

Ah of course! thanks

On a slightly unrelated note, I have been told that the separation of J-coupled peaks increases as B0 increases. Please could you suggest why this is, again, as ppm is unrelated to B0

 

[Dale]

Sadly not :( Is it possible to explain why it is more complicated without going into it?

I had to think a bit about this. Basically, what makes the Fourier transform easy is that it is separable. A 2D FFT is just a bunch of 1D FFT in one direction followed by a bunch of 1D FFT in the other direction. But this relies on the data being laid out in a nice rectilinear grid in k space.

But the corresponding "gradient" for spectroscopy is just time, which you cannot turn off or rewind. So if you do spatial encoding and spectroscopic encoding you wind up going diagonally through k space which complicates things.

 

[Dale]

Ah of course! thanks

On a slightly unrelated note, I have been told that the separation of J-coupled peaks increases as B0 increases. Please could you suggest why this is, again, as ppm is unrelated to B0

I think that this is purely due to SNR, but I am more on the imaging side than the spectroscopy side. So I am not certain.

 

[DrClaude]

On a slightly unrelated note, I have been told that the separation of J-coupled peaks increases as B0 increases. Please could you suggest why this is, again, as ppm is unrelated to B0

Be careful here. The chemical shift is independent of B₀, but the calculation of ppm depends on the operating frequency, as it is basically
$$
\frac{\nu_\mathrm{sample} - \nu_0}{\nu_0} \times 10^6
$$
Since the transition frequency due to the chemical shift ($\nu_\mathrm{sample} - \nu_0$) and the reference frequency have the same variation with respect to B₀, the field cancels out and the chemical shift, expressed in ppm, is independent of B₀.

However, the J-coupling comes from spin-spin interaction, which is independent of B₀. Since it induces a fixed shift in $\nu_\mathrm{sample}$, in terms of ppm it decreases as B₀ increases.:
$$J\ \mathrm{Hz} = \delta\ \mathrm{ppm} \times \nu_0\ \mathrm{MHz}$$

 

[BobP]

I think that this is purely due to SNR, but I am more on the imaging side than the spectroscopy side. So I am not certain.

I see. thanks

 

[BobP]

Be careful here. The chemical shift is independent of B₀, but the calculation of ppm depends on the operating frequency, as it is basically
$$
\frac{\nu_\mathrm{sample} - \nu_0}{\nu_0} \times 10^6
$$
Since the transition frequency due to the chemical shift ($\nu_\mathrm{sample} - \nu_0$) and the reference frequency have the same variation with respect to B₀, the field cancels out and the chemical shift, expressed in ppm, is independent of B₀.

However, the J-coupling comes from spin-spin interaction, which is independent of B₀. Since it induces a fixed shift in $\nu_\mathrm{sample}$, in terms of ppm it decreases as B₀ increases.:
$$J\ \mathrm{Hz} = \delta\ \mathrm{ppm} \times \nu_0\ \mathrm{MHz}$$

thank you. To clarify, did you mean However, the J-coupling comes from spin-spin interaction, which is DEPENDANT on B₀.

 

[DrClaude]

To clarify, did you mean However, the J-coupling comes from spin-spin interaction, which is DEPENDANT on B₀.

No. The spin-spin coupling is intrinsic to the molecule, thus is independent of any external field.I've stolen the following at http://orgchem.colorado.edu/Spectroscopy/nmrtheory/splitting.html

http://orgchem.colorado.edu/Spectroscopy/nmrtheory/figures/splittingsum.gif

 

[BobP]

No. The spin-spin coupling is intrinsic to the molecule, thus is independent of any external field.I've stolen the following at http://orgchem.colorado.edu/Spectroscopy/nmrtheory/splitting.html

http://orgchem.colorado.edu/Spectroscopy/nmrtheory/figures/splittingsum.gif

thanks