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

  • Context: Graduate 
  • Thread starter Thread starter BobP
  • Start date Start date
  • Tags Tags
    Nmr Spectroscopy
Click For Summary

Discussion Overview

The discussion focuses on the principles of NMR spectroscopy, particularly the effects of magnetic field strength (B0) on chemical shifts and J-coupling. Participants explore the relationship between frequency, peak separation in ppm, and the implications for signal localization in 3D space.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • Some participants assert that increasing frequency improves resolution but does not change the separation of peaks in ppm.
  • There is a suggestion that higher signal-to-noise ratio (SNR) at higher field strengths allows for better peak separation.
  • Participants discuss the complications of using frequency-encoded gradients for spatial localization in NMR, with some questioning the applicability to solid-state NMR.
  • One participant notes that the separation of J-coupled peaks increases with B0, prompting inquiries about the underlying reasons.
  • It is mentioned that while chemical shifts are independent of B0, the calculation of ppm is affected by the operating frequency.
  • Some participants express uncertainty about the relationship between J-coupling and B0, with differing views on whether J-coupling is dependent on B0.
  • There is a discussion about the mathematical complexities involved in spatial and spectroscopic encoding in NMR.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the relationship between J-coupling and B0, with some asserting independence while others question this view. The discussion remains unresolved regarding the implications of frequency on peak separation and the use of frequency-encoded gradients.

Contextual Notes

Participants reference specific technical concepts such as signal-to-noise ratio, shimming, and k-space, indicating a reliance on these definitions and assumptions. The discussion also highlights the complexity of mathematical relationships in NMR without resolving them.

BobP
Messages
70
Reaction score
1
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
 
Physics news on Phys.org
BobP said:
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).

BobP said:
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?
 
DrClaude said:
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?
DrClaude said:
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
 
Last edited:
BobP said:
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.
BobP said:
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?
 
DaleSpam said:
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

DaleSpam said:
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?
 
BobP said:
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.
 
DaleSpam said:
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
 
BobP said:
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.
 
BobP said:
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.
 
  • #10
BobP said:
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.:
$$J\ \mathrm{Hz} = \delta\ \mathrm{ppm} \times \nu_0\ \mathrm{MHz}$$
 
  • Like
Likes   Reactions: BobP
  • #11
DaleSpam said:
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
 
  • #12
DrClaude said:
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 dependent on B0.
 
  • #13
BobP said:
To clarify, did you mean However, the J-coupling comes from spin-spin interaction, which is dependent 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
 
  • Like
Likes   Reactions: BobP
  • #14
DrClaude said:
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
 

Similar threads

Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 5 ·
Replies
5
Views
4K