I have 1D NOESY spectra from which I need to find out the value of the NOE (%). Is it enough to simply measure the intensity of the resulting peaks or is it necessary to subtract it from the intensity of the standard spectrum?

  • 3
    $\begingroup$ What does your 1D NOESY look like? Is it one big peak (the proton that was irradiated) plus several small peaks (which often have a different sign compared to the original peak)? $\endgroup$ Commented Oct 15, 2023 at 20:45
  • 1
    $\begingroup$ Percentage NOE enhancements are very rarely used nowadays -- this refers to an older version of the 1D NOE where the irradiated proton is saturated and then a 1D spectrum recorded. You have to compare this against a normal 1D where the proton isn't saturated -- this is where the term 'enhancement' comes from. But modern 1D NOE experiments are transient NOE experiments which selectively excite the target proton, and any other peaks that appear arise solely from the NOE. There is no 'enhancement' to be measured here, only the presence/absence of a peak. $\endgroup$ Commented Oct 15, 2023 at 20:49
  • 1
    $\begingroup$ (Just for completeness' sake: of course, you can get quantitative information from transient NOE experiments, in particular, by repeating the experiment with multiple values of d8 — but I assume that is not what OP is looking for.) $\endgroup$ Commented Oct 15, 2023 at 20:53

1 Answer 1


From documentation (https://chemnmrlab.uchicago.edu/2021/03/10/1d-noe_1/):

Magnetization transfer by NOE between spins can be quantified. NOESY with D8 of just 10 msec (very short) yields almost no magnetization transfer, so the integral of the excited peak in this 1D spectrum can be used for quantitative reference when compared to peaks in spectra acquired with the same parameters (but longer D8 values).

So the denominator is the signal strength of the proton where magnetization originates (the peak you have to pick before running the experiment). You divide, not subtract.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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