While acquiring some NMRs I find that there is a consistent pattern in that protons attached to sp2 carbons tend to have slightly smaller integrals than protons attached to sp3 carbons:

spectrum of compound

(peaks in aromatic region are from a para-disubstituted benzene ring; peak at 4.92 ppm is from a trisubstituted alkene; 4.50 is a NH; the remainder are aliphatic CH2's or CH3's.)

I calibrated the total sum of integrals to 21; one can see from the spectrum that the aromatic and olefinic peaks are slightly smaller than integral values, and the others are slightly larger. Obviously it doesn't hinder the analysis of the spectrum, but it was an interesting trend (the same was observed in multiple spectra).

Am I reading too much into this, or is there a reason behind it? I strongly suspect there is something - perhaps related to the marginally slower relaxation of these protons, since there are no geminal protons (only vicinal protons)?

If it is of relevance - a 60° pulse is used (pulse program zg60). The number of scans ns is 16, and the relaxation delay d1 is 1 second. I'm happy to provide any other necessary acquisition parameters, or to provide the full structure of the compound if it's necessary.

P/S I found something in Findeisen & Berger's 50 and More Essential NMR Experiments, in which they mention that the integral of an aromatic proton in strychnine is smaller than expected as it has the longest relaxation time. Even though the authors used ns = 1, too short a delay between the receiver gain adjust and the scan was used, which led to incomplete relaxation and hence a smaller integral.

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    $\begingroup$ Indeed I think all that's going on is that 1 second is too short a relaxation delay; bump it up to 5 s or maybe 10 s for safety and the integration should be essentially perfect, bar random noise. For quantitative NMR standards, it's pretty common to use at least ~5 s. See this Sigma-Aldrich brochure (pdf) for some typical relaxation delays for reference. I once had a very slowly relaxing aryl proton in a zwitterion, but with 25 s relaxation delay, it too integrated perfectly. $\endgroup$ Oct 12, 2017 at 13:31
  • $\begingroup$ Just so I can fix my answer below, can you give the duration of the recorded FID, np x dw? $\endgroup$
    – Karl
    Oct 7, 2022 at 18:10
  • $\begingroup$ @Karl not totally sure, but I'd expect it to be around 2 or 3 s (probably 16k complex points, sw ~ 20 ppm = 8 kHz (I think this was acquired at 400 MHz), so dwell time between complex points ~ 125 µs) $\endgroup$ Oct 7, 2022 at 19:58

1 Answer 1


That's to be expected: Less protons as spatially close neighbours, less dipolar coupling, slower relaxation.

For a relaxation/recycle delay of $\pu{1 s}$ and $T_1$ of perhaps $\pu{3 s}$, the Ernst angle is already only $44^\circ$, so your measurement is far from optimal anyway. Use a shorter pulse or longer recycle delay.

At one second, your are anyway dangerously close to generating echoes which overlay the signal from further scans, distorting the accumulated spectrum.

  • $\begingroup$ d1 is the delay between end of the acquisition and next scan. For the Ernst angle, der relevant delay includes the acquisition, so the number of 44° I give above is wrong. $\endgroup$
    – Karl
    Oct 7, 2022 at 18:08

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