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I've been given this NMR along with the IR and Mass Spec and assigned the task of figuring out the unknown compound. I know that it is cyclohexanol and by analysing the NMR I know that the peak at around 2.8 is the OH and the peak around 3.6 is the deshielded CH. However I can't pair up the other protons and the peaks between 1-2 ppm. I know that the difference between the peaks/protons is their vicinity to the OH group.

My questions is: Is there any way to do the integral of the points lying between 1-2 ppm. This will be useful as I think that the peak at around 1.3 ppm represents more protons than the others. Also could the peaks be different due axial and equatorial protons resonating at different frequencies, or would the temperature

Or is there a better way to pair up the peaks between 1-2 ppm and the remaining protons in the cyclic.

enter image description here

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    $\begingroup$ Yes you can measure integrals; resort to the NMR visualising software of your choice. (No, the problem is non-trivial so it cannot be precisely done by hand unless you want to cut out slabs of paper and weigh them.) Also it is most likely that axial and equatorial protons give different signals. $\endgroup$
    – Jan
    Commented Oct 27, 2015 at 9:20
  • $\begingroup$ (Unfortunately, I cannot link you to the exact page, but note that SDBS has an entry for cyclohexanol which contains the relevant assignments. Check the $400~\mathrm{MHz}$ spectrum presented there.) $\endgroup$
    – Jan
    Commented Oct 27, 2015 at 13:49

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To answer your first question about measuring the integrals for the peaks between 1 and 2 ppm: If you seriously need to do a very accurate integral of these peaks without having access to NMR processing software (ie you've just been given the spectrum as a handout), you can resort to one of the methods of choice from the good 'ol days.

  • Print your spectrum out on a nice big piece of paper, get a nice sharp pair of scissors and cut out your peaks and weight them (a 4 or 5 figure balance may be required). These method gives incredibly accurate integrals

  • Print out your spectrum on graph/grid paper and count the grids. Again, very accurate, depending on the scale of grid you choose to print on.

As for 'pairing up' the remainder of the ring protons, this is not a trivial exercise, made even less so with the dispersion of the spectrum you are working with. You are quite right in your assessment. The protons closer to the -OH group will be at higher chemical shifts; equatorial and axial protons will be non-equivalent (equatorial at higher chemical shifts); and the whole fluxional ring dynamic will be temperature, concentration, solvent and field dependent. So, probably it is not worth trying to pair up any of these protons. You could (and hopefully would, if you are a thorough researcher) assign all of your peaks with the assistance of 2D COSY, HSQC/HMBC methods.

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    $\begingroup$ Geeze you all are spoiled by high tech. Going really old school. Look at the peak heights. Crude but good enough for 1-2 ppm region. Key is that bonds are similar enough that peaks will be the same width. So height is proportional to area. From left to right I see 2, 2, 1, 4, 1. You know it has to add to 10. $\endgroup$
    – MaxW
    Commented Oct 28, 2015 at 7:26
  • $\begingroup$ I must’ve stopped looking at peak heights sometime when my NMR software gave me a different integral by factor two, but I’ll agree it can be helpful @MaxW. $\endgroup$
    – Jan
    Commented Oct 28, 2015 at 8:13
  • $\begingroup$ Considering that the $400~\mathrm{MHz}$ spectrum on SDBS looks exactly like the image in the question, I’ll assume some TAs were lazy and copied. Assuming SDBS is correct, we get distinct peaks for axial and equatorial protons. Otherwise I agree; $+1$. $\endgroup$
    – Jan
    Commented Oct 28, 2015 at 8:14
  • $\begingroup$ I don't understand the "otherwise". I assure you it is 2,2,1,4,1. You have to understand that the molecule isn't planar, nor is it static relative to the time that the NMR scan takes. // The Diophantine solution with peak heights only works for small numbers of atoms with fairly well separated lines. $\endgroup$
    – MaxW
    Commented Oct 28, 2015 at 8:28
  • $\begingroup$ I just did the integrals, it seems to work by just printing it on graph paper and measuring the heights. Thanks for the help. $\endgroup$
    – Jonathan
    Commented Oct 28, 2015 at 10:29
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Calculating the spectrum (APFD cc-pVDZ NMR GIAO) with -OH in equatorial position gives the following signals:

  • H-1 axial 3.3
  • H-2 axial 1.1, equatorial 1.8
  • H-3 axial 1.1, equatorial 1.6
  • H-4 axial 1.0, equatorial 1.4

That would fit the spectrum decently but it also would mean that flipping through the conformations is surprisingly slow.

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  • $\begingroup$ Welcome. It might help to format the calculated shifts into a list to make it easier to read. Did you perform the calculations yourself? $\endgroup$
    – Geoff Hutchison
    Commented Jan 3, 2023 at 15:28
  • $\begingroup$ On its own, I don't think these shifts provide any evidence for slow ring flipping. At RT ring flips should be fast, and you haven't evaluated this possibility to see if it fits better...? I think to be thorough, one would need to calculate shifts for the other conformer, then simulate the resulting spectrum for a variety of exchange rate constants (not done with a typical QC programme, you need specific software which deal with NMR / spin dynamics Hamiltonians), and see which rate constant (or range thereof) best matches the experimental spectrum. $\endgroup$
    – orthocresol
    Commented Jan 3, 2023 at 15:38

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