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I read about splitting patterns in proton NMR and found that this is due to spin-spin coupling, where reference to Pascal's triangle can explain the splitting patterns found in doublet, triplet and quartet.

My book mentions this later:

There are also no splitting patterns caused by the spin-spin coupling, i.e. interaction of neighbouring protons in the 1-H atoms within molecules.

Advanced Chemistry for You, Third Edition (Ryan)

Why do you not find splitting patterns in Carbon-13 NMR?

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  • $\begingroup$ There is splitting in 13C. Think about solvent signals or compounds with CF bonds. $\endgroup$
    – DSVA
    Dec 3 '17 at 23:24
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In 1H NMR, spin-spin coupling leading to multiplet structure most commonly arises due to coupling to other 1H nuclei. This is most often taught in terms of neighbouring 1H nuclei having "up" and "down" spins, both of which lead to the nucleus of interest experiencing a slightly different magnetic field, and hence having a slightly different chemical shift. While this is only a simplification, it is probably sufficient for the purposes of this question.

In 13C NMR, though, proton decoupling is often employed. As the name suggests, this removes all effects of coupling to protons (i.e. 1H). How is this achieved? Essentially, while the signal from the 13C nuclei is being recorded, the NMR spectrometer simultaneously and continually irradiates the 1H nuclei with energy. This means that each 1H nucleus is rapidly undergoing "spin-flips" between up and down spin, and therefore, what the 13C nucleus sees is actually a magnetic field that is very rapidly oscillating between the two possible values. What is actually observed in the NMR spectrum is an average of the two possible values, which is equivalent to what would be seen if there was no coupling to begin with.

Why do we do this, though?

  • One concern is sensitivity: when peaks are split due to spin-spin coupling, the intensity of the peak is decreased. Since 13C has a naturally very low abundance, it is of interest to prevent the peak intensity from being spread out due to coupling.
  • Another concern is simplicity: carbon spectra which have C–H coupling can often be extremely difficult to interpret. For an example I refer you to Glenn Facey's blogpost where an inset of a proton-coupled spectrum of toluene is shown.
  • A subtle point that is often neglected is the fact that the decoupling process also enhances the intensity of 13C peaks via the nuclear Overhauser effect.

All of these factors make proton-decoupled carbon spectra – or in NMR lingo, 13C{1H} spectra – the preferred experiment.


If you're interested in coupling to nuclei other than 1H:

  • coupling to 13C is all but invisible, considering the low natural abundance of 13C. Recall that in a 13C spectrum you are only observing 1% of all the carbons present. How many of those carbons are coupled to another 13C nucleus? Or in other words, what are the chances of finding two 13C nuclei next to each other? Pretty low is the answer – so much so that an NMR experiment designed to measure C–C couplings was given the acronym INADEQUATE.

  • coupling to other abundant spin-1/2 nuclei is in fact very often seen. Two simple examples are 19F and 31P, both of which are close to 100% natural abundance – in compounds containing F or P, multiplet structure will be observed in the carbon spectrum. Here's the spectrum of 1-chloro-2-fluoroethane, $\ce{FCH2CH2Cl}$, courtesy of SDBS. The two peaks on the left are from the carbon next to fluorine; coupling to fluorine leads to a 1:1 doublet being observed. Likewise, the two peaks on the right are from the carbon next to chlorine. It is still coupled to fluorine, but because of the greater distance, the coupling constant is smaller. carbon spectrum of 1-chloro-2-fluoroethane There is also probably a wealth of examples in organometallic chemistry (for example 103Rh is also spin-1/2), but I don't currently have the energy to find a good example.

  • coupling to quadrupolar nuclei (nuclei with spin > 1/2) can be seen, depending on the symmetry of the molecule and on some other physical parameters. For example, since deuterium has spin 1, the carbon peak in $\ce{CDCl3}$ is observed as a 1:1:1 triplet (a familiar sight to organic chemists). But this is not always the case: for a more in-depth explanation see, e.g. Symmetry and NMR linewidths of quadrupolar nuclei

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  • $\begingroup$ You mentioned that proton decoupling is by averaging the values where the ¹H are up and down spin. However, doesn’t irradiating it only ensures that all of the ¹H is up spin? (where it opposes the magnetic field from what i understand) How does this method ensure that we have a signal where all ¹H is down spin? (where it has an enhancing effect on the external magnetic field) $\endgroup$ Aug 7 at 14:50
  • $\begingroup$ @ImJustKawaii It's not possible to place all spins in the higher-energy state through continuous irradiation. The explanation for this requires some slightly more complicated theory (see: Einstein coefficients), but the short answer is that the absorption (i.e. excitation) is balanced by stimulated emission (i.e. decay). That is, for every spin you flip from down to up, there will also be a spin you flip from up to down. Eventually the up and down populations are equalised. $\endgroup$
    – orthocresol
    Aug 7 at 15:10
  • $\begingroup$ If it’s not possible to place all spins in the higher or lower energy state, what then is the two possible values that we average to remove coupling effect? and what happens at these two values? $\endgroup$ Aug 7 at 15:19
  • $\begingroup$ @ImJustKawaii The chemical shift of the 13C nucleus depends on whether the 1H is spin up or down. So if the 1H is continually flip-flopping between different states, the 13C chemical shift you see is the average of two different chemical shifts, i.e. a chemical shift in the middle. I thought I described this part already in my answer; what exactly are you confused about? [Note that this description is rather simplified, but it works. QM is needed for a full explanation.] $\endgroup$
    – orthocresol
    Aug 7 at 15:30
  • $\begingroup$ I think my confusion stems from what two values we are averaging. Initially, I was thinking that to remove the coupling effect we average the chemical shift values of ¹³C when all the protons oppose the external magnetic field and when all the protons enhance the external magnetic field. However, you then mentioned that it’s not possible for all protons to be in the higher energy state (opposing the field I think) by irradiation. So I am now confused as to what two values we are averaging and how irradiation achieves this two values. $\endgroup$ Aug 7 at 16:08

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