I've noticed that as the chain length of n-alkanes increase, their number of carbon environments tend to 5 peaks, and no more, no less. Why is this? I've looked on multiple NMR (C13 NMR, I might add) databases and they show that there are 5 carbon environments. The outer carbons are all distinct environments like you'd normally expect, but the inner part of it (e.g. the innermost carbons in tridecane, C13H28) is all one environment. I hope somebody could shed some light on it. It's stumped both me and my chemistry teachers.
2 Answers
The environments are the same only to within the limits of the spectrometer's resolution and other experimental details; remember that every measurement comes with uncertainties, some of which are baked into the very fabric of the universe.
If you had an "infinitely precise" NMR spectrometer, all the peaks would be distinct (except when symmetry demands they be the same). In fact, every single possible compound would be uniquely identifiable by its atoms' specific shifts, shared with no others. In principle, you'd be able to assign a carbon peak at 123.6207426[...]964109454164[...] ppm to riboflavin, whereas a peak at 123.6207426[...]964109454165[...] ppm would indicate monobrominated hexabenzocoronene intercalated into a strand of DNA, and so on. Of course this is just a thought experiment and would never be possible in practice, even though it would make characterisation of complex molecules and mixtures a dream...
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$\begingroup$ Yes, I thought that may be possible. However, is that the only reason, do you think? Would there be another reason that it works too, apart from simply the spectrometer only able to be so precise? And why 5 peaks? Why not four, or six? Do you know why the number of peaks doesn't keep on increasing even after, say, dozens and dozens in the carbon chain? Thank you for the comment; it really helped. I'm just wondering if that's the only explanation for it. $\endgroup$ Feb 26, 2018 at 6:44
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$\begingroup$ I'm afraid that trying to quantify at what point the signals will start to overlap is probably quite hard from theoretical considerations alone. It makes sense for it to happen though; the further apart things are, the weaker they interact with each other. You might find it interesting to check the 13C NMR of a sequence of 1-functionalized alkanes, e.g. 1-hexanol, 1-heptanol, 1-octanol, 1-nonanol, 1-decanol, .... See, for example, here. $\endgroup$ Feb 27, 2018 at 9:13
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1$\begingroup$ Specifically, to get such precise peak positions, you would need to have a lifetime of the excited state of the spin system in the order of days. I.e. a T2 of many hours. That's not a limitation of the spectometer, but of the very sample itself. $\endgroup$– KarlOct 13 at 10:29
There is a very simple anwer: these methylene groups are practically chemically equivalent. It is impossible to make a specific reaction with one of them. Chemically equivalent hydrogens have the same chemical shift, and that's it.
Additionally, when two protons with similar CS couple, they create a higher order spectrum instead of two doubletts. The closer the CS are, the more the outer peaks get suppressed and the inner coalesce. Classic transition from AB to AA', the roof effect becomes so pronounced that you only see one peak, perhaps with a funny shape.
With 13C+decoupling, you seem to be slightly better off because there are no interactions (line splitting) visible. Still each line has an intrinsic bandwidth, because the relaxation times are finite, and the magnetic field has a finite homogenity. And the temperature distribution over the sample has a finite homogenity. Etc.
And of course you have in fact still interaction with surrounding protons, only they cancel out in the spectrum. Doesn't mean they're not there. Because otherwise you'd get relaxation times in the order of days with decoupling, which is not the case.