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.
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...
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.