nmr spectroscopy - Splitting of multiplets in ¹H NMR spectra?

I'm currently taking VCE (Victorian Certificate of Education) Chemistry classes, and we're currently studying the interpretation of spectra produced by Hydrogen NMR (Nuclear Magnetic Resonance) spectroscopy.

When studying the spectra of High Resolution ¹H NMR, the peaks representing the different Hydrogen environments are split into multiplets based on the protons surrounding these environments. There has been a considerable amount of confusion in my classes over the actual principles / rules of thumb on how to calculate the multiplets for a particular environment of a known chemical (i.e.; known structue), based off the 'n + 1' rule.
(i.e.; an environment with n neighbours will be split into n+1 multiplets).

We are absolute on the principles that;

• Peaks of a particular Hydrogen environment are not split by neighbouring protons in equivalent environments.


• OH does not, and is not split by, it's neighbouring environments.

However, immense confusion arised over whether the following principle was correct.

• Peaks of a particular Hydrogen environment will only be split by the protons in neighbouring environments once for each type of neighbouring environment.
  e.g. The middle "CH₂" environment in "CH₃–CH₂–CH₃" will only have 4 peaks;
  Although it has 6 neighbouring protons, they are two lots of the same environment (CH₃).

As a class, we found numerous examples from different text books and sources that provide examples of ¹H NMR spectra which did not clarify the matter; Some considered all neighbouring protons as neighbours, others discriminated on the repeated neighbouring environments. For example, the CH₂ in CH₃–CH₂–CH₃ was sometimes split into 4 peaks or 7 peaks, depending on the source.

Many Chemistry teachers contradicted each other on the matter.
There was repeated self corrections made by the teachers, such that now nobody really knows whether this principle is correct or not.

So, is there anybody that has the correct information on the matter?
Is there a reasonable explanation behind this strange lack of correlation,
or is there a common misconception about multiplet splitting?

Ultimately; How many multiplets should the CH₂ in CH₃–CH₂–CH₃ have?

(Note that if there is a complicated explanation that I am currently only at Year 12 VCE level, so links to resources I can pursue would be extremely helpful! It's been established that VCAA (an authority for the education system in Australia) ensures that the chemicals featured in the exams for NMR analysis will not be of a structure so as to allow the ambiguity above.)

Answer

Which nuclei you should consider as neighboring atoms for the multiplet splitting depends on the size of the coupling constant. The coupling constant are usually named after the number of bonds between the coupling nuclei, so a coupling between two hydrogens that are three bonds apart from each other would be a $\mathrm{^3J_{HH}}$-coupling. In an alkyl chain like CH₃CH₂ the coupling between the CH2 and the CH3 hydrogens would be a $\mathrm{^3J}$-coupling that is about 7 Hz large. Couplings over more than three bonds are usually not observed, because they are smaller than the linewidth of the signals in your NMR spectrum. The exception to that you're likely to encounter are couplings along C=C double bonds, where you can even see $\mathrm{^4J}$-couplings or more.

So the simplified rule would be that couplings along three bonds are visible, couplings along more than three bonds are only visible when there is at least one C=C double bond along the way.

The concepts of chemical and magnetic equivalence are essential to understanding how different multiplicities arise. Chemical equivalence means there exists a symmetry operation that exchanges those nuclei. Magnetically equivalent nuclei additionally need to have identical couplings to all other spins in the molecule. Magnetically equivalent nuclei don't couple to each other.

In propane the hydrogens of both CH₃ groups are three bonds away from the hydrogens of the CH₂ group, so the coupling between those is visible in the NMR spectrum. The hydrogens of each CH₃ group are magnetically equivalent, due to the fast rotation along the C–C bond, and the two CH₃ groups should also be magnetically equivalant. So you have 6 magnetically equivalent hydrogens that couple to your CH₂ hydrogens. The result of that is a splitting into a septet (7).

The OH-group is an interesting exception, as you would expect it to lead to a visible coupling on hydrogens connected to the same carbon, but you don't observe that under most conditions. The reason is that the OH is acidic enough that the hydrogen exchanges quickly with the solvent, so the hydrogen dissasociates and associates quickly. This happens too fast for NMR, so the other nuclei only see the average OH-hydrogen. This eliminates the coupling to the OH, and it is also the reason why the OH-signal is often very broad or even completely gone in NMR spectra.