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.)


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.


In addition to mad scientists answer I would like to explain what the quantum mechanical phenomenon is that leads to a J-Coupling. The interesting part about the J-coupling for chemists is that this coupling is a proof for the presence of an electron bond between the two coupling nuclei.

The coupling is based on two principles: a Fermi contact between the nucleus and an electron, the Pauli exclusion principle that two electrons must not have the same set of quantum numbers. In this case if one electron spin is up, then the other electron in this bonding orbital must be down. The other electron couples then to the second nucleus by another Fermi contact.

The chain of couplings is therefore:

Nucleus 1 ---Fermi contact---> electron 1 ----Pauli exclusion principle----> electron 2 ---Fermi contact----> Nucleus 2

This total chain is the effect which is called J-coupling. In a handwaving fashion one could say that through the first nucleus feels through the coupling if the second nucleus is up or down. This is shifting the energy levels a bit which can be then found in the spectrum as a shift in the resonance position. As the energy conservation principle holds one line is shifted to lower energies and the other by the same amount to higher energy leading to a doublet.

The OH proton is actually split as well by the J-coupling, however the line width of the OH resonance is bigger than the coupling constant, so the doublet is smeared out.

  • $\begingroup$ One could also talk about Hund's rules to further refine the description. $\endgroup$ – CHM Apr 26 '12 at 21:12
  • 1
    $\begingroup$ @CHM No that is not correct. Hund's rule gives guidance how to occupy orbitals in multi-electron atoms. The first rule is the Pauli principle. Other rules like having electron repulsion do not apply here. Furthermore the orbital I am talking about is an molecular orbital, because it describes the bond, not an atomic orbital. $\endgroup$ – GorillaPatch Apr 29 '12 at 19:40

While the above answers quite clearly explain the theory, I thought I would provide some clarification and citations:

If it is indeed propane you were taking the spectrum of, then you would get a heptet (7 peaks, in the standard pascals triangle pattern). Lets make another example though R-CH2-CH2-CH2-R'

If R = R' then you get a quintet (5 peaks) in a pascal's triangle. However, if R is not the same as R' then you will get a triplet of triplets, that is, three distinct triplets with the same J-coupling. This may or may not be easy to discern, depending on what R and R' are.

Here are some useful sources: These two papers can give you some great examples to examine, and are useful enough I always keep them on my USB key. NMR Chemical Shifts of Common Laboratory Solvents as Trace Impurities; Hugo E. Gottlieb, Vadim Kotlyar, and Abraham Nudelman; J. Org. Chem. 1997, 62, 7512-7515

NMR Chemical Shifts of Trace Impurities: Common Laboratory Solvents, Organics, and Gases in Deuterated Solvents Relevant to the Organometallic Chemist; Gregory R. Fulmer,Alexander J. M. Miller, Nathaniel H. Sherden, Hugo E. Gottlieb, Abraham Nudelman, Brian M. Stoltz, John E. Bercaw, and Karen I. Goldberg; Organometallics 2010, 29, 2176–2179

  • $\begingroup$ Is J-coupling equivalent or directly proportional to "chemical shift"? i.e. the x axis on a spectra? $\endgroup$ – Anti Earth Apr 30 '12 at 23:56
  • $\begingroup$ No, shift is the position of the peak, measured in ppm, J-coupling is the distance between two coupled peaks, i.e. how far the two peaks in a doublet are from one another, measured in Hz. $\endgroup$ – Canageek May 2 '12 at 5:29
  • $\begingroup$ Oh! So when you say "two distinct triplets with the same J coupling", there are two visible sets of triplets (obviously with the same distances between multiplets) or only 1 visible set of triplets (because the two sets have the same J coupling and so appear as 1 triplet)? Thanks (Sorry, still trying to articulate these ideas into the spectra) $\endgroup$ – Anti Earth May 2 '12 at 10:50
  • $\begingroup$ You will have 6 peaks, organized into two sets of three, forming 1:2:1 triplets. The distance between the center peak and the two flanking ones will be the same in both cases. $\endgroup$ – Canageek May 3 '12 at 17:48
  • $\begingroup$ @Canageek you write "...then you will get a triplet of triplets, that is, two distinct triplets..." was that a typo and should read "...you will get a doublet of triplets..."? $\endgroup$ – Reto Höhener Nov 1 '15 at 22:31

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