If you have $\ce{CH3-\color{\red}{CH2}-CH2Br}$

Then the $\ce{CH2}$ (in red) would have a peak splitting of $12$ because $(n+1)(m+1)$ where $n$ for the $\ce{CH3}$ is $(3+1=4)$ and m for $\ce{CH2}$ is $(2+1=3)$ so its $4\times3=12$


enter image description here

In the general case, a signal will be split into $(n + 1) \times (m + 1)$ peaks for an $\ce{H}$ atom that is coupled to a set of $n~\ce{H}$ atoms with one coupling constant, and to a set of $m~\ce{H}$ atoms with another coupling constant.

How can we determine differences in the coupling constant by just looking at the structure? Otherwise how do I know when to apply $n+1$ OR $(n+1)(m+1)$?

But in the other textbook it says:


The $\ce{CH2}$ (in red) has a peak splitting of $6$ because $(n+1)$ where $n=5$, $(5+1=6)$ and the $n$ is the $3~\ce{H}$ on $\ce{CH3}$ plus the $2~\ce{H}$ on $\ce{CH2}$.


enter image description here

Which is the correct method? Or are both right? and why?

Alternatively symmetric adjacent hydrogen groups:


Is the $\ce{CH2}$ split to 4 peaks due to symmetric adjacent hydrogen environments? Due to $n+1$, where $n=3$ so $(3+1=4)$. I read that it should be split into 7 peaks - but where are they getting 7 from - is it $n+1$, where $n=6$ so $(6+1=7)$. For symmetric adjacent groups do you allow $n$ to be the sum of all hydrogens adjacent/or not?

  • 1
    $\begingroup$ I am not an expert, but I remember the first rule to be exact, while the second rule is an approximation (meaning the neighbour hydrogens are almost indistinguishable). Given the right nmr, should should be able to resolve the 12 signals. In the (last) symmetric case you have six indistinguishable neighbour hydrogens, hence $n=6$ and a septett splitting. But I am far away from being an expert... (Very nice question, btw.) $\endgroup$ May 22, 2014 at 10:36

2 Answers 2


Both are correct to some degree.

The first book predicting 12 lines is being very theoretical. In principle with narrow enough line widths 12 lines would be observed. However, due to coupling constants (J) being similar from the beta protons to the alpha and gamma protons, in practice with reasonable resolution, only 6 lines are observed (the first book is actually explaining this in fine print).

enter image description here
(source: ucalgary.ca)

Credit: Univ. Calgary


The second text is merely reporting the observed number of peaks, not predicting the splitting pattern you would obtain from this molecule.

So, this is not a case of one being an exact rule, and one being an approximate rule. It is a case of one text describing how to determine the expected splitting pattern (which would be a quartet of triplets, or 12 peaks) and the other simply reporting how many peaks are observed (six peaks). If you understand the method explained in the first text, you can very easily rationalise why the expected 12 peak qt appears as a 6 line signal; if you merely follow the observations described in the second text, you will fail to understand complicated splitting patterns that provide valuable information about molecular structure in 'real' molecules. (No disrespect to C3H7X text book examples intended).

Both of these systems would commonly be referred to as a A3M2X2 system, representing $\ce{CH3CH2CH2X}$. To predict the splitting of the M2 group, simply consider the effect of each coupled group in turn. The A3 group will cause the M2 protons group to split into a quartet. Each of those quartet signals is then split due to coupling to the X2 protons, so each quartet signal becomes a triplet. this gives you the 12 lines, as so elegantly described in the first text.

However, if you then consider that the coupling constants between M and A, (JMA) and M and X (JMX) are roughly equivalent (let's say roughly 7.5Hz), then you get overlap of some of the 12 lines, and the resulting spectrum appears, roughly, as 6 lines. On a lower field instrument, or with poor resolution, this may be indeed what you see (first spectrum below, simulated at 200MHz). However at high field with good resolution (without having to resolution enhance the data), you can clearly see that there are in fact 12 lines, and the coupling constants are not quite equal (second spectrum, simulated at 800MHz). In fact, coupling constants are closer to JMA=7.8, and JMX=7.0.

Low field 1H spectrum of A3M2X2 spin system Simulated Low Field Spectrum of A3M2X2 Spin System (Simulated at 200MHz, 0.5Hz linewidth)

High field 1H spectrum of A3M2X2 spin system Simulated High Field Spectrum of A3M2X2 Spin System (Simulated at 800MHz, 0.3Hz linewidth)

As an added note, in both molecules, $\ce{CH3-CH2-CH2OH}$ and $\ce{CH3CH2CH2Br}$, the central $\ce{CH2}$ is prochiral. A prochiral centre is defined as one that will become chiral if a single substituent is changed. Prochirality is a very important property to consider, as it is very common for protons at a prochiral centre to be chemically and magnetically non-equivalent if there is a chiral centre elsewhere in the molecule. This often gives rise to methylene protons of ethyl sidechains appearing as non-equivalent groups in NMR.


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