Chemistry-analytical-analysing the n+1 rule-multiplicity molecular structure-complex H NMR splitting

Multiplicities- NMR
For the questions that the VCE course will ask of students the n+1 rule is sufficient, but, it is worth noting that this is a very simplified rule and a basic attempt is offered below to try and show a little of the complexity that lies behind the NMR spectroscopy. Each neighbour that is not chemically equivalent to the proton producing the signal splits the proton's signal
When a given nucleus has its signal split by two or more sets of nuclei the peaks are separated by a distance called the coupling constant, and given the symbol J. If neighbouring nuclei are chemically different, then each peak in the signal is separated by different J values and the n+1 rule fails to predict the entire splitting pattern. No coupled hydrogens One coupled hydrogen Two coupled hydrogens Three coupled hydrogens
Splitting is multiplicative. Take the example of 1,1,2 trichloropropane. Its H NMR is shown on the right. The hydrogen shown in the red has it signal split into 8 peaks or in other words a doublet of quartets. This doublet of quartets occurs when the green hydrogens, as pictured on the right, split the red signal into a quartet then the blue hydrogen further splits this quartet in a doublet of quartets having 8 peaks. In other words (n+1) X (n+1) = 2 X 4 = 8 peaks.
Sometimes the resolution of the spectrum is not high enough to show the extra peaks and appears as though the n+1 rule works nicely. See below. The middle CH₂ group is surrounded by two CH₂ that are not chemically equivalent hence the splitting should produce 9 peaks (3 X 3 =9). The spectrum on the other hand shows 5peaks that conveniently fit into out naive n+1 rule. A closer examination of the spectrum at a higher resolution reveals the 9 peaks.

Take pentane for example. The H NMR spectrum agrees with the n+1 rule, eventhough the blue CH₂ has two groups of hydrogens on each side that are chemically different. This is because the difference in the coupling constant of each group, the red CH₃ and the green CH₂ is so small that it is irrelevant and they appear as if they are five similar hydrogens producing 6 peaks.
1) Draw the structure of 1-chloropentane 2) How many peaks will the signal from the hydrogens on the 3rd carbon be split into? solution
3) Consider the molecule CH₂BrCH₂CH₂Cl a) How many sets of peaks will be present in the H NMR spectrum? b) Describe these sets of peaks