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October 28, 2007
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NMR N+1 Rule and # of peaks?

Using the N+1 Rule for finding multiplicities, I found that the number of adjacent protons for my problem came out to be 7, which meant that there would be 8 peaks.
My question is, is there a limit on how many peaks there are in NMR? My textbook only gives me examples of signals up to 7 peaks, so I'm unsure.

Here's what I'm working with: (CH3)2CHCH(CH3)2
Member since:
July 04, 2007
Total points:
156,484 (Level 7)

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Your N+1 will usually apply to hydrogen, but when it comes to the number of peaks, there probably is no limit. For instance, coupling from sulfur or phosphorous atoms results in a tremendous amount of complexity.

Then, you can get long range coupling through double bonds, that will give coupling.

At your probable level, just consider the N+1 as a general guide in the absence of other complications.

But realize that in a publication, there are reasons that a ppm range is given, with the description of multiplet.
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• Member since:
November 03, 2007
Total points:
82,762 (Level 7)
More than that! Each CH has got two kindsof hydrogen neighbour; CH and (CH3)2. These will have different coupling constants (why not, after all?)

So the CH proton will be split into seven peaks by the (CH3)2 protons.

I don't know whether you will see each of the seven peaks being split by the other CH, or whether this splitting won't show up because the two CH groups are identical. Ask your professor.

Anyway, you will either get 7 peaks, or 7 pairs of peaks.
• by Skip
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July 03, 2009
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5,695 (Level 5)
Be careful about which splitting H's are equivalent and which are not. The tertiary H on this symmetric molecule will not be split by the equivalent tertiary H next to it. So this compound should give a clean isopropyl group signal (at high resolution), with an upfield doublet (area 12) and a slightly downfield, clean septet (area 2).