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Since the nuclear spin value of $\ce{^{14}N}$ is $I = 1$, we should expect a coupling between $\ce{^1H}$ and $\ce{^{14}N}$. Hence, the proton peak in a primary amine $(\ce{-NH2})$ should split in a triplet ($2 × 1 × 1 + 1 = 3$).

But generally that is not observed. Is this because of lower gyromagnetic ratio of $\ce{^{14}N}$ that leads to very less population difference between ground and excited state, or is there some other reason behind it?

andselisk
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Deepta Chattapadhyay
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  • Although not exactly the same question, my answer here addresses this issue, in particular the last paragraph, although one would likely need to read the whole thing. – orthocresol Apr 20 '19 at 09:57
  • The highly symmetric ammonium cation does in fact display ¹H-¹⁴N coupling, forming a nice sharp 1:1:1 triplet in the ¹H-NMR spectrum, as shown here. Once the source of much confusion, before an amazing a-ha moment! – Nicolau Saker Neto Apr 20 '19 at 11:16

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