# Why ¹H–¹⁴N coupling is not observed in ¹H NMR

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?

• 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. Apr 20 '19 at 9:57
• 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! Apr 20 '19 at 11:16