In my high school chemistry class, we learned that NMR only works for nuclei with odd mass number. Hence, we often use the $\ce{^1{H}}$ and $\ce{^{13}{C}}$ spectra. However, on this page about NMR I read, it says that $\ce{^{14}{N}}$ is also used, but with a mass number of 14, this evidently violates what we learned in class.

What am I missing here? Is the "odd mass number" thing just a simplification? If so, what's the actual criteria for whether or not something can be used for NMR?

  • 4
    $\begingroup$ A better way of saying this is that a nuclei with a non-zero spin is NMR active. This means that the nuclei must have an odd number of neutrons and/or protons. N-14 has a spin of 1, so it is NMR active. $\endgroup$
    – AChem
    Apr 24, 2022 at 2:45
  • $\begingroup$ See this link. This answers your question in detail. www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/… $\endgroup$
    – AChem
    Apr 24, 2022 at 2:47
  • 3
    $\begingroup$ An odd mass/nucleon number guarantees one of proton or neutron number is odd as well (odd + even = odd). Therefore the mandatory NMR condition of the nonzero nucleus spin is met. nitrogen-14 as the double odd nucleus (odd + odd = even) meets this condition by 2 ways. BTW nitrogen-14 is it the heaviest (2H, 6Li, 10B, 14N) stable odd-odd nucleus (with one exception of metastable isomer tantalum-180m1) $\endgroup$
    – Poutnik
    Apr 24, 2022 at 5:51
  • $\begingroup$ For an even simpler example, $^2\ce{H}$ NMR is also possible $\endgroup$
    – Andrew
    Apr 24, 2022 at 11:45
  • $\begingroup$ Your high school chemistry class was wrong. It's not NMR active if it has an even number of protons and an even number of neutrons. Otherwise, active. Having an even mass number does not imply this condition, but having an odd mass number immediately implies the opposite. $\endgroup$
    – Zhe
    Apr 25, 2022 at 15:10


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.