Why does $\ce{CH2NH2}$ have lower priority than $\ce{CN}$? The carbon on the first molecule is bonded to (N,H,H) while the carbon on the second molecule is just bonded to (N). I can't seem to find this scenario anywhere online.


1 Answer 1


Each multiple bond is considered to be equivalent to an individual bond. So $\ce{-CN}$ has the same priority as (N,N,N), making it higher than (N,H,H). Read more about it here.

  • $\begingroup$ Thanks for the quick reply. Also, I have one more question. In the case of C≡C-H vs C(CH3)3, why does the first one have higher priority than the second? They are both the same two branches out, but then the first one becomes (H) vs the second one which is (H,H,H,H,H,H,H,H,H). If two branches contain the same elements but just a different number of them, does the shorter one automatically get priority? Thanks. $\endgroup$
    – Jon
    Jul 30, 2018 at 20:18
  • 2
    $\begingroup$ Because the terminal carbon is considered as being equivalently bonded to (C,C,H), while the methyl groups are all (H,H,H). $\endgroup$
    – ringo
    Jul 30, 2018 at 20:28
  • 2
    $\begingroup$ Important to note that the additional N's on CN are phantom. Treat them with the same priority as N, but compared to a real N, it has lower priority. $\endgroup$
    – Zhe
    Jul 30, 2018 at 20:48

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