Why is bond length $(a)$ greater than that of $(b)$?


Is this because $\ce{-NH}$ is electron donating group (+M) and this results in a higher electron density in the ring causing repulsion?

  • 3
    $\begingroup$ Are you sure? The second structure tautomerises to phenol(almost 100%), so the C-O bond becomes single, which should cause its bond length to be more. $\endgroup$ May 3, 2020 at 7:36
  • $\begingroup$ Firstly the tautomers of both of them, I think are aromatic. b becomes phenol, which is of course aromatic while a has a lone pair of nitrogen which makes it aromatic. So both of them have bond lengths greater than the regular length expected. But, when the resonance structures are made for a, you would find greater contribution of structures with single C-O bond than those with double bond. So, this may be a possible reason for that. $\endgroup$
    – KaBe2003
    May 3, 2020 at 7:40
  • 1
    $\begingroup$ @user93057 The first structure's tautomer is non-aromatic, since it still has an sp3 carbon. $\endgroup$ May 3, 2020 at 7:43
  • $\begingroup$ @NikhilAnand thanks for your point $\endgroup$
    – KaBe2003
    May 3, 2020 at 7:44
  • $\begingroup$ Tautomerisn should be let apart in such a discussion. The question does compare those two structures and not other molecules. The comments above are just related but dangerous as for they somehow push to treat tautomers as they are limiting forms. $\endgroup$
    – Alchimista
    May 3, 2020 at 9:09

1 Answer 1


So, although I stated in the comments that the second structure is very unstable and converts to phenol, as @Alchimista pointed out, tautomeric structures are different compounds and we need to solve the problem based on the compounds that are given to us, regardless of how stable they are.

However, resonance is operative in both of the above compounds. A compound such as this exists as a resonance hybrid, so we need to analyse the bond lengths in all the canonical structures and judge their relative contributions to the resonance hybrid.

Read this extract:

As noted above, we can more accurately describe the bonding in a molecule or polyatomic ion using the (weighted) average of its resonance structures. One model for estimating bond orders and charges in a compound is to simply take the average of those values from all (important) contributing resonance structures.

Note that here 'weighted' refers to giving resonance structures which are more stable more weight.

Try drawing out the resonance structures of both the compounds given above. Alternate resonance structures will have more contribution in the first as opposed to the second, as resonance in the second structure leads to an incomplete octet on the carbon which becomes a $\ce{C+}$, while in the first, only an $\ce{N+}$ results which is still a complete octet. So resonance structures in the first are more stable, and thus, the single bond character of the CO bond in the first structure increases more than in the second, and that causes its bond length to increase. So, the bond length of (a) is greater.

References: Resonance Structures, Chemlibre texts

  • $\begingroup$ There is no such thing as a more stable resonance structure. See the comment of Martin - マーチン in chemistry.stackexchange.com/questions/133010/…. $\endgroup$ May 3, 2020 at 16:08
  • $\begingroup$ Alright, what word do I use to indicate structures that 'have more contribution to the hybrid'? $\endgroup$ May 4, 2020 at 1:21
  • 1
    $\begingroup$ You can probably used "favored"! $\endgroup$ May 4, 2020 at 5:24

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