Which compound has larger $\ce{C-N}$ bond length, 4-aminobenzaldehyde (1) or aniline (2)?


I think it's the second compound. $\ce{-CHO}$ is an electron withdrawing group. This means the electron density of the ring decreases. So the carbon attached to nitrogen becomes more electronegative. Hence its p-character decreases and the $\ce{C-N}$ bond length decreases.


You think right, but for the wrong reason. The inductive effect is not that great and does not quite reach that far. It is the effect of conjugated $\pi$-system that really matters. See, we may draw an alternative resonance structure for I with minus on O, plus on N, and rearranged double bonds (quinone-like ring instead of aromatic). This means the C-N bond in I is kind of "partially double" (i.e., has a noticeable $\pi$ component), hence it is shorter.

Resonance structures

Indeed, there is some conjugation between the aromatic ring and nitrogen's lone pair in unsubstituted aniline as well. That's what makes aniline a weaker base than ammonia, and its aromatic ring more reactive towards electrophiles than benzene. It is just that the $\ce{-CH=O}$ group in para position makes that conjugation stronger yet, as we may see qualitatively by drawing the resonance structures, or (supposedly) quantitatively by means of quantum chemistry.

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