Why is ethanoic acid more acidic than phthalimide? $\mathrm{p}K_\mathrm{a}$ of phthalimide is 8.3 (source), while that of ethanoic acid is 4.76 (source).

Applying the logic found here would suggest the opposite. This is because both conjugate bases delocalize their negative charge, but while ethanoic acid has two oxygens to delocalize its charge among, phthalimide has two oxygens and a nitrogen (and a benzene ring); it seems that the quality of the delocalization is the same, but the quantity favors phthalimide (as the answers to the question put it). As for the other factors that affect acidity (inductive effects, hybridization, atom size/electronegativity, formal charge, etc.) none seem to come into play.


1 Answer 1


The deprotonated phthalimide anion is indeed highly delocalized, but the neutral molecule with the N-H proton reattached retains most of that delocalization. Protonating acetate ion back to acetic acid causes a much greater loss of delocalization.

  • 2
    $\begingroup$ I don't get this. Even the neutral acetic acid molecule shows delocalisation, doesn't it? $\endgroup$ Apr 9, 2018 at 2:48
  • $\begingroup$ Yes, that's another point I was thinking about, nitrogen donates it's electron density much more eagerly then oxygen, so energy gain between delocalisation of ftalimide and it's anion is lower. $\endgroup$
    – Mithoron
    Apr 9, 2018 at 17:27
  • $\begingroup$ I asked a TA about this, and they explained it the following way: in both cases, when the hydrogen is taken away, the "new" electrons go into an sp2 orbital. This is because the protonated forms of both molecules already have resonance, and therefore both already have filled p orbitals. The "new" sp2 electrons don't participate in resonance, so the electronegativity difference between nitrogen and oxygen makes the difference. Since oxygen is more electronegative, the deprotonated form of ethanoic acid is actually more stable. $\endgroup$ Apr 10, 2018 at 22:45
  • $\begingroup$ Ok, can whoever downvoted please provide an alternative, better answer? Thank you. $\endgroup$ Jul 14, 2018 at 12:17

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