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Consider the following order of acidity based on pKa values:-

ethanoic acid > phthalimide >phenol >water >acetamide

Looking towards similar questions on stack exchange such as this , the recurring theme is that both quality(the electronegativities of the atoms on which charge is delocalized)as well as quantity of resonance(the number of resonating structures) should be seen when determining the stability of conjugate base and quality is given precedence over quantity. However, this doesn't seem to work here. Let me categorize the conjugate bases of the aforementioned compounds:-

  1. Ethanoic Acid- Delocalisation over 2 oxygen atoms
  2. Phtalimide- Delocalisation over 2 oxygen atoms and 1 nitrogen atom
  3. Phenol-Delocalisation over 1 oxygen atom and 3 carbon atoms
  4. Water-Localised over 1 oxygen atom
  5. Acetamide- Delocalisation over 1 oxygen atom and 1 nitrogen atom

Now, based on the linked question, we get ethanoic acid > phenol > water where we have seen quality of resonance first then quantity. However, this doesn't work for phtalimide which should be placed above carboxylic acid and also acetamide which should be placed above water and possibly even phenol.

So how would you explain this order?

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    $\begingroup$ I think you have this whole thing backwards. From the experimental acidity, we give these rule-of-thumb explanations, which fit to varying degrees. You shouldn't expect the other way around to work in general. New chemistry students get too caught up in these rules and overestimate their explanatory power and precision. $\endgroup$ Jun 11 at 6:42
  • $\begingroup$ @NicolauSakerNeto Agreed, but there must be some explanation given for experimental results, right? Is the explanation in this case too advanced(like based on the schridinger equation) or can it be explained by basic principles? $\endgroup$
    – user121185
    Jun 11 at 6:53
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    $\begingroup$ Indeed. The accurate computation of substance acidities purely from first principles, especially outside of the gas phase, is still a matter of active research. $\endgroup$ Jun 11 at 7:24
  • $\begingroup$ chemistry.stackexchange.com/questions/94870/… $\endgroup$
    – Mithoron
    Jun 11 at 12:03
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    $\begingroup$ I assume it the matter of stronger C=O then C=N $\endgroup$
    – Mithoron
    Jun 11 at 12:41

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