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The question requires me to identify the decreasing order of the acidic strength of the following functional group. But I'm afraid I don't know how to distinguish one from the other since all the hydroxyl groups would be in resonance. What would be the next step in this case? I felt like the answer should start with (i) as there is a carbonyl carbon near and the delocalisation would be stabilised. enter image description here

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  • $\begingroup$ The (ii) OH has a seperate delocalisation and is not connected to the others so it will probably be the most acidic. Delocalisation hinders the acidity of (i). So, according to me, the order should be ii>iii>i. I may be incorrect though. $\endgroup$
    – Aditya Roychowdhury
    May 1, 2021 at 9:58
  • $\begingroup$ I don't understand what you mean by separate delocalization. Wouldn't (iii) OH exhibit this as well? $\endgroup$
    – ljm
    May 1, 2021 at 10:22
  • $\begingroup$ There are only two resonance structures for (ii) a keto and an enol. It would not be possible to draw extended delocalisation structures for ii. But the resonance of i and iii is connected. In other words two seperate electron clouds exist. An extended one for i and iii and another cloud in ii. $\endgroup$
    – Aditya Roychowdhury
    May 1, 2021 at 10:30
  • $\begingroup$ @Aditya Roychowdhury: The order is $iii \gt i \gt ii$. More the resonance structure of conjugaste base, more the acidity. $\endgroup$
    – Mathew Mahindaratne
    May 1, 2021 at 15:44
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    $\begingroup$ Well, as far as I'm concerned, only important thing about structure you present is that it's a non aromatic tautomer of polyphenol, which makes all this a pointless exercise. $\endgroup$
    – Mithoron
    May 1, 2021 at 19:19

1 Answer 1

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This question is, at best, flawed. The most acidic protons on the structure provided are not any of the labeled OH protons. Rather, the most acidic position is the one I have drawn a rectangle around below.

enter image description here

Deprotonation at this position produces an anion with a resonance contributor in which both rings are aromatic. The circa 150 kJ/mol resonance stabilization for each ring will vastly outweigh any delocalization that might occur for the oxyanions formed by deprotonating any of the OH groups.

enter image description here

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