This reaction (http://orgsyn.org/demo.aspx?prep=CV1P0476) gives quinizarin as the major product in about 70% yield.

However, when the same reaction is conducted with phenol insteas of p-chlorophenol, the product we get is not 1-hydroxyantraquinone (as would be expected if we only knew the above reaction), but phenolphthalein.

Why does phenolphtalein form instead of 1-hydroxyantraquinone? Would 1-hydroxyanthraquinone form if we added only 1 mol of phenol instead of 2 in the phenolphthalein synthesis?


1 Answer 1


In the Org. Syn. link you provided phthalic anhydride 1 and p-chlorophenol 2 in 95% H2SO4 and boric acid at 200oC undergo condensation and cyclization. Phthalic anhydride 1 under acid catalysis reacts with p-chlorophenol 2 at the ortho position as shown below to form ketoacid 3. Chlorine is a weak o,p-director and boric acid, I believe, is complexing with the phenolic hydroxyl and the ketone to reduce bond rotation and aid in cyclization to form anthroquinone 4. There is an aqueous work-up at this point followed by treatment with hot aqueous KOH which forms quinizarin 5 by nucleophilic aromatic substitution which is aided by the adjacent keto group.

In the reaction of phthalic anhydride 1 with phenol 7 in the presence of sulfuric acid, (presumably no boric acid), cyclization is repressed because of a lack of electron density at the phenol meta-position in 6. In addition, protonation of the ketone in 6 gives extended conjugation with the phenolic hydroxyl. Preferrable addition of phenol to the ketone of 6 leads to 8 and on to phenolphthalein 9.

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  • $\begingroup$ from 4 to 5 , Cl para to hydroxyl group is undergoing nucleophillic aromatic substitution. Aryl halides are known to undergo nucleophillic aromatic substitution if nitro group is at ortho or para position. While this us not the case a mechanistic apparoch would help. $\endgroup$ Commented Dec 10, 2019 at 14:13
  • $\begingroup$ There is a ketone ortho to the Cl, which is the reason why the nucleophilic substitution takes place, or am I wrong? $\endgroup$
    – FusRoDah
    Commented Dec 10, 2019 at 15:50
  • $\begingroup$ @FusRoDah; You are correct! The base conditions are fairly vigorous. $\endgroup$
    – user55119
    Commented Dec 10, 2019 at 19:20
  • $\begingroup$ @Chakravarthy Kalyan: The mechanism of OH for Cl substitution is straightforward. I'll entertain a better suggestion. $\endgroup$
    – user55119
    Commented Dec 10, 2019 at 19:23

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