Stoichiometry of the reaction between horseradish peroxidase and p-cresol

Over a wide range of pH horseradish peroxidase compound I can be reduced quantitatively via compound II to the native enzyme by only 1 molar equivalent of p-cresol. Since 2 molar equivalents of electrons are required for the single turnover of the enzymatic cycle, p-cresol behaves as a 2-electron reductant. With p-cresol and compound I in a 1:1 ratio compound II and p-methylphenoxy radicals are obtained in the transient state. Compound II is then reduced to the native enzyme. A possible explanation for the facile reduction of compound II involves reaction with the dimerization product of these radicals, 1/2 molar equivalent of 2,2'-dihydroxy-5,5'-dimethylbiphenyl. If only 1/2 molar equivalent of p-cresol is present, than at high pH the reduction stops at compound II. The major steady state peroxidase oxidation product of p-cresol (with p-cresol in large excess compared to the enzyme concentration) is Pummerer's ketone. Pummerer's ketone is only reactive at pH values greater than about 9 where significant amounts of the enol can be formed via the enolate anion. Therefore, in alkaline solution it is reactive with compound I, but not with compound II, which is converted into an unreactive basic form. These results indicate that Pummerer's ketone cannot be the intermediate free radical product responsible for reducing compound II in the single turnover experiments. It is postulated that Pummerer's ketone is formed only in the steady state by the reaction of the p-methylphenoxy radical with excess p-cresol.