Abstract

The reactions of anthracene with lead tetra-acetate in benzene, benzene–pyridine, benzene–cyclohexane, and chloroform are shown to give cis- and trans-9,10-diacetoxy-9,10-dihydroanthracene [(1) and (2)]. In benzene containing increasing amounts of methanol these products are gradually replaced by cis- and trans-9,10-dimethoxy-9,10-dihydroanthracene [(3) and (4)] and cis- and trans-9-acetoxy-10-methoxy-9,10-dihydroanthracene [(5) and (6)], with a preference for the trans-isomers. The conclusion is drawn that in non-protic solvents the reaction involves an intermediate carbonium ion, thus giving equimolecular amounts of (1) and (2), whereas in the presence of methanol the methoxy-group is transferred by the lead atom. The second substituent group is introduced by an SN2 displacement.

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