Abstract
Cleavage reaction with metallic sodium in liquid ammonia was carried out on O-methyldomesticine (III) and only one kind of phenolic base was obtained as the degradation product, isolated as its hydrochloride of needles, m.p. 255-256° (decomp.). This substance did not possess a methylenedioxy group and had only one methoxyl, its analytical values agreeing with the formula of C18H19O2N⋅HCl. The hydrochloride, m.p. 233-234°(decomp.), of its O-methylated compound possessed two methoxyls and its analytical values agreed with C19H21O2N⋅HCl, there being no evidence for the presence of phenolic hydroxyl in its infrared spectrum. It follows, therefore, that the cleaved base is monomethoxymonohydroxy-aporphine (IV) and its methyl ether would be dimethoxyaporphine (V).
Hofmann degradation of (V) by the route shown in Chart 1, followed by oxidation and decarboxylation afforded dimethoxyphenanthrene (XI) which was identified through admixture and comparison of infrared spectra to be 3, 6-dimethoxyphenanthrene (XII). Consequently, the cleaved phenolic base (IV) obtained by the cleavage of O-methyldomesticine with metallic sodium in liquid ammonia is 2-hydroxy-10-methoxyaporphine (XIII) and its O-methylated compound (V) is 2, 10-dimethoxyaporphine (XIV).
It is considered from these results that the methylenedioxy group in (III) is cleaved at 9-position by this reaction to produce a fresh hydroxyl in 10-position and the methoxyl in 1-positron is completely liberated. In general, the methoxyl bonded to the benzene ring is known not to change by cleavage reaction with alkali metal in liquid ammonia at a comparatively low temperature or undergoes demethylation in some cases to produce a new phenolic hydroxyl. Therefore, the loss of an aromatic methoxyl group by this cleavage reaction observed in the present series of experiments is an interesting new phenomenon.
