Cleavage reaction of O-methyldomesticine (I) with lithium in liquid ammonia afforded two kinds of decomposition bases, A and B. The base A was obtained as its hydrochloride of needles, m.p. 256°(decomp.), and the base was found to be identical with the hydrochloride of 2-methoxy-10-hydroxyaporphine (II) obtained by the action of sodium on (I) in liquid ammonia. Consequently, this reaction was also found to effect complete liberation of the methoxyl group in 1-position, indicating a rather interesting abnormal reaction to take place.
The base B (needles, m.p. 239°; C
17H
21O
2N) was found to be a monohydroxy-oxohexahydroaporphine. Wolff-Kischner reduction of this base gave a reduction product of the ketone (C
17H
23ON; [α]
D19+161.7°(MeOH), m.p. 210-212°) and its O-methyl ether (m.p. 57-59°; [α]
D20+148°(MeOH)). These substances were found to be identical with the synthesized (+)-10-hydroxy-1, 2, 3, 3a, 11b, 11c-hexahydroaporphine (VIIIa), m.p. 206-207°; [α]
D25+160.7°, and its O-methyl ether (IXa), m.p. 65-66°; [α]
D18.6+158.6°, by mixed melting point determination and infrared absorption spectrum (in Nujol or chloroform). The base B was therefore determined as 2-oxo-10-hydroxy-1, 2, 3, 3a, 11b, 11c-hexahydroaporphine (VI).
The same lithium-liquid ammonia reaction was carried out on the base A (II) and formation of the ketone compound (VI) was also noticed in this case. This has revealed that this cleavage reaction of O-methyldomesticine (I) first produces the intermediate compound (II) which undergoes reduction to form (VI).
As one of the model experiments of this reaction, lithium-liquid ammonia reaction was carried out on 2-methoxybiphenyl (X) and it was found that the benzene ring carrying the methoxyl group submitted to partial hydrogenation.
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