Abstract
AbstractSodium borohydride reduction of 3‐methyl‐2,3‐dioxo‐7,8‐benzo‐3‐aza[3.3.3]propellan‐6‐one (1b) gave 2,4‐dioxo‐3‐methyl‐7,8‐benzo‐3‐aza[3.3.3]propellan‐6‐ol, while lithium aluminum hydride reduction gave 3‐methyl‐7,8‐benzo‐3‐aza[3.3.3]propellan‐6‐ol, which on oxidation, gave the corresponding ketone. This ketone formed the corresponding thioketal upon reaction with 1,2‐ethanedithiol. Raney nickel desulfurization of the thioketal provided 3‐methyl‐6,7‐benzo‐3‐aza[3.3.3]propellane. The same compound was also obtained in poor yield by forming the thioketal of Ib followed by lithium aluminum hydride reduction and Raney nickel desulfurization of the product. Desulfurization of the thioketal of Ib gave 2,4‐dioxo‐6,7‐Benzo‐3‐aza[3.3.3]propellane.
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