Abstract

2-Bromoadamantane (2-BrAd) reacted in liquid ammonia under irradiation with diphenylphosphide (Ph(2)P(-)) ions whereas 2-chloroadamantane (2-ClAd) did not under the same experimental conditions. However, 2-ClAd yielded 2-(trimethylstannyl)adamantane in its photostimulated reaction with trimethylstannyl (Me(3)Sn(-)) ions. The compound 1-ClAd yielded the substitution product in a photostimulated slow reaction when the nucleophile is Ph(2)P(-) ion; the reaction occurs faster with the nucleophile Me(3)Sn(-) ion. All these reactions can be explained by the S(RN)1 mechanism as they did not occur in the dark and were inhibited by p-dinitrobenzene when photostimulated. In competition experiments, 1-haloadamantane showed more reactivity than 2-haloadamantane. Either with Ph(2)P(-) or Me(3)Sn(-) ions, 1-BrAd is 1.4 times more reactive than 2-BrAd while 1-ClAd is 12 times more reactive than 2-ClAd with Me(3)Sn(-) ions. In the photostimulated reaction of 1,2-dichloroadamantane (7) with Ph(2)P(-) the monosubstitution products 1-adamantyldiphenylphosphine (64%) and 2-adamantyldiphenylphosphine (15%) were formed, isolated as the oxides. From these results, it appears that when 7 receives an electron, the 1-position fragments ca. four times faster than the 2-position. The disubstitution product was not formed with Ph(2)P(-) ions, but when 7 reacted with a nucleophile having less steric bulk such as a Me(3)Sn(-) ion, the 2-chloro-1-(trimethylstannyl)adamantane and the disubstitution product 1,2-bis(trimethylstannyl)adamantane were formed. The formation of these products is explained in terms of the different rates of the electron transfer reactions of the radical anion intermediates.

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