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Efficient Regio- and Stereoselective Formation of Azocan-2-ones via 8-Endo Cyclization of α-Carbamoyl Radicals

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journal contribution
posted on 24.02.2010, 00:00 by Xinqiang Fang, Kun Liu, Chaozhong Li
The iodine-atom-transfer 8-endo cyclization of α-carbamoyl radicals was investigated experimentally and theoretically. With the aid of Mg(ClO4)2 and a bis(oxazoline) ligand, N-ethoxycarbonyl-substituted N-(pent-4-enyl)-2-iodoalkanamides underwent 8-endo cyclization leading to the formation of only the corresponding 3,5-trans-substituted azocan-2-ones in excellent yields. Similarly, the BF3·OEt2/H2O-promoted reactions of N-ethoxycarbonyl-N-(2-allylaryl)-2-iodoalkanamides afforded exclusively the benzazocanone products with a 3,5-cis configuration in high yields. The bidentate chelation of substrate radicals not only significantly improved the efficiency of cyclization but also resulted in the change of stereochemistry of azocanone products from 3,8-trans to 3,8-cis. Theoretical calculations at the UB3LYP/6-31G* level revealed that the cyclization of N-carbonyl-substituted α-carbamoyl radicals occurs via the E-conformational transition states without the presence of a Lewis acid. On the other hand, the cyclization proceeds via the Z-conformational transition states when the substrates form the bidentate chelation with a Lewis acid. In both cases, the 8-endo cyclization is always fundamentally preferred over the corresponding 7-exo cyclization. The complexed radicals having the more rigid conformations also allow the better stereochemical control in the iodine-atom-abstraction step. To further understand the reactivity of α-carbamoyl radicals, the competition between the 8-endo and 5-exo cyclization was also studied. The results demonstrated that the 8-endo cyclization is of comparable rate to the corresponding 5-exo cyclization for α-carbamoyl radicals with fixed Z-conformational transition states. As a comparison, the 8-endo mode is fundamentally preferred over the 5-exo mode in the cyclization of NH-amide substrates because the latter requires the Z-conformational transition states, whereas the former proceeds via the more stable E-conformational transition states.