Abstract

Elimination from 3α-chloro-3β-methyl-(1) and 3β-chloro-3α-methyl-5α-cholestane (2) to give 3-methylene-5α-cholestane (exo-olefin) and 3-methyl-5α-cholest-2-ene (endo-olefin) has been investigated in ButOK–ButOH solution. The exo:endo olefin ratio is affected by the ButOK concentration but does not depend significantly upon the conformation of the chlorine substituent. This also holds for the reaction rates since the endo-olefin is formed at 100° only 1.7 fold faster from (1)(axial chlorine) than from (2)(equatorial chlorine). In contrast, elimination from 3α-chloro-5α-cholestane (3) is ca. 500 fold faster than from the equatorial isomer (4). The small difference between (1) and (2) in the rate of formation of the endo-olefin is due to compensation between enthalpy and entropy factors. A similar situation is also observed when the exo-olefin forming reactions from (1) and (2) are considered. The kinetic data and the activation parameters support the hypothesis that both olefins are formed from (2) in an anticoplanar process involving a skew-boat conformation.

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