Journal of the American Chemical Society p. 7403 - 7407 (1992)
Update date:2022-08-11
Topics:
Thibblin, Alf
Sidhu, Harvinder
Solvolysis of 1,1-diphenyl-1-chloroethane (1-Cl) in acetonitrile at 25°C provides the elimination product 1,1-diphenylethene (3). Addition of water or methanol to the acetonitrile increases the rate of elimination and gives rise to a second product, 1,1-diphenyl-l-hydroxyethane (1-OH) or 1,1-diphenyl-l-methoxyethane (1-OMe). The deuteriated analogue, 1,1- diphenyl-l-chloro[2,2,2-2H3]ethane (d-1-Cl), reacts slower. The overall kinetic deuterium isotope effect was found to decrease from (k12H + k13H)/(k12D + k13D) = 1.73 in pure acetonitrile to 1.34 in 20 vol % methanol in acetonitrile. This isotope effect is composed of a substitution isotope effect k12H/k12D and an elimination isotope effect k13H/k13D, which simultaneously change from 0.84 to 0.96 and from 2.2 to 3.2, respectively, when the methanol concentration is increased from 1.96 to 9.09 vol %. These results indicate a branched mechanism involving an almost rate-limiting formation of a common carbocationic intermediate. The acid-catalyzed solvolysis of 1,1-diphenyl-l-phenoxyethane (1-OPh) in acetonitrile containing 0.4 vol % aqueous 2 M sulfuric acid quickly provides the alcohol 1-OH, k12/k13 ≈ 11, which in a subsequent fast reaction yields the olefin 3 as the final product. In contrast, the solvolysis of the chloride 1-Cl in 0.4 vol % water in acetonitrile yields almost exclusively olefin 3, k12/k12 = 0.003. The results strongly indicate that the leaving chloride anion is involved in the elimination process and that the common intermediate is the contact ion pair. The isotope effect of the acid-catalyzed elimination of the alcohol 1-OH in 0.4 vol % water in acetonitrile is k13H/k13D = 5.3, which increases to k13H/k13D = 6.5 in 25 vol % acetonitrile in water. The possibility of an ion-molecule pair intermediate in the acid-catalyzed solvolysis is discussed.
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