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Solutions to Problems 105 63. (a) There are three (C9, C13, C14): (b) CH₃O H H NCH₃ 13 H H 9 14 NCH₃ (c) For each of these stereocenters, the groups are drawn out as far as needed to reach the first point of difference. C9: C CH₂ H d C C Flip so that the lowest priority C * N b a group d (the H) is in the back. CH * C = S b a C13: Treat the benzene carbon as HC CH any carbon with a double bond: C * C C HC CH becomes CH₂ CH C CH₂ CH₂ C C HC CH and is C CH₂ CH₂ N therefore highest in priority over all the other groups. a a C Rotate d in back: = S again. b c d b c C14: C C H C CH Careful! The C on the right (attached to C C * N N, C, H) takes precedence over the C on C the left (attached to 3C's). At the first CH₂ C point of difference, N (circled) > C. d b a b a C III S again. c c Dextromethorphan therefore has the (9S, 13S, 14S) configuration.