1 (126)

Prévia do material em texto

96 Chapter 5 STEREOISOMERS C(OCH₃)₂ is the issue. The relevant comparison is C O , and the C CH₃ second one is higher by virtue of having carbon atoms attached to both oxygens in the functional group. Answer is R. 42. Letters correspond to compounds in Problem 33. Consider only (b), (e), (j), and (m), the ones that are chiral. Since you are given a choice of which configuration to draw at each stereocenter, draw the one that has the lowest priority group going "away" from you-on a hashed bond-to make assignment of R and S easier. a Br (b) (e) * C d H * C d BrCH₂ b c CH₃ H CH₃ c bCH₂CH₃ R R (j) b CH₂NHCH₃ Notice that CH₂N beats the ring, which is ranked as CC₃, because N>C at the * C d HO first point of difference. c H OH a HO S c (m) CH₂OH CH₂OH d C * H Evaluate each * C b O CHCO > chiral carbon H CH H separately: HO a C S CH₂OH a HOCH O and b H d c S 43. CH₃ 2nd (because The (S) enantiomer (note priorities) of 0 atom) c b 3rd H b C 1st C III (S) CH₃ a a Conversely, (-) -carvone has the R configuration.