1 (96)

Prévia do material em texto

82 Chapter 5 STEREOISOMERS Procedure: Highest priority is Cl, and lowest is H. A priority choice between the two outlined groups needs to be made, however. We write the groups out side-by-side: b b CH CH₃ CH₂ CH₃ H CH₃ H First point of difference Group 1 Group 2 Start at the carbon atom labeled "a" in each group. In both groups, the highest priority atom attached to car- bon "a" is oxygen, the second highest is carbon, and the last is hydrogen. No difference is found, so move out to the highest priority atom on carbon "a" for the next comparison: Follow the arrow and move to the oxy- gen (not carbon atom "b"-oxygen is higher priority, so it is evaluated first). Both groups 1 and 2 have iden- tical CH₂'s attached directly to O, so they are still tied. Turning to the second largest group attached to "a," we look to carbon "b," and see that here the tie can be broken: In group 1, this atom is attached to two carbons and one hydrogen, whereas in group 2, it is attached to just one carbon and two hydrogens. Group 1 therefore is higher in priority than group 2 at atom "b," the first point of difference. With priorities done, the groups can all be labeled and R or S assigned: b b C III C d (d in back) c a c a Clockwise = R If, on the other hand, the oxygen in group 1 were attached to an H instead of a C, that would become the first point of difference, and group 2 would be higher in priority. The differences at the "b" carbons would become irrelevant. Lower Higher C CH CH₃ VS. H CH₃ H Group 1 Group 2 The above three examples have each been designed to contain a "trick"-an unusual feature that is not often encountered but illustrates the application of the rules in detail. Most chapter and exam problems you see will not be "tricky." However, by understanding the toughies, proper application of the procedure be- comes quicker and simpler in all cases, because you've now seen what to do when things get complicated.