1559Tch0570-9810/22/0520:19.Pag972 ⊕ 72.Chapter 5 STEREOISOMERS a point of difference.If no difference is found,then examine the second highest priority group.and It is actually easier to do than it is to describe.so let's analyze three examples Example:Determine R-S designation for CCls H- CHa CHaBr Procedure:H is obviously lowest priority:the other three need to be compared.Write them out side-by-side: @ -C1 一H一C一H -CH:Br to atomic number on each carbon.In -CCl it is Cl:in .All three are different therefore priorities can be assiged imme that ther 0 ⊕ the first Cl on -CCl and Br is "higger ☒ redraw the molecule with the highest priority group designated"a. becomes (d in back) CH.CH2B Counterclockwise S Erample:Determine R-S designation for H :The sr is a member of a ring.but the isnot really different.The H is lowest up is ethyl:that's iu eally just the quence of ring atoms attached to the stere start at the ringC(CHand move around the ring in the othercoo enter.For one group,start at the ring -CHz-CH3. -CH-C(CHa)-etc..and -C(CHa)-CH,-ctc.72 • Chapter 5 STEREOISOMERS a point of difference. If no difference is found, then examine the second highest priority group, and so on. It is actually easier to do than it is to describe, so let’s analyze three examples. Example: Determine R–S designation for Procedure: H is obviously lowest priority; the other three need to be compared. Write them out side-by-side: Identify the highest priority atom according to atomic number on each carbon. In OCCl3, it is Cl; in OCH2Br, it is Br; and in OCH3, it is H. All three are different; therefore priorities can be assigned imme￾diately. Br
Cl
H, therefore OCH2Br
OCCl3
OCH3. The fact that there are three Cl’s on OCCl3 and only one Br on OCH2Br is irrelevant. The first point of difference is the Br on OCH2Br vs. the first Cl on OCCl3, and Br is “bigger” than Cl. Once the first point of difference is identified, nothing else matters. With priorities now assigned, we can redraw the molecule with the highest priority group designated “a,” the second “b,” the third “c,” and the lowest “d”: Example: Determine R–S designation for Procedure: The stereocenter is a member of a ring, but the procedure is not really different. The H is lowest priority, and the other three groups have to be compared. The problem is interpreting what it means to have the stereocenter in a ring. One group is ethyl; that’s obvious. The other two “groups” are really just the se￾quence of ring atoms attached to the stereocenter. For one group, start at the ring OCH2 and move around the ring; for the other, start at the ring OC(CH3)2, and move around the ring in the other direction. So, we com￾pare the “groups” OCH2OCH3, OCH2OC(CH3)2Oetc., and OC(CH3)2OCH2Oetc. C CH2CH3 (CH3)2C H2C H C CH2Br H CH3 CCl3 becomes C
d c a b C c a b (d in back) Counterclockwise
S C Cl Cl Cl –CCl3 C Br H H –CH2Br C H H H –CH3 H C CH2Br CH3 CCl3 1559T_ch05_70-98 10/22/05 20:19 Page 72