Solutions_Manual_for_Organic_Chemistry_6th_Ed 111

Solutions_Manual_for_Organic_Chemistry_6th_Ed 111 - 6-20...

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Unformatted text preview: 6-20 continued H3C _ (d) c I� ,� H" � CH3 -: SH F Br •• CH3CH2 : Br: .. H .. H H 1 + : Br: .. - c::::> H H CH3 + + I CH} CH2CH} (e) CH3 � : - �S �/ � H ,Br . H3CO , � R � ,� , H + r .B• . • ;··· c::::> (f) H- �: H I H �I �� - H C � ID C;CI ---II"� + :CJ : 6-2 1 (a) The best leaving groups are the weakest bases. Bromide ion is so weak it is not considered at all basic; it is an excellent leaving group. Fluoride is moderately basic, by far the most basic of the halides. It is a terrible leaving group. Bromide is many orders of magnitude better than fluoride in leaving group ability. (b) inverted, but still named S H"� "'" : CH} . F : H ,I �� CH} I OCH3 I + : Br: .. transition state (c) As noted on the structure above, the configuration is inverted even though the designations o f the configuration for both the starting material and the product are S; the oxygen of the product has a lower priority than the bromine it replaces. Refer to the solution to problem 5-38, p. 97, for the caution about con fusing absolute configuration with the designation of configuration. Cd) The result is perfectly consistent with the SN2 mechanis m. Even though both the reactant and the product have the S designation, the con figuration has been inverted: the nomenclature priority of fluorine changes from second (after bromine) in the reactant to first (before oxygen) in the product. While the designation may be misleading, the structure shows with certainty that an inversion has occurred. 104 ...
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This note was uploaded on 02/27/2010 for the course CHEM 140 taught by Professor Wade during the Spring '10 term at Whitman.

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