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7-65
tert-Butyl esters [RCO₂C(CH3)3] are converted into carboxylic acids
(RCO₂H) by reaction with trifluoroacetic acid, a reaction useful in pro-
tein synthesis (Section 26-7). Assign E,Z designation to the double
bonds of both reactant and product in the following scheme, and explain
why there is an apparent change in double-bond stereochemistry:
H
C-OCH3
C-OCH3
CF3CO2H
C=C
+ H2C=C(CH3)2
H3C
C-OC(CH3)3
C-OH
0-
H3C

7.65 Treatment of the tert-butyl ester with trifluoroacetic acid cleaves the -OC(CH3)3 group and
replaces it with an -OH group, which has a lower priority than the -OCH3 group on the
upper carbon and the -OC(CH3)3 group that was removed. The result is a change in the E,Z
designation around the double bond without breaking any of the bonds attached to the
double-bond carbons.
O
Low H
Low
C-OCH3 Low
H
C-OCH3 High
CF3COOH
C=C
High H3C
-OC(CH3)3 High
High H3C
C-OH Low
O=

Answer 1

✨ 最佳解答 ✨

Ann

原文翻譯
反應物的丙醇酯會被轉變成醋酸基團
請寫下反應物和生成物是E還是Z
並說明為何有這樣的變化
答案
E Z是用雙鍵上分子量的大小去比的
如果低分子量都在同一側 則為Z
而為什麼會又Z變成E
是因為不用斷掉雙鍵