21.119 The following synthetic step was utilized as part of a synthesis of the polycyclic natural product haouamine B.15 This step represents both an electrophilic aromatic substitution and a Michael addition. In this reaction, the function of the first reagent (triflic anhydride) is to activate the Michael acceptor that is present in the starting compound (rendering it even more electrophilic), so that it can undergo an intramolecular Michael reaction. Draw a mechanism for this reaction and explain the observed stereochemistry of the newly formed chiral center. OMe MeO MeO F3C-S-O-S-CF3 Pyridine MeO OMe H OMe ZIR O=S=0 CF3
21.119 The following synthetic step was utilized as part of a synthesis of the polycyclic natural product haouamine B.15 This step represents both an electrophilic aromatic substitution and a Michael addition. In this reaction, the function of the first reagent (triflic anhydride) is to activate the Michael acceptor that is present in the starting compound (rendering it even more electrophilic), so that it can undergo an intramolecular Michael reaction. Draw a mechanism for this reaction and explain the observed stereochemistry of the newly formed chiral center. OMe MeO MeO F3C-S-O-S-CF3 Pyridine MeO OMe H OMe ZIR O=S=0 CF3
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Transcribed Image Text:21.119 The following synthetic step was utilized as part of a synthesis of the polycyclic natural
product haouamine B.15 This step represents both an electrophilic aromatic substitution and a
Michael addition. In this reaction, the function of the first reagent (triflic anhydride) is to activate
the Michael acceptor that is present in the starting compound (rendering it even more
electrophilic), so that it can undergo an intramolecular Michael reaction. Draw a mechanism for
this reaction and explain the observed stereochemistry of the newly formed chiral center.
OMe
MeO
MeO
F3C-S-O-S-CF3
Pyridine
OMe
MeO
OMe
O=S=0
5px
R
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