1. (a) Explain why compound A is optically active while compound B is optically inactive at room temperature. •. .. P. PHCH2 H. Me Me А В (b) If optically active compound A is dissolved in water containing a small amount of H3O*, its optical rotation rapidly decreases in magnitude, eventually becoming 0. Provide an explanation for this observation, using structures and chemical reactions.

Organic Chemistry
8th Edition
ISBN:9781305580350
Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote
Publisher:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote
Chapter19: Enolate Anions And Enamines
Section: Chapter Questions
Problem 19.58P
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1. (a) Explain why compound A is optically active while compound B is optically inactive at room temperature.
•.
..
P.
PHCH2
H.
Me
Me
А
В
(b) If optically active compound A is dissolved in water containing a small amount of H3O*, its optical
rotation rapidly decreases in magnitude, eventually becoming 0. Provide an explanation for this
observation, using structures and chemical reactions.
Transcribed Image Text:1. (a) Explain why compound A is optically active while compound B is optically inactive at room temperature. •. .. P. PHCH2 H. Me Me А В (b) If optically active compound A is dissolved in water containing a small amount of H3O*, its optical rotation rapidly decreases in magnitude, eventually becoming 0. Provide an explanation for this observation, using structures and chemical reactions.
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