Chapter 7 : Stereochemistry |
1. Given that (S)-bromobutane has a specific rotation of +23.1o and (R)-bromobutane has a specific rotation of -23.1o, what is the optical purity and % composition of a mixture whose specific rotation was found to be +18.4o?
The positive sign indicates that the (S)-isomer is in excess.
Optical purity, %
= 100 [a]mixture / [a]pure
sample
= 100 (+18.4) / +23.1o
= 80% this indicates a 80% excess of S over R!
The 20% leftover, which is optically inactive, must be equal amounts of both (R)- and (S)-bromobutane. The excess 90% is all S so there is a total of 10% (R) and 90% (S).
© Dr. Ian Hunt, Department of Chemistry |