Answer to Question #123684 in General Chemistry for 250300183747

Question

Consider that (S)-bromobutane has a specific rotation of +23.1degree
and (R)-bromobutane has a specific rotation of -23.1degree
.
a. Determine the optical purity of a racemic mixture
b. Which isomer is dominant and what is the optical purity of a mixture, of
(R)- and (S)-bromobutane, whose specific rotation was found to be -9.2degree?
c. What is the percent composition of the mixture?

Answer 1

a)The specific rotation, [a], of the racemate is expected to be 0, since the effect of one enantiomer cancel the other out, molecule for molecule.

Optical purity, % = 100 [a]_mixture / [a]_{pure sample}

                           = 100 (0) / +23.1^o

                           = 0%

b) The negative sign tells indicates that the R enantiomer is the dominant one.

Optical purity, % = 100 [a]_mixture / [a]_{pure sample}

                           = 100 (-9.2) / -23.1^o

                           = 40%  it means 40% excess oc R over S

c)The 60% leftover, which is optically inactive, must be equal amounts of both (R)- and (S)-bromobutane. The excess 40% is all R so there is a total of 70% (R) and 30% (S)