Chapter 7 : Stereochemistry

Consider that (S)-bromobutane has a specific rotation of +23.1^o and (R)-bromobutane has a specific rotation of -23.1^o
 

Question:  Determine the optical purity of a racemic mixture.

Answer:  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%

Question:  Determine the enantiomeric excess of the racemic mixture.

Answer:  You would expect [R] = [S] = 50%.

ee%  =  100 ([R]-[S]) / ([R]+[S])
         = 100 (50-50) / (50+50)
         = 0%

Let's consider something a bit harder......

Question: 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 -9.2^o?

Answer: 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%  this indicates a 40% excess of R over S!

Question: What is the percent composition of the mixture?

Answer:  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).

Try a similar problem.

© Dr. Ian Hunt, Department of Chemistry