Question

Given that observed rotation for (S)-2-bromobutane is +23.1degrees and that for (R)-2-bromobutane -23.1degrees, if the specific rotation of a sample of 2-bromobutane is -9.2degrees, what is the percentage of each stereoisomer in the sample?

Answer 1

The mixture is 70 % (`R`) and 30 % (`S`).

Explanation:

The negative sign tells us that the (`R`) enantiomer is the dominant one.

The formula for optical purity is

`"Optical purity" = [α]_"mixture"/[α ]_"pure sample" × 100 %`

∴ `"Optical purity" = "-9.2 °"/"-23.1 °" × 100 % = 39.8 %`

The other 60.2 % must consist of equal amounts of the (`R`) and (`S`) isomers, or 30.1 % of each.

Hence the amount of (`S`) is 30 %. and the amount of (`R`) is (39.8 + 30.1) % = 70 %