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?

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Answer 1 · 2016-02-09T02:17:03

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

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 %