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 %