Question

The specific rotation of $(R) (-) 2- $ bromooctane is $ -\,40 $ . What is the percentage composition of mixture of enantiomers of $ 2- $ bromooctane whose rotation is $ +20 $ ?

• A. The mixture has 25% R and 75% S
• B. The mixture has 20% R and 80% S
• C. The mixture has 50% R and 50% S
• D. The mixture has 75% R and 25% S

Answer 1

Answer: (A)

Specific rotation of $(R) (-) 2$-bromooctane is $ =-\,40 $
Specific rotation of $(S) (+) 2$-bromooctane is $ =+40 $
Let the mixture contains $ x \%R$ and $ (100-x) \ % S$ form.
$ \frac{x\times (-40)+(100-x)(+40)}{100}=20 $
$ x=25 $ and $ S=100-x=75 $
Hence, the mixture has $25\% R$ and $75\% S$.