Question

The enantiomeric excess of a mixture of 2-bromobutane with an observed specific rotation of $+{18.5}^{\circ }$. The specific rotation of (R)-2-bromobutane is $-{23.1}^{\circ }$ and (S)-2-bromobutane is $+{23.1}^{\circ }$.
How many percent of each enantiomer is present in the mixture?

• A. (S)-2-bromobutane - 90%
  (R)-2-bromobutane - 10%
• B. (S)-2-bromobutane - 10%
  (R)-2-bromobutane - 90%
• C. (S)-2-bromobutane - 20%
  (R)-2-bromobutane - 80%
• D. (S)-2-bromobutane - 80%
  (R)-2-bromobutane - 20%

Answer 1

The correct option is A (S)-2-bromobutane - 90%

(R)-2-bromobutane - 10%
The specific rotation of mixture of enantiomers is positive value so the mixture contains excess of S-2-bromobutane.

$\%\text{Enantiomeric excess}=\frac{+18.5}{+23.1}\times 100\%$

$\%\text{Enantiomeric excess}=80\%$

Thus, 80% of S-2-bromobutane is excess in the given mixture.
Therefore, other 20% have racemic mixture of S-2-bromobutane and R-2-bromobutane.
So within those 20%, there is 10% of R and additional 10% of the S enantiomer.
Hence, the percent of S enantiomer is 90% and percent of R enantiomer is 10%