Question

Calculate the enantiomeric excess of a mixture of 2-bomobutane with an observed specific rotation of $+{18.5}^{\circ }$. The specific rotation of (R)-2-bomobutane is $-{23.1}^{\circ }$ and (S)-2-bomobutane is $+{23.1}^{\circ }$.

• A. 80%
• B. 90%
• C. 75%
• D. 59%

Answer 1

The correct option is A 80%

The specific rotation of mixture of enantiomers is positive value so the mixture contains excess of S-2-bromobutane.

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

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