Question

A compound is having an observed rotation of +10.5. The rotation for the pure enantiomer is calculated as +14.
The % optical purity of the compound is:

• A. $100\%$
• B. $33.3\%$
• C. $75\%$
• D. Cannot be predicted.

Answer 1

The correct option is C $75\%$

Optical purity is defined as the ratio of observed rotation to the rotation of pure enantiomer.
$\%\text{Optical purity}=\frac{\text{Observed optical rotation}}{\text{Optical rotation of pure enantiomer}}\times 100\%$

$\%\text{Optical purity}=\frac{10.5}{14}\times 100=75\%$