Question

The specific rotation of an optically pure compound is +75.3 (c 1.0 in CHCl₃) at 20 °C. A synthetic sample of the same compound showed a specific rotation of +66.3 (c 1.0 in CHCl₃) at 20 °C. The enantiomeric excess (ee) of the synthetic sample is _______ %.
(rounded off to the nearest integer)

Answer 1

Correct Answer: 88

The enantiomeric excess (ee) can be calculated using the following formula:

\(ee = \frac{\text{Observed specific rotation}}{\text{Specific rotation of optically pure compound}} \times 100\)

Given:

• Observed specific rotation = +66.3
• Specific rotation of the optically pure compound = +75.3

Substitute the values:

\(ee = \frac{66.3}{75.3} \times 100 \approx 88\)

Thus, the enantiomeric excess (ee) of the synthetic sample is 88%.