If a sample of the pure R enantiomer of a molecule has a specific rotation of - 40° and your mixture of that molecule has an observed optical rotation of + 22°, what is the % ee of your sample?

Question

Quizplus · Accepted Answer

% ee = (observed/specific) × 100.% ee = (22/40) × 100; 0.55 × 100 = 55 %. Since the pure R rotates the polarized light counterclockwise or levorotatory (which is determined by the rotation being negative) and your observed rotation is positive, that means your sample has more S than R since S is positive in this case.

If a Sample of the Pure R Enantiomer of a Molecule