A-Level Structural isomerism and stereoisomerism: worked solution

Question

It is found that the value obtained for optical purity matches the enantiomeric excess exactly. Taking (+)-P and (-)-P as your example, explain why these two quantities come out to the same numerical value.

(+)-P and (-)-P are enantiomers, so they have specific rotations of equal magnitude but opposite sign. In a mixture the observed rotation is the sum of the two contributions, so equal amounts of (+) and (-) (the racemic part) cancel exactly and contribute nothing. The net rotation therefore comes only from the excess of one enantiomer. Optical purity = observed rotation / rotation of the pure enantiomer = (fraction of excess enantiomer) = (mol(+) - mol(-))/(mol(+) + mol(-)), which is exactly the definition of enantiomeric excess. Hence the two quantities are numerically identical.