An optically active liquid mixture consists of a pure dextrorotatory () enantiomer and its correspon — Isomerism and Stereochemistry Chemistry Question
Question
An optically active liquid mixture consists of a pure dextrorotatory ($d$) enantiomer and its corresponding pure levorotatory ($l$) enantiomer. If the specific rotation of the pure '$d$' isomer is $+30^\circ$, and the observed specific rotation of the contaminated mixture is exactly $+22.5^\circ$, what is the calculated optical purity (enantiomeric excess, ee) of the final mixture?
💡 Solution & Explanation
Optical Purity (Enantiomeric Excess, ee) is defined mathematically as the ratio of the observed specific rotation of the mixture to the specific rotation of the pure major enantiomer, multiplied by 100. Therefore, $ee = ( \alpha_{obs} / \alpha_{pure} ) \times 100$. Substituting the known values: $ee = (22.5^\circ / 30^\circ) \times 100 = 0.75 \times 100 = 75\%$.