A mixture of the two enantiomers of a compound is prepared. The pure (+) enantiomer has [alpha] = +66.0 in the solvent used. The mixture shows [alpha] = +39.6 under the same conditions. The percentage of the (+) enantiomer in the mixture is

JEE Main levelSingle correct optionDifficulty 2 of 3st122

A mixture of the two enantiomers of a compound is prepared. The pure (+) enantiomer has $[α] = + 66.0$ in the solvent used. The mixture shows $[α] = + 39.6$ under the same conditions. The percentage of the (+) enantiomer in the mixture is

Optical purity (= ee) = observed

[α]

[α]_{p u re}

39.6/66.0 = 0.600

, so ee = 60%. %(+) = 50 + ee/2 = 80%, %(-) = 20%. The value 60% is the ee, not the percentage of the (+) form.