See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Concept: Enantiomeric excess (ee) measures the excess of one enantiomer over the other, expressed as a percentage of the total mixture. Formula: ee = [(moles or mass of major enantiomer - moles or mass of minor enantiomer) / (total mass of both enantiomers)] × 100 Since both enantiomers are the same compound (2-butanol) with the same molar mass, we can use masses directly. Step 1: Identify the masses. - Mass of (+) 2-butanol = 6 g (major enantiomer) - Mass of (-) 2-butanol = 4 g (minor enantiomer) - Total mass = 6 + 4 = 10 g Step 2: Apply the enantiomeric excess formula. ee = [(6 - 4) / (6 + 4)] × 100 ee = [2 / 10] × 100 ee = 20% Why other options fail: - (a) 10%: This would result if the difference were 1g out of 10g, which is incorrect. - (c) 40%: This would result if someone incorrectly used (6-4)/5 × 100 or some other wrong denominator. - (d) 33%: This would result if someone incorrectly calculated 2/6 × 100 (using only the major enantiomer mass as denominator). The correct calculation gives ee = (6-4)/(6+4) × 100 = 2/10 × 100 = 20%. Therefore, the correct answer is B.