See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: The observed specific rotation of a mixture depends on the enantiomeric excess (ee) and the composition of the sample. A racemic mixture has a specific rotation of 0 degrees because equal amounts of (R) and (S) enantiomers cancel each other out. Step 2 - Calculate the ratio: The observed rotation is +6.76 degrees, and the pure (S) enantiomer gives +13.52 degrees. The ratio is 6.76/13.52 = 0.50, meaning the enantiomeric excess (ee) is 50%. This means 50% of the sample is pure (S)-2-butanol in excess. Step 3 - Interpreting ee of 50%: An ee of 50% means that 50% of the sample contributes net rotation (as pure S), and the other 50% contributes zero net rotation. The 50% that contributes zero rotation must be a racemic mixture (equal R and S), since a racemic mixture has [α] = 0. Step 4 - Composition breakdown: If the sample is 50% pure (S) and 50% racemic mixture, then: - The 50% racemic portion contains 25% (S) and 25% (R). - Total (S) = 50% + 25% = 75% (S); Total (R) = 25% (R). - Net rotation = 0.75 × (+13.52) + 0.25 × (−13.52) = 10.14 − 3.38 = +6.76 degrees. ✓ This confirms the composition is 50% (S) and 50% racemic. Step 5 - Why other options fail: - (a) 50% (S), 50% impurity: An impurity with zero rotation would give [α] = 0.50 × 13.52 = +6.76, which is mathematically consistent, but distillation purifies the compound; an impurity scenario is unlikely and not the best chemical interpretation. - (b) 50% (S), 50% (R): This would be a racemic mixture giving [α] = 0, not +6.76. - (d) Some other mixture: This is vague and not the most precise conclusion given the clean 50% ee result. - Option (c) correctly describes the mixture as 50% (S) enantiomer combined with 50% racemic mixture, which mathematically and conceptually accounts for the observed +6.76 degrees rotation. Therefore, the correct answer is C.