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5.4 4. The specific rotation of pure (R)-2-butanol is -13.5°. What % of a mixture of the two 12.5 forms is (S)-2-butanol if the specific rotation of this mixture is -5.4° ? enantiomeric (A) 40% (B) 30% (C) 60% (D) 70% (E) None of the above

5.4 4. The specific rotation of pure (R)-2-butanol is -13.5°. What % of a mixture of the two 12.5 forms is (S)-2-butanol if the specific rotation of this mixture is -5.4° ? enantiomeric (A) 40% (B) 30% (C) 60% (D) 70% (E) None of the above

Organic Chemistry
Organic Chemistry
8th Edition
ISBN:9781305580350
Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote
Publisher:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote
Chapter16: Aldehydes And Ketones
Section: Chapter Questions
Problem 16.71P
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**Question 4:** The specific rotation of pure (R)-2-butanol is +13.5°. What percentage of a mixture of the two enantiomeric forms is (S)-2-butanol if the specific rotation of this mixture is +5.4°? Options: - (A) 40% - (B) 30% - (C) 60% - (D) 70% - (E) None of the above **Solution:** Given: - Specific rotation of pure (R)-2-butanol = +13.5° - Specific rotation of the mixture = +5.4° To find the percentage of (S)-2-butanol in the mixture: - Assume the percentage of (R)-2-butanol is \( x \). - Then the percentage of (S)-2-butanol is \( 100 - x \). Using the formula for calculating specific rotation of the mixture: \[ [\alpha]_{mixture} = x \cdot [\alpha]_{R-2-butanol} + (100 - x) \cdot [\alpha]_{S-2-butanol} \] Given: \[ +5.4 = x \cdot (+13.5) + (100 - x) \cdot (-13.5) \] Calculate to find \( x \). The handwritten solution indicates: \[ \frac{5.4}{13.5} = \frac{2}{5} = 0.4 = 40\% \] Therefore, the correct answer is (A) 40%.
Transcribed Image Text:**Question 4:** The specific rotation of pure (R)-2-butanol is +13.5°. What percentage of a mixture of the two enantiomeric forms is (S)-2-butanol if the specific rotation of this mixture is +5.4°? Options: - (A) 40% - (B) 30% - (C) 60% - (D) 70% - (E) None of the above **Solution:** Given: - Specific rotation of pure (R)-2-butanol = +13.5° - Specific rotation of the mixture = +5.4° To find the percentage of (S)-2-butanol in the mixture: - Assume the percentage of (R)-2-butanol is \( x \). - Then the percentage of (S)-2-butanol is \( 100 - x \). Using the formula for calculating specific rotation of the mixture: \[ [\alpha]_{mixture} = x \cdot [\alpha]_{R-2-butanol} + (100 - x) \cdot [\alpha]_{S-2-butanol} \] Given: \[ +5.4 = x \cdot (+13.5) + (100 - x) \cdot (-13.5) \] Calculate to find \( x \). The handwritten solution indicates: \[ \frac{5.4}{13.5} = \frac{2}{5} = 0.4 = 40\% \] Therefore, the correct answer is (A) 40%.
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