How is this answer found? **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%.

Enantiomeric excess (%ee) = 100 x (observed rotation)/rotation of pure enantiomer %ee = %Major…

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

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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