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8) How much of the S-enantiomer is present in 10 g of a mixture which has an enantiomeric excess of 20% of the R-enantiomer? A) 2 g B) 3 g C) 4 g D) 5 g E) 6 g

8) How much of the S-enantiomer is present in 10 g of a mixture which has an enantiomeric excess of 20% of the R-enantiomer? A) 2 g B) 3 g C) 4 g D) 5 g E) 6 g

Organic Chemistry: A Guided Inquiry
Organic Chemistry: A Guided Inquiry
2nd Edition
ISBN:9780618974122
Author:Andrei Straumanis
Publisher:Andrei Straumanis
Chapter6: Alkanes & Alkenes
Section: Chapter Questions
Problem 25E
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### Enantiomer Calculations #### Question 8 **Problem Statement:** How much of the S-enantiomer is present in 10 g of a mixture which has an enantiomeric excess of 20% of the R-enantiomer? - **Options:** - A) 2 g - B) 3 g - C) 4 g - D) 5 g - E) 6 g **Explanation:** - Given data: Total mixture mass = 10 g, Enantiomeric excess (ee) = 20% of R-enantiomer - Enantiomeric excess (ee) is defined as the difference between the percentages of the two enantiomers. - If ee = 20%, then 80% of the mixture is composed of an equal mixture of R and S enantiomers, and 20% is the excess amount of the R-enantiomer. - Let \( R \) be the amount of R-enantiomer and \( S \) be the amount of S-enantiomer. - The amount of R-enantiomer in the 10 g mixture: \( R \) = \( 5g + \frac{(20/100) \times 10g}{2} \) = 6 g - Therefore, the amount of S-enantiomer in the 10 g mixture: \( S \) = 10 - 6 = 4 g - Hence, the correct answer is C) 4 g. #### Question 9 **Problem Statement:** A solution of 0.2 g/ml of a pure R-enantiomer in a 1 dm polarimeter sample tube rotates the plane of polarized light by +5°. What is the rotation observed on this solution in a 2 dm polarimeter sample tube? - **Options:** - A) 0 - B) -5 - C) 10 - D) -10 - E) 5 **Explanation:** - Given data: concentration = 0.2 g/ml, tube length = 1 dm, observed rotation = +5° - Optical rotation is directly proportional to the path length of the polarimeter tube. - New tube length: 2 dm - Rotation observed: \( +5° \times \frac{2 \,
Transcribed Image Text:### Enantiomer Calculations #### Question 8 **Problem Statement:** How much of the S-enantiomer is present in 10 g of a mixture which has an enantiomeric excess of 20% of the R-enantiomer? - **Options:** - A) 2 g - B) 3 g - C) 4 g - D) 5 g - E) 6 g **Explanation:** - Given data: Total mixture mass = 10 g, Enantiomeric excess (ee) = 20% of R-enantiomer - Enantiomeric excess (ee) is defined as the difference between the percentages of the two enantiomers. - If ee = 20%, then 80% of the mixture is composed of an equal mixture of R and S enantiomers, and 20% is the excess amount of the R-enantiomer. - Let \( R \) be the amount of R-enantiomer and \( S \) be the amount of S-enantiomer. - The amount of R-enantiomer in the 10 g mixture: \( R \) = \( 5g + \frac{(20/100) \times 10g}{2} \) = 6 g - Therefore, the amount of S-enantiomer in the 10 g mixture: \( S \) = 10 - 6 = 4 g - Hence, the correct answer is C) 4 g. #### Question 9 **Problem Statement:** A solution of 0.2 g/ml of a pure R-enantiomer in a 1 dm polarimeter sample tube rotates the plane of polarized light by +5°. What is the rotation observed on this solution in a 2 dm polarimeter sample tube? - **Options:** - A) 0 - B) -5 - C) 10 - D) -10 - E) 5 **Explanation:** - Given data: concentration = 0.2 g/ml, tube length = 1 dm, observed rotation = +5° - Optical rotation is directly proportional to the path length of the polarimeter tube. - New tube length: 2 dm - Rotation observed: \( +5° \times \frac{2 \,
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