1. If we have mostly (+) 2-butanol with a specific rotation of +9.72°. We compar
ID: 494319 • Letter: 1
Question
1. If we have mostly (+) 2-butanol with a specific rotation of +9.72°. We compare this rotation with the +13.5° rotation of the pure (+) enantiomer. Calculate % enantiomer excess of the mixture? What percentage of the mixture is the (+) enantiomer and what percentage is the (-) enantiomer?2. Suppose a solution of a pure chiral molecule has a specific rotation of -25°. What is the specific rotation of a solution that is 70% of the (-) enantiomer and 25% of the (+) enantiomer?
Please help me solve these! 1. If we have mostly (+) 2-butanol with a specific rotation of +9.72°. We compare this rotation with the +13.5° rotation of the pure (+) enantiomer. Calculate % enantiomer excess of the mixture? What percentage of the mixture is the (+) enantiomer and what percentage is the (-) enantiomer?
2. Suppose a solution of a pure chiral molecule has a specific rotation of -25°. What is the specific rotation of a solution that is 70% of the (-) enantiomer and 25% of the (+) enantiomer?
Please help me solve these! 1. If we have mostly (+) 2-butanol with a specific rotation of +9.72°. We compare this rotation with the +13.5° rotation of the pure (+) enantiomer. Calculate % enantiomer excess of the mixture? What percentage of the mixture is the (+) enantiomer and what percentage is the (-) enantiomer?
Please help me solve these!
Explanation / Answer
Observed purity = 100*(observed rotation/rotation of pure sample)=100*9.72/13.5= 72%
enantiometric excess= optical purity = 100* Mod*(d-l)/(d+l)
where d-l= 72 and d+l=100 on solving d= 86% and l= 14%. +ve is 86% and -ve is 14%.
2. Specific rotation =0.7*(-25)+0.25*25 = -11.25
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