1. The optical rotation of a sample of 2-butanol is measured to be obs = -0.35º.

Question

1. The optical rotation of a sample of 2-butanol is measured to be obs = -0.35º. The specific rotation for pure (+)-2-butanol is []D = 13.52° ml/g dm. If the cell path length was 0.6 dm and the concentration of 2-butanol in the sample was 0.15 g/ml, calculate the specific rotation and the percentage of the (+) and (-) isomers in your sample.
2. You have a sample that you know is composed of 21% R isomer and 79% S isomer. What is the % optical purity of the sample? If you know that the specific rotation for the pure S isomer is [] = +75°, what would you expect the observed specific rotation for your sample to be?

Any help would be appreciated!!

Explanation / Answer

1. In this question, you'll have to use the following equation:

[]obs = D / c*l

obs = 0.35 / 0.6*0.15 = 3.89 this is the obs adjusted for the concentration and pathleght.

EE = obs/D = 3.88 / 13.52 = 0.2870 x 100 = 28.70%

Now, If 100% = M + N where M is the major isomer (+) and N the minor (-), and

EE = M - N then:

M = (100 + EE)/2 and N = (100 - EE)/2

M = (100 + 28.70)/2 = 64.35%

N = (100 - 28.70)/2 = 35.65%

2. We use the same equationns of above:

EE = M - N

EE = 79 - 21 = 58%

EE = obs/D ----> obs = EE x D

obs = (58/100) * 0.75

obs = 0.435 °

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