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The optical rotation of a sample of 2-butanol is measured to be αobs = -0.35º. The specific rotation for pure (+)-2-butanol is [α]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.

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Final answer:

The specific rotation can be calculated using the formula [α]D = αobs/(c*l). To determine the percentage of the (+) and (-) isomers in the sample, compare the observed rotation with the specific rotation of the pure (+)-2-butanol.

Step-by-step explanation:

The specific rotation is a characteristic physical property of chiral compounds. It is measured in degrees and represents the rotation of the plane-polarized light by an enantiomer. The specific rotation can be calculated using the formula [α]D = αobs/(c*l), where [α]D is the specific rotation, αobs is the observed rotation, c is the concentration, and l is the path length. In this case, the specific rotation for pure (+)-2-butanol is given as [α]D = 13.52° ml/g dm.

To calculate the specific rotation for the sample, we can substitute the given values into the formula: [α]D = -0.35°/ (0.15 g/ml * 0.6 dm). After performing the calculation, the specific rotation of the sample is obtained.

To determine the percentage of the (+) and (-) isomers in the sample, we need to compare the observed rotation with the specific rotation of the pure (+)-2-butanol. If the observed rotation is equal to the specific rotation, it means the sample only contains the (+) isomer. If the observed rotation is the negative value of the specific rotation, it means the sample only contains the (-) isomer. However, if the observed rotation is different from the specific rotation, it means the sample contains a mixture of both isomers.

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