Question

The specific rotation of optically pure 2-butanol has a specific rotation of +13.52 degree. A synthesized and purified sample of 2-butanol has the observed specific rotation of +6.76 degrees. The correct statement based on this observation is

• A. The sample is completely racemized
• B. 25% of the sample is racemic
• C. 50% of the sample is racemic
• D. 6.76% of the sample is racemic

Answer 1

Answer: (C)

50% of the sample is racemic

Answer 2

The enantiomeric excess (ee) is calculated as the ratio of the observed specific rotation to the specific rotation of the pure enantiomer, multiplied by 100%. For the given sample,

ee = (+6.76 / +13.52) * 100% = 50%.

An enantiomeric excess of 50% means that 50% of the sample is the excess enantiomer, and the remaining (100% - 50%) = 50% of the sample is a racemic mixture.