The specific rotation of optically pure (R)-2-bromobutane is –112.00. A given sample of  2 bromobutane exhibited a specific rotation of –82.88

Q. The specific rotation of optically pure (R)-2-bromobutane is –112.00. A given sample of  2-bromobutane exhibited a specific rotation of –82.88. The percentage of (S)-(+)-enantiomer present in this sample is

Ans: The specific rotation of optically pure (R)-2-bromobutane is –112.00. A given sample of  2-bromobutane exhibited a specific rotation of –82.88. The percentage of (S)-(+)-enantiomer present in this sample is 13.

Solution:

Given, Specific rotation of optically pure (R)-2-bromobutane = -112.00.

The specific rotation of the sample mixture = -82.88.

Using, ee = observed rotation / actual rotation x 100 ⇒

ee = -82.88 / -112.00 x 100 ⇒ ee = 74% (R).

Percentage of Racemic Mixture = 100 – R% = 100 – 74 % = 26%.⇒

26% = 13%(S)+13%(R).