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).
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