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

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answer is 13.

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Detailed Solution

x100×112+100−x100×(−112)=−82.88⇒x=13

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