Pure (S)-2-butanol has a specific rotation of | vedclass

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(C) Optical purity is calculated as: $	ext{Optical purity} = \frac{	ext{Observed specific rotation}}{	ext{Specific rotation of pure enantiomer}}

Vedclass · Accepted Answer

$50\% \,(S), 50\%$ racemic mixture

Vedclass · Answer

(C) Optical purity is calculated as: $	ext{Optical purity} = \frac{	ext{Observed specific rotation}}{	ext{Specific rotation of pure enantiomer}} 	imes 100$. Given: $	ext{Observed rotation} = +6.76^{\circ}$,$	ext{Pure rotation} = +13.52^{\circ}$. $	ext{Optical purity} = \frac{6.76}{13.52} 	imes 100 = 50\%$. This means $50\%$ of the sample is the pure $(S)$-enantiomer,and the remaining $50\%$ is a racemic mixture (which contains equal amounts of $(R)$ and $(S)$ enantiomers and is optically inactive).

Pure $(S)-2-$butanol has a specific rotation of $+13.52^{\circ}$. $A$ sample of $2-$butanol prepared in the lab and purified by distillation has a calculated specific rotation of $+6.76^{\circ}$. What can you conclude about the composition?

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