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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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?

Similar Questions

Which one of the following shows optical activity?

How many chiral carbon atoms are present in $2,3,4$-trichloropentane?

The number of meso forms of the given compound $CH_3-CH(OH)-CH(OH)-CH(OH)-CH_3$ is

How many optically active stereoisomers are possible for butan-$2,3$-diol?

Which of the following is not chiral?

Vedclass Test Series

Exam Paper Generator

Online Exam Module