Subject: Organic Chemistry · Chapter: General Organic Chemistry · Topic: Stereochemistry R S E Z · Exam: CSIR-NET JUNE 2011 · Marks: 2 · Difficulty: Medium
An optically active compound enriched with $R$-enantiomer (60% ee) exhibited $[\alpha]_{D}+90^{\circ}$. If the $[\alpha]_{D}$ value of the sample is $-135^{\circ}$, the ratio of R and S enantiomers would be
(a)$\mathrm{R}: \mathrm{S}=1: 19$
(b)$\mathrm{R}: \mathrm{S}=19: 1$
(c)$\mathrm{R}: \mathrm{S}=1: 9$
(d)$\mathrm{R}: \mathrm{S}=9: 1$
Answer
Answer (as printed): A
Explanation
$[\alpha]_D = (\alpha_D / \text{ee}) \times 100$. $[\alpha]_D$ of R enantiomer $= (+90/60)\times100 = 150^{\circ}$. $[\alpha]_D$ of S enantiomer $=-150^{\circ}$. Now, ee of S enantiomer $=(-135/150)\times100=90\%$. Racemic part will be 10%: 5% R and 5% S isomer. So, total R = 5%; total S = 95%. Ratio R:S = 5:95 = 1:19.