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: A ✓ checked by 4AB · confidence high
Explanation
[α] of the pure enantiomer = 90/0.60 = 150°. A sample with −135° has 90 % ee of S: S = 95 %, R = 5 %, i.e. R : S = 1 : 19.