Question

What is the percent composition of a mixture of $\left(S\right)-(+)-2-butanol,[\pm ]\frac{25}{D}=+{13.52}^0$, and $\left(R\right)-(-)-2-butanol,[\pm ]\frac{25}{D}=-{13.52}^0$, with a specific rotation $[\pm ]\frac{25}{D}=+{.76}^0$:

• A. $75\%\left(R\right)25\%\left(S\right)$
• B. $25\%\left(R\right)75\%\left(S\right)$
• C. $50\%\left(R\right)50\%\left(S\right)$
• D. $67\%\left(R\right)33\%\left(S\right)$

Answer 1

Answer: (C)

$50\%\left(R\right)50\%\left(S\right)$

Rotaion for $S(+)$ is $+{13.52}^o$ and for $R(-)$ is $-{13.52}^o$ given rotation of mixture is $+{0.76}^o$

Let mixture R be x and of S be $100-x$

$x\left({13.52}^o\right)+(100-x)(-{13.52}^o)={76}^o$

${27.04}^{o}x-{1352}^o={0.76}^o$

${27.04}^{o}x={1352.76}^o$

$x=50$

So mixture has $50$ of $S(+)$ & $50$% of $R(-)$