# SN1, SN2 and enantiomeric excess

When (R)-2-bromobutane is heated with water, the $$\mathrm{S_N1}$$ substitution proceeds twice as fast as $$\mathrm{S_N2}$$. Calculate the appropriate enantiomeric excess and the specific rotation of the product of the mixture. The specific rotation of (R)-2-butanol is +13.5.

I din't understand how to interpret '$$\mathrm{S_N1}$$ proceeds twice as fast as $$\mathrm{S_N2}$$' mathematically.

• Basically, if 100% conversion occurs, you have $\frac{2}{3}$ of molecules converted by SN1 and other $\frac{1}{3}$ of molecules converted by SN2. That means, you have $\frac{2}{3}$ of $(S)$-isomers and other $\frac{1}{3}$ of $(R)$-isomers. – Mathew Mahindaratne May 5 '19 at 22:09

In this case, the absolute configuration remains same (as R) for $$\mathrm{S_N1}$$ and changes (into S) for $$\mathrm{S_N2}.$$ This is because the priority order of non substituted groups (wrt $$\ce{C2})$$ remains the same before and after substitution of $$\ce{Br}$$ by $$\ce{OH}.$$
Let the initial mole fraction of (R)-2-bromobutane be 1. After the reaction, let the mole fractions of R- and S-2-bromobutane be $$x$$ and $$1-x$$, respectively.
According to question condition, as $$\mathrm{S_N1}$$ proceeds twice as fast as $$\mathrm{S_N2},$$ we get
$$x = 2(1 - x) \quad\implies\quad x = 2/3$$
$$\frac{2}{3}\cdot (+13.5) + \frac{1}{3}\cdot (-13.5) = +4.5$$