SN1, SN2 and enantiomeric excess


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> 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.
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I din't understand how to interpret '$\mathrm{S\_N1}$ proceeds twice as fast as $\mathrm{S\_N2}$' mathematically. 




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


$$x = 2(1 - x) \quad\implies\quad x = 2/3$$


So we have excess of *R*-enantiomer, and the specific rotation is


$$\frac{2}{3}\cdot (+13.5) + \frac{1}{3}\cdot (-13.5) = +4.5$$


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