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.

  • 1
    $\begingroup$ 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. $\endgroup$ – 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

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

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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