Optically pure (+)-2-iodooctane is dissolved in acetone containing a very large excess of radioactive iodide, I^*-. Every substitution of I by I^*…

Let each substrate molecule undergo substitution with pseudo-first-order rate constant $k$. Because the labelled iodide is in huge excess, a molecule that has reacted once is almost certainly labelled, so the unlabelled fraction is $e^{-kt}$ and the label half-time $\ln2/k=20.0\ \mathrm{min}$ gives $k=0.03466\ \mathrm{min^{-1}}$. Each substitution inverts the configuration. If $R$ and $S$ are the mole fractions of the two enantiomers, then $d(R-S)/dt=-2k(R-S)$ because every event moves a molecule from one enantiomer to the other and changes the difference by two units. Hence the optical rotation, proportional to $R-S$, decays as $e^{-2kt}$, i.e. the racemisation rate constant is twice the substitution rate constant. Setting $e^{-2kt}=0.100$ gives $t=\ln10/(2k)=2.3026/(0.06931)=33.22\ \mathrm{min}$. Equivalently $t=\tfrac12(20.0)\log_2 10=33.22\ \mathrm{min}$.