= Solution
The kink indicates that \b[two pathways with different leaving-group sensitivities compete]: one dominates for good leaving groups and another for poorer ones. A single straight-line free-energy relationship is insufficient. The additional hydroxide-order information can be accommodated by rapid phenol <deprotonation>,
$$
\mathrm{C+OH^-\rightleftharpoons C^-+H_2O},\qquad
[\mathrm{C^-}]=K_{\rm OH}[\mathrm C][\mathrm{OH^-}].
$$
Two rates are then possible. A route with one <hydroxide> equivalent in its overall kinetic dependence contributes $k_2[\mathrm C][\mathrm{OH^-}]$. Attack of a \b[further <hydroxide>] on phenoxide <ester> contributes
$$
k_{\rm add}[\mathrm{C^-}][\mathrm{OH^-}]
=k_{\rm add}K_{\rm OH}[\mathrm C][\mathrm{OH^-}]^2.
$$
Thus $k_3=k_{\rm add}K_{\rm OH}$. The third-order term need not involve a simultaneous collision of three molecules: sequential acid-base equilibration followed by bimolecular attack produces it.
Back to article page
