Question #44837

The oxidation of 1-phenylpropan-1-ol by \ce{KMnO4} proceeds via two competing kinetic pathways: Pathway I: Targeted oxidation of the secondary alcohol to propiophenone. Pathway II: Oxidative cleavage of the alkyl side-chain to yield benzoic acid.

According to the Bell-Evans-Polanyi principle, the activation energies ($E_a$, in kJ/mol) for these pathways are linearly dependent on the effective reduction potential ($E_{red}$, in V) of the permanganate half-cell: $E_{a,I} = 60 - 10 E_{red}$ $E_{a,II} = 105 - 40 E_{red}$

The dominant reduction half-reaction in the mildly acidic regime is: \ce{MnO4^- + 4H^+ + 3e^- -> MnO2(s) + 2H2O} \quad $E^\circ = 1.70 \text{ V}$

Assuming standard concentration for \ce{MnO4^-} ($1 \text{ M}$) and $T = 298 \text{ K}$ (where $\frac{2.303 RT}{F} = 0.06 \text{ V}$), at what pH does the kinetic preference switch from oxidative cleavage to targeted alcohol oxidation, and what is the major product at pH = 4.0?