Write the equation for the oxidation of propene by \(\mathrm{KMnO}_{4}\) solution at \(\mathrm{pH}=7.0 .\) Show any intermediates. Assume cold and dilute conditions.

Write the chemical formulas of the reactants and products. Reactants: Propene (C\(_3\)H\(_6\)) and potassium permanganate (\(\mathrm{KMnO}_{4}\)) Products: Propanol (C\(_3\)H\(_7\)OH) and manganese dioxide (MnO\(_2\)) Unbalanced equation: C\(_3\)H\(_6\) (g) + \(\mathrm{KMnO}_{4}\) (aq) → C\(_3\)H\(_7\)OH (l) + MnO\(_2\) (s) + KOH (aq)

Balance the equation by ensuring that the number of atoms for each element is the same on both sides of the equation. Balanced equation: C\(_3\)H\(_6\) (g) + \(\mathrm{KMnO}_{4}\) (aq) + H\(_2\)O (l) → C\(_3\)H\(_7\)OH (l) + MnO\(_2\) (s) + KOH (aq)

In cold, dilute conditions, potassium permanganate is reduced to form manganese dioxide, while the propene is oxidized to form propanol. This can be divided into two half-reactions: one for the oxidation of propene and another for the reduction of potassium permanganate. Oxidation half-reaction: C\(_3\)H\(_6\) (g) + OH\(^-\) (aq) → C\(_3\)H\(_7\)OH (l) + e\(^-\) Reduction half-reaction: \(\mathrm{MnO}_{4}^-\) (aq) + 2H\(_2\)O (l) + 3e\(^-\) → MnO\(_2\) (s) + 4OH\(^-\) (aq)

Multiply the half-reactions by the appropriate factors to equalize the number of electrons transferred. Then, add the two half-reactions together, canceling out any species that appear on both sides. Oxidation: 3(C\(_3\)H\(_6\) (g) + OH\(^-\) (aq) → C\(_3\)H\(_7\)OH (l) + e\(^-\)) Reduction: (MnO\(_4^-\) (aq) + 2H\(_2\)O (l) + 3e\(^-\) → MnO\(_2\) (s) + 4OH\(^-\) (aq)) Final balanced equation: 3C\(_3\)H\(_6\) (g) + 3OH\(^-\) (aq) + \(\mathrm{MnO}_4^-\) (aq) + 2H\(_2\)O (l) → 3C\(_3\)H\(_7\)OH (l) + MnO\(_2\) (s) + 4OH\(^-\) (aq) + KOH (aq). Thus, the balanced equation for the oxidation of propene by \(\mathrm{KMnO}_{4}\) at \(\mathrm{pH}=7.0\) is: 3C\(_3\)H\(_6\) (g) + 3OH\(^-\) (aq) + \(\mathrm{KMnO}_4\) (aq) + 2H\(_2\)O (l) → 3C\(_3\)H\(_7\)OH (l) + MnO\(_2\) (s) + KOH (aq).