Explanation

(c)
Aromatic compounds which have$= or \equiv$ bond in the side chain decolourise acidic/ alkaline ${KMnO}_{4}$. Benzene does not delcolourise the acidic/alkaline ${KMnO}_{4}$due to the delocalization of $\pi$-$\ electrons$.
While propene decolourized the alkaline ${KMnO}_{4}$ due to the presence of = bond. Ferrous ammonium sulphate and oxalic acid decolourized the ${KMnO}_{4}$ in acidic medium.
$2\ {KMnO}_{4} + 10{FeSO}_{4} + 8H_{2}{SO}_{4} \longrightarrow$
$\ \ \ \ \ \ \ \ \ \ \ \ K_{2}{SO}_{4} + 2{MnSO}_{4} + 5{Fe}_{2}\left( {SO}_{4} \right)_{3} + 8H_{2}O$
$2\ {KMnO}_{4} + 5C_{2}H_{2}O_{4} + 3H_{2}{SO}_{4} \longrightarrow$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ K_{2}{SO}_{4} + 2{MnSO}_{4} + 10{CO}_{2} + 8H_{2}O$