The sequence of reagents which convert p-methyl aniline to p-methyl benzoic acid are

$\mathrm{KMnO}_4 / \mathrm{H}^{+} ; \mathrm{NaNO}_2+\mathrm{HCl} ; \mathrm{Cu} / \mathrm{HCl}$
$\mathrm{NaNO}_2+\mathrm{HCl} / 273 \mathrm{~K} ; \mathrm{Cu} / \mathrm{HCl} ; \mathrm{KMnO}_4 / \mathrm{H}^{+}$
$\mathrm{NaNO}_2+\mathrm{HCl} / 273 \mathrm{~K} ; \mathrm{CuCN} / \mathrm{KCN} ; \mathrm{H}_3 \mathrm{O}^{+}$
$\mathrm{NaNO}_2+\mathrm{HCl} / 285 \mathrm{~K} ; \mathrm{KCN} ; \mathrm{H}_3 \mathrm{O}^{+}$

Asked in: JEE-TOPICTESTS-CHEMISTRY