What are the products obtained by the oxidation of aniline with (i) chromic acid and (ii) alkaline \(\mathrm{KMnO}_{4} ?\)

Aniline (C\(_6\)H\(_5\)NH\(_2)\) is an aromatic amine consisting of a phenyl ring (C\(_6\)H\(_5\)-) attached to an amino group (-NH\(_2\)). In general, the amino group is a reducing agent which can be oxidized in various ways depending on the conditions and the oxidizing agents used.

In this first part of the exercise, aniline reacts with chromic acid (H\(_2\)CrO\(_4\)) as the oxidizing agent. Chromic acid is a strong oxidizing agent that mainly oxidizes the amino group (-NH\(_2\)) into a nitro group (-NO\(_2\)). The oxidation process involves losing electrons for the nitrogen atom and replacing the hydrogen atoms with oxygen atoms. The overall reaction is as follows: C\(_6\)H\(_5\)NH\(_2\) + H\(_2\)CrO\(_4\) -> C\(_6\)H\(_5\)NO\(_2\) + Cr(OH)\(_3\) + H\(_2\)O So, the product obtained by the oxidation of aniline with chromic acid is nitrobenzene (C\(_6\)H\(_5\)NO\(_2\)).

In this second part of the exercise, aniline reacts with alkaline \(\mathrm{KMnO}_{4}\) as the oxidizing agent. Alkaline \(\mathrm{KMnO}_{4}\) is another strong oxidizing agent which promotes cleavage of the benzene ring and converts the amino group (-NH\(_2\)) into a carboxylic group (-COOH). The overall reaction is as follows: 3 C\(_6\)H\(_5\)NH\(_2\) + 2 KMnO\(_4\) + 4 KOH -> 3 C\(_6\)H\(_4\)(COOK)COOH + 2 MnO\(_2\) + 4 KOH + 4 H\(_2\)O After acidic hydrolysis (adding HCl, for instance): 3 C\(_6\)H\(_4\)(COOH)\(_2\) + 6 KCl -> 3 C\(_6\)H\(_4\)(COOH)\(_2\) + 6 KCl So, the product obtained by the oxidation of aniline with alkaline \(\mathrm{KMnO}_{4}\) is 1,4-benzoquinone (C\(_6\)H\(_4\)(COOH)\(_2\)). In conclusion, the products obtained by the oxidation of aniline are: (i) Nitrobenzene (C\(_6\)H\(_5\)NO\(_2\)) using chromic acid, (ii) 1,4-Benzoquinone (C\(_6\)H\(_4\)(COOH)\(_2\)) using alkaline \(\mathrm{KMnO}_{4}\).