See image — Practical Organic Chemistry and Purification Chemistry Question

💡 Solution & Explanation

Concept: Bromobenzene (Ph-Br) is an aryl halide, while benzyl bromide (PhCH2Br) is a benzylic halide. These two compounds differ significantly in their reactivity toward various reagents. Step 1 - Test with aq. KOH followed by Na (option a): Benzyl bromide (benzylic halide) undergoes hydrolysis with aq. KOH readily to give benzyl alcohol (PhCH2OH), which then reacts with Na to liberate H2 gas. Bromobenzene (aryl halide) does not undergo nucleophilic substitution with aq. KOH under normal conditions due to the strong C(sp2)-Br bond and the absence of a suitable leaving group mechanism. Thus, aq. KOH + Na can differentiate them. This test works. Step 2 - Test with AgNO3 (option b): Benzyl bromide is a benzylic halide and ionizes readily in polar solvents to give a stable benzylic carbocation, producing a cream/pale yellow precipitate of AgBr with AgNO3 solution. Bromobenzene, being an aryl halide, does not ionize easily and gives no precipitate (or a very slow/no reaction) with AgNO3 under normal conditions. Thus, AgNO3 differentiates them. This test works. Step 3 - Test with KMnO4 (option c): Benzyl bromide contains a benzylic CH2 group that can be oxidized by KMnO4. The benzylic position is oxidized, and KMnO4 (purple) gets decolorized or a brown precipitate of MnO2 forms. Bromobenzene lacks an oxidizable benzylic position and does not decolorize KMnO4 under mild conditions. Thus, KMnO4 differentiates them. This test works. Step 4 - Conclusion: Since all three tests — (a) aq. KOH then Na, (b) AgNO3, and (c) KMnO4 — can individually differentiate bromobenzene from benzyl bromide, the correct answer is (d) All these. Therefore, the correct answer is D.