Match List I with List II
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List I Reaction of benzene diazonium chloride |
List II Products |
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Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-IV, B-I, C-III, D-II
Reactions of benzenediazonium chloride fall into two groups. In replacement reactions the –\( \ce{N2+} \) group leaves as nitrogen gas and is replaced by another group: Cl, Br or CN with Cu(I) salts (Sandmeyer), I with KI, OH with warm water, H with hypophosphorous acid or ethanol. In coupling reactions the –N=N– group is kept: the diazonium ion attacks the para position of phenol or aniline and gives a coloured azo compound.
(A) with H3PO2 — This matches (IV) benzene. Hypophosphorous acid reduces the diazonium salt to the arene.
\( \displaystyle \ce{C6H5N2^+Cl^- + H3PO2 + H2O -> C6H6 + N2 + H3PO3 + HCl} \)
(B) with KI — This matches (I) iodobenzene. Iodine is not easily put into the ring directly, but iodide ion replaces the diazonium group readily.
\( \displaystyle \ce{C6H5N2^+Cl^- + KI -> C6H5I + KCl + N2} \)
(C) with warm H2O — This matches (III) phenol. If the solution is allowed to warm up to 283 K, the salt is hydrolysed.
\( \displaystyle \ce{C6H5N2^+Cl^- + H2O -> C6H5OH + N2 + HCl} \)
(D) with aniline — This matches (II) p-aminoazobenzene. The diazonium ion couples at the para position of aniline (electrophilic substitution), and the –N=N– group stays in the product.
\( \displaystyle \ce{C6H5N2^+Cl^- + H-C6H4-NH2 -> C6H5-N=N-C6H4-NH2 + HCl} \)