Q.In the reaction (benzenediazonium chloride) + Cu/HCl → X + N₂, X is ________ a) benzene (plain ring) b) chlorobenzene (ring with one Cl) c) 1,2-dichlorobenzene (ring with two ortho Cl) d) 1,3,5-trichlorobenzene (ring with three Cl)

Five named routes build a haloarene, each keyed to a specific product. Direct halogenation: benzene + Cl2 over a Lewis-acid catalyst (FeCl3) gives chlorobenzene + HCl -- an electrophilic aromatic substitution. Sandmeyer / Gattermann reaction (from benzenediazonium chloride): warming aqueous benzenediazonium chloride with Cu2Cl2/HCl (Sandmeyer) or with copper powder/HCl (Gattermann) gives chlorobenzene + N2 -- the diazonium group is displaced by the halide, catalysed by copper in either metallic or cuprous-salt form. Iodobenzene: benzenediazonium chloride warmed with aqueous KI gives iodobenzene + N2 + KCl directly, with no copper catalyst needed. Balz-Schiemann reaction (fluorobenzene): benzenediazonium chloride + fluoroboric acid (HBF4) first gives a stable diazonium fluoroborate salt, which only decomposes to fluorobenzene + BF3 + N2 on heating -- fluorine is the one halo …