On adding $KI$ to benzenediazonium chloride, the product obtained is

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benzene

1-4-diiodobenzene

iodobenzene

1,3,5-triiodobenzene

answer is C.

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Detailed Solution

Treating with $KI$ produces iodobenzene. When benzene diazonium chloride reacts with potassium iodide, iodobenzene is obtained. Diazonium group is replaced with −I.

$C_{6} H_{5} - N_{2}^{+} {Cl}^{-} + KI \overset{Δ}{⟶} C_{6} H_{5} - I + N_{2} ↑ + KCl$

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