Sandmeyer's reaction converts $ArN_2^+X^-$ to $ArCl$ using

A$NaOH$

B$Br_2/FeBr_3$

C$Zn/HCl$

D$Cu_2Cl_2/HCl$

Answer & Solution

Correct answer: D. $Cu_2Cl_2/HCl$

**Sandmeyer:** cuprous halide is the key reagent — $Cu_2Cl_2/HCl \to ArCl$, $Cu_2Br_2/HBr \to ArBr$, $CuCN/KCN \to ArCN$. Mechanism is radical (single-electron transfer from Cu(I)).

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**Sandmeyer's reaction** converts $ArN_2^+X^-$ to $ArCl$ using

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Sandmeyer's reaction converts $ArN_2^+X^-$ to $ArCl$ using