An organic reaction used to convert an aryl diazonium salt into an aryl halide using a copper(I) halide catalyst is called:

Correct — A. The Sandmeyer reaction is the one defined by the catalyst the question names: an aryl diazonium salt treated with a copper(I) halide in the matching acid, with nitrogen leaving as gas.

ArN₂⁺Cl⁻ + CuCl → ArCl + N₂
ArN₂⁺Br⁻ + CuBr → ArBr + N₂

The same route with CuCN puts a cyanide group on the ring, which is why Sandmeyer is taught as the general way of replacing the −N₂⁺ group left behind by an amine. That is option (a).

Why the others are wrong

• (b)
  Balz-Schiemann reaction — The Balz-Schiemann reaction makes an aryl fluoride, by isolating the diazonium tetrafluoroborate and heating it. No copper is involved anywhere in it.
• (c)
  Finkelstein reaction — The Finkelstein reaction is a halide exchange on an alkyl halide, using NaI in dry acetone to give the iodide. It begins from no diazonium salt at all.
• (d)
  Gattermann reaction — The Gattermann reaction reaches the same aryl halide but with copper powder and HX in place of the copper(I) salt. The question specifies a copper(I) halide, which names Sandmeyer.

Diazonium chemistry is the classroom route from aniline to almost any substituted benzene.

Aniline with NaNO₂ and HCl at 0-5 °C gives benzenediazonium chloride. That step is diazotisation, and the low temperature matters because the salt decomposes on warming.

The −N₂⁺ group is then swapped for Cl, Br, CN, I, F or OH, and each swap carries its own name. Learn which reagent goes with which name, because SSC asks the name from the reagent.

Key facts

• The Sandmeyer reaction converts an aryl diazonium salt into an aryl chloride, bromide or cyanide using CuCl, CuBr or CuCN.
• Benzenediazonium chloride is prepared from aniline with NaNO2 and HCl at 0-5 °C.
• The Balz-Schiemann reaction gives an aryl fluoride by heating the diazonium tetrafluoroborate.
• The Finkelstein reaction exchanges halogen on an alkyl halide using NaI in dry acetone.

Common traps

• Choosing Gattermann because it also yields an aryl halide, when the question names a copper(I) halide catalyst.
• Choosing Finkelstein on the word halide alone, when it acts on alkyl halides and needs no diazonium salt.

No directly related past PYQ was found.