Question

Which of the following is a Sandmeyer reaction?
A. $2{C_6}{H_5}Cl\xrightarrow{{C{u_2}C{l_2}/HCl}}{C_6}{H_6} + C{l_2}$
B. ${C_6}{H_5}{N_2}Cl\xrightarrow{{C{u_2}C{l_2}/HCl}}{C_6}{H_5}Cl + {N_2}$
C. ${C_6}{H_5}OH\xrightarrow{{ZnDust}}{C_6}{H_6} + ZnO$
D. ${C_6}{H_5}N{O_2} + 6\left[ H \right]\xrightarrow{{Sn/HCl}}{C_6}{H_5}N{H_2} + 2{H_2}O$

Answer 1

Hint- Many Sandmeyer Reactions proceed under copper(I) catalysis, while the Sandmeyer-type reactions with thiols, water and potassium iodide don't require catalysis.

Complete answer:
Sandmeyer reaction is a kind of radical-nucleophilic fragrant substitution reaction. It is a helpful device by which an amino group on a sweet-smelling ring is supplanted with various substituents. During the Sandmeyer reaction, the amino group is changed over into a diazonium salt that can be changed into different functional groups utilizing a catalyst. The system includes a few stages, including the generation of intermediary compounds.
It is an organic reaction used to change over an aryl diazonium salt to an aryl halide utilizing a copper(I) halide catalyst. The system starts with a solitary electron move (SET) from the copper to the diazonium to frame a neutral daiso radical and copper(II) halide. The diazo radical at that point discharges an atom of nitrogen gas to frame an aryl radical. The aryl radical responds with the copper(II) halide to recover the copper(I) halide catalyst and yield the last aryl halide product.
The Sandmeyer reaction is the reaction between Copper(I)-Halogens and Cyanide and Diazonium compounds. Diazonium compounds can be prepared using amines and nitrous acid (from sodium nitrite and Hydrochloric acid). Copper(I) acts as a catalyst.

Thus, option B is the correct answer.

Note: The Sandmeyer Reaction is a significant change in sweet-smelling science, since it can bring about some substitution patterns that are not feasible by direct substitution.