Question

Benzene diazonium salt in the presence of Cu/HBr will give bromobenzene (C\(_6\)H\(_5\)Br) and release nitrogen gas (N\(_2\)). This reaction is known as:

Hint:

The Sandmeyer reaction is used to introduce halogens into aromatic rings using a diazonium salt and copper halides.

Answer 1

The reaction described is known as the
Sandmeyer Reaction. In the Sandmeyer reaction, a benzene diazonium salt reacts with copper(I) halides (such as CuBr or CuCl) in the presence of HBr or HCl to substitute the diazonium group with a halide, producing an aromatic halide and releasing nitrogen gas. Step 1: Write the general equation for the Sandmeyer Reaction.
The general reaction for the Sandmeyer reaction is: \[ \text{C}_6\text{H}_5\text{N}_2^+ \text{Cl}^- \xrightarrow{\text{Cu}/\text{HBr}} \text{C}_6\text{H}_5\text{Br} + \text{N}_2 \] In this case, the diazonium salt (C\(_6\)H\(_5\)N\(_2\)^+Cl^-) reacts with CuBr and HBr to produce bromobenzene and nitrogen gas.
Step 2: Conclusion.
Thus, the reaction where benzene diazonium salt in the presence of Cu/HBr gives bromobenzene and releases nitrogen gas is known as the Sandmeyer Reaction.