Step 1: Recall the concept of Sandmeyer’s reaction
Sandmeyer’s reaction involves the conversion of an aryl diazonium salt into an aryl halide using cuprous salts. It is mainly applicable for introducing Cl, Br, and CN groups on an aromatic ring. The general form is:
\[
C_6H_5N_2^+Cl^- \xrightarrow{CuCl} C_6H_5Cl, \quad
C_6H_5N_2^+Br^- \xrightarrow{CuBr} C_6H_5Br.
\]
This reaction proceeds smoothly with
CuCl and
CuBr but not with CuF or CuI.
Step 2: Analyze each compound
I.
Fluorobenzene (C6H5F): Cannot be prepared by Sandmeyer’s reaction because CuF is unstable. It is prepared by the
Balz–Schiemann reaction (using BF₄⁻ salts).
II.
Chlorobenzene (C6H5Cl): Formed by Sandmeyer’s reaction using CuCl.
III.
Bromobenzene (C6H5Br): Formed by Sandmeyer’s reaction using CuBr.
IV.
Iodobenzene (C6H5I): Formed by simply treating diazonium salt with KI (not Sandmeyer’s reaction since no copper salt is used).
V.
Atodobenzene (C6H5At): Not practical and not prepared by Sandmeyer’s method.
Step 3: Count the applicable halobenzenes
Only
Chlorobenzene (II) and
Bromobenzene (III) are prepared by Sandmeyer’s reaction.
Final answer
2