Identify the standard reaction sequence involved

In questions of this type, the sequence is typically:

$\text{Aniline} \xrightarrow[0-5^\circ C]{NaNO_2/HCl} \text{benzene diazonium chloride} \xrightarrow{HBF_4,\ \Delta} \text{fluorobenzene}$

The first step is diazotization of aniline.

Formation of compound A

Aniline reacts with nitrous acid generated in situ from $NaNO_2 + HCl$ at low temperature $(0-5^\circ C)$ to form the diazonium salt:

$C_6H_5NH_2 \xrightarrow[0-5^\circ C]{NaNO_2/HCl} C_6H_5N_2^+Cl^-$

So,

$A = \text{benzene diazonium chloride}$

Formation of compound B

Benzene diazonium chloride on treatment through the Balz–Schiemann reaction gives fluorobenzene:

$C_6H_5N_2^+Cl^- \xrightarrow{HBF_4} C_6H_5N_2^+BF_4^- \xrightarrow{\Delta} C_6H_5F + N_2 + BF_3$

Thus,

$B = \text{fluorobenzene}$

Match with options

• A: nitrobenzene and fluorobenzene $\rightarrow$ incorrect
• B: phenol and benzene $\rightarrow$ incorrect
• C: benzene diazonium chloride and fluorobenzene $\rightarrow$ correct
• D: nitrobenzene and chlorobenzene $\rightarrow$ incorrect

Final answer

Therefore, the compounds $A$ and $B$ are:

$\boxed{\text{benzene diazonium chloride and fluorobenzene}}$