Identify the reliable laboratory method for preparing fluorobenzene

Aryl fluorides are not usually prepared by simple nucleophilic substitution on haloarenes, because the $C-F$ bond is very strong and direct fluorination with $F_2$ is too violent and uncontrolled.

The standard laboratory method is the Balz–Schiemann reaction:

• Convert aniline to benzene diazonium salt by diazotisation.
• Treat with $HBF_4$ to form benzene diazonium tetrafluoroborate.
• On heating, it decomposes to give fluorobenzene.

$C_6H_5NH_2 \xrightarrow[0-5^\circ C]{NaNO_2/HCl} C_6H_5N_2^+Cl^-$ $C_6H_5N_2^+Cl^- + HBF_4 \rightarrow C_6H_5N_2^+BF_4^-$ $C_6H_5N_2^+BF_4^- \xrightarrow{\Delta} C_6H_5F + BF_3 + N_2$

Evaluate each option

A: by heating phenol with HF and KF

• Phenol does not get converted directly to fluorobenzene by this method.
• The $-OH$ group in phenol is not replaced by $F$ under these conditions.
• Incorrect.

B: from aniline by diazotisation followed by heating the diazonium salt with $HBF_4$

• This is exactly the Balz–Schiemann reaction, the standard laboratory preparation of fluorobenzene.
• Correct.

C: by direct fluorination of benzene with $F_2$ gas

• Direct fluorination is highly reactive, explosive, and not a controlled laboratory method for preparing fluorobenzene.
• Incorrect.

D: by reacting bromobenzene with NaF solution

• Haloarenes do not undergo such simple halogen exchange with $NaF$ solution.
• Incorrect.

Final answer

The correct option is: $\boxed{\text{B}}$

Comparison with stored correct answer

Stored correct answer = B.

My derived answer also = B.

Therefore, they agree.