Fluorobenzene $$(C_6H_5F)$$ can be synthesized in the laboratory

The laboratory synthesis of fluorobenzene relies on converting an aromatic amine into an arenediazonium salt and then replacing the diazonium group by fluorine. This two-step sequence is known as the Balz-Schiemann reaction.

Step 1 : Diazotisation of aniline
Aniline is treated with cold $$NaNO_2$$ and $$HCl$$ (0-5 °C) to give benzenediazonium chloride:
$$C_6H_5NH_2 + NaNO_2 + 2HCl \rightarrow C_6H_5N_2^+Cl^- + NaCl + 2H_2O$$

Step 2 : Formation of diazonium tetrafluoroborate and thermal decomposition
The diazonium chloride is reacted with aqueous $$HBF_4$$ to precipitate benzenediazonium tetrafluoroborate:
$$C_6H_5N_2^+Cl^- + HBF_4 \rightarrow C_6H_5N_2^+BF_4^- + HCl$$
On gentle heating (≈ 60-70 °C) the solid diazonium salt decomposes, losing $$N_2$$ and forming fluorobenzene:
$$C_6H_5N_2^+BF_4^- \xrightarrow{\;\Delta\;} C_6H_5F + BF_3 + N_2$$

Thus fluorobenzene is obtained smoothly and in good yield. This exactly matches Option B.

Why the other options are unsuitable

Option A : Phenol + HF / KF — Aryl-OH groups cannot be replaced by F⁻ under such conditions; the C-O bond in phenol is too strong and fluoride is a very poor nucleophile toward sp² carbon.

Option C : Direct fluorination with $$F_2$$ — Molecular fluorine attacks benzene violently, giving a mixture of polyfluorinated and degraded products; controlled mono-fluorination is not feasible in the lab.

Option D : Bromobenzene + NaF (halogen-exchange) — The Finkelstein reaction works only for aliphatic halides. Aryl halides do not undergo nucleophilic substitution with $$F^-$$ because the aryl C-Br bond is resistant to SNAr or SN1/SN2 pathways under these conditions.

Therefore, the only practical laboratory route is the Balz-Schiemann reaction described in Option B.

Final answer: Option B which is: from aniline by diazotisation followed by heating the diazonium salt with $$HBF_4$$