The synthesis of Alkyl fluoride is best carried through:

The correct answer is Option (2) → Swarts Reaction

Alkyl fluorides cannot be made well by direct fluorination, which is too violent. They are made by halogen exchange. Two such exchange reactions are known:

• Swarts reaction: an alkyl chloride or bromide is heated with a metallic fluoride such as AgF, \( \ce{Hg2F2} \), \( \ce{CoF2} \) or \( \ce{SbF3} \), giving the alkyl fluoride.

      \( \displaystyle \ce{CH3Br + AgF -> CH3F + AgBr} \)

• Finkelstein reaction: an alkyl chloride or bromide is treated with NaI in dry acetone, giving the alkyl iodide.

So alkyl fluorides are best made by the Swarts reaction.

Why the other options are incorrect

• Option (1) Free radicals — Incorrect. Free radical halogenation of an alkane gives a mixture of products, and with fluorine the reaction is explosive:

      \( \displaystyle \ce{CH4 + Cl2} \xrightarrow{h\nu} \ce{CH3Cl + HCl} \)

• Option (3) Sandmeyer Reaction — Incorrect. It makes aryl chlorides and bromides from diazonium salts with Cu(I) halides, not alkyl fluorides:

      \( \displaystyle \ce{C6H5N2+Cl^-} \xrightarrow{\ce{Cu2Cl2}/\ce{HCl}} \ce{C6H5Cl + N2} \)

• Option (4) Finkelstein Reaction — Incorrect. It gives alkyl iodides; it works because NaCl or NaBr is insoluble in dry acetone and precipitates:

      \( \displaystyle \ce{CH3CH2Br + NaI} \xrightarrow{\text{dry acetone}} \ce{CH3CH2I + NaBr v} \)

NCERT (2026-27 edition):
Halogen Exchange – …

Note: The same/similar question has been asked in different years by NTA as per the details given below: