The synthesis of alkyl fluoride is best achieved through

The correct answer is Option (1) → 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 (2) Finkelstein reaction — Incorrect. It gives alkyl iodides; it works because NaCl or NaBr is insoluble in dry acetone and precipitates:

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

• Option (3) Free radical Fluorination — Incorrect. Direct fluorination of an alkane is too violent (explosive) to control, and free radical halogenation gives a mixture of products.

• Option (4) Hydrogen fluoride — Incorrect. The reactivity of halogen acids with alcohols falls in the order HI > HBr > HCl, and HF is still less reactive, so alkyl fluorides are not made this way; they are made by halogen exchange.

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: