The synthesis of alkyl fluorides is best accomplished by:

We begin by recalling the well-known named reactions that are commonly taught in organic chemistry for the replacement of halogen atoms.

First, the Swarts reaction is a halogen-exchange reaction which specifically converts an alkyl chloride, bromide or iodide into the corresponding alkyl fluoride. The general form of the reaction is written as

$$R{-}X \;+\; AgF \;\longrightarrow\; R{-}F \;+\; AgX,$$

where $$R$$ represents any alkyl group and $$X$$ can be $$Cl$$, $$Br$$ or $$I$$. Silver fluoride, antimony trifluoride $$\big(SbF_3\big)$$ or other metal fluorides are typically used because fluorine itself is too reactive to be handled safely in most laboratory situations. Thus, Swarts reaction is deliberately designed to give alkyl fluorides smoothly and selectively.

Next, we examine free-radical fluorination. Although it can, in principle, introduce a fluorine atom into an alkane, it proceeds via a chain-radical mechanism that is extremely vigorous. Side reactions, over-fluorination and even explosions are frequent. Therefore, it is not the method “best accomplished” for controlled synthesis of specific alkyl fluorides.

Sandmeyer’s reaction is employed for converting diazonium salts into chlorides, bromides or cyanides, not fluorides. Hence Sandmeyer’s reaction does not suit the present need.

Finkelstein reaction is another halogen-exchange reaction, but it is used for interconverting alkyl chlorides or bromides and alkyl iodides, typically with sodium iodide in acetone:

$$R{-}Cl \;+\; NaI \;\longrightarrow\; R{-}I \;+\; NaCl.$$

This reaction does not generate fluorides either.

Comparing all four possibilities, only the Swarts reaction is both selective and practical for preparing alkyl fluorides from other alkyl halides. Therefore, we conclude that the best method for synthesising alkyl fluorides is indeed the Swarts reaction.

Hence, the correct answer is Option A.