The synthesis of alkyl fluoride is best accomplished by the following reaction
\(CH_3Br + AgF \longrightarrow CH_3F + AgBr\)
The reaction is termed as:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Swarts reaction
Haloalkanes are prepared by several named reactions, grouped by what they do:
Halogen exchange: Finkelstein reaction (R–Cl/R–Br + NaI in dry acetone → R–I) and Swarts reaction (R–Cl/R–Br + a metallic fluoride such as AgF, \( \ce{Hg2F2} \), \( \ce{CoF2} \) or \( \ce{SbF3} \) → R–F).
Coupling with sodium in dry ether: Wurtz (two alkyl halides → alkane), Fittig (two aryl halides → biaryl) and Wurtz–Fittig (alkyl + aryl halide → alkylarene).
Alkyl fluorides cannot be made well by direct fluorination, which is too violent, so they are made by heating an alkyl chloride or bromide with a metallic fluoride. The given reaction exchanges Br for F using AgF, so it is the Swarts reaction:
\( \ce{CH3Br + AgF -> CH3F + AgBr} \)
Why the other options are incorrect
Option (2) Finkelstein reaction — Incorrect. Also a halogen exchange, but it gives alkyl iodides with NaI in dry acetone; NaCl or NaBr precipitates and drives the reaction forward.
\( \ce{CH3Br + NaI} \xrightarrow{\text{dry acetone}} \ce{CH3I + NaBr} \)
Option (3) Wurtz reaction — Incorrect. Two alkyl halides couple with sodium in dry ether to form a higher alkane; no fluoride is formed.
\( \ce{2CH3Br + 2Na} \xrightarrow{\text{dry ether}} \ce{CH3-CH3 + 2NaBr} \)
Option (4) Fittig reaction — Incorrect. Two aryl halides couple with sodium in dry ether to give a biaryl.
\( \ce{2C6H5Cl + 2Na} \xrightarrow{\text{dry ether}} \ce{C6H5-C6H5 + 2NaCl} \)
NCERT (2026-27 edition):
Halogen Exchange – …
