Which of the following pairs of alkyl halides are best prepared by the halogen exchange method?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Alkyl fluorides and alkyl iodides
In halogen exchange, an alkyl chloride or bromide is the starting material, and its halogen is swapped for another halogen. Two exchange reactions are used, and they give the two halides that are hard to make in other ways:
Alkyl iodides — Finkelstein reaction: the alkyl chloride or bromide is treated with NaI in dry acetone. NaCl or NaBr is precipitated, which drives the reaction forward.
\( \displaystyle \ce{R-Cl + NaI} \xrightarrow{\text{dry acetone}} \ce{R-I + NaCl v} \)
Alkyl fluorides — Swarts reaction: the alkyl chloride or bromide is heated with a metallic fluoride such as AgF, Hg2F2, CoF2 or SbF3.
\( \displaystyle \ce{2R-Br + Hg2F2} \xrightarrow{\Delta} \ce{2R-F + Hg2Br2} \)
Alkyl chlorides and bromides are the starting halides in both reactions; they are made from alcohols, alkanes or alkenes.
NCERT (2026-27 edition):
Summary – “…Aryl halides are prepared by electrophilic substitution to arenes. Fluorides and iodides are best prepared by halogen exchange method.”
6.4.3 Halogen Exchange – “Alkyl iodides are often prepared by the reaction of alkyl chlorides/ bromides with NaI in dry acetone. This reaction is known as Finkelstein reaction.” … “The synthesis of alkyl fluorides is best accomplished by heating an alkyl chloride/bromide in the presence of a metallic fluoride such as AgF, Hg2F2, CoF2 or SbF3. The reaction is termed as Swarts reaction.”