Alkyl chlorides are converted to alkyl iodides by treating with NaI i…

The correct answer is Option (3) → Finkelstein reaction.

Haloalkanes are converted into one another, or coupled, by a small family of named reactions:

• Halogen exchange: the Finkelstein reaction gives alkyl iodides from alkyl chlorides or bromides with NaI in dry acetone; the Swarts reaction gives alkyl fluorides with a metallic fluoride such as AgF, \( \ce{Hg2F2} \), \( \ce{CoF2} \) or \( \ce{SbF3} \).

• From diazonium salts: the Sandmeyer reaction gives aryl chlorides or bromides with \( \ce{Cu2Cl2} \)/HCl or \( \ce{Cu2Br2} \)/HBr; its look-alike, the Gattermann reaction, uses copper powder with HCl or HBr:

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

• Coupling with sodium: the Wurtz reaction joins two alkyl groups with sodium in dry ether.

In the Finkelstein reaction, NaI is soluble in dry acetone but NaCl (or NaBr) is not. It precipitates out, and by Le Chatelier’s principle this drives the reaction forward:

    \( \displaystyle \ce{R-Cl + NaI} \xrightarrow{\text{dry acetone}} \ce{R-I + NaCl v} \)

So the conversion of alkyl chlorides to alkyl iodides with NaI in dry acetone is the Finkelstein reaction.

Why the other options are incorrect

• Option (1) Swarts reaction. — Incorrect. It prepares alkyl fluorides, using metallic fluorides:

      \( \displaystyle \ce{H3C-Br + AgF -> H3C-F + AgBr} \)

• Option (2) Sandmeyer’s reaction. — Incorrect. It starts from a diazonium salt and gives an aryl halide:

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

• Option (4) Wurtz reaction. — Incorrect. It joins two alkyl groups to form an alkane; no halogen exchange takes place:

      \( \displaystyle \ce{2R-X + 2Na} \xrightarrow{\text{dry ether}} \ce{R-R + 2NaX} \)

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: