Match List-I with List-II
| Reaction | Name of Reaction |
(A)  | (I) Finkelstein Reaction |
| (B) CH3–Br + AgF → CH3–F + AgBr | (II) Swarts Reaction |
| (C) CH3–Cl + NaI → CH3–I + NaCl | (III) Fittig Reaction |
(D)  | (IV) Sandmeyer’s Reaction |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (IV), (B) - (II), (C) - (I), (D) - (III)
The four named reactions belong to three families, and each is recognised by its starting material and reagent:
Halogen exchange: Finkelstein (alkyl chloride or bromide + NaI in dry acetone gives an alkyl iodide) and Swarts (alkyl chloride or bromide + a metallic fluoride gives an alkyl fluoride).
From diazonium salts: Sandmeyer (\( \ce{Cu2Cl2} \)/HCl or \( \ce{Cu2Br2} \)/HBr) and its relative Gattermann (copper powder with HCl or HBr) replace the –\( \ce{N2^+} \) group by Cl or Br.
Coupling with sodium in dry ether: Wurtz joins two alkyl groups, Fittig joins two aryl groups, and Wurtz–Fittig joins an alkyl and an aryl group.
Each item of the list:
(A) Reaction A — This matches (IV) Sandmeyer’s Reaction. A benzenediazonium salt treated with cuprous chloride or cuprous bromide gives the aryl halide, and nitrogen gas is released.
\( \displaystyle \ce{C6H5N2^+X^-} \xrightarrow{\ce{Cu2X2}} \ce{C6H5X + N2} \)
(B) \( \ce{CH3-Br} \) + AgF → \( \ce{CH3-F} \) + AgBr — This matches (II) Swarts Reaction. An alkyl chloride or bromide heated with a metallic fluoride such as AgF, \( \ce{Hg2F2} \), \( \ce{CoF2} \) or \( \ce{SbF3} \) gives an alkyl fluoride.
\( \displaystyle \ce{CH3-Br + AgF -> CH3-F + AgBr} \)
(C) \( \ce{CH3-Cl} \) + NaI → \( \ce{CH3-I} \) + NaCl — This matches (I) Finkelstein Reaction. An alkyl chloride or bromide reacts with NaI in dry acetone to give an alkyl iodide. The NaCl formed is insoluble in acetone and precipitates, which drives the reaction forward.
\( \displaystyle \ce{CH3-Cl + NaI} \xrightarrow{\text{dry acetone}} \ce{CH3-I + NaCl v} \)
(D) Reaction D — This matches (III) Fittig Reaction. Two molecules of an aryl halide react with sodium in dry ether, and the two aryl groups join to form a diaryl.
\( \displaystyle \ce{2C6H5X + 2Na} \xrightarrow{\text{Dry ether}} \ce{C6H5-C6H5 + 2NaX} \)
Shortcut: The pair (A)–(IV) appears only in Option (3).