Match List-I with List-II :
| List-I | List-II |
| (A) Swarts Reaction | (I) C6H5NH2 + NaNO2 + HX + Cu2X2 → C6H5X + N2 |
| (B) Finkelstein reaction | (II) 2RX + 2Na → R – R + 2NaX |
| (C) Sandmeyer’s reaction | (III) RX + AgF → R – F + AgX |
| (D) Wurtz reaction | (IV) RX + NaI → R – I + NaX |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
Each named reaction is recognised by its starting material and reagent. Swarts and Finkelstein reactions are halogen exchange reactions, Sandmeyer’s reaction replaces the diazonium group of an arenediazonium salt, and Wurtz reaction joins two alkyl groups with sodium.
(A) Swarts Reaction — This matches (III) RX + AgF → R – F + AgX. A metallic fluoride replaces the halogen by fluorine.
\( \displaystyle \ce{H3C-Br + AgF -> H3C-F + AgBr} \)
(B) Finkelstein reaction — This matches (IV) RX + NaI → R – I + NaX. NaI in dry acetone gives the alkyl iodide.
\( \displaystyle \ce{R-X + NaI} \xrightarrow{\text{dry acetone}} \ce{R-I + NaX}\quad(\ce{X = Cl, Br}) \)
(C) Sandmeyer’s reaction — This matches (I) C6H5NH2 + NaNO2 + HX + Cu2X2 → C6H5X + N2. The item combines two steps: aniline is first diazotised, and cuprous halide then replaces the –\( \ce{N2^{+}} \) group by X.
\( \displaystyle \ce{C6H5NH2 + NaNO2 + 2HX} \xrightarrow{273\text{-}278\ \mathrm{K}} \ce{C6H5N2+X- + NaX + 2H2O} \)
\( \displaystyle \ce{C6H5N2+X-} \xrightarrow{\ce{Cu2X2}} \ce{C6H5X + N2} \)
(D) Wurtz reaction — This matches (II) 2RX + 2Na → R – R + 2NaX. Sodium in dry ether joins two alkyl groups.
\( \displaystyle \ce{2CH3CH2Br + 2Na} \xrightarrow{\text{dry ether}} \ce{CH3CH2-CH2CH3 + 2NaBr} \)
Shortcut: (A)–(III) appears only in Option (4).