Match List-I (reagents) with List-II (transformation)List-IList-IIRea…

The correct answer is Option (2) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)

• (A) \( \ce{NaNO2} \) + HCl + \( \ce{H2O} \) (273-278 K) — This matches (III) C₆H₅-NH₂ → C₆H₅N₂⁺Cl⁻. Sodium nitrite and HCl produce nitrous acid in the mixture, which converts a primary aromatic amine into its diazonium salt. The low temperature is needed because the diazonium salt decomposes on warming.

      \( \displaystyle \ce{C6H5NH2 + NaNO2 + 2HCl} \xrightarrow{\text{273-278 K}} \ce{C6H5N2^+Cl^- + NaCl + 2H2O} \)

• (B) \( \ce{H2} \)/Ni + \( \ce{C2H5OH} \) — This matches (I) RCN → RCH₂NH₂. Catalytic hydrogenation adds two molecules of hydrogen across the C≡N bond of a nitrile and gives a primary amine with the same number of carbon atoms.

      \( \displaystyle \ce{R-C#N + 2H2} \xrightarrow{\ce{Ni}\text{, }\ce{C2H5OH}} \ce{R-CH2-NH2} \)

• (C) \( \ce{LiAlH4} \) — This matches (IV) RCONH₂ → RCH₂NH₂. Lithium aluminium hydride reduces the C=O group of an amide to \( \ce{CH2} \), so the amide becomes an amine without losing a carbon atom.

      \( \displaystyle \ce{R-CONH2} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{LiAlH4}} \ce{R-CH2-NH2} \)

• (D) \( \ce{HNO3} \) + \( \ce{H2SO4} \), 288K — This matches (II) C₆H₅NH₂ → p- Nitroaniline (major). Direct nitration of aniline with the nitrating mixture gives p-nitroaniline as the main product (about 51%). In the strongly acidic mixture part of the aniline is protonated to the anilinium ion, which is meta directing, so a large amount of m-nitroaniline (about 47%) also forms.

      \( \displaystyle \ce{C6H5NH2 + HNO3} \xrightarrow{\ce{H2SO4}\text{, 288 K}} \ce{O2N-C6H4-NH2 + H2O} \)

  The para isomer is the main product (51%); the anilinium ion, \( \ce{C6H5NH3^{+}} \), formed in the acid is meta directing, which accounts for 47% m-nitroaniline (and 2% ortho).

Shortcut: The pair (A)–(III) appears in Options (2) and (4), and (B)–(I) then leaves only Option (2).