Match List-I with List-II List-I List-II (A) Gatterman's reaction (I)…

The correct answer is Option (1) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

Each named reaction is known by its starting material, its reagent and its product. Two of the four start from a benzene diazonium salt, two start from an amine or an amide:

• From diazonium salts: replacement of \( \ce{-N2+} \) by halogen (Sandmeyer with Cu(I) salts, Gattermann with copper powder and HX), and coupling with phenol or aniline, in which the \( \ce{-N=N-} \) link is kept.

• From amines and amides: the carbylamine reaction (primary amine + \( \ce{CHCl3} \) + KOH) and the Hofmann bromamide degradation (amide + \( \ce{Br2} \) + NaOH).

• (A) Gatterman's reaction — This matches (III) Aryl Halide. A diazonium salt with copper powder and HCl or HBr gives an aryl halide.

      \( \displaystyle \ce{C6H5N2^+Cl^- + HBr} \xrightarrow{\ce{Cu}} \ce{C6H5Br + N2 + HCl} \)

• (B) Carbylamine reaction — This matches (I) Alkyl isocyanide. A primary amine heated with chloroform and ethanolic KOH gives a foul-smelling isocyanide.

      \( \displaystyle \ce{R-NH2 + CHCl3 + 3KOH} \xrightarrow{\text{heat}} \ce{R-NC + 3KCl + 3H2O} \)

• (C) Hofmann bromamide reaction — This matches (IV) Primary amine. An amide with bromine and sodium hydroxide gives a primary amine with one carbon atom less than the amide.

      \( \displaystyle \ce{R-CONH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O} \)

• (D). Coupling reaction — This matches (II) Azo Compound. A diazonium salt couples with phenol in alkaline solution at the para position, keeping the nitrogen atoms as an azo (–N=N–) link; the product is the orange dye p-hydroxyazobenzene.

      \( \displaystyle \ce{C6H5N2+Cl^- + C6H5OH} \xrightarrow{\ce{OH-}} \ce{C6H5-N=N-C6H4-OH + HCl} \)

Shortcut: (A)-(III) appears in Options (1), (2) and (4); of these, only Option (1) also has (C)-(IV).