To solve the matching problem, we need to understand the chemical reactions between the reactants in List-I and their corresponding products in List-II. Let's analyze each pair:
(A) Phenol, Zn/Δ: The reaction of phenol with zinc dust upon heating leads to the removal of the hydroxyl group, yielding benzene as the product. Thus, (A) matches with (III).
(B) Phenol, CHCl3, NaOH, HCl: This is the Reimer-Tiemann reaction, where phenol reacts with chloroform in the presence of a base and then acid, giving salicylaldehyde. Thus, (B) matches with (I).
(C) Phenol, CO2, NaOH, HCl: This is the Kolvbe-Schmitt reaction, where phenol reacts with carbon dioxide in the presence of a base to form salicylic acid. Thus, (C) matches with (II).
(D) Phenol, Conc. HNO3: Nitration of phenol using concentrated nitric acid produces picric acid. Thus, (D) matches with (IV).
Based on the above explanation, the correct matching of List-I with List-II is:
(A) - (III)
(B) - (I)
(C) - (II)
(D) - (IV)
Therefore, the correct answer is: (A)-(III), (B)-(I), (C)-(II), (D)-(IV).
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Approach Solution -2
(A) Phenol, Zn/$\Delta$: When phenol is treated with zinc dust and heated, it gets reduced to benzene. Thus, (A) corresponds to (III) Benzene. (B) Phenol, CHCl$_3$, NaOH, HCl (Reimer-Tiemann Reaction): Phenol undergoes the Reimer-Tiemann reaction with chloroform and aqueous sodium hydroxide, forming salicylaldehyde. Thus, (B) corresponds to (I) Salicylaldehyde. (C) Phenol, CO$_2$, NaOH, HCl (Kolbe-Schmitt Reaction): Phenol reacts with carbon dioxide under high pressure in the presence of sodium hydroxide, followed by acidification, to form salicylic acid. Thus, (C) corresponds to (II) Salicylic acid. (D) Phenol, Conc. HNO$_3$: Phenol reacts with concentrated nitric acid to form picric acid (2,4,6-trinitrophenol). Thus, (D) corresponds to (IV) Picric acid. Conclusion: The correct matching is (A)-(III), (B)-(I), (C)-(II), (D)-(IV).
