Match the List-I with List-II.List-IList-IIConversion ofReagent used(…

The correct answer is Option (3) → (A) - (IV), (B) - (III), (C) - (II), (D) - (I)

Each conversion is a standard reaction of phenols or alcohols, recognised by what happens to the ring or to the –OH group:

• (A) phenol to salicylic acid — This matches (IV) NaOH, CO₂, H⁺. This is Kolbe’s reaction: NaOH forms the phenoxide ion, which is more reactive than phenol towards the weak electrophile \( \ce{CO2} \); acidification then gives the acid.

      \( \displaystyle \ce{C6H5OH + NaOH -> C6H5ONa + H2O} \)

      \( \displaystyle \ce{C6H5ONa} \xrightarrow[\text{(ii) }\ce{H^+}]{\text{(i) }\ce{CO2}\text{, 400 K, 4-7 atm}} \underset{\text{salicylic acid}}{\ce{HOC6H4COOH}} \)

• (B) alcohol to alkene — This matches (III) Conc. H₂SO₄. Heating an alcohol with concentrated \( \ce{H2SO4} \) removes a molecule of water (acid-catalysed dehydration).

      \( \displaystyle \ce{C2H5OH} \xrightarrow{\text{conc. }\ce{H2SO4}\text{, 443 K}} \ce{CH2=CH2 + H2O} \)

• (C) phenol to picric acid — This matches (II) Conc. HNO₃. The strongly activating –OH group allows nitration at all three ortho/para positions, giving 2,4,6-trinitrophenol (picric acid).

      \( \displaystyle \ce{C6H5OH + 3HNO3} \xrightarrow{\text{conc.}} \underset{\text{picric acid}}{\ce{C6H2(NO2)3OH}} + \ce{3H2O} \)

• (D) phenol to benzene — This matches (I) Zn dust. Heating phenol with zinc dust removes the oxygen (reduction).

      \( \displaystyle \ce{C6H5OH + Zn -> C6H6 + ZnO} \)

Shortcut: The pair (A)–(IV) appears only in Option (3).