Match List-I (Name of reaction)with List-II (End product)
| List-I | List-II |
| Name of reaction | End product |
| (A) Reimer Tiemann reaction | (I) Aspirin |
| (B) Nitration of phenol by conc. HNO3 | (II) Salicylic acid |
| (C) Acetylation of salicylic acid | (III) Salicylaldehyde |
| (D) Kolbe reaction | (IV) Picric Acid |
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)
All four are standard reactions of phenol or its derivatives, and each one puts a different group on the ring:
(A) Reimer Tiemann reaction — This matches (III) Salicylaldehyde. Phenol with chloroform and aqueous NaOH puts a –\( \ce{CHCl2} \) group at the ortho position; it is hydrolysed and acidified to 2-hydroxybenzaldehyde.
\( \displaystyle \ce{C6H5OH} \xrightarrow[\text{(ii) }\ce{H+}]{\text{(i) }\ce{CHCl3}\text{ + aq. }\ce{NaOH}} \text{salicylaldehyde} \)
(B) Nitration of phenol by conc. HNO3 — This matches (IV) Picric Acid. Concentrated nitric acid nitrates all three ortho and para positions; dilute nitric acid gives only 2- and 4-nitrophenol.
\( \displaystyle \ce{C6H5OH + 3HNO3} \xrightarrow{\text{conc.}} \underset{\text{2,4,6-trinitrophenol}}{\ce{C6H2(NO2)3OH}} + \ce{3H2O} \)
(C) Acetylation of salicylic acid — This matches (I) Aspirin. Acetic anhydride acetylates the –OH group of salicylic acid, giving acetylsalicylic acid.
\( \displaystyle \ce{HOC6H4COOH + (CH3CO)2O} \xrightarrow{\ce{H+}} \underset{\text{aspirin}}{\ce{CH3COOC6H4COOH}} + \ce{CH3COOH} \)
(D) Kolbe reaction — This matches (II) Salicylic acid. Sodium phenoxide is attacked by CO2 at the ortho position, and acidification gives 2-hydroxybenzoic acid.
\( \displaystyle \ce{C6H5ONa} \xrightarrow[\text{(ii) }\ce{H+}]{\text{(i) }\ce{CO2}} \text{salicylic acid} \)
Shortcut: The pair (A)–(III), Reimer–Tiemann giving salicylaldehyde, appears only in Option (4).