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JEE MAIN - Chemistry (2024 - 1st February Evening Shift - No. 8)

Match List - I with List - II.

List I (Reactants) List II (Product)
(A) Phenol, Zn/Δ (I) Salicylaldehyde
(B) Phenol, CHCl3, NaOH, HCl (II) Salicylic acid
(C) Phenol, CO2, NaOH, HCl (III) Benzene
(D) Phenol, Conc. HNO3 (IV) Picric acid

Choose the correct answer from the options given below :
(A)-(IV), (B)-(I), (C)-(II), (D)-(III)
(A)-(III), (B)-(I), (C)-(II), (D)-(IV)
(A)-(IV), (B)-(II), (C)-(I), (D)-(III)
(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Explanation

Let's analyze each reaction given in List I to identify the correct product from List II.

(A) Phenol, Zn/Δ: When phenol is reacted with zinc dust (Zn) upon heating (Δ), it undergoes a reduction reaction known as the Clemmensen reduction, which results in the removal of the oxygen from the hydroxyl group (-OH) attached to the benzene ring, thus forming benzene. Hence, (A) corresponds to (III) Benzene.

(B) Phenol, CHCl3, NaOH, HCl: This reaction is known as the Reimer-Tiemann reaction. Phenol, when reacted with chloroform (CHCl3) in the presence of an aqueous base (NaOH), followed by acidification with HCl, forms salicylaldehyde. Therefore, (B) corresponds to (I) Salicylaldehyde.

(C) Phenol, CO2, NaOH, HCl: This is the Kolbe-Schmitt reaction. In this reaction, phenol reacts with carbon dioxide (CO2) under pressure at high temperatures in the presence of sodium hydroxide (NaOH), followed by acidification with HCl, to yield salicylic acid. Thus, (C) corresponds to (II) Salicylic acid.

(D) Phenol, Conc. HNO3: The nitration of phenol with concentrated nitric acid (HNO3) produces picric acid (2,4,6-trinitrophenol). So, (D) corresponds to (IV) Picric acid.

Based on the above reactions, the correct matching would be:

(A)-(III) Benzene
(B)-(I) Salicylaldehyde
(C)-(II) Salicylic acid
(D)-(IV) Picric acid

Therefore, the correct answer is Option B: (A)-(III), (B)-(I), (C)-(II), (D)-(IV).

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