In the following reaction, identify the product D.

The correct option is m-Nitrobenzoic acid.

Let's analyze the reaction sequence step-by-step as shown in the provided reaction scheme starting from phenol:
C₆H₅OH → A → B → C → D

Step 1: Conversion of Phenol (C₆H₅OH) to A
When phenol is heated with zinc (Zn) dust, it undergoes a reduction reaction. The zinc dust removes the oxygen atom from the hydroxyl (-OH) group of phenol to form zinc oxide (ZnO) and benzene.
Therefore, intermediate A is benzene (C₆H₆).

Step 2: Conversion of A to B (Friedel-Crafts Alkylation)
Benzene (A) reacts with methyl chloride (CH₃Cl) in the presence of anhydrous aluminium chloride (AlCl₃), which acts as a Lewis acid catalyst. The AlCl₃ facilitates the formation of a methyl electrophile (CH₃⁺) that attacks the benzene ring to yield toluene.
Therefore, intermediate B is toluene (C₆H₅CH₃).

Step 3: Conversion of B to C (Side-Chain Oxidation)
Toluene (B) is treated with a strong oxidizing agent, potassium dichromate in acidic medium (K₂Cr₂O₇ + H₂SO₄). The benzylic carbon of the methyl side-chain is oxidized completely to a carboxylic acid (-COOH) group.
Therefore, intermediate C is benzoic acid (C₆H₅COOH).

Step 4: Conversion of C to D (Nitration)
Benzoic acid (C) is reacted with a nitrating mixture containing concentrated sulfuric acid and nitric acid (H₂SO₄ + HNO₃). The carboxylic acid group (-COOH) already present on the ring is an electron-withdrawing group, making it meta-directing. Consequently, the incoming nitro group (-NO₂) is directed to the meta-position on the benzene ring.
Therefore, the final product D is m-nitrobenzoic acid.