Reaction Pathway Analysis

The question asks to identify the final product (D) of a multi-step chemical reaction starting from Phenol.

Step 1: Phenol to Intermediate A

Phenol ($C_6H_5OH$) is treated with Zinc dust ($Zn$). This is a reduction reaction where the hydroxyl group is removed.

Reaction: $C_6H_5OH \xrightarrow{Zn\ dust} C_6H_6$

Intermediate A is Benzene ($C_6H_6$).

Step 2: Intermediate A to Intermediate B

Benzene ($A$) undergoes Friedel-Crafts alkylation with methyl chloride ($CH_3Cl$) in the presence of anhydrous aluminum chloride ($AlCl_3$).

Reaction: $C_6H_6 + CH_3Cl \xrightarrow{anhy.\ AlCl_3} C_6H_5CH_3 + HCl$

Intermediate B is Toluene ($C_6H_5CH_3$).

Step 3: Intermediate B to Intermediate C

Toluene ($B$) is oxidized using a strong oxidizing agent, potassium dichromate ($K_2Cr_2O_7$) in acidic medium ($H_2SO_4$). This oxidizes the methyl group to a carboxyl group.

Reaction: $C_6H_5CH_3 \xrightarrow{K_2Cr_2O_7\ +\ H_2SO_4} C_6H_5COOH$

Intermediate C is Benzoic acid ($C_6H_5COOH$).

Step 4: Intermediate C to Product D

Benzoic acid ($C$) undergoes nitration using a mixture of concentrated sulfuric acid ($H_2SO_4$) and concentrated nitric acid ($HNO_3$). The carboxyl group ($-COOH$) is an electron-withdrawing group and acts as a meta-director.

Reaction: $C_6H_5COOH + HNO_3 \xrightarrow{H_2SO_4} m\text{-}NO_2C_6H_4COOH + H_2O$

The nitro group ($NO_2$) is directed to the meta position relative to the $-COOH$ group.

Product D is m-Nitrobenzoic acid.