See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Starting material: 4-nitrotoluene (CH3 at C1, NO2 at C4). Step 2 - Reaction with Br2/Fe (electrophilic aromatic bromination): The NO2 group is a meta-director and the CH3 group is an ortho/para-director. In 4-nitrotoluene, the two groups are in a 1,4-relationship. The CH3 is ortho/para-directing and NO2 is meta-directing. The combined directing effects place incoming Br ortho to CH3 and meta to NO2, which corresponds to position 2 (ortho to CH3 at C1, meta to NO2 at C4). Thus product (A) is 2-bromo-4-nitrotoluene (CH3 at C1, Br at C2, NO2 at C4). Step 3 - Reaction with Sn/HCl (reduction of nitro group to amine): The NO2 group is reduced to NH2, giving (B) = 2-bromo-4-aminotoluene, i.e., 4-amino-2-bromotoluene (CH3 at C1, Br at C2, NH2 at C4). Step 4 - Reaction with NaNO2+HCl (diazotization): The NH2 group is converted to a diazonium salt N2+Cl-, giving (C) = 2-bromo-4-(diazonium)toluene diazonium chloride (CH3 at C1, Br at C2, N2+Cl- at C4). Step 5 - Reaction with H3PO2 (hypophosphorous acid, reductive removal of diazonium group - Sandmeyer-type replacement with H): The diazonium group at C4 is replaced by H, giving (D) = 2-bromotoluene (CH3 at C1, Br at C2). Step 6 - Reaction with KMnO4/HO- (alkaline permanganate oxidation of methyl group to carboxylic acid): The CH3 group at C1 is oxidized to COOH, giving (E) = 2-bromobenzoic acid (COOH at C1, Br at C2), which is ortho-bromobenzoic acid. This matches option (c): benzene ring with CO2H at position 1 and Br at position 2 (ortho). Why other options fail: - (a) para-bromobenzoic acid would require Br at C4, but Br was introduced at C2 (ortho to CH3) and the diazonium (at C4) was replaced by H. - (b) meta-bromobenzoic acid: Br is not at meta position relative to COOH. - (d) terephthalic acid: no second methyl group was present; the Br at C2 is not oxidized by KMnO4 under these conditions. Therefore, the correct answer is C.