See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is 2-bromotoluene (ortho-bromotoluene), which has a methyl group (CH3) and a bromine (Br) on adjacent carbons of a benzene ring. Step 2 - Step 1 of reaction (Mg/Ether): Treatment of an aryl bromide with magnesium in dry ether forms a Grignard reagent, specifically ortho-tolylmagnesium bromide (2-methylphenylmagnesium bromide). The C-Br bond is converted to C-MgBr. Step 3 - Step 2 of reaction (H3O+): Protonation (hydrolysis) of the Grignard reagent with aqueous acid (H3O+) replaces the C-MgBr bond with a C-H bond, regenerating the arene. This gives toluene (methylbenzene). Wait - but step 3 involves KMnO4/OH-, so let's reconsider the sequence. Step 4 - Reconsidering: Steps 1 and 2 together (Mg/Ether then H3O+) convert ArBr -> ArMgBr -> ArH. This means the Br is replaced by H, giving toluene (methylbenzene). Then step 3 (KMnO4/OH-) oxidizes the methyl group on the benzene ring. KMnO4 under basic conditions oxidizes an alkyl side chain on an aromatic ring to a carboxylate (-COO- K+). Step 4 (H+) acidifies the carboxylate to give the carboxylic acid (-COOH). Step 5 - Final product: The methyl group of toluene is oxidized by KMnO4/OH- followed by H+ workup to give benzoic acid (benzene ring with one COOH group), which corresponds to option (c). Step 6 - Why other options fail: - Option (a): 2-hydroxybenzoic acid would require introduction of an OH group, which does not occur here. - Option (b): ortho-cresol would require replacement of Br with OH, which is not what Mg/Ether + H3O+ does. - Option (d): Benzene would result if the methyl group were also removed, but KMnO4 oxidation of toluene gives benzoic acid, not benzene. Therefore, the correct answer is C.