See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: When aryl methyl ethers (ArOCH3) are treated with excess HI under heating, the O-CH3 bond is cleaved (not the Ar-O bond). This is because the Ar-O bond has partial double bond character due to resonance with the aromatic ring, making it much stronger and resistant to cleavage. HI cleaves the weaker O-CH3 bond instead, producing the phenol (ArOH) and CH3I. Step 1: Identify the starting material. The substrate is 1,2,3-trimethoxybenzene, which has three OCH3 groups attached to the benzene ring at the 1-, 2-, and 3-positions. Step 2: Apply the reaction with excess HI and heat. For each ArOCH3 group, HI cleaves the methyl-oxygen bond: ArOCH3 + HI → ArOH + CH3I. Since HI is in excess, all three OCH3 groups are cleaved. Step 3: The product is 1,2,3-trihydroxybenzene (pyrogallol), which has three OH groups on the benzene ring at positions 1, 2, and 3. Option (a) depicts a benzene ring with multiple OH groups consistent with the cleavage of all methoxy groups to give phenolic OH groups. Why other options fail: - Option (b): HI does not substitute the methoxy groups with iodine on the ring; electrophilic aromatic substitution of OCH3 by I is not what occurs here. - Option (c): Reduction to give CH3 groups does not occur under these conditions; HI cleaves O-CH3 to give ArOH + CH3I, not ArCH3. - Option (d): Formation of an O-I species (hypoiodite-like product) is not a standard product of ArOCH3 + HI reaction. Step 4: The major product is 1,2,3-benzenetriol (pyrogallol), matching option (a). Therefore, the correct answer is A.