See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Nitration of phenol (the starting material shown is benzene with an OH group, i.e., phenol) using HNO3/H2SO4 gives a mixture of ortho-nitrophenol (o-nitrophenol) and para-nitrophenol (p-nitrophenol). The question identifies two products: (A) more volatile and (B) less volatile. Step 1: Identify the reaction. Phenol undergoes electrophilic aromatic substitution with HNO3/H2SO4 to give predominantly ortho- and para-nitrophenol. Step 2: Distinguish volatility. o-Nitrophenol (2-nitrophenol) forms an intramolecular hydrogen bond between the OH group and the adjacent NO2 group. This intramolecular H-bonding reduces intermolecular association, lowering the boiling point and making it MORE volatile (steam distillable). p-Nitrophenol (4-nitrophenol) forms intermolecular hydrogen bonds, leading to stronger intermolecular forces, a higher boiling point, and making it LESS volatile. Step 3: Assign products. Product (A) = more volatile = o-nitrophenol (2-nitrophenol), which corresponds to option (a). Product (B) = less volatile = p-nitrophenol (4-nitrophenol), which corresponds to option (c). Step 4: Why other options fail. Option (b) meta-nitrophenol is not a major product of phenol nitration because OH is an ortho/para director. Option (c) para-nitrophenol is the less volatile product (B), not (A). Option (d) nitrobenzene would require a different starting material. Therefore, the correct answer is A.