See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: In electrophilic aromatic substitution (EAS), the directing effects of substituents on each aromatic ring determine the regiochemistry of nitration. In phenyl benzoate (Ph-C(=O)-O-Ph), there are two rings: the acyl (benzoyl) ring and the phenoxy ring. Step 1 - Analyze the acyl ring (benzoyl side): The carbonyl group C(=O) attached directly to the ring is a strong electron-withdrawing group (EWG) via resonance and induction. EWGs are meta-directors and deactivate the ring. This ring is strongly deactivated toward EAS. Step 2 - Analyze the phenoxy ring (the O-Ph side): The oxygen of the ester linkage (-O-) is attached directly to this ring. Although oxygen is electronegative, when directly bonded to an aromatic ring it acts as an ortho/para director through resonance donation of lone pairs into the ring. This ring is activated (relative to the acyl ring) and the -O- group directs incoming electrophile to ortho and para positions. Step 3 - Determine which ring reacts preferentially: The phenoxy ring is more electron-rich (activated by the -O- group) compared to the benzoyl ring (deactivated by the C=O group). Therefore, nitration occurs preferentially on the phenoxy ring. Step 4 - Determine regiochemistry on the phenoxy ring: The -O- substituent on the phenoxy ring is an ortho/para director. Between ortho and para products, para is generally the major product due to steric reasons (less steric hindrance at para position). Step 5 - Identify the product: Nitration at the para position of the phenoxy ring gives phenyl 4-nitrobenzoate... wait, more precisely it gives the product where the phenoxy ring has NO2 at para position: this corresponds to option (d), which shows the acyl ring unsubstituted and the phenoxy (O-connected) ring with NO2 at the para position. Why other options fail: - (a) and (b): These show NO2 on the benzoyl (acyl) ring, which is deactivated and a meta-director. This ring is less reactive, so nitration there is disfavored. - (c): Shows NO2 at the ortho position of the phenoxy ring; while ortho is possible, para is sterically preferred and is the major product. - (d): Correctly shows NO2 at the para position of the phenoxy ring, consistent with ortho/para direction by the -O- group and preference for para. Therefore, the correct answer is D.