See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the two rings and their substituents: The compound is 4-acetyl-4'-methoxybiphenyl. The left ring carries an electron-withdrawing acetyl group (–COCH3), and the right ring carries an electron-donating methoxy group (–OCH3). Step 2 - Determine directing effects on each ring: • Left ring: The –COCH3 group is a meta-director and deactivator. It withdraws electron density from the ring via resonance and induction, making the ring less reactive toward electrophilic aromatic substitution (EAS). It directs incoming electrophiles to the meta position relative to itself. • Right ring: The –OCH3 group is an ortho/para-director and strong activator. It donates electron density into the ring via resonance (+M effect), making the ring highly reactive toward EAS. It directs incoming electrophiles to the ortho and para positions relative to itself. Step 3 - Identify which ring is attacked preferentially: The right ring (bearing –OCH3) is far more electron-rich and reactive than the left ring (bearing –COCH3). Therefore, the nitronium ion preferentially attacks the right ring. Step 4 - Identify the preferred position on the right ring: The –OCH3 group at the para position of the right ring (relative to the biphenyl bond) directs to its ortho and para positions. The para position is already occupied by –OCH3, so attack at para is blocked. The ortho positions to –OCH3 are available. Additionally, the biphenyl bond itself provides some electron donation to the ring. Position D is ortho to the biphenyl linkage and ortho to the –OCH3 group (i.e., it is activated by both the OCH3 ortho/para direction and the ring-to-ring conjugation), making it the most electron-rich and most activated position on the molecule. Step 5 - Why other positions fail: • Position A: On the left (deactivated, COCH3-bearing) ring — less reactive overall. • Position B: Also on the left ring — same deactivation issue. • Position C: On the right ring but less activated than D due to positional geometry relative to both –OCH3 and the biphenyl bond. • Position D: Ortho to both the biphenyl bond (which transmits electron density from the OCH3-activated right ring back) and in the ortho/para direction of –OCH3, making it the most nucleophilic carbon and the fastest site of attack. Therefore, the correct answer is D.