See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: The nitrosonium ion (NO+) is an electrophile that undergoes electrophilic aromatic substitution (EAS) with aromatic compounds. The rate of EAS depends on the electron density of the aromatic ring — electron-donating groups (EDGs) activate the ring (increase rate) and electron-withdrawing groups (EWGs) deactivate the ring (decrease rate). Step 1: Analyze each substituent's electronic effect on the benzene ring. (a) Benzene — no substituent, baseline reactivity. (b) tert-Butylbenzene — tert-butyl is an alkyl group, which is an EDG via hyperconjugation/induction; activates the ring, making it MORE reactive than benzene. (c) Acetophenone (PhCOCH3) — the carbonyl group (C=O) is a strong EWG via both induction and resonance (it withdraws electron density from the ring through resonance, placing positive charge on the ring carbons). This strongly deactivates the ring, making it the LEAST reactive. (d) Methoxybenzene (anisole, PhOMe) — the OMe group is a strong EDG via resonance (lone pairs donated into the ring); strongly activates the ring, making it MORE reactive than benzene. Step 2: Rank reactivity toward the electrophile NO+. Most reactive: OMe (d) > tert-Bu (b) > benzene (a) > COCH3 (c) Least reactive. Step 3: Why other options fail. (a) Benzene is baseline — more reactive than acetophenone but not the slowest. (b) tert-Butylbenzene is activated — not the slowest. (d) Methoxybenzene is strongly activated — not the slowest. The acetophenone carbonyl group is a meta-director and strong deactivator, making option (c) the slowest to react with the electrophilic NO+. Therefore, the correct answer is C.