See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the starting material: The starting material is 1,4-dibromocyclohex-2-ene (Br at C1 and C4 of a cyclohexene ring). Step 2 – Reaction with CH3NHCH3 (2 molar equivalents) to give (A): Each C–Br undergoes nucleophilic substitution with dimethylamine (CH3NHCH3). With 2 moles of dimethylamine, both bromines are replaced, giving 1,4-bis(dimethylamino)cyclohex-2-ene as product (A). Step 3 – Treatment of (A) with BaO to give (B): BaO is a base that promotes elimination. The bis-dimethylamino compound undergoes double elimination (E2 or E1cb) across both positions, losing two equivalents of dimethylamine and introducing additional unsaturation. The product (B) is benzene. The two eliminations aromatize the six-membered ring, giving benzene. Step 4 – Reaction of (B) with CH3I (2 molar equivalents) to give (C): Benzene itself does not react with CH3I under normal conditions; re-examining the sequence suggests that (B) retains amine functionality. More precisely, (A) is 1,4-bis(dimethylamino)cyclohex-2-ene; BaO deprotonates or assists in generating the diene but does NOT fully aromatize at this stage — rather, (B) is still the bis-amine with a diene system. Then CH3I (2 equivalents) quaternizes both nitrogen atoms, giving the bis-quaternary ammonium salt (C): 1,4-bis(trimethylammonio)cyclohex-2-ene diiodide. Step 5 – Treatment of (C) with Ag2O/H2O (Hofmann elimination conditions) to give (D) + (CH3)3N: Ag2O converts iodide to hydroxide (wet silver oxide), generating the quaternary ammonium hydroxide. Hofmann elimination then occurs at both ammonium centers. Each trimethylammonium group undergoes beta-elimination, expelling trimethylamine [(CH3)3N] and generating a double bond. Two successive Hofmann eliminations from the 1,4-positions of cyclohexene ring introduce two new double bonds, and together with the existing double bond in the ring, full aromatization to benzene occurs. Step 6 – Product (D) is benzene: The sequence is a classic route using Hofmann exhaustive methylation/elimination to aromatize a cyclohexene derivative, ultimately yielding benzene as the aromatic product (D), which matches option (b). Why other options fail: - (a) cross-conjugated cyclohexadiene: insufficient elimination, not the final aromatic product. - (c) 1,4-benzoquinone: would require oxidation, not present in these conditions. - (d) p-cresol: no source of methyl or hydroxyl on the ring in this sequence. Therefore, the correct answer is B.