See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is 4-chlorobenzyl bromide (Cl on one end of para-phenylene, CH2Br on the other end). Step 2 - Formation of Grignard reagent (A): When 4-chlorobenzyl bromide is treated with Mg in dry Et2O, a Grignard reagent forms. Mg selectively inserts into the C–Br bond of the benzylic bromide (C–Br is more reactive than the aryl C–Cl bond toward Mg). Therefore, A = 4-ClC6H4CH2MgBr (4-chlorobenzylmagnesium bromide). The aryl C–Cl bond remains intact because aryl chlorides react much more slowly with Mg under these mild conditions compared to benzylic/alkyl bromides. Step 3 - Reaction with CH3CHO: The Grignard reagent (A) acts as a nucleophile and attacks the carbonyl carbon of acetaldehyde (CH3CHO). The benzyl carbanion (4-ClC6H4CH2–) attacks CH3CHO to give, after workup with H3O+, the alcohol: 4-ClC6H4–CH2–CH(OH)–CH3. Step 4 - Workup with H3O+: Protonation of the magnesium alkoxide intermediate gives the free alcohol B = 1-(4-chlorophenyl)propan-2-ol, i.e., Cl-[para-phenylene]-CH2CH(OH)CH3. Step 5 - Elimination of other options: (a) Incorrect: This would require the Grignard to have reacted at the aryl-Cl position while retaining the CH2Br, which does not happen. (c) Incorrect: This implies reaction at both positions of the ring, which is not what occurs under these selective conditions. (d) Incorrect: This would be an elimination product, not a Grignard addition product. Option (b) matches exactly the product of benzylic Grignard addition to acetaldehyde with intact aryl chloride. Therefore, the correct answer is B.