See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is ethylbenzene (benzene with a -CH2CH3 group). Step 2 - First reaction with AlCl3: AlCl3 is a Lewis acid catalyst used in Friedel-Crafts alkylation. However, the reagent shown going into the first arrow is not explicitly shown in the crop, but AlCl3 is the catalyst. Given the context and that the product is used in the next step, the first step likely involves a Friedel-Crafts acylation (with an acyl chloride or similar) to give product (A) — a ketone intermediate (acylation product). Looking at option (a), it shows a triethylbenzene with a ketone (=O), suggesting (A) is a propiophenone-type or polysubstituted aryl ketone. Step 3 - Second reaction with triethylene glycol and heat: Triethylene glycol (HOCH2CH2OCH2CH2OCH2CH2OH) at high temperature is a high-boiling solvent used for Wolff-Kishner-type or Clemmensen-type reduction conditions. More specifically, heating a carbonyl compound (ketone) with a diol under acidic or basic conditions can effect reduction. However, triethylene glycol with heat (without hydrazine or KOH) is actually used as a high-boiling solvent for thermal reactions. In this context, the combination suggests a Wolff-Kishner reduction variant or simply that the ketone C=O is reduced to CH2 (deoxygenation). The ketone -CO- between two ethyl-bearing carbons gets reduced to -CH2-, effectively giving a fully alkylated ethylbenzene product. Step 4 - Determine regiochemistry of product (A): Friedel-Crafts acylation of ethylbenzene would place the acyl group ortho or para to the ethyl group (directing). The major product (A) at 80% would be para-substituted or a diethyl-substituted aryl ketone depending on the reagent. Given the final product (B) is 1,3,5-triethylbenzene (option b), the intermediate (A) must be a ketone that upon reduction gives the 1,3,5-pattern. Step 5 - Why 1,3,5-triethylbenzene (option b): The Friedel-Crafts reaction on ethylbenzene can give a diethyl-substituted benzene ketone. Reduction of the carbonyl (C=O → CH2) via heating in triethylene glycol (acting as solvent for a modified Huang Minlon / Wolff-Kishner) yields 1,3,5-triethylbenzene. The 1,3,5-arrangement is thermodynamically favored and consistent with para-selectivity in successive electrophilic substitutions. Options (a) still retains the ketone so it is not fully reduced. Option (c) is 1,2,3-triethylbenzene which would require ortho-ortho substitution, sterically disfavored. Option (d) shows isopropyl substituents rather than ethyl groups. Step 6 - Elimination of other options: (a) retains =O, so reduction did not occur — eliminated. (c) 1,2,3-pattern is sterically unfavorable for electrophilic substitution — eliminated. (d) shows branched (isopropyl) groups, not ethyl groups — eliminated. (b) 1,3,5-triethylbenzene is the correct symmetric product of para-directed Friedel-Crafts followed by carbonyl reduction. Therefore, the correct answer is B.