See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the reagents and reaction sequence: The starting materials are benzene and a compound that appears to be 2-chloro-1-butene (CH2=CCl-CH2CH3) or more likely methacryloyl chloride-type compound. Looking at the structure carefully: it is 2-chlorobut-1-ene (CH2=C(Cl)-CH2CH3). Under AlCl3/H2O conditions, this undergoes Friedel-Crafts acylation or alkylation with benzene. Step 2 - Friedel-Crafts reaction (A): The acid chloride or alkyl chloride reacts with benzene under AlCl3. The structure shown is CH2=C(Cl)-CH2CH3, which is an allylic/vinylic chloride. With AlCl3, this can generate a carbocation. However, looking at the product options and the second step (Wolff-Kishner reduction with N2H4, HO-, heat), product A must be a ketone. So the first step is a Friedel-Crafts acylation. The reagent is likely CH3CH2-CO-Cl (propanoyl chloride, i.e., ethyl acid chloride) giving propiophenone (1-phenyl-1-propanone, PhCO-CH2CH3) as product A. Step 3 - Wolff-Kishner reduction (A -> B): The second step uses N2H4 (hydrazine) and HO- with heat, which is the Wolff-Kishner reduction. This converts a ketone (C=O) to a methylene group (CH2), i.e., it reduces the carbonyl to a CH2 without changing the carbon skeleton. Propiophenone (Ph-CO-CH2CH3) under Wolff-Kishner gives Ph-CH2-CH2-CH3, which is n-propylbenzene (propylbenzene). Step 4 - Verify option B: Propylbenzene (n-propylbenzene) matches option (b), a benzene ring with a propyl (-CH2CH2CH3) group. Step 5 - Why other options fail: (a) Ethylbenzene would result from reduction of acetophenone (PhCOCH3), but our acyl group has 3 carbons total, giving propylbenzene not ethylbenzene. (c) and (d) are alcohols, which would result from incomplete reduction or a different reaction type, not Wolff-Kishner. Therefore, the correct answer is B.