See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Friedel-Crafts Acylation In a Friedel-Crafts acylation, an acid chloride reacts with an aromatic ring in the presence of AlCl3 (a Lewis acid catalyst) to introduce an acyl group onto the ring. Step 1: Identify the reagents. The acid chloride is 4-chlorobenzoyl chloride (Cl-C(=O)-C6H4-Cl, where the ring bears a Cl at the para position relative to the carbonyl). The aromatic substrate is benzene (C6H6). AlCl3 is the Lewis acid catalyst. Step 2: Determine the reaction. AlCl3 activates the acid chloride by coordinating with the chlorine of the acyl chloride, generating an acylium ion: [4-ClC6H4-C≡O]+. This electrophilic acylium ion attacks benzene in an electrophilic aromatic substitution (EAS) reaction. Step 3: Determine the product. The acylium ion from 4-chlorobenzoyl chloride is (4-chlorophenyl)oxocarbenium. It attacks benzene to give (4-chlorophenyl)(phenyl)methanone, i.e., 4-chlorobenzophenone. This is a diaryl ketone where one aryl group is an unsubstituted phenyl and the other is a para-chlorophenyl, connected through a C=O. Step 4: Match with options. (a) Shows a biphenyl-type structure with C(=O)Cl still present — this would imply a Friedel-Crafts coupling rather than acylation losing Cl from carbonyl incorrectly. Incorrect. (b) Shows (phenyl)(2-chlorophenyl)methanone — ortho-chlorobenzophenone. The starting acid chloride has Cl at para position, not ortho. Incorrect. (c) Shows (phenyl)(4-chlorophenyl)methanone — para-chlorobenzophenone. This matches exactly: benzene is acylated by 4-chlorobenzoyl chloride to give 4-chlorobenzophenone. Correct. (d) Shows benzophenone (Ph-CO-Ph) with no chlorine — loss of the para-Cl is not expected under these conditions. Incorrect. Therefore, the correct answer is C.