See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the reagent: The second reactant is beta,beta-dimethylacrylic acid (3-methylbut-2-enoic acid): (CH3)2C=CH-COOH. This is an alpha,beta-unsaturated carboxylic acid. Step 2 - Role of AlCl3: AlCl3 is a Lewis acid catalyst used in Friedel-Crafts reactions. With an unsaturated carboxylic acid, AlCl3 can coordinate with the carbonyl oxygen and promote formation of an acylium-type or carbocation intermediate. Step 3 - Mechanism with beta,beta-dimethylacrylic acid and AlCl3: AlCl3 complexes with the carboxylic acid. The protonation/activation of the double bond or the carboxyl group generates a tertiary carbocation at the beta-carbon (the (CH3)2C+ end), because the dimethyl-substituted carbon forms the more stable tertiary carbocation. This is a classical Friedel-Crafts alkylation via a carbocation intermediate rather than acylation. Step 4 - Carbocation formation: (CH3)2C=CH-COOH + AlCl3 -> [(CH3)2C(+)-CH2-COOH] (tertiary carbocation at the isobutenyl terminus via protonation of the double bond, or via ionization). The tertiary carbocation (CH3)2C(+)- attacks benzene in electrophilic aromatic substitution. Step 5 - Product formation: The tertiary carbocation (CH3)2C(+)-CH2-COOH attacks the benzene ring, giving Ph-C(CH3)2-CH2-COOH after loss of proton. This is 3-methyl-3-phenylbutanoic acid, which corresponds to option (c): a benzene ring bearing a -C(CH3)2-CH2-CO2H group. Step 6 - Why other options fail: - Option (a) would require a different carbon skeleton with isopropyl and COOH on the same carbon attached to phenyl; does not match the expected carbocation pathway. - Option (b) would be the Friedel-Crafts acylation product (a ketone); acylation would give Ph-CO-CH=C(CH3)2, but the high yield (82%) and the known chemistry of beta,beta-dimethylacrylic acid with AlCl3 favor alkylation via the tertiary carbocation, not acylation. - Option (d) is tert-butylbenzene, which would require loss of CO2 as well; while decarboxylation can occur, the observed 82% yield product retains the carboxylic acid function as in option (c). Step 7 - Conclusion: The reaction proceeds via Friedel-Crafts alkylation where AlCl3 promotes protonation across the double bond to give the tertiary carbocation (CH3)2C(+)-CH2-COOH, which then alkylates benzene to give Ph-C(CH3)2-CH2-COOH (option c). Therefore, the correct answer is C.