See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Aldol Condensation (first arrow): Acetophenone (Ph-CO-CH3) reacts with NaOH as base and p-tolualdehyde (4-MeC6H4-CHO) in an aldol condensation. The methyl group of acetophenone is deprotonated to give an enolate, which attacks the aldehyde carbonyl of p-tolualdehyde. After dehydration (-H2O, Heat), the product is an alpha,beta-unsaturated ketone (chalcone-type): (E)-1-phenyl-3-(4-methylphenyl)prop-2-en-1-one, i.e., Ph-CO-CH=CH-(4-MeC6H4). The double bond forms between the alpha carbon of acetophenone and the carbon from the aldehyde; the aryl group of the aldehyde (4-MeC6H4) ends up on the beta carbon (far from the carbonyl). Step 2 - Conjugate (1,4) Addition with lithium enolate of cyclohex-2-en-1-ol: The reagent shown is the lithium enolate of cyclohex-2-en-1-one (drawn as OLi on the cyclohexene ring). This is equivalent to the lithium enolate of cyclohexanone acting as a nucleophile. In a 1,4 (Michael) addition, the enolate carbon of cyclohexanone attacks the beta carbon of the chalcone (the carbon bearing the 4-methylphenyl group). After aqueous workup (H2O), the enolate is protonated and the ketone is restored. Step 3 - Regiochemical outcome: The Michael addition places the cyclohexanone carbon at the beta position of the enone. The beta carbon of Ph-CO-CH=CH-(4-MeC6H4) bears the 4-MeC6H4 group. After 1,4-addition by the cyclohexanone enolate at this beta carbon, the product is: 2-[CH(4-MeC6H4)-CH2-CO-Ph]cyclohexan-1-one. That is, cyclohexanone at C2 has a substituent consisting of -CH(4-MeC6H4)-CH2-C(=O)-Ph. Step 4 - Match to options: This saturated 1,4-adduct with 4-MeC6H4 on the carbon directly attached to the cyclohexanone ring and Ph-CO on the terminal end corresponds to option (c): 2-(3-oxo-1-(4-methylphenyl)-3-phenylpropyl)cyclohexan-1-one. Why other options fail: - (a) and (d) show unsaturated (vinylogous/conjugated) products - these would require no Michael addition or incomplete reaction; the 1,4-addition gives a saturated chain. - (b) has phenyl on the carbon attached to cyclohexanone and 4-MeC6H4 on the carbonyl end - this would require 1,2-addition or reversed regiochemistry, which is not favored. - (c) correctly has 4-MeC6H4 on the carbon alpha to cyclohexanone and Ph-C=O at the end, consistent with Michael addition at the beta carbon of the chalcone. Therefore, the correct answer is C.