See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: This reaction involves a base-catalyzed crossed aldol condensation (specifically a conjugate addition / Michael-type aldol) between two alpha,beta-unsaturated aldehydes under heating conditions in ethanol solvent. Step 1: Identify the reactants. - Reactant 1: Ph–CH=CH–CHO (cinnamaldehyde, an alpha,beta-unsaturated aldehyde with no alpha-hydrogens on the aldehyde carbon side that are easily enolizable, but it has a beta-carbon activated by the phenyl group) - Reactant 2: CH3–CH=CH–CHO (crotonaldehyde, an alpha,beta-unsaturated aldehyde with an alpha-methyl group) Step 2: Determine the reactive site under base conditions. Under base (EtOH, heat), the alpha-carbon of crotonaldehyde (the CH3 group adjacent to the conjugated system) is deprotonated to form a resonance-stabilized carbanion/enolate. This nucleophilic carbon attacks the beta-carbon of cinnamaldehyde in a Michael (1,4-conjugate addition) fashion, followed by aldol condensation and dehydration. Step 3: Trace the bond formation. - The enolate of crotonaldehyde attacks the beta-carbon of cinnamaldehyde (the carbon alpha to Ph). - After the Michael addition and subsequent intramolecular aldol condensation with dehydration, the chain is extended by two carbons with formation of a new C=C double bond. - The product extends the conjugated polyene chain: Ph is connected through three CH=CH units to CHO. - This gives Ph–(CH=CH)3–CHO. Step 4: Count the vinyl units. - Cinnamaldehyde contributes: Ph–CH=CH–CHO (one CH=CH unit between Ph and CHO) - Crotonaldehyde contributes: CH3–CH=CH–CHO (one CH=CH unit) - The condensation merges the two molecules: the methyl end of crotonaldehyde couples with the beta-carbon of cinnamaldehyde, and after dehydration, two additional CH=CH units are incorporated, giving Ph–(CH=CH)3–CHO (three conjugated double bonds between Ph and CHO). Step 5: Why other options fail. - Option (a) Ph–(CH=CH)2–CHO: This would result from adding only one additional vinyl unit, which does not account for the full condensation product of both reactants. - Option (c) Ph–(CH=CH)4–CHO: This would require incorporation of four vinyl units, which would need three reactant molecules to combine, not two. - Option (d) Ph–CH=CH–CH=CH–CH3: This lacks the aldehyde group and is not consistent with a product retaining the CHO functionality from an aldol condensation. Therefore, the correct answer is B.