KCET 2026 Chemistry Solved Paper

Concept:This is a crossed aldol condensation, also called Claisen-Schmidt reaction, between benzaldehyde and acetophenone.Explanation:Benzaldehyde has no $\alpha$-hydrogen, so it cannot form an enolate ion.Acetophenone has $\alpha$-hydrogen atoms, so the base $\mathrm{OH}^{-}$ removes one to form the enolate ion:$$\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COCH}_{3} + \mathrm{OH}^{-} \rightleftharpoons \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COCH}_{2}^{-} + \mathrm{H}_{2}\mathrm{O}$$This enolate attacks the electrophilic carbonyl carbon of benzaldehyde, forming an alkoxide intermediate:$$\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CHO} + \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COCH}_{2}^{-} \rightarrow \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CH}(\mathrm{O}^{-})\mathrm{CH}_{2}\mathrm{COC}_{6}\mathrm{H}_{5}$$The alkoxide ion takes a proton from water to give a $\beta$-hydroxy ketone:$$\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CH}(\mathrm{O}^{-})\mathrm{CH}_{2}\mathrm{COC}_{6}\mathrm{H}_{5} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CH}(\mathrm{OH})\mathrm{CH}_{2}\mathrm{COC}_{6}\mathrm{H}_{5} + \mathrm{OH}^{-}$$This $\beta$-hydroxy ketone readily undergoes dehydration to form a conjugated $\alpha,\beta$-unsaturated ketone:$$\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CH}(\mathrm{OH})\mathrm{CH}_{2}\mathrm{COC}_{6}\mathrm{H}_{5} \xrightarrow{-\mathrm{H}_{2}\mathrm{O}} \mathrm{C}_{6}\mathrm{H}_{5}-\mathrm{CH}=\mathrm{CH}-\mathrm{CO}-\mathrm{C}_{6}\mathrm{H}_{5}$$The product is 1,3-diphenylprop-2-en-1-one, commonly called chalcone.Answer:The major product $A$ is chalcone, $\mathrm{C}_{6}\mathrm{H}_{5}-\mathrm{CH}=\mathrm{CH}-\mathrm{CO}-\mathrm{C}_{6}\mathrm{H}_{5}$, which matches option D.