Which one of the following is \textbf{not} formed when acetone ($\mathrm{CH}{}_3\mathrm{COCH}{}_3$) reacts with 2-pentanone ($\mathrm{CH}{}_3\mathrm{COCH}{}_2\mathrm{CH}{}_2\mathrm{CH}{}_3$) in the presence of dilute $\mathrm{NaOH}$ followed by heating?

Solution & Explanation

\textbf{Answer:} (B) $\mathrm{CH}{}_3-\mathrm{CO}-\mathrm{C}(\mathrm{CH}{}_3)=\mathrm{CH}-\mathrm{CH}{}_3$. \textbf{Solution:} Both ketones have $\alpha$-hydrogens, so dilute $\mathrm{NaOH}$ with heat gives self- and cross-aldol condensation products ($\alpha ,\beta$-unsaturated ketones). Valid products arise only from the enolate of one ketone attacking the carbonyl of either ketone: acetone self-condensation gives mesityl oxide (A), and the acetone/2-pentanone cross combinations give (C) and (D). The structure in (B) cannot be assembled from any legitimate enolate-carbonyl pairing of these two ketones, so it is not formed.