Consider the following reactions of $\mathrm{X}$ ($\mathrm{C}{}_2\mathrm{H}{}_6\mathrm{O}$): $\mathrm{X}\underset{573\mathrm{K}}{\overset{\mathrm{Cu}}{\to }}\mathrm{A}$ $\mathrm{A}\underset{\mathrm{\Delta }}{\overset{\text{[Ag(NH3)2]+, OH-}}{\to }}\mathrm{silver}\mathrm{mirror}$ $\mathrm{A}\underset{\mathrm{aldol}}{\overset{\mathrm{OH}{}^{-},\mathrm{\Delta }}{\to }}\mathrm{Y}$ $\mathrm{A}\stackrel{\text{H2N-NH-CO-NH2}}{\to }\mathrm{Z}$ Identify $\mathrm{A}$, $\mathrm{X}$, $\mathrm{Y}$ and $\mathrm{Z}$.

Solution & Explanation

\textbf{Answer:} (C) A = ethanal, X = ethanol, Y = but-2-enal, Z = semicarbazone. \textbf{Solution:} Catalytic dehydrogenation over $\mathrm{Cu}$ at 573 K converts a primary alcohol to an aldehyde, so $\mathrm{X}$ = ethanol ($\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{OH}$) and $\mathrm{A}$ = ethanal ($\mathrm{CH}{}_3\mathrm{CHO}$); the positive Tollens' test (silver mirror) confirms $\mathrm{A}$ is an aldehyde. Base-catalysed aldol of ethanal followed by heating gives $\mathrm{CH}{}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}$ (but-2-enal, $\mathrm{Y}$). Ethanal with semicarbazide ($\mathrm{H}{}_2\mathrm{N}-\mathrm{NH}-\mathrm{CO}-\mathrm{NH}{}_2$) gives the semicarbazone $\mathrm{CH}{}_3-\mathrm{CH}=\mathrm{N}-\mathrm{NH}-\mathrm{CO}-\mathrm{NH}{}_2$ ($\mathrm{Z}$).