When the vapours of a secondary alcohol are passed over heated copper at $573$K, the product formed is:

Solution & Explanation

\textbf{Answer:} (C) a ketone \textbf{Solution:} Passing alcohol vapours over heated copper at $573$K brings about catalytic dehydrogenation. A secondary alcohol loses two hydrogen atoms (one from $-\mathrm{OH}$ and one from the carbinol carbon) to give a ketone, e.g. $(\mathrm{CH}{}_3){}_2\mathrm{CHOH}\underset{573\text{K}}{\overset{\text{Cu}}{\to }}(\mathrm{CH}{}_3){}_2\mathrm{C}=\mathrm{O}+\mathrm{H}{}_2$. (A primary alcohol would give an aldehyde and a tertiary alcohol would undergo dehydration to an alkene under these conditions.) Hence a secondary alcohol gives a ketone.