JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Hydrocarbons

The clue chain pins down $A$ one step at a time. First, $A$ gives a precipitate with $\mathrm{A}{\mathrm{g}}_2\mathrm{O}$ (ammoniacal silver): only a terminal alkyne, with its acidic $\equiv \mathrm{C}-\mathrm{H}$, forms an insoluble silver acetylide, so $A$ has to be a terminal alkyne. Next, $A$ undergoes $\mathrm{H}{\mathrm{g}}^{2+}/{\mathrm{H}}^{+}$ hydration (Markovnikov) to the carbonyl $B$, which $\mathrm{NaB}{\mathrm{H}}_4$ then reduces to the alcohol $C$. Finally $C$ is treated with $\mathrm{ZnC}{\mathrm{l}}_2$/conc. $\mathrm{HCl}$, the Lucas reagent: turbidity appearing within about five minutes is the signature of a secondary alcohol. For $C$ to be secondary, its precursor $B$ must be a ketone, i.e. the hydration of $A$ delivered the oxygen onto an internal carbon. Acetylene, $\mathrm{CH}\equiv \mathrm{CH}$, would hydrate to acetaldehyde and finish as ethanol, a primary alcohol that gives no quick Lucas turbidity; but-$2$-yne is internal and would give no silver-acetylide precipitate at all. Propyne, $\mathrm{C}{\mathrm{H}}_3-\mathrm{C}\equiv \mathrm{CH}$, hydrates to acetone, which $\mathrm{NaB}{\mathrm{H}}_4$ reduces to propan-$2$-ol — a secondary alcohol that turns the Lucas reagent turbid in a few minutes — and its terminal $\equiv \mathrm{C}-\mathrm{H}$ accounts for the $\mathrm{A}{\mathrm{g}}_2\mathrm{O}$ precipitate. Hence $A$ is $\mathrm{C}{\mathrm{H}}_3-\mathrm{C}\equiv \mathrm{CH}$.