Q.Explain why ethyne is more acidic than ethene, which is in turn more acidic than ethane, and describe a chemical test that distinguishes a terminal alkyne from the other two.

The carbon bearing the hydrogen being removed is $sp$ hybridised in ethyne ($50\%$ $s$-character), $sp^2$ hybridised in ethene ($33\%$ $s$-character), and $sp^3$ hybridised in ethane ($25\%$ $s$-character). An $s$ orbital holds its electron density, on average, much closer to the positively charged nucleus than a $p$ orbital does, so a hybrid orbital with more $s$-character behaves as though it is more electronegative and stabilises a lone pair of electrons (as left behind on the carbanion after deprotonation) more effectively. Because acidity is set by how well the conjugate base (the carbanion) is stabilised, acidity therefore runs in the same order as $s$-character: $sp$ (ethyne, $\text{p}K_a \approx 25$) $>$ $sp^2$ (ethene, $\approx 44$) $>$ $sp^3$ (ethane, $\approx 50$). A simple test that exploits this is the reaction of a TERMINAL alkyne (one with the triple bond at the very end of the chain, carrying an acidic $\equiv\text{C--H}$) with ammoniacal silver nitrate solution: the acidic hydrogen is displaced and a white precipitate of silver acetylide, $\text{RC}{\equiv}\text{CAg}$, forms. Neither an alkene nor …

Identify the hybridisation of the carbon bearing the relevant hydrogen in each case, rank by $s$-character, and connect higher $s$-character to better stabilisation of the resulting carbanion (hence higher acidity); recall the specific reagent (am …

Reversing the acidity order (assuming more p-character, not more s-character, stabilises the carbanion); applying the silver-nitrate test to a NON-terminal (internal) alk …