Q.Starting from 1,2-dibromopropane, outline the steps needed to prepare propyne.

1,2-Dibromopropane, $\text{CH}_3\text{CHBrCH}_2\text{Br}$, is a vicinal dihalide and needs TWO successive elimination steps to reach the triple bond. The first elimination, using hot alcoholic KOH, removes one molecule of $\text{HBr}$ across two adjacent carbons to give the vinyl bromide (haloalkene) intermediate, $\text{CH}_3\text{CBr=CH}_2$ (2-bromopropene) -- an ordinary dehydrohalogenation, exactly as used to make an alkene from a mono-halide. This first elimination is comparatively easy because the starting $\text{C--Br}$ bonds are on ordinary $sp^3$ carbons. The SECOND elimination -- removing the remaining $\text{HBr}$ from the vinyl bromide to form the triple bond -- is intrinsically harder, because it requires deprotonating a vinylic hydrogen (on an $sp^2$ carbon, which holds it more tightly and does not lose it easily to a weaker base), so a stronger base than …

Recognise that a dihalide needs elimination TWICE to reach a triple bond, and that the second elimination (removing a vinylic hydrogen) needs a str …

Trying to remove both HBr molecules in a single step with only alcoholic KOH (it is not strong enough to complete the second, vinylic elimination); forgetting that a GEMINAL dihalide (both halogens on the same carbon) also …