Q.Why sodamide is used in second step to remove HX from alkenyl halide in place of alcoholic KOH?

Step 1. Recall the two-step route to an alkyne. Section 15.3.2 makes an alkyne from a vicinal dihalide in two eliminations: alcoholic KOH removes the first HX to give an alkenyl (vinylic) halide, and a second elimination removes the second HX to give the alkyne.

Step 2. Explain why the second step needs a different, stronger base. In the alkenyl halide intermediate, the hydrogen that must now be removed sits on an sp2-hybridised (vinylic) carbon rather than an ordinary sp3 alkyl carbon. Vinylic C-H bonds are much stronger and far less prone to base-mediated elimination than alkyl C-H bonds, so the comparatively mild alcoholic KOH that worked for the FIRST elimination is simply not strong enough to remove this second, much less reactive hydrogen. …

Comparing the reactivity of a vinylic (sp2) C-H toward base-mediated elimination against an ordinary alkyl C-H, and mat …

• Assuming any strong base would do for both steps -- the point is specifically that alcoholic KOH, though sufficient for the FIRST (alkyl) elimination, is too weak for the SECOND (vinylic) one. …