Markovnikov addition of excess HBr to terminal alkyne

Steps:

• Propyne (CH3C≡CH) undergoes electrophilic addition with HBr.
• First HBr adds Markovnikov, forming vinyl bromide CH3CBr=CH2.
• Second HBr adds to the alkene, again Markovnikov, yielding geminal dibromide CH3CBr2CH3.
• Excess HBr drives complete double addition.

Why D is correct:

• Double Markovnikov addition to terminal alkynes with excess HX produces geminal dihalides (CH3CBr2CH3), as per alkyne reactivity rules.

Why the others are wrong:

• A: Forms from one equivalent of HBr only, not excess.
• B: Isopropyl bromide from propene + HBr, not alkyne.
• C: 1,2-dibromide from anti-Markovnikov or alkene addition.

Final answer: D