Which of the following reactions will yield $$2, 2-$$dibromopropane?

The reaction required is an electrophilic addition of two moles of $$HBr$$ to an unsymmetrical alkyne or alkene. For an alkyne, the first mole of $$HBr$$ gives a vinyl bromide; the second mole adds to the double bond formed in the first step. In the absence of peroxides, each addition follows Markovnikov’s rule: the hydrogen attaches to the carbon that already has more hydrogens and the bromine attaches to the other carbon.

Case A: $$CH_3\!-\!C\equiv CH$$ (propyne) + $$2\,HBr$$
  Step 1 (first mole of $$HBr$$):
  H attaches to the terminal carbon, Br to the internal carbon, giving $$CH_3\!-\!C(Br)=CH_2$$ (2-bromoprop-1-ene).
  Step 2 (second mole of $$HBr$$):
  H again attaches to the terminal carbon of the new double bond, Br to the same internal carbon, yielding $$CH_3\!-\!CBr_2\!-\!CH_3$$, i.e. $$2,2$$-dibromopropane.

Thus Option A furnishes the required geminal dihalide.

Case B: $$CH_3CH=CHBr$$ + $$HBr$$ → Markovnikov addition places H on $$CH_3$$ carbon and Br on the vinylic carbon that already bears Br, giving $$CH_3CH_2CHBr_2$$ (1,1-dibromopropane), not the desired product.

Case C: $$CH\equiv CH$$ + $$2\,HBr$$ → successively gives $$CH_2=CHBr$$ and then $$CH_3CHBr_2$$ (1,1-dibromoethane), which has only two carbon atoms.

Case D: $$CH_3CH=CH_2$$ + $$HBr$$ → gives the mono-bromide $$CH_3CHBrCH_3$$ (2-bromopropane), containing only one bromine atom.

Only Option A produces $$2,2$$-dibromopropane.

Answer: Option A which is: $$CH_3\!-\!C\equiv CH + 2\,HBr \longrightarrow CH_3\!-\!CBr_2\!-\!CH_3\; (2,2\text{-dibromopropane})$$