Step 1: First reaction:
Heating a secondary alkyl halide with alcoholic KOH favours elimination over substitution. Removing H and Br from adjacent carbons gives the alkene $CH_3CH=CH_2$ (propene).
Step 2: Second reaction:
HBr with no peroxide means the ionic mechanism. $H^+$ adds first to the terminal carbon, giving the secondary carbocation $CH_3-CH^+-CH_3$, which is more stable than the primary one.
Step 3: Capture by bromide:
$Br^-$ attacks the cation and produces $CH_3CHBrCH_3$, which is 2-bromopropane.
Step 4: Eliminate the rest:
1-Bromopropane needs the radical (peroxide) route. Bromoethane and $CBr_4$ have the wrong carbon count. Option (C) stands.
Final Answer:
Option (C), 2-bromopropane.
\[ \boxed{\text{2-Bromopropane}} \]
