Concept:Dehydrohalogenation of an alkyl halide gives an alkene, which then undergoes Markovnikov addition of HBr.Explanation:The starting compound is 2-bromopropane: CH3​−CH(Br)−CH3​.On heating with alcoholic KOH, β-elimination occurs to give propene, which is product A.CH3​−CH(Br)−CH3​Alc. KOH, Δ​CH3​−CH=CH2​Propene then reacts with HBr in the absence of peroxide.According to Markovnikov's rule, the H atom adds to the carbon of the double bond that already has more hydrogen atoms.So, H adds to the terminal carbon, and Br adds to the middle carbon, regenerating 2-bromopropane.CH3​−CH=CH2​+HBr⟶CH3​−CH(Br)−CH3​Therefore, the major product B is 2-bromopropane.Answer:C. 2-Bromopropane
