Elimination followed by halogen addition

Steps:

• 2-Bromopropane undergoes dehydrohalogenation with ethanolic NaOH under reflux to form propene (CH₃CH=CH₂) via E2 mechanism.
• Propene, as compound X, reacts with Br₂ at room temperature to add across the double bond, yielding 1,2-dibromopropane (CH₃CHBrCH₂Br).
• Aqueous NaOH promotes substitution to alcohol, not elimination.
• HBr addition to propene regenerates bromopropane, not dibromide.

Why D is correct:

• Ethanolic NaOH favors E2 elimination to alkene; Br₂ undergoes electrophilic addition to form vicinal dibromide per Markovnikov's rule.

Why the others are wrong:

• A: Aqueous NaOH gives substitution to propan-2-ol; HBr adds to alkene but yields monobromide.
• B: Ethanolic NaOH gives propene, but HBr addition reforms 2-bromopropane.
• C: Aqueous NaOH gives alcohol, not alkene for Br₂ addition.

Final answer: D