Two-step synthesis: elimination to alkene, then Br₂ addition

Steps:

• Start with 1-bromopropane (CH₃CH₂CH₂Br), an alkyl halide.
• Eliminate HBr using a base (e.g., alcoholic KOH) to form propene (CH₃CH=CH₂).
• Add Br₂ across the propene double bond via electrophilic addition, yielding 1,2-dibromopropane (CH₃CHBrCH₂Br).

Why B is correct:

• Elimination (E2 mechanism) removes HBr from adjacent carbons to create a C=C bond; addition of Br₂ follows Markovnikov's rule but yields the symmetric vicinal dibromide from propene.

Why the others are wrong:

• A: Addition of Br₂ requires an alkene, but 1-bromopropane lacks unsaturation; substitution wouldn't add a second Br.
• C: Hydrolysis converts Br to OH (forming propan-1-ol); elimination then gives propene, but no second Br is introduced.
• D: Substitution replaces the existing Br (e.g., with another nucleophile), preventing formation of a dibromide.

Final answer: B