Elimination Reaction in Hot Ethanolic NaOH

Steps:

• Hot ethanolic NaOH with secondary alkyl halide favors E2 elimination over substitution.
• 2-Bromopropane (CH₃CHBrCH₃) undergoes dehydrohalogenation.
• OH⁻ abstracts β-hydrogen from adjacent carbon, Br⁻ leaves.
• Forms propene (CH₃CH=CH₂) as the major product.

Why D is correct:

• E2 mechanism in alcoholic KOH/NaOH yields alkenes from alkyl halides, per standard organic reaction conditions.

Why the others are wrong:

• A: Propan-1-ol requires primary halide and aqueous NaOH for SN2 substitution.
• B: Propan-2-ol is minor SN1 substitution product, but elimination dominates in ethanol.
• C: 2-Hydroxypropane names propan-2-ol, which is substitution product, not major here.

Final answer: D