Elimination favored in ethanolic NaOH

Steps:

• Ethanolic NaOH acts as a strong base in alcohol solvent, promoting elimination over substitution.
• 2-Bromopropane is a secondary alkyl halide, prone to E2 mechanism.
• In E2, beta-hydrogen is abstracted, and Br leaves, forming a double bond.
• The product is propene (CH3-CH=CH2) via dehydrohalogenation.

Why D is correct:

• Alcoholic NaOH induces E2 elimination per Zaitsev's rule, yielding the most stable alkene, propene, as the major product.

Why the others are wrong:

• A: Propan-1-ol requires SN2 on primary halide; secondary favors elimination.
• B: Propan-2-ol from SN1 substitution, but ethanolic conditions suppress nucleophilic attack.
• C: 2-Hydroxypropane names propan-2-ol, incorrect for elimination pathway.

Final answer: D