Ethanolic NaOH promotes elimination in 2°/3° haloalkanes but substitution in unhindered 1° ones

Steps:

• Classify options: A (primary, beta-branched), B (tertiary), C (secondary), D (primary, unhindered).
• Recall ethanolic NaOH mechanism: 2°/3° favor E2 elimination forming alkenes; unhindered 1° favor SN2 substitution forming ethers.
• Note beta-branching in 1° haloalkanes sterically hinders SN2, allowing E2 competition.
• Select D as unhindered primary, which resists elimination.

Why D is correct:

• 1-Bromobutane is an unhindered primary haloalkane; by SN2 mechanism, it forms 1-ethoxybutane via substitution, not alkene via elimination.

Why the others are wrong:

• A: Beta-branched primary; hindrance favors E2 to 2-methylpropene.
• B: Tertiary; E2 rapidly forms 2-methylpropene.
• C: Secondary; E2 forms but-1-ene and but-2-ene.

Final answer: D