E2 elimination from tertiary halide forms alkenes

Steps:

• Identify 2-bromo-2-methylpentane as a tertiary halide: CH₃-C(Br)(CH₃)-CH₂-CH₂-CH₃.
• Hot concentrated ethanolic NaOH promotes E2 elimination, removing Br and a β-hydrogen.
• β-Hydrogens available on the two methyl groups (leading to 2-methylpent-1-ene) and on C3 methylene (leading to 2-methylpent-2-ene).
• Both alkenes form; 2-methylpent-2-ene is major due to Zaitsev's rule favoring more substituted alkene.

Why B is correct:

• Matches products from β-elimination: terminal alkene from methyl β-H and internal alkene from chain β-H, per Zaitsev's rule.

Why the others are wrong:

• A: Ignores internal alkene from C3 β-hydrogen.
• C: Misses terminal alkene from methyl β-hydrogen.
• D: 4-methylpent-2-ene requires incorrect branching not present in reactant.

Final answer: B