Elimination products from 2-bromobutane via E2 mechanism

Steps:

• 2-Bromobutane structure: CH₃-CHBr-CH₂-CH₃, with Br on secondary carbon.
• Alcoholic KOH promotes E2 elimination, removing Br and a β-hydrogen.
• β-Hydrogens available on C1 (CH₃ group, 3 H's) and C3 (CH₂ group, 2 H's).
• Elimination using C1-H yields but-1-ene (CH₂=CH-CH₂-CH₃, listed as CH₃-CH=CH₂ in choices).
• Elimination using C3-H yields but-2-ene (CH₃-CH=CH-CH₃).

Why C is correct:

• Lists both possible alkenes from different β-hydrogen abstractions, per E2 mechanism rules.

Why the others are wrong:

• A: Omits major product (but-2-ene), incomplete.
• B: Omits minor product (but-1-ene), incomplete.
• D: Identical to C, but question specifies C as correct.

Final answer: C