Elimination reaction yields isomeric alkenes from 2-bromobutane (noted as propene in query, but matches standard for 3 products)

Steps:

• 2-Bromobutane (CH₃CHBrCH₂CH₃) undergoes E2 elimination with ethanolic NaOH, removing HBr.
• Beta-hydrogens on C1 yield but-1-ene (CH₂=CHCH₂CH₃).
• Beta-hydrogens on C3 yield but-2-ene (CH₃CH=CHCH₃), with E/Z stereoisomers.
• No other alkenes form; substitution products (ethers/alcohols) are not hydrocarbons.

Why C is correct:

• Three distinct C₄H₈ hydrocarbons: but-1-ene (structural isomer), (E)-but-2-ene, (Z)-but-2-ene (stereoisomers), per Zaitsev's rule favoring internal alkene.

Why the others are wrong:

• A: Ignores multiple beta positions and stereoisomerism, assuming only one product.
• B: Counts but-2-ene as one (ignoring E/Z) plus but-1-ene, missing stereoisomers.
• D: No fourth isomer; 2-methylpropene requires different structure.

Final answer: C