Elimination reaction in hot ethanolic NaOH

Steps:

• Hot ethanolic NaOH favors E2 elimination over substitution for primary alkyl halides like bromoethane (likely intended, as bromomethane lacks beta hydrogens).
• The reaction removes HBr from the alkyl halide: CH3CH2Br + OH⁻ → CH2=CH2 + H2O + Br⁻.
• Ethene (C2H4) forms as the alkene product.
• Ethene is a colorless, gaseous alkene at room temperature.

Why C is correct:

• Ethene is the gaseous product of dehydrohalogenation, matching the colorless gas description per organic elimination mechanisms.

Why the others are wrong:

• A: 1,2-Dibromoethane is a liquid, not a gas, and requires different reactants.
• B: Ethanol is a liquid byproduct in substitution, not elimination.
• D: Hydrogen bromide reacts with NaOH to form NaBr and water, not evolving as a gas.

Final answer: C