Final answer:
Conditions that promote elimination reactions in haloalkanes include the use of strong bases, heat, and steric hindrance. Sulfuric acid catalyzes the elimination in alcohols by protonating the OH group, and the poor nucleophilicity of its conjugate base, [HSO4]-, favors elimination over substitution.
Step-by-step explanation:
Haloalkanes can undergo either elimination reactions or substitution when reacting with NaOH, but certain conditions favor elimination. Specifically, elimination is promoted by steric hindrance at the substrate or the presence of a bulky base, the use of stronger bases, and heat. When it comes to alcohols, the O of alcohols (R-OH) is protonated by sulfuric acid (H₂SO₄) in water solution to facilitate elimination. The resulting conjugate base [HSO4]- is a poor nucleophile, which encourages elimination over substitution. Additionally, in the presence of a strong acid catalyst, water molecules act as bases to remove a β-hydrogen atom, leading to the formation of an alkene. For haloalkanes, the catalyst is typically a strong base, which helps remove a halide ion and a neighboring hydrogen to form a double bond.