When 2-bromobutane reacts with $\mathrm{KOH}$ in dry ethanol, the major product is but-2-ene. If the ethanol is progressively replaced by water, the proportion of butan-2-ol rises. The best explanation is

1. Awater converts the E2 mechanism into E1, which always gives substitution only
2. Bwater drives the equilibrium $\mathrm{EtO}{}^{-}+\mathrm{H}{}_2\mathrm{O}\rightleftharpoons \mathrm{EtOH}+\mathrm{OH}{}^{-}$ to the left to give less ethoxide, and solvated hydroxide acts less as a base, so SN2 competes better
3. Cwater is a polar aprotic solvent, which favours elimination less than SN2
4. Dwater increases the steric hindrance at the alpha carbon of 2-bromobutane