Substitution Versus Elimination

Whether an alkyl halide undergoes substitution or elimination depends on the reagent and medium. In aqueous potassium hydroxide the hydroxide ion is heavily hydrated and behaves mainly as a nucleophile, attacking the carbon bearing bromine to give an alcohol by substitution. In alcoholic potassium hydroxide some ethoxide ion is generated, and the less-solvated, more basic species abstracts a beta-hydrogen, promoting elimination to give an alkene. The bromide-leaving option is wrong because the leaving group departs in both solvents; only the product differs. The temperature option is incorrect because the change is due to reagent behaviour, not to a large difference in boiling-driven temperature. The potassium oxidation-state option is wrong because potassium remains in its +1 state throughout and plays no redox role. The deciding factor is the relative solvation and basicity of the attacking species: water heavily solvates hydroxide and blunts its basicity so it prefers carbon attack, whereas in ethanol the partially formed ethoxide is a stronger, less-hindered base that prefers to grab a beta-hydrogen. This same competition between substitution and elimination is central to predicting products throughout haloalkane chemistry. This matches the NCERT contrast between aqueous and alcoholic KOH. Sanity check: nucleophilic substitution gives a C-OH bond while base-driven elimination gives a C=C bond, exactly the two observed products.