1-chloropropane can react with aqueous sodium hydroxide to form propan-1-ol. However, when 1-chloropropane reacts with ethanolic sodium hydroxide instead, a different type of reaction occurs and propene is formed. Explain why changing the solvent changes the type of reaction that takes place, and describe the difference between the two reactions.

1-chloropropane reacts with aqueous NaOH via a nucleophilic substitution (S_N2). The hydroxide ion (OH⁻) is a strong nucleophile and attacks the electrophilic carbon bearing the chlorine, displacing Cl⁻ and giving propan‑1‑ol.

In ethanolic NaOH the solvent is polar protic but less able to stabilise the transition state for substitution. The hydroxide ion behaves mainly as a strong base. It abstracts a β‑hydrogen from the 2‑carbon of 1‑chloropropane, while the C–Cl bond breaks to give a double bond. This is an E2 elimination, producing propene.

Thus the change of solvent changes the reaction from a nucleophilic substitution to a base‑promoted elimination, giving an alcohol in water and an alkene in ethanol.