Assertion (A): 1-Chloropropane gives mainly propan-1-ol with aqueous KOH but mainly propene with alcoholic KOH. Reason (R): Alcoholic KOH contains the alkoxide ion $C X_{2} H X_{5} O X^{-}$ , a stronger base than $OH X^{-}$ in water, which favours E2 elimination.

Answer: A

In water,

OH X^{-}

is heavily solvated and acts mainly as a nucleophile (SN2), giving the alcohol. In ethanol the equilibrium produces

C X_{2} H X_{5} O X^{-}

, a stronger base, so E2 elimination to propene dominates. Both statements are true and R explains A.