Step 1: Identify the reagent type.
Alcoholic KOH provides $OH^-$ ions in a less polar (alcohol) medium. In this environment $OH^-$ acts as a strong base rather than a nucleophile, favouring elimination over substitution.
Step 2: Identify the substrate.
n-Butyl chloride is $CH_3CH_2CH_2CH_2Cl$ (1-chlorobutane), a primary alkyl halide. The $\alpha$-carbon bears the $-Cl$ group, and the $\beta$-carbon is $-CH_2$-adjacent to it.
Step 3: Describe the elimination mechanism.
The base abstracts a $\beta$-hydrogen while the C-Cl bond breaks simultaneously (E2 mechanism). This forms a double bond between the $\alpha$ and $\beta$ carbons and releases $Cl^-$.
Step 4: State the product.
But-1-ene ($CH_3CH_2CH=CH_2$) is formed as the product along with KCl and water: $CH_3CH_2CH_2CH_2Cl + KOH \xrightarrow{\text{alc, } \Delta} CH_3CH_2CH=CH_2 + KCl + H_2O$.