Write the mechanism of the following reaction:
$n-BuBr + KCN \xrightarrow{EtOH/H_2O} n-BuCN$

(N/A) The given reaction is:
$n-BuBr + KCN \xrightarrow{EtOH/H_2O} n-BuCN$
This reaction proceeds via an $S_{N}2$ mechanism.
$1$. The nucleophile $CN^{-}$ attacks the electrophilic carbon atom of $n-butyl$ bromide $(n-BuBr)$ from the side opposite to the leaving group $(Br^{-})$.
$2$. This results in the formation of a transition state where the $C-CN$ bond is partially formed and the $C-Br$ bond is partially broken.
$3$. Finally,the $Br^{-}$ ion leaves,resulting in the formation of $n-butyl$ cyanide $(n-BuCN)$.
$4$. The $CN^{-}$ ion is an ambident nucleophile,but in the presence of $KCN$,the attack occurs primarily through the carbon atom due to the higher nucleophilicity of the carbon center.