$tert$-Butylbromide reacts with $aq.$ $NaOH$ by $S_N1$ mechanism while $n$-butylbromide reacts by $S_N2$ mechanism. Why?

(N/A) $tert$-Butylbromide is a tertiary $(3^{\circ})$ alkyl halide. The carbocation formed after the loss of the leaving group $(Br^-)$ is a tertiary carbocation,which is highly stable due to the inductive effect and hyperconjugation of the three methyl groups. Thus,it prefers the $S_N1$ mechanism.
$n$-Butylbromide is a primary $(1^{\circ})$ alkyl halide. The formation of a primary carbocation is energetically unfavorable due to its low stability. Furthermore,the primary carbon is less sterically hindered,allowing the nucleophile $(OH^-)$ to attack from the back side,which is the characteristic step of the $S_N2$ mechanism.