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

Tert-butylbromide undergoes substitution by S_N1 mechanism because it is able to form a stable carbocation in the first step after cleavage of the halide group. The carbocation then reacts with the nucleophile OH^-. On the other hand, primary halide n-butylbromide cannot form a stable carbocation so it undergoes S_N2 mechanism, which is a one-step substitution which involves attack of OH^- and simultaneous leaving of X^- to form n-butyl alcohol.