tert-Butylbromide reacts with aq. NaOH by `S_(N)1` mechanism while n butylbromide reacts by `S_(N)2` mechanism. Why?

Explanation

tert-Butyl readily loses `Br^(-)` ion to form stable tert-butyl carbocation. Therefore, it undergoes reaction by `S_(N)1` mechanism which occurs in two steps. In the first step, tert-butyl carbocation is formed. This step is slow and hence is the rate determining step of the reaction. In the second step, the tert-butyl carbocation is readily attacked by `OH^(-)` ion to form tert-butyl alcohol.

On the other hand, n-butyl bromide does not undergo ionization to produce n-butyl carbocation because it is not stable. Therefore, it prefers to undergo reaction by `S_(N)2` mechanism which occurs in one step through a transition state involving nucleophilic attack of the `OH^(-)` ion from the back side with simultaneous expulsion of `Br^(-)` ion from the front side.