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

Explanation

In general, the `S_(N)1` reaction proceeds through the formation of carbocation. The tert-butyl bromide readily laus `Br^(-)` ion to form table `3^(@)` carbocation.
Therefore, it reacts with aqueous KOH by `S_(N)1` mechanism as:
`CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-Br underset(underset("slow")(-Br))overset("Ionisation")to underset(underset("(stable)")("tert-butyl carbocation"))(CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C^(+)))`
`CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C^(+)) underset("fast")overset(OH^(-))to underset("tert-Butyl alcohol")(CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-OH)`
On the other hand, n-butyl bromide does not undergo ionization to form n-butyl carbocation (`1^(@)`) 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 `OH^(-)` lon from the back side with simultaneous expulsion of `Br^(-)` ion from the front side.

`S_(N)1`mechanism follows the reactivity order as `3^(@) gt 2^(@) gt 1^(@)` while `S_(N)2` mechanism follows the reactivity order as `1^(@) gt 2^(@) gt 3^(@)`
Therefore, tert-butylbromide (`3^(@)`) reacts by `S_(N)1` mechanism while n-butylbromide (`1^(@)`) reacts by `S_(N)2` mechanism