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

Explanation

Tert-butylbromide reacts with aq. NaOH as follows :
`undersetunderset((3^(@)" alkyl halide"))("Tert-butylbromide")(H_(3)C-undersetunderset(CH_(3))(|)oversetoverset(CH_(3))(|)(C)-Br)overset(S_(N)1)rarrundersetunderset(("more stable"))(3^(@)" carbocation")(H_(3)C-undersetunderset(CH_(3))(|)oversetoverset(CH_(3))(|)(C^(+))+Br^(-))`
`H_(3)C-undersetunderset(CH_(3))(|)oversetoverset(CH_(3))(|)(C^(+))+OH^(-)rarrH_(3)C-undersetunderset(CH_(3))(|)oversetoverset(CH_(3))(|)(C)-OH`
tert-butylbromide when treated with aq. NaOH, it forms tert-carbocation which is more stable intermediate. This intermediate is further attacked by `OH^(-)` ion.
As tert-carbocation is highly stable, so tert-butylbromide follows `S_(N)1` mechanism. In case of n-butylbromide, primary carbocation is formed which is least stable, so it does not follows `S_(N)1` mechanism.
Here, steric hindrance is very less, so it follows `S_(N)2` mechanism. In `S_(N)2` mechanism, `OH^(-)` will attack from backside and a transition state is formed.

The leaving group is then pushed off and the product is formed.