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

Explanation

Tert. butyl bromide reacts with aq. NaOH as follows
`underset((3^(@) " alkyl halide"))underset("tert Butylbromide")(H_(3)C-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-Br)overset(S_(N)1)tounderset(("more stable"))underset(3^(@)"carbocation")(H_(3)C-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C^(+))+Br^(-))`
`H_(3)C-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C^(+))+OH^(-)toH_(3)C-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-OH`
tert. butyl bromide when treated with aq. NaOH, it forms tert. corbocation which is more
stable intermediate. This intermediate is further attacked by `""^(-)OH` ion.
As tert. carbocation is highly stable so tert butylbromide follow `S_(N)1` mechanism.
In case of n-nutylbromide,primary carbocation is formed which is least stable so, it does not follow `S_(N)1` mechanism. Here, stearic hindrance is very less so, it follow `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 the eopposite side and the product is formed,