$C{H}_{3 } C{H}_3$
$| |$
$H_3—C—Br \stackrel{S_{N}1}{\to } H_{3}C—C^{+} + B{r}^{-}$
$| |$
$C{H}_3 C{H}_3$
$tert Butlbromide 3° carbocation$
$(3°alkl halide) (more stable)$
$$
$C{H}_{3 } C{H}_3$
$| |$
$H_{3}C—C^{+} + O{H}^{-} \to H_{3}C—C—OH$
$| |$
$C{H}_3 C{H}_3$

tert. butyl bromide when treated with aq, NaOH, il 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$1mechanism. Here, stearic hindrance is very less so, it follow S${}_N$2 mechanism. In

S${}_N$2  mechanism, ${}_{-}OH$ will atlack from backside and a transilion state is formed,

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