Explain the mechanism of `S_N1` reaction taking 2-bromo-2-methyl propane (t-butyl bromide)

Explanation

Tertiary butyl bromide. When treated with aqueous KOH, tertiary butyl alcohol is formed.
`(CH_(3))_(3)C-Br+KOH rarr (CH_(3))C-OH+KBr`.
The `S_(N)`. reaction mechanism involves two steps.
step (i) : Formation of Carbonation.
Tertiary butyl bromide undergoes ionization to form tertiary carbo cation of bromide ion.
`CH_(3)-overset(CH_(3))overset("| ")underset(CH_(3))underset("| ")("C ")-Br overset(" Slow ")rarr CH_(3)-overset(CH_(3))overset("| ")underset(CH_(3))underset("| ")("C ")+Br^(-)`
This is the slow step is hence is a rate determining step.
Step - (ii) : Attack of nuclephile on carbocation.
The nucleophile `OH^(-)` attacks the positive centre of carbocation from any one of the direction forming the final product i.e., tertiary butyal alcohol. `CH_(3)-overset(CH_(3))overset("| ")underset(CH_(3))underset("| ")(C^(+))+OH^(-) overset(" Fast ")rarr CH_(3)-overset(CH_(3))overset("| ")underset(CH_(3))underset("| ")("C ")-OH`