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

Explanation

Detailed Answer:
Tertiary butyl bromide. When treated with aqueous KOH, tertiary butyl alcohol is formed.
`(CH_(3))_(3)C-Br+KOH to (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)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-Br overset("Slow")to CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C+Br^(-)`
This step 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 butyl alcohol.
`CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C^(+))+OH^(-) overset("Fast")to CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-OH`