Write a note on `S_(N)1` reaction.

Explanation

The reaction in which the rate is independent of concentration of nuclephile and depends only on the concentration of substrate is called `S_(N)1` reaction. Thus `S_(N)1` reaction follows first order kinetics.
`(CH_(3))_(3)C-Br+""^(-)OH to(CH_(3))_(3)C-OH+Br^(-)`
Rate `alpha[(CH_(3))_(3)C-Br]`
The `S_(N)1` reaction takes place in two steps :
Step - 1 : The polarised C - Br bond undergoes cleavage slowly to form a carbocation and a bromide ion. The carbocation thus formed is then attacked by nucleophile in step - II to complete the substitution reaction.
`CH_(3)-overset(CH_(3))overset("| ")underset(CH_(3))underset("| ")("C ")-Br overset(" Slow ")hArr CH_(3)-overset(CH_(3))overset("| ")underset(CH_(3))underset("| ")(C^(o+))+Br^(Θ)`
The step - 1 is the slow and reversible and thus it is a rate determining step. The energy required for breaking of C - Br bond is obtained through solvantion of halide ion with the proton of a protic solvent. The rate of reaction depends only on concentration of alkyl halide and not on the concentration of `""^(-)H` ion because slow step involves only the alkyl halide.
Step - 2 : The carbocation being planar is attacted from either side by nucleophile to form a final product. It is a fast step.
`CH_(3)-overset(CH_(3))overset("| ")underset(CH_(3))underset("| ")(C^(o+))+""^(ө)OH overset(" Fast ")rarr CH_(3)- overset(CH_(3))overset("| ")underset(CH_(3))underset("| ")("C ")-OH`
In case if the substrate is optically active, the `S_(N)1` reaction results in partial racemisation. Due to planar structure of carbocation, the nucleophile attacks from either side to give enantiomers. The attack from backside is more favourable and thus the enantiomers are not obtained in equal amount.

For the given alkyl group the reactivity of alkyl halides follow the order.
`R-I gt R-Br gt R-Cl gt R-F`