Question

in the reaction of (CH3)3CCl with Oh- there is no formation of (CH3)3C-Oh?​

Answer 1

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Answer 2

The given reaction follows SN1 mechanism

Explanation:

• " S N " denotes "nucleophilic substitution", and the "1" along with SN stands for the rate-determining step in the reaction which is unimolecular.
• First step is the dissociation of C-Cl bond which is a slow process and is therefore the rate determining step in the reaction. A carbocation is thus formed. It is given by:

         $(CH_3)_3CCl \rightarrow (CH_3)_3C^+ + Cl^-$

• In the second step , nucleophile $OH^-$ attacks the carbocation intermediate formed in the first rate determining step to give the product:

       $(CH_3)_3C^+ + OH^-\rightarrow (CH_3)_3C-OH$