Chemistry, asked by alan0296, 4 hours ago

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

Answers

Answered by Dhruwsonwani
0

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Answered by sushmaa1912
0

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

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