Chapter 27, Problem 86SAE

Interpretation Introduction

Interpretation:

The mechanism for the reaction between ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_{\text{3}}\text{CBr}$ and ${\text{CH}}_{\text{3}}\text{OH}$ should be drawn.

Concept introduction:

Nucleophilic substitution reaction $\left({\text{S}}_{\text{N}}1\right)$ is a two-step substitution reaction. In the first step removal of the leaving group and formation of the carbocation intermediate occurs. The second step involves the formation of the product by the attack of the incoming nucleophile on the carbocation intermediate. It is a unimolecular substitution reaction. A mixture of both retention and inversion product is obtained as the incoming nucleophile can attack from either side of the carbocation.

Elimination reaction is the type of reaction in which two substituents attached to the adjacent carbon atoms are removed from the reactant molecule to form the product. In $\text{E2}$ elimination reactions, the two substituents are removed in a single step and a transition state is formed. Both the substituents should be at $180°\text{C}$ to each other in $\text{E2}$ elimination reaction. The rate of $\text{E2}$ elimination reaction depends upon both the reactant molecules and therefore, follows second order kinetics. Generally, strong bases favor $\text{E2}$ elimination reactions.