question_answer

Tertiary alkyi halides are practically inert to substitution by\[{{S}_{N}}2\]mechanism because of     AIEEE  Solved  Paper-2005

A) steric hindrance               

B) inductive effect

C)                        instability          

D)        insolubility

Correct Answer: A

Solution :

In\[{{S}_{N}}2\]reaction, nucleophile and alkyl halide react in one step. Thus, tertiary carbon is under steric hindrance thus reaction does not take place until \[(CBr)\]bond breaks \[R-\underset{\begin{smallmatrix}  | \\  R \end{smallmatrix}}{\overset{\begin{smallmatrix}  R \\  | \end{smallmatrix}}{\mathop{C}}}\,-Br\xrightarrow{Slow}\,R-\underset{\begin{smallmatrix}  | \\  R \end{smallmatrix}}{\overset{\begin{smallmatrix}  R \\  | \end{smallmatrix}}{\mathop{{{C}^{\oplus }}}}}\,+B{{r}^{-}}\] which is then\[{{S}_{N}}1\]reaction. Alternate Solution Back side attack takes place in\[{{S}_{N}}2\]reaction but it is not possible in tertiary alkyl halide because of steric hinderence. Thus, it is practically inert to substitution by\[{{S}_{N}}2\]mechanism.