question_answer

Tertiary alkyl halides are practically inert towards substitution by\[{{S}_{N}}2\]mechanism because of

A)  steric hindrance

B)  inductive effect

C)  instability

D)  insolubility

Correct Answer: A

Solution :

                 In\[{{S}_{N}}2\]reaction, nucleophile and alkyi halide react in one step. Here, it is evident that 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.