In the mechanism for the reaction of HBr with t-butyl alcohol, pick out the correct statement.

• A

  Formation of protonated alcohol in a slow step
• Correct answer
  B

  Formation of $(\mathrm{C}{\mathrm{H}}_3{)}_3{\mathrm{C}}^{+}$ is a slow step
• C

  Formation of $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{C}\mathrm{B}\mathrm{r}$ from $(\mathrm{C}{\mathrm{H}}_3{)}_3{\mathrm{C}}^{+}$ is a slow step
• D

  Formation of $(\mathrm{C}{\mathrm{H}}_3{)}_3{\mathrm{C}}^{+}$ is a fast step

Solution

The mechanism of this reaction is as follows
${{\text{(CH}}_{\text{3}}\text{)}}_{\text{3}}\text{COH}\text{+}\text{HBr}\rightleftharpoons {{\text{(CH}}_{\text{3}}\text{)}}_{\text{3}}\text{C}\stackrel{\text{+}}{\text{O}}{\text{H}}_{\text{2}}\text{+}{\text{Br}}^{-}$ (fast)

Hence, formation on carbocation $(\mathrm{C}{\mathrm{H}}_3{)}_3{\mathrm{C}}^{+}$ is slow step (R.D.S.).

Hence, the correct answer is option: 'B'.