Question

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
• B. Formation of $\left(\right.CH_{3}\left(\left.\right)_{3}C^{+}$ is a slow step
• C. Formation of $\left(\right.CH_{3}\left(\left.\right)_{3}CBr$ from $\left(\right.CH_{3}\left(\left.\right)_{3}C^{+}$ is a slow step
• D. Formation of $\left(\right.CH_{3}\left(\left.\right)_{3}C^{+}$ is a fast step

Answer 1

Answer: (B)

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

Hence, formation on carbocation $\left(\right.CH_{3}\left(\left.\right)_{3}C^{+}$ is slow step (R.D.S.).