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

Options

A. Formation of protonated alcohol in a slow step
B. Formation of `(CH_3)_3C^(+)` is a slow step
C. Formation of `(CH_3)_3CBr` from `(CH_3)_3C^+` is a slow step
D. Formation of `(CH_3)_3C^+` is a fast step

Answer

To solve the question regarding the mechanism for the reaction of HBr with t-butyl alcohol, we need to analyze the steps involved in the mechanism and identify the correct statement.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants in this reaction are t-butyl alcohol (tert-butanol, \( \text{(CH}_3)_3\text{COH} \)) and HBr.

2. **Protonation of Alcohol**:
- The first step involves the protonation of t-butyl alcohol by HBr. This can be represented as:
\[ \text{(CH}_3)_3\text{COH} + \text{H}^+ \rightarrow \text{(CH}_3)_3\text{C(OH}_2^+\text{)} \]
- This step is fast because it is an acid-base reaction.

3. **Formation of Carbocation**:
- The next step is the formation of a carbocation from the protonated alcohol. This step can be represented as:
\[ \text{(CH}_3)_3\text{C(OH}_2^+\text{)} \rightarrow \text{(CH}_3)_3\text{C}^+ + \text{H}_2\text{O} \]
- This step is slow and is the rate-determining step of the reaction.

4. **Nucleophilic Attack by Bromide Ion**:
- In the final step, the bromide ion (\( \text{Br}^- \)) attacks the carbocation to form the product:
\[ \text{(CH}_3)_3\text{C}^+ + \text{Br}^- \rightarrow \text{(CH}_3)_3\text{CBr} \]
- This step is fast.

5. **Conclusion**:
- From the analysis, we can conclude that the correct statement is that the formation of the carbocation (\( \text{(CH}_3)_3\text{C}^+ \)) is a slow step and is the rate-determining step of the reaction.

### Correct Statement:
- The formation of \( \text{(CH}_3)_3\text{C}^+ \) is a slow step.