In the reaction: `CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-OH+HCl rarr CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-Cl`
the rate-determining step is:

Options

A. `CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-OH+H_(3)O^(+)hArrCH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-underset(o+)overset(H)overset(|)O-H-H_(2)O`
B.
C.
D. all steps take place with equal ease

Explanation

To determine the rate-determining step in the reaction of the given alcohol with HCl, we can follow these steps:

### Step 1: Identify the Reactants and Products
The reactant is a tertiary alcohol (specifically, 2-methyl-2-propanol, also known as tert-butanol) and the product is a tertiary alkyl chloride (tert-butyl chloride). The reaction can be represented as:
\[ \text{(CH}_3)_3\text{C-OH} + \text{HCl} \rightarrow \text{(CH}_3)_3\text{C-Cl} + \text{H}_2\text{O} \]

### Step 2: Understand the Mechanism of the Reaction
The reaction proceeds via a two-step mechanism:
1. Protonation of the alcohol to form a better leaving group.
2. Formation of a carbocation followed by nucleophilic attack by chloride ion.

### Step 3: Protonation of the Alcohol
In the first step, the hydroxyl group (-OH) of the alcohol is protonated by HCl. This forms an oxonium ion, which is more reactive and can lead to the formation of a carbocation:
\[ \text{(CH}_3)_3\text{C-OH} + \text{H}^+ \rightarrow \text{(CH}_3)_3\text{C-OH}_2^+ \]

### Step 4: Formation of Carbocation
The oxonium ion formed in the previous step is unstable and loses a water molecule, resulting in the formation of a tertiary carbocation:
\[ \text{(CH}_3)_3\text{C-OH}_2^+ \rightarrow \text{(CH}_3)_3\text{C}^+ + \text{H}_2\text{O} \]

### Step 5: Nucleophilic Attack
In the final step, the chloride ion (Cl⁻) attacks the carbocation to form the final product, tert-butyl chloride:
\[ \text{(CH}_3)_3\text{C}^+ + \text{Cl}^- \rightarrow \text{(CH}_3)_3\text{C-Cl} \]

### Step 6: Identify the Rate-Determining Step
The slowest step in this mechanism is the formation of the carbocation, as it involves the breaking of a bond and the formation of a highly reactive intermediate. Therefore, the rate-determining step is:
\[ \text{Formation of the carbocation} \]

### Final Answer
The rate-determining step in the reaction is the formation of the carbocation from the protonated alcohol.

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