Consider the reaction,
`(CH_(3))_(3)C-underset(CH_(3))underset("| ")"CH"-OH+"Conc. HCl"rarrX`.
The product (X) is

Options

A. `(CH_(3))_(3)"CCH"(CH_(3))Cl`
B. `(CH_(3))_(2)"CCl"CH(CH_(3))_(2)`
C. Mixture of both A and B
D. `(CH_(3))_(2)CHCl(Cl)(CH_(3))_(2)`

Answer

To solve the given reaction, we will analyze the structure of the reactant and the mechanism of the reaction with concentrated HCl.

### Step-by-Step Solution:

1. **Identify the Reactant Structure**:
The reactant is (CH₃)₃C-CH₂OH, which can be represented as:
```
CH₃
|
CH₃-C-CH₂-OH
|
CH₃
```
This is a tertiary alcohol (due to the presence of the -OH group on a carbon that is bonded to three other carbon atoms).

2. **Protonation of the Alcohol**:
When concentrated HCl is added, the -OH group of the alcohol gets protonated to form a better leaving group (water):
```
CH₃
|
CH₃-C-CH₂-OH + H⁺ → CH₃
|
CH₃-C-CH₂-OH₂⁺
|
CH₃
```

3. **Formation of Carbocation**:
The protonated alcohol (oxonium ion) can lose a water molecule (H₂O), leading to the formation of a carbocation:
```
CH₃
|
CH₃-C⁺-CH₂ + H₂O
|
CH₃
```

4. **Carbocation Rearrangement**:
The carbocation formed is a tertiary carbocation, which is stable. However, to further stabilize it, a methyl group (CH₃) can shift from the adjacent carbon to the carbocation, resulting in a more stable carbocation:
```
CH₃
|
CH₃-C-CH₃⁺
|
CH₂
```

5. **Nucleophilic Attack by Chloride Ion**:
The chloride ion (Cl⁻) from HCl will now attack the positively charged carbon to form the final product:
```
CH₃
|
CH₃-C-CH₃
|
CH₂-Cl
```

6. **Final Product**:
The final product (X) is 2-chloro-2-methylpropane, which can be represented as:
```
CH₃
|
CH₃-C-CH₂-Cl
|
CH₃
```

### Final Answer:
The product (X) is 2-chloro-2-methylpropane.