Which of the following alcohols will yield the corresponding alkyl chloride on reaction with concentrated HCl at room temperature ?

Options

A. `CH_(3)CH_(2)-CH_(2)-OH`
B. `CH_(3)CH_(2)-underset(CH_(3))underset(|)CH-OH`
C. `CH_(3)CH_(2)-underset(CH_(3))underset(|)CH-CH_(2)OH`
D. `CH_(3)CH_(2)-underset(CH_(3))underset(|) overset(CH_(3))overset(|)C-OH`

Explanation

To determine which alcohol will yield the corresponding alkyl chloride upon reaction with concentrated HCl at room temperature, we need to analyze the reaction mechanism involved. The reaction of alcohols with concentrated HCl can proceed via two mechanisms: SN1 and SN2.

### Step-by-Step Solution:

1. **Identify the Reaction Type**:
- The reaction between alcohol and concentrated HCl is a nucleophilic substitution reaction. We need to determine whether it will follow the SN1 or SN2 mechanism.

2. **Analyze SN2 Mechanism**:
- In an SN2 reaction, the nucleophile attacks the carbon atom at the same time as the leaving group (OH) departs. However, OH- is a poor leaving group, making SN2 less favorable for alcohols.

3. **Analyze SN1 Mechanism**:
- In an SN1 reaction, the first step involves the protonation of the alcohol by HCl, forming a better leaving group (water). The reaction can be represented as:
\[ RCH_2OH + HCl \rightarrow RCH_2OH_2^+ + Cl^- \]
- The protonated alcohol (RCH2OH2+) can then lose water to form a carbocation.

4. **Stability of Carbocations**:
- The stability of the carbocation formed is crucial for the SN1 mechanism. The order of stability is tertiary > secondary > primary.
- A tertiary carbocation (3°) is more stable than a secondary (2°) or primary (1°) carbocation.

5. **Identify the Alcohols**:
- Examine the structure of the given alcohols to determine whether they can form primary, secondary, or tertiary carbocations upon protonation.
- Count the number of carbon atoms attached to the carbon bearing the hydroxyl group (OH):
- **Primary (1°)**: 1 carbon attached.
- **Secondary (2°)**: 2 carbons attached.
- **Tertiary (3°)**: 3 carbons attached.

6. **Determine the Suitable Alcohol**:
- The alcohol that can form a tertiary carbocation will be the one that can effectively yield the corresponding alkyl chloride upon reaction with concentrated HCl at room temperature.
- From the analysis, if an alcohol has a tertiary structure, it will undergo the SN1 mechanism efficiently.

7. **Conclusion**:
- The correct option is the alcohol that forms a tertiary carbocation, as it will yield the corresponding alkyl chloride via the SN1 mechanism.