Predict the compound which will undergo `S_(N)2` reaction faster:

Options

A. `CH_(3)-CH_(2)-CH_(2)-Cl`
B. `H_(3)C-overset(Cl)overset(|)C-CH_(3)`
C. `CH_(3)-CH_(2)-CH_(2)-l`
D. `H_(3)C-overset(l)overset(|)CH-CH_(3)`

Answer

To determine which compound will undergo an \( S_N2 \) reaction faster, we need to consider two main factors: the degree of the alkyl halide and the quality of the leaving group.

### Step-by-Step Solution:

1. **Identify the Types of Alkyl Halides**:
- Alkyl halides can be classified as primary (1°), secondary (2°), or tertiary (3°). The reactivity in \( S_N2 \) reactions follows the order: 1° > 2° > 3°.
- In this case, we have two compounds to compare:
- Option 1: \( CH_3CH_2CH_2Cl \) (1° alkyl halide)
- Option 3: \( CH_3CH_2CH_2I \) (1° alkyl halide)

2. **Evaluate the Leaving Groups**:
- The leaving group is crucial in determining the rate of the \( S_N2 \) reaction. A better leaving group will facilitate a faster reaction.
- In this case, we have:
- Chlorine (Cl) in Option 1
- Iodine (I) in Option 3
- Iodine is a better leaving group than chlorine because it is larger and can stabilize the negative charge better once it leaves.

3. **Combine the Factors**:
- Both compounds are primary alkyl halides, which means they have the same steric hindrance and will react at similar rates based on their structure.
- However, since Option 3 has iodine as the leaving group, it will react faster than Option 1, which has chlorine.

4. **Conclusion**:
- Therefore, the compound that will undergo the \( S_N2 \) reaction faster is **Option 3: \( CH_3CH_2CH_2I \)**.