`S_(N)2` reaction will be fastest in:

Options

A. `CH_(3)Br`
B. `CH_(3)Cl`
C. `CH_(3)CH_(2)Cl`
D. `(CH_(3))_(2)CHCl`

Answer

To determine which compound will undergo an \( S_N2 \) reaction the fastest, we need to consider two main factors: steric hindrance and the quality of the leaving group. The \( S_N2 \) mechanism involves a nucleophile attacking a carbon atom from the opposite side of the leaving group, leading to a transition state. Here's how we can analyze the options step by step:

### Step 1: Understand the \( S_N2 \) Mechanism
The \( S_N2 \) reaction is characterized by a single concerted step where the nucleophile attacks the carbon atom while the leaving group departs. The reaction rate depends on:
- The steric hindrance around the carbon atom (less steric hindrance leads to faster reactions).
- The ability of the leaving group to depart (better leaving groups lead to faster reactions).

### Step 2: Evaluate Steric Hindrance
- **Methyl carbon** (no alkyl groups attached) has the least steric hindrance.
- **Primary carbon** (one alkyl group attached) has moderate steric hindrance.
- **Secondary carbon** (two alkyl groups attached) has more steric hindrance.
- **Tertiary carbon** (three alkyl groups attached) has the most steric hindrance and does not undergo \( S_N2 \) reactions effectively.

### Step 3: Assess the Leaving Group
Common leaving groups in \( S_N2 \) reactions include:
- Bromide (Br\(^-\))
- Chloride (Cl\(^-\))
- Iodide (I\(^-\))

Among these, bromide is generally a better leaving group than chloride due to its larger size and the stability of the bromide ion.

### Step 4: Analyze the Given Options
Let’s analyze the options provided in the question:

1. **Option A**: Methyl halide (e.g., CH\(_3\)Br)
2. **Option B**: Primary halide (e.g., CH\(_3\)CH\(_2\)Br)
3. **Option C**: Secondary halide (e.g., CH\(_3\)CH(Br)CH\(_2\)CH\(_3\))
4. **Option D**: Tertiary halide (e.g., (CH\(_3\))\(_3\)CBr)

- **Option A** (Methyl halide) has no steric hindrance and a good leaving group (Br\(^-\)).
- **Option B** (Primary halide) has low steric hindrance and a good leaving group (Br\(^-\)).
- **Option C** (Secondary halide) has more steric hindrance and a good leaving group (Br\(^-\)).
- **Option D** (Tertiary halide) has the highest steric hindrance and does not favor \( S_N2 \).

### Step 5: Conclusion
Based on the analysis, **Option A** (Methyl halide with Br\(^-\)) will undergo the \( S_N2 \) reaction the fastest due to minimal steric hindrance and the presence of a good leaving group.

### Final Answer
The \( S_N2 \) reaction will be fastest in **Option A** (Methyl halide with Br\(^-\)).

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