When the concentration of nucleophile is reduced to half, the rate of `S_(N^(2))` reaction is decreased by

Options

A. 3 times
B. 0.5 times
C. 2 times
D. 6 times

Answer

To solve the problem, we need to analyze how the rate of an \( S_N2 \) reaction changes when the concentration of the nucleophile is reduced to half.

### Step-by-Step Solution:

1. **Understand the Rate Law for \( S_N2 \) Reactions**:
The rate of an \( S_N2 \) reaction is given by the equation:
\[ \text{Rate} = k [\text{RX}]^1 [\text{Nucleophile}]^1 \]
Here, \( k \) is the rate constant, \( [\text{RX}] \) is the concentration of the substrate, and \( [\text{Nucleophile}] \) is the concentration of the nucleophile.

2. **Initial Rate Calculation**:
Let’s denote the initial concentration of the nucleophile as \( [\text{Nucleophile}] = [N] \). Therefore, the initial rate \( R \) can be expressed as:
\[ R = k [\text{RX}] [N] \]

3. **Change in Nucleophile Concentration**:
When the concentration of the nucleophile is reduced to half, the new concentration becomes:
\[ [\text{Nucleophile}] = \frac{[N]}{2} \]

4. **New Rate Calculation**:
The new rate \( R' \) with the reduced nucleophile concentration is:
\[ R' = k [\text{RX}] \left(\frac{[N]}{2}\right) \]
Simplifying this, we get:
\[ R' = k [\text{RX}] \frac{[N]}{2} = \frac{1}{2} \cdot k [\text{RX}] [N] \]
Thus, we can express \( R' \) in terms of the original rate \( R \):
\[ R' = \frac{1}{2} R \]

5. **Conclusion**:
Therefore, when the concentration of the nucleophile is reduced to half, the rate of the \( S_N2 \) reaction decreases to half of its original value.

### Final Answer:
The rate of the \( S_N2 \) reaction is decreased by half.