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 halved.

### Step-by-Step Solution:

1. **Understand the Rate Law for \( S_N2 \) Reactions**:
The rate of an \( S_N2 \) reaction can be expressed by the rate law:
\[ \text{Rate} = k [\text{R-X}] [\text{Nucleophile}] \]
where:
- \( k \) is the rate constant,
- \( [\text{R-X}] \) is the concentration of the substrate,
- \( [\text{Nucleophile}] \) is the concentration of the nucleophile.

2. **Identify the Relationship Between Rate and Nucleophile Concentration**:
From the rate law, we see that the rate of the reaction is directly proportional to the concentration of the nucleophile. This means if the concentration of the nucleophile changes, the rate will change in the same proportion.

3. **Halve the Concentration of the Nucleophile**:
If the concentration of the nucleophile is reduced to half, we can denote the new concentration as:
\[ [\text{Nucleophile}]' = \frac{1}{2} [\text{Nucleophile}] \]

4. **Calculate the New Rate**:
Substituting the new concentration into the rate law gives us:
\[ \text{Rate}' = k [\text{R-X}] \left(\frac{1}{2} [\text{Nucleophile}]\right) = \frac{1}{2} k [\text{R-X}] [\text{Nucleophile}] \]
This shows that the new rate \( \text{Rate}' \) is half of the original rate.

5. **Conclusion**:
Therefore, when the concentration of the nucleophile is reduced to half, the rate of the \( S_N2 \) reaction is also decreased by half.

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