When the concentration of alkyl halide is triple and concentration of `overset(Θ)OH` is reduced to half, the rate of `S_(N^(2))` reaction increased by :

Options

A. 3 times
B. 1.5 times
C. 2 times
D. 6 times

Answer

To solve the problem, we need to analyze how the changes in concentrations of the alkyl halide and the hydroxide ion (OH⁻) affect the rate of an SN2 reaction.

### Step-by-Step Solution:

1. **Understand the Rate Law for SN2 Reactions**:
The rate of an SN2 reaction is given by the rate law:
\[ \text{Rate} = k [\text{Alkyl Halide}] [\text{Nucleophile}] \]
where \( k \) is the rate constant, \([\text{Alkyl Halide}]\) is the concentration of the alkyl halide, and \([\text{Nucleophile}]\) is the concentration of the nucleophile (in this case, OH⁻).

2. **Define Initial Concentrations**:
Let:
- Initial concentration of alkyl halide = \([R-X]\)
- Initial concentration of OH⁻ = \([OH^-]\)

3. **Apply the Changes in Concentration**:
According to the problem:
- The concentration of alkyl halide is tripled: \([R-X] = 3[R-X]\)
- The concentration of OH⁻ is reduced to half: \([OH^-] = \frac{1}{2}[OH^-]\)

4. **Substitute into the Rate Law**:
The new rate of reaction can be expressed as:
\[ \text{New Rate} = k [3(R-X)] \left[\frac{1}{2}(OH^-)\right] \]
Simplifying this gives:
\[ \text{New Rate} = k \cdot 3[R-X] \cdot \frac{1}{2}[OH^-] = \frac{3}{2} k [R-X][OH^-] \]

5. **Compare with the Original Rate**:
The original rate is:
\[ \text{Original Rate} = k [R-X][OH^-] \]
Therefore, the ratio of the new rate to the original rate is:
\[ \frac{\text{New Rate}}{\text{Original Rate}} = \frac{\frac{3}{2} k [R-X][OH^-]}{k [R-X][OH^-]} = \frac{3}{2} \]

6. **Conclusion**:
The rate of the SN2 reaction increases by a factor of \( \frac{3}{2} \) or 1.5 times.

### Final Answer:
The rate of the SN2 reaction increased by **1.5 times**.