A secondary alkyl halide `(A)` hydrolyzes with alkali `(B)` in aqueous medium ismultaneously via `S_(N)1` and `S_(N)2` pathways with rate constants `k_(1)` and `k_(2)`, respectively. form kinetic data, it was found that a plot of `= (-1)/([A])(d[A])/(dt)` vs `[B]` is straight line with a slope equal to `2.7 xx 10^(-4) L mol^(-1) min^(-1)` and intercept equal to `1.02 xx 10^(-3)`. Minimum initial concentration of `[A] = 0.2 M` and `[B]`, i.e., `[overset(ɵ)(OH)] = 0.5 M`.
The initial rate of consumption is isopropyl chloride (in `M min^(-1))` is
(a)`0.2 xx 10^(-3)`
(b)`3 xx 10^(-5)`
(c)`2.31 xx 10^(-4)`
(d)`2.31 xx 10^(-3)`

Options

A. `a. 0.2 xx 10^(-3)`
B. `b. 3 xx 10^(-5)`
C. `c 2.31 xx 10^(-4)`
D. `d. 2.31 xx 10^(-3)`

Explanation

To solve the problem, we need to analyze the hydrolysis of the secondary alkyl halide (isopropyl chloride) through both SN1 and SN2 mechanisms. We will derive the initial rate of consumption of isopropyl chloride based on the given data.

### Step-by-step Solution:

1. **Identify the Mechanisms**:
- For the SN1 mechanism, the rate of reaction is given by:
\[ -\frac{d[A]}{dt} = k_1 [A] \]
- For the SN2 mechanism, the rate of reaction is given by:
\[ -\frac{d[A]}{dt} = k_2 [A][B] \]

2. **Combine the Rate Expressions**:
- The overall rate of consumption of isopropyl chloride can be expressed as:
\[ -\frac{d[A]}{dt} = k_1 [A] + k_2 [A][B] \]
- Factoring out \([A]\):
\[ -\frac{d[A]}{dt} = [A](k_1 + k_2 [B]) \]

3. **Use Given Data**:
- The slope of the plot is given as \(2.7 \times 10^{-4} \, L \, mol^{-1} \, min^{-1}\) and the intercept is \(1.02 \times 10^{-3}\).
- The minimum initial concentrations are \([A] = 0.2 \, M\) and \([B] = 0.5 \, M\).

4. **Set Up the Rate Equation**:
- From the slope and intercept, we can identify:
\[ k_1 = \text{intercept} = 1.02 \times 10^{-3} \, min^{-1} \]
\[ k_2 = \text{slope} = 2.7 \times 10^{-4} \, L \, mol^{-1} \, min^{-1} \]

5. **Calculate the Initial Rate**:
- Substitute the values into the rate equation:
\[ -\frac{d[A]}{dt} = [A](k_1 + k_2 [B]) \]
\[ -\frac{d[A]}{dt} = 0.2 \left(1.02 \times 10^{-3} + 2.7 \times 10^{-4} \times 0.5\right) \]
- Calculate \(k_2 [B]\):
\[ k_2 [B] = 2.7 \times 10^{-4} \times 0.5 = 1.35 \times 10^{-4} \]
- Now substitute back:
\[ -\frac{d[A]}{dt} = 0.2 \left(1.02 \times 10^{-3} + 1.35 \times 10^{-4}\right) \]
\[ -\frac{d[A]}{dt} = 0.2 \left(1.155 \times 10^{-3}\right) \]
\[ -\frac{d[A]}{dt} = 2.31 \times 10^{-4} \, M \, min^{-1} \]

### Final Answer:
The initial rate of consumption of isopropyl chloride is:
\[ \boxed{2.31 \times 10^{-4} \, M \, min^{-1}} \]