A partially racemised (+2) 2-Bromooctane on reaction with aq. NaOH in acetone gives an alcolhol with 80% inversion and 20% racemisation .
Find out the rate expression of the reaction.

Options

A. Rate = `K_1[R-X][barOH]`
B. Rate = `K_2[R-X]`
C. Rate = `K_1[R-X]+K_2[R-X][barOH]`
D. None of these

Answer

To solve the problem, we need to analyze the reaction of partially racemized (+2) 2-Bromooctane with aqueous NaOH in acetone, which results in an alcohol with 80% inversion and 20% racemization.

### Step-by-Step Solution:

1. **Identify the Reaction Type**:
The reaction involves a nucleophilic substitution where the bromine atom (a good leaving group) is replaced by a hydroxide ion (OH-) from NaOH. This indicates that both SN1 and SN2 mechanisms are likely involved due to the presence of both inversion and racemization.

2. **Understanding Inversion and Racemization**:
- **Inversion** (80%): This is characteristic of the SN2 mechanism, where the nucleophile attacks the substrate leading to a complete inversion of configuration.
- **Racemization** (20%): This is characteristic of the SN1 mechanism, where the formation of a carbocation intermediate allows for attack from either side, leading to a racemic mixture.

3. **Rate Laws for SN1 and SN2**:
- For the **SN1 mechanism**, the rate law is given by:
\[ \text{Rate}_{\text{SN1}} = k_1 [\text{RX}] \]
where \( k_1 \) is the rate constant for the SN1 reaction and \([\text{RX}]\) is the concentration of the substrate.

- For the **SN2 mechanism**, the rate law is given by:
\[ \text{Rate}_{\text{SN2}} = k_2 [\text{RX}][\text{OH}^-] \]
where \( k_2 \) is the rate constant for the SN2 reaction and \([\text{OH}^-]\) is the concentration of the nucleophile.

4. **Combining the Rate Laws**:
Since both mechanisms are contributing to the overall reaction, we can combine the rate laws:
\[ \text{Total Rate} = \text{Rate}_{\text{SN1}} + \text{Rate}_{\text{SN2}} \]
Therefore, the overall rate expression becomes:
\[ \text{Total Rate} = k_1 [\text{RX}] + k_2 [\text{RX}][\text{OH}^-] \]

5. **Final Rate Expression**:
The final rate expression for the reaction is:
\[ \text{Rate} = k_1 [\text{RX}] + k_2 [\text{RX}][\text{OH}^-] \]