The reaction of (S) - 2 - bromobutane with `OH^-` to produce (R) - butan -2 - ol will be

Options

A. first order in 2 - bromobutane only
B. first order in `OH^-` only
C. first order in 2 - bromobutane and first order in `OH^-`
D. second order in `OH^-`

Answer

To solve the problem of the reaction of (S)-2-bromobutane with OH⁻ to produce (R)-butan-2-ol, we need to analyze the reaction mechanism and determine the order of the reaction. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Reactants and Products
- The reactant is (S)-2-bromobutane, and the product is (R)-butan-2-ol.
- The reaction involves the nucleophile OH⁻ attacking the carbon atom bonded to the bromine atom.

**Hint:** Always start by clearly identifying the reactants and products in the reaction.

### Step 2: Determine the Configuration of the Reactant
- (S)-2-bromobutane has a specific stereochemistry. The bromine atom is attached to the second carbon, which is a chiral center.
- To determine the configuration, we assign priorities to the groups attached to the chiral carbon based on atomic number.

**Hint:** Use the Cahn-Ingold-Prelog priority rules to assign priorities to the substituents around the chiral center.

### Step 3: Analyze the Reaction Mechanism
- The reaction of (S)-2-bromobutane with OH⁻ can proceed via two mechanisms: SN1 or SN2.
- In an SN2 reaction, the nucleophile attacks the substrate from the opposite side of the leaving group (Br), resulting in inversion of configuration.
- In an SN1 reaction, the leaving group departs first, forming a carbocation, and then the nucleophile attacks, which can lead to racemization.

**Hint:** Consider the stereochemical outcome of the reaction to determine whether it is SN1 or SN2.

### Step 4: Determine the Stereochemical Outcome
- Since the reaction converts (S)-2-bromobutane to (R)-butan-2-ol, this indicates that there is an inversion of configuration.
- This inversion is characteristic of an SN2 mechanism, where the nucleophile attacks the backside of the carbon atom, leading to a direct conversion from one stereoisomer to another.

**Hint:** Look for clues in the product's stereochemistry to confirm the type of mechanism.

### Step 5: Determine the Order of the Reaction
- In an SN2 reaction, the rate of reaction depends on the concentration of both the substrate (2-bromobutane) and the nucleophile (OH⁻).
- Therefore, the reaction is bimolecular, meaning it is first order with respect to each reactant.

**Hint:** Recall that the rate law for an SN2 reaction involves both the substrate and nucleophile.

### Conclusion
The reaction of (S)-2-bromobutane with OH⁻ to produce (R)-butan-2-ol is an SN2 reaction, and the order of the reaction is first order in 2-bromobutane and first order in OH⁻, making it a total of second order.

### Final Answer
The reaction is an **SN2 reaction** with **second order** kinetics.