A dextro- rotatory optically active alkyl halide undergoes hydrolysis by `S_(N) 2` mechanism. The resulting alcohol is `:`

Options

A. Dextorotatory
B. Laevorotatory
C. Optically inactive due to racemization
D. May be dexture or laevorotatory

Explanation

To solve the question regarding the hydrolysis of a dextro-rotatory optically active alkyl halide undergoing an SN2 mechanism, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the SN2 Mechanism**:
- The SN2 (Substitution Nucleophilic Bimolecular) mechanism involves a nucleophile attacking the electrophilic carbon atom from the opposite side of the leaving group. This results in an inversion of configuration at the carbon center.

2. **Identifying the Reactant**:
- The reactant in this case is a dextro-rotatory alkyl halide. Dextrorotatory means that it rotates plane-polarized light to the right (clockwise).

3. **Nucleophilic Attack**:
- In the SN2 mechanism, the nucleophile (in this case, hydroxide ion, OH-) attacks the carbon atom that is bonded to the halogen (the leaving group). This attack occurs from the back side of the carbon atom.

4. **Inversion of Configuration**:
- Due to the back-side attack, the configuration of the carbon atom is inverted. If the original alkyl halide was dextrorotatory (D form), the product after the nucleophilic substitution will be levorotatory (L form) because of this inversion.

5. **Resulting Product**:
- The product formed from the hydrolysis of the dextrorotatory alkyl halide through the SN2 mechanism will be an alcohol. Since the configuration has inverted, the resulting alcohol will be levorotatory.

6. **Conclusion**:
- Therefore, the resulting alcohol from the hydrolysis of the dextro-rotatory optically active alkyl halide via the SN2 mechanism is levorotatory.

### Final Answer:
The resulting alcohol is **levorotatory**.