`underset("Specific rotation" = +50^(@))(Ph-CH(OH)CH_(3)) overset(SOCl_(2))(rarr) underset(Cl)underset(|)(PhCH) - CH_(3)`
Which of the following acts as a nucelphilic?

Options

A. `CI^(o-)`
B. `overset(o-)(O)H`
C. `SO_(2)`
D. None

Explanation

To solve the question regarding which species acts as a nucleophile in the reaction of phenyl-2-propanol with thionyl chloride (SOCl2), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants and Reaction Type**:
- The reactant is phenyl-2-propanol (Ph-CH(OH)CH3).
- The reagent is thionyl chloride (SOCl2).
- This is a nucleophilic substitution reaction where an alcohol is converted into an alkyl halide.

2. **Understanding the Mechanism**:
- SOCl2 reacts with the alcohol to convert the hydroxyl (-OH) group into a better leaving group.
- The oxygen atom in the -OH group has lone pairs of electrons that can interact with the sulfur atom in SOCl2.

3. **Formation of the Intermediate**:
- The lone pair of electrons on the oxygen atom attacks the sulfur atom, leading to the formation of an intermediate where the oxygen carries a positive charge and sulfur carries a partial positive charge.
- This results in the formation of a chlorosulfite intermediate.

4. **Leaving Group Formation**:
- The intermediate undergoes a rearrangement where the chlorine atom from SOCl2 leaves as a chloride ion (Cl-), and the oxygen returns its electrons to form a double bond with sulfur, releasing sulfur dioxide (SO2) and hydrogen chloride (HCl).

5. **Nucleophilic Attack**:
- The chloride ion (Cl-) that was generated acts as a nucleophile.
- It attacks the positively charged carbon atom that was originally bonded to the hydroxyl group, leading to the formation of the final product (Ph-CHCl-CH3).

6. **Conclusion**:
- In this reaction, the nucleophile is the chloride ion (Cl-), which seeks out the positively charged carbon atom to form the alkyl halide.

### Final Answer:
The nucleophile in this reaction is **chloride ion (Cl-)**.