`PhCH_(2)-overset(O)overset(||)C-Cl+CH_(3)CH_(2)SH to X, X` is

Options

A. `Ph-overset(O)overset(||)C-S-underset(CH_(3))underset(|)(CH)-CH_(3)`
B. `Ph-CH_(2)-overset(O)overset(||)C-S-CH_(2)-CH_(3)`
C. `CH_(3)-CH_(2)-CH_(2)-overset(O)overset(||)C-S-Ph`
D. `CH_(3)-underset(CH_(3))underset(|)(CH)-overset(O)overset(||)C-S-Ph`

Answer

To solve the given question, we need to analyze the reaction step by step. The reactants are PhCH₂-C(=O)Cl (a phenyl group attached to a carbonyl group with a chlorine atom) and CH₃CH₂SH (ethyl mercaptan). We want to determine the product X.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The first reactant is PhCH₂-C(=O)Cl, which is a phenyl group (Ph) attached to a methylene group (CH₂) and a carbonyl (C=O) with a chlorine (Cl) as a leaving group.
- The second reactant is CH₃CH₂SH, which is ethyl mercaptan (a thiol).

2. **Nucleophilic Attack**:
- The sulfur atom in the thiol (CH₃CH₂SH) has lone pairs and acts as a nucleophile. It will attack the carbon atom of the carbonyl group (C=O) in the first reactant.
- This attack leads to the formation of a tetrahedral intermediate where the carbonyl oxygen will have a negative charge.

3. **Rearrangement of Bonds**:
- The negative charge on the oxygen will push the electrons back to form a double bond between carbon and oxygen, while the chlorine atom (Cl) will act as a leaving group and depart.
- This results in the formation of a new product where the sulfur atom is now bonded to the carbon that was part of the carbonyl group.

4. **Final Product Formation**:
- After the chlorine leaves, we are left with a product that has the structure: Ph-CH₂-C(=O)-S-CH₂-CH₃.
- This product contains a phenyl group, a methylene group, a carbonyl group, a sulfur atom, and an ethyl group.

5. **Conclusion**:
- The final product X can be represented as Ph-CH₂-C(=O)-S-CH₂-CH₃, which corresponds to a thioester.

### Final Answer:
The product X is Ph-CH₂-C(=O)-S-CH₂-CH₃.