Identify the product in the reaction `PhC-=CMe overset(H_3O^(+),Hg^(2+)?)rarr`

Options

A. `PhCOCH_2CH_3`
B. `PhCH_2CH_2CHO`
C. `PhCOCOMe`
D. `PhCH_2COCH_3`

Answer

To identify the product of the reaction `PhC≡CMe` with `H3O^+` and `Hg^2+`, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactants are phenylacetylene (`PhC≡CMe`), hydronium ion (`H3O^+`), and mercuric ion (`Hg^2+`).

2. **Understand the Reaction Type**: This reaction is a hydrolysis of an alkyne. The presence of `Hg^2+` indicates that we are likely dealing with a Markovnikov addition, where the more substituted carbon will bond with the hydroxyl group.

3. **Protonation of the Alkyne**: The triple bond in the alkyne will first react with the proton from `H3O^+`. This results in the formation of a carbocation. The protonation occurs at the less substituted carbon, leading to the formation of a carbocation at the more substituted carbon.

- **Carbocation Formation**:
\[ PhC≡CMe + H^+ \rightarrow PhC^+(H)CMe \]

4. **Nucleophilic Attack by Water**: Water, acting as a nucleophile, will attack the carbocation. The lone pair of electrons from the water molecule will bond with the positively charged carbon.

- **Formation of an Alcohol**:
\[ PhC^+(H)CMe + H2O \rightarrow PhC(OH)CMe^+ \]

5. **Deprotonation**: The positively charged alcohol will lose a proton to form a stable product. This step involves the removal of a proton from the hydroxyl group.

- **Final Product Formation**:
\[ PhC(OH)CMe^+ \rightarrow PhC=O + CH2Me \]

6. **Identify the Product**: The final product after the hydrolysis will be a ketone. Specifically, the product is `PhC(=O)CH2Me`, which is an aromatic ketone.

### Final Answer:
The product of the reaction is `PhC(=O)CH2Me`.