`Ph -ch_(2)-ch=ch_(2)overset(dil H_(2)SO_(4))rarrX,`
Identify product `'X'` is `:`

Options

A. `Ph-CH_(2)-CH_(2)-CH_(2)-OH`
B.
C.
D. `Ph-CH_(2)-OH`

Explanation

To solve the question regarding the reaction of the compound \( \text{Ph-CH}_2\text{-CH}=\text{CH}_2 \) with dilute \( \text{H}_2\text{SO}_4 \), we will follow these steps:

### Step 1: Identify the Compound
The compound given is \( \text{Ph-CH}_2\text{-CH}=\text{CH}_2 \), which consists of a phenyl group attached to a propene structure.

### Step 2: Understand the Reaction Conditions
The reaction is carried out in the presence of dilute sulfuric acid (\( \text{H}_2\text{SO}_4 \)). Dilute sulfuric acid acts as an acid catalyst and can protonate the alkene, leading to the formation of a carbocation.

### Step 3: Protonation of the Alkene
In the presence of dilute \( \text{H}_2\text{SO}_4 \), the double bond in the alkene will undergo protonation. The double bond between \( \text{CH} \) and \( \text{CH}_2 \) will break, and a proton (\( \text{H}^+ \)) from the acid will add to one of the carbon atoms of the double bond, forming a carbocation.

### Step 4: Formation of Carbocation
The protonation leads to the formation of a more stable carbocation. The structure after protonation will be \( \text{Ph-CH}_2\text{-C}^+(H)\text{-CH}_3 \) (where \( C^+ \) is the positively charged carbon).

### Step 5: Nucleophilic Attack by Water
Next, the hydroxide ion (\( \text{OH}^- \)) from the dissociation of sulfuric acid will act as a nucleophile and attack the carbocation. This will lead to the formation of an alcohol.

### Step 6: Final Product Formation
The final product after the nucleophilic attack will be \( \text{Ph-CH}_2\text{-CH(OH)-CH}_3 \). This compound is an alcohol, specifically phenylpropan-2-ol.

### Conclusion
Thus, the product \( X \) formed from the reaction of \( \text{Ph-CH}_2\text{-CH}=\text{CH}_2 \) with dilute \( \text{H}_2\text{SO}_4 \) is phenylpropan-2-ol.

### Final Answer
The product \( X \) is \( \text{Ph-CH}_2\text{-CH(OH)-CH}_3 \).

---