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

Answer

To solve the question, we need to identify the product 'X' formed from the reaction of phenyl-CH2-CH=CH2 with dilute H2SO4. Here’s the step-by-step solution:

### Step 1: Understand the Reactants
We start with phenyl-CH2-CH=CH2. This compound has a phenyl group (C6H5) attached to a CH2 group, which is further connected to a double bond (CH=CH2).

### Step 2: Reaction with Dilute H2SO4
When this compound reacts with dilute H2SO4, the acid will dissociate to provide H⁺ ions. The presence of water in dilute H2SO4 also plays a crucial role in the reaction.

### Step 3: Markovnikov's Rule
According to Markovnikov's rule, when adding H⁺ to an alkene, the hydrogen will add to the carbon with the greater number of hydrogen atoms. In our case, the double bond is between CH2 and CH. The hydrogen will add to the CH2 carbon, leading to the formation of a more stable carbocation.

### Step 4: Formation of Carbocation
After the addition of H⁺ to the double bond, we form a carbocation. The positive charge will be on the carbon that is more stable, which in this case is the benzylic carbon (the carbon adjacent to the phenyl group). The structure now looks like this:

- pH-CH⁺-CH2-CH3

### Step 5: Nucleophilic Attack by Water
Next, water (H2O) acts as a nucleophile and attacks the positively charged carbon. The oxygen of water has lone pairs of electrons, which it can donate to form a bond with the carbocation.

### Step 6: Deprotonation
After the nucleophilic attack, we have an intermediate with a positive charge on the oxygen. This intermediate will lose a proton (H⁺) to stabilize the molecule.

### Step 7: Final Product
The final product after the deprotonation will be:

- pH-CH(OH)-CH2-CH3

This compound is phenyl-CH(OH)-CH2-CH3, which is phenylpropan-2-ol.

### Conclusion
Thus, the product 'X' formed in the reaction is phenylpropan-2-ol.

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