Predict the correct intermediate and product in the following reaction.
`H_(2)-C=CH overset(H_(2)O H_(2)SO_(2)) underset(H_(2)SO_(4)) to underset((A)) ("intermediante") to underset((B)) ("product")`

Options

A. `A:H_(3) C- underset(SO_(4)) underset(|) (C)=CH_(2) B: H_(3)C- underset(O) underset(||)(C)-CH_(3)`
B. `A:H_(3) C- underset(OH) underset(|) (C)=CH_(2) B: H_(3)C- underset(O) underset(||)(C)-CH_(3)`
C. `A:H_(3) C- underset(OH) underset(|)(C)=CH_(2) B: H_(3)C- underset(SO_(4)) underset(||)(C)-CH_(2)`
D. `A:H_(3) C- underset(O) underset(|) (C)=CH_(2) B: H_(3)C-C-=H`

Explanation

To solve the problem, we need to predict the correct intermediate and product for the reaction involving H2C=CH2 (ethylene) with H2O and H2SO4. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Reactants
The reactants in the reaction are ethylene (H2C=CH2) and water (H2O) in the presence of sulfuric acid (H2SO4).

**Hint:** Look for the functional groups in the reactants that will participate in the reaction.

### Step 2: Understand the Reaction Mechanism
In the presence of sulfuric acid, water can dissociate into H+ and OH-. The H+ ion will protonate the double bond of ethylene, leading to the formation of a carbocation intermediate.

**Hint:** Remember that the addition of H+ to a double bond generates a carbocation.

### Step 3: Formation of the Carbocation Intermediate
When H2C=CH2 is protonated, the double bond breaks, and a carbocation is formed. The more stable carbocation will be formed at the carbon that can accommodate the positive charge better. In this case, the intermediate can be represented as CH3-C^+(CH2)-OH, where the positive charge is on the more substituted carbon.

**Hint:** Consider the stability of carbocations; more substituted carbocations are generally more stable.

### Step 4: Nucleophilic Attack by Water
The hydroxide ion (OH-) from water will attack the positively charged carbon in the carbocation intermediate, resulting in the formation of an alcohol.

**Hint:** Look for how nucleophiles interact with electrophiles in the reaction.

### Step 5: Identify the Final Product
After the nucleophilic attack, the product will be an alcohol. In this case, the product formed will be CH3-CH(OH)-CH3, which is isopropanol (or propan-2-ol).

**Hint:** Check the structure of the product to ensure it matches the expected outcome based on the reactants.

### Summary of the Intermediate and Product
- **Intermediate (A):** The carbocation formed after the addition of H+ to the double bond.
- **Product (B):** The final product after the nucleophilic attack by water, which is isopropanol (CH3-CH(OH)-CH3).

### Final Answer
- Intermediate (A): CH3-C^+(CH2)-OH (carbocation)
- Product (B): CH3-CH(OH)-CH3 (isopropanol)