`CH_(3)CH= CH_(2) overset(H//H_(2)O)(to)` major product is

Options

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

Explanation

To solve the question regarding the reaction of propene (CH₃CH=CH₂) with water (H₂O) in the presence of an acid (H⁺), we will follow these steps:

### Step 1: Identify the Reactants
The reactant is propene, which is an alkene with the molecular formula CH₃CH=CH₂. It will react with water in the presence of an acid.

**Hint:** Recognize that alkenes react with water in the presence of acid to form alcohols.

### Step 2: Understand the Mechanism of Hydration
When propene reacts with water in the presence of an acid, the first step involves the protonation of the alkene. The double bond (C=C) acts as a nucleophile and attacks a proton (H⁺) from the acid, leading to the formation of a carbocation.

**Hint:** Remember that the double bond in alkenes can act as a nucleophile and will interact with protons to form a carbocation.

### Step 3: Formation of Carbocation
The protonation of propene can lead to two possible carbocations:
1. A secondary carbocation (more stable) at the second carbon (CH₃-CH⁺-CH₂).
2. A primary carbocation (less stable) at the first carbon (CH₃-CH₂-CH₂⁺).

The secondary carbocation is more stable due to the presence of two alkyl groups that can donate electron density.

**Hint:** Stability of carbocations is key; secondary carbocations are more stable than primary ones due to hyperconjugation and inductive effects.

### Step 4: Nucleophilic Attack by Water
The more stable secondary carbocation will then be attacked by a water molecule. The oxygen atom in water has lone pairs that can donate electrons to the positively charged carbon, forming an oxonium ion (CH₃-CH(OH₂)-CH₂⁺).

**Hint:** The nucleophile (water) attacks the carbocation, leading to the formation of an oxonium ion.

### Step 5: Deprotonation to Form Alcohol
The oxonium ion (CH₃-CH(OH₂)-CH₂⁺) will lose a proton (H⁺) to form the final product, which is an alcohol. The resulting product is isopropanol (CH₃-CHOH-CH₃).

**Hint:** The final step involves the loss of a proton from the oxonium ion to yield the alcohol.

### Conclusion
The major product of the reaction of propene with water in the presence of acid is isopropanol (CH₃-CHOH-CH₃).

### Final Answer
The major product is **isopropanol (CH₃-CHOH-CH₃)**.