Which of the following alkene in acid catalysed hydration form `2-`methyl propan `-2-ol` ?

Options

A. `(CH_3)_(2)-C=CH_(2)`
B. `CH_(3)-CH=CH_(2)`
C. `CH_(3)-CH=CH-CH_(3)`
D. `CH_(3)-CH_(2)-CH=CH_(2)`

Answer

To determine which alkene forms 2-methylpropan-2-ol upon acid-catalyzed hydration, we will analyze each option step by step.

### Step 1: Understand the Reaction
In acid-catalyzed hydration of alkenes, water (H₂O) is added across the double bond. According to Markovnikov's rule, the hydrogen atom (H⁺) from the acid will attach to the carbon atom of the double bond that has more hydrogen atoms already attached, while the hydroxyl group (OH⁻) will attach to the carbon with fewer hydrogen atoms.

### Step 2: Analyze Each Option

**Option A: CH3-C(=CH2)-CH3**
- Structure: CH3-CH=CH2
- Markovnikov addition: H⁺ goes to the carbon with 2 H atoms (the first carbon), and OH⁻ goes to the carbon with no H atoms (the second carbon).
- Product: CH3-C(OH)(CH3)-CH3 (2-methylpropan-2-ol)
- This is the required product.

**Option B: CH3-CH=CH2**
- Structure: CH3-CH=CH2
- Markovnikov addition: H⁺ goes to the second carbon (which has more H atoms), and OH⁻ goes to the first carbon.
- Product: CH3-CHOH-CH3 (2-propanol)
- This is not the required product.

**Option C: CH3-CH=CH-CH3**
- Structure: CH3-CH=CH-CH3
- Markovnikov addition: H⁺ can go to either of the two carbons with equal H atoms (symmetrical), and OH⁻ goes to the other.
- Product: CH3-CH2-CHOH-CH3 (2-butanol)
- This is not the required product.

**Option D: CH3-CH2-CH=CH2**
- Structure: CH3-CH2-CH=CH2
- Markovnikov addition: H⁺ goes to the second carbon (which has more H atoms), and OH⁻ goes to the first carbon.
- Product: CH3-CH2-CHOH-CH3 (2-butanol)
- This is not the required product.

### Conclusion
The only alkene that forms 2-methylpropan-2-ol upon acid-catalyzed hydration is **Option A: CH3-C(=CH2)-CH3**.

### Final Answer
**Option A: CH3-C(=CH2)-CH3 forms 2-methylpropan-2-ol.**