Consider the given reaction
`CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-CH_(2)-OH overset("conc." H_(2)SO_(4))underset(Delta)to "Alkene(major)"`
Identify alkene.

Options

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

Answer

To identify the major alkene formed from the given reaction of the alcohol with concentrated H₂SO₄, we can follow these steps:

### Step 1: Identify the structure of the alcohol
The given alcohol is 2-methyl-2-butanol, which can be represented as:
```
CH3
|
CH3-C-CH2-OH
|
CH3
```
This structure shows that the alcohol has a hydroxyl (-OH) group attached to a carbon that is also bonded to three other carbon groups.

### Step 2: Protonation of the alcohol
When the alcohol reacts with concentrated H₂SO₄, the hydroxyl group (-OH) gets protonated to form a better leaving group, water (H₂O):
```
CH3
|
CH3-C-CH2-OH2^+
|
CH3
```
This protonation increases the electrophilicity of the carbon attached to the -OH group.

### Step 3: Formation of carbocation
The protonated alcohol (now a better leaving group) can lose water (H₂O) to form a carbocation:
```
CH3
|
CH3-C^+-CH2
|
CH3
```
In this case, the carbocation is formed on the tertiary carbon, which is more stable due to the presence of three alkyl groups.

### Step 4: Rearrangement of carbocation (if necessary)
In this case, the carbocation is already tertiary, which is the most stable form. No rearrangement is needed.

### Step 5: Deprotonation to form the alkene
The next step involves the removal of a proton (H⁺) from the adjacent carbon to form the double bond:
1. If the H⁺ is removed from the carbon adjacent to the carbocation, the alkene formed will be:
```
CH3
|
CH3-C=CH2
|
CH3
```
2. Alternatively, if the H⁺ is removed from the other adjacent carbon, we would get:
```
CH3
|
CH2=C-CH3
|
CH3
```

### Step 6: Determine the major product using Zaitsev's rule
According to Zaitsev's rule, the more substituted alkene is favored. In this case:
- The first alkene has two alkyl groups attached to the double bond (more substituted).
- The second alkene has only one alkyl group attached to the double bond.

Thus, the major alkene product will be:
```
CH3
|
CH3-C=CH2
|
CH3
```

### Final Answer
The major alkene formed from the reaction is **2-methylpropene**.

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