The product of the reaction of `(CH_3)_3 "CCH"(OH)CH_3` with conc. `H_2SO_4` is

Options

A. `(CH_3)_3CH=CH_2`
B. `(CH_3)_2C = C(CH_3)_2`
C. `CH_3CH_2CH_2CH=CH_2`
D. `CH_3CH_2CH=CHCH_3`

Answer

To determine the product of the reaction of `(CH₃)₃CCH(OH)CH₃` with concentrated `H₂SO₄`, we can follow these steps:

### Step 1: Identify the Structure
The compound `(CH₃)₃CCH(OH)CH₃` can be represented as:
- A tert-butyl group `(CH₃)₃C-`
- A hydroxyl group `-CH(OH)-`
- A methyl group `-CH₃`

This compound can be rewritten as:
```
CH₃
|
CH₃-C-CH(OH)-CH₃
|
CH₃
```

### Step 2: Protonation of the Hydroxyl Group
When this compound is treated with concentrated sulfuric acid (`H₂SO₄`), the hydroxyl group (`-OH`) gets protonated by the acid:
```
CH₃
|
CH₃-C-CH(OH₂⁺)-CH₃
|
CH₃
```
This creates a positively charged intermediate.

### Step 3: Formation of Carbocation
The protonation of the hydroxyl group leads to the formation of a carbocation. Since we have a tertiary carbocation here, it is relatively stable:
```
CH₃
|
CH₃-C⁺-CH₃
|
CH₃
```

### Step 4: Rearrangement (if applicable)
In this case, there is no need for rearrangement since the carbocation is already tertiary and stable.

### Step 5: Elimination Reaction
Upon heating, the intermediate undergoes an elimination reaction (E1 mechanism), where a molecule of water (`H₂O`) is lost, leading to the formation of a double bond:
```
CH₃
|
CH₃-C=CH-CH₃
|
CH₃
```

### Step 6: Final Product
The final product of the reaction is:
```
(CH₃)₂C=CH(CH₃)
```
This can also be written as:
```
(CH₃)₂C=C(CH₃)₂
```
This corresponds to 2-methyl-2-butene.

### Conclusion
The product of the reaction of `(CH₃)₃CCH(OH)CH₃` with concentrated `H₂SO₄` is 2-methyl-2-butene.