The major product in the dehydrohalogenation of 3 - Bromo - 2, 2 - dimethylbutane is

Options

A. 3, 3 - dimethyl but -1- ene
B. 2, 3 - dimethylbut -1- ene
C. 2, 3 - dimethyl but -2- ene
D. 4 - methylpent -2- ene

Answer

To determine the major product in the dehydrohalogenation of 3-bromo-2,2-dimethylbutane, we can follow these steps:

### Step 1: Identify the Structure of 3-Bromo-2,2-Dimethylbutane
3-Bromo-2,2-dimethylbutane has the following structure:
- The main chain is a butane (4 carbon atoms).
- There are two methyl groups attached to the second carbon.
- A bromine atom is attached to the third carbon.

The structure can be represented as:
```
CH3
|
CH3 - C - CH - Br
|
CH3
```

### Step 2: Understand the Dehydrohalogenation Reaction
Dehydrohalogenation is the elimination of a hydrogen halide (in this case, HBr) from an alkyl halide. This reaction typically requires a strong base, such as sodium ethoxide (C2H5ONa).

### Step 3: Formation of Carbocation
In the presence of a strong base, the bromine atom will leave, resulting in the formation of a carbocation. Since the bromine is attached to a secondary carbon (the third carbon), the reaction will proceed via an E1 mechanism:
- The bromine (Br) leaves, forming a secondary carbocation.

### Step 4: Carbocation Rearrangement
The secondary carbocation can undergo rearrangement to form a more stable tertiary carbocation. In this case, a methyl group from the adjacent carbon can shift to the carbocation:
```
CH3
|
CH3 - C - C+
|
CH3
```
This rearrangement leads to a tertiary carbocation:
```
CH3
|
CH3 - C - CH3
|
C+
```

### Step 5: Elimination of Proton
From the tertiary carbocation, a proton (H+) will be eliminated to form a double bond, resulting in the formation of an alkene. The elimination can occur from the adjacent carbon:
```
CH3
|
CH3 - C = C
|
CH3
```

### Step 6: Determine the Major Product
The major product will be the alkene with the most substituted double bond, which is known as the Zaitsev's rule. In this case, the major product is:
- **2,3-Dimethylbut-2-ene** (the double bond is between the second and third carbons).

### Final Answer
The major product in the dehydrohalogenation of 3-bromo-2,2-dimethylbutane is **2,3-dimethylbut-2-ene**.

---