Neopentyl bormide, undergoes dehydrohalogenation to give alkenes even though it has no `beta-H`. This is due to

Options

A. `E_(2)` mechanism
B. Rearrangement of carbocations by `E_(1)` mechanism
C. `E_(1)cB` mechanism
D. `E_(4)` mechanism

Answer

To solve the question regarding neopentyl bromide undergoing dehydrohalogenation to give alkenes despite having no beta-hydrogen, we can break down the process step by step.

### Step-by-Step Solution:

1. **Understanding Neopentyl Bromide Structure**:
Neopentyl bromide (C5H11Br) is structured as follows:
\[ \text{(CH}_3\text{)}_4\text{CBr} \]
This means that the carbon bonded to bromine (the alpha carbon) has no hydrogen atoms on the adjacent (beta) carbons.

2. **Dehydrohalogenation Process**:
Dehydrohalogenation typically involves the elimination of a hydrogen halide (in this case, HBr) to form an alkene. However, neopentyl bromide lacks beta-hydrogens, which complicates the process.

3. **Formation of Carbocation**:
When bromine leaves, it forms a carbocation at the alpha carbon. The leaving of bromine generates a positive charge on the alpha carbon:
\[ \text{(CH}_3\text{)}_4\text{C}^+ \]

4. **Carbocation Stability**:
The positive charge on the primary carbocation is not stable. To stabilize the carbocation, a rearrangement occurs. In this case, a methyl shift can happen where a methyl group from a neighboring carbon moves to the positively charged carbon:
\[ \text{(CH}_3\text{)}_3\text{C}^+ \]
This results in a more stable tertiary carbocation.

5. **Elimination of Hydrogen**:
After the formation of the tertiary carbocation, a hydrogen atom from one of the adjacent carbons (which now has beta-hydrogens due to the rearrangement) is eliminated. This leads to the formation of a double bond:
\[ \text{(CH}_3\text{)}_2\text{C}=\text{C}(\text{CH}_3) \]
This is the alkene product formed from neopentyl bromide.

6. **Conclusion**:
The reason neopentyl bromide can undergo dehydrohalogenation despite having no beta-hydrogens is due to the rearrangement of the carbocation to a more stable tertiary form, which then allows for the elimination of hydrogen to form an alkene.

### Final Answer:
Neopentyl bromide undergoes dehydrohalogenation to give alkenes due to the rearrangement of the carbocation to a more stable tertiary carbocation, allowing for the elimination of hydrogen.