`CH_(3)-underset(CH_(3))underset(|)(C)H-CH=CH_(2)+HBr rarr` (product) which is predominate , X is -

Options

A. `CH_(3)-underset(CH_(3))underset(|)(CH)-CH_(2)CH_(2)Br`
B. `CH_(3)-underset(CH_(3))underset(|)overset(Br)overset(|)(C)-CH_(2)-CH_(3)`
C. `CH_(3)-underset(CH_(3))underset(|)(CH)-underset(Br)underset(|)(CH)-CH_(3)`
D. None is correct

Answer

To solve the question regarding the reaction of 3-methylbut-1-ene with HBr, we need to determine the predominant product formed during the reaction. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Reactant
The reactant is 3-methylbut-1-ene, which can be represented as:
\[ \text{CH}_3-\text{C}(\text{CH}_3)=\text{CH}-\text{CH}_2 \]

### Step 2: Understand the Reaction with HBr
When HBr reacts with alkenes, the hydrogen (H) from HBr will add to one of the carbon atoms involved in the double bond, while the bromine (Br) will add to the other carbon. The addition follows Markovnikov's rule, which states that the more substituted carbon will receive the halide (Br) to form the more stable carbocation.

### Step 3: Form the Carbocation
1. The double bond between the carbon atoms can break, leading to the formation of a carbocation.
2. In this case, the hydrogen will add to the carbon with fewer alkyl substituents (the terminal carbon), and the bromine will add to the carbon with more alkyl substituents (the more substituted carbon).
3. This will lead to the formation of a secondary carbocation at the more substituted carbon.

### Step 4: Determine Carbocation Stability
- A secondary carbocation (2°) is formed initially.
- However, a more stable tertiary carbocation (3°) can be formed through a hydride shift (migration of a hydrogen atom from an adjacent carbon).
- This shift will convert the secondary carbocation into a tertiary carbocation, which is more stable.

### Step 5: Attack of Bromide Ion
Once the more stable tertiary carbocation is formed, the bromide ion (Br⁻) will attack this positively charged carbon, leading to the formation of the final product.

### Step 6: Write the Predominant Product
The predominant product will be:
\[ \text{CH}_3-\text{C}(\text{Br})(\text{CH}_3)-\text{CH}_2 \]
This indicates that bromine is attached to the tertiary carbon, and the structure is more stable.

### Conclusion
The major product formed from the reaction of 3-methylbut-1-ene with HBr is the one where bromine is attached to the tertiary carbon. This is the predominant product.

### Final Answer
The predominant product, X, is:
\[ \text{CH}_3-\text{C}(\text{Br})(\text{CH}_3)-\text{CH}_2 \]