The product formed in the following reaction is `CH_3CH(CH_3)CH=CH_2+HBrrarr` Product

Options

A. `(CH_3)_2CHCH(Br)CH_3`
B. `(CH_3)_2CHCH_2CH_2Br`
C. `(CH_3)_2C(Br)CH_2CH_3`
D. `CH_3CH(CH_3)CH(Br)CH_2CH_3`

Explanation

To determine the product formed in the reaction of 2-methylbutene (CH₃CH(CH₃)CH=CH₂) with HBr, we will follow these steps:

### Step 1: Identify the Reactants
The reactant is 2-methylbutene, which has the structure:
```
CH3
|
CH3-CH-CH=CH2
```
We are reacting this with HBr.

### Step 2: Understand the Mechanism
The reaction of alkenes with HBr follows the Markovnikov's rule, which states that when HX (where X is a halogen) adds to an alkene, the hydrogen atom will attach to the carbon with the greater number of hydrogen atoms already attached (the more substituted carbon).

### Step 3: Protonation of the Alkene
In our case, the double bond between the carbons will react with HBr. The hydrogen (H⁺) from HBr will add to the carbon atom that has more hydrogen atoms (the terminal carbon in this case).

### Step 4: Formation of the Carbocation
After the addition of H⁺, a carbocation will form. The possible carbocation structures are:
1. A secondary carbocation at the carbon adjacent to the double bond:
```
CH3
|
CH3-CH-CH2⁺
```
2. A primary carbocation at the terminal carbon:
```
CH3-CH(CH3)-CH2-CH2⁺
```

### Step 5: Nucleophilic Attack by Bromide Ion
The bromide ion (Br⁻) will then attack the carbocation. Since the secondary carbocation is more stable than the primary, the bromide will preferentially attack the secondary carbocation.

### Step 6: Write the Final Product
The final product of the reaction will be:
```
CH3
|
CH3-CH-CH2-Br
```
This corresponds to 2-bromo-2-methylbutane.

### Final Answer
The product formed in the reaction of 2-methylbutene with HBr is **2-bromo-2-methylbutane**.

---