The reaction of `C_(6) H_(5) CH=CHCH_(3)` with HBr produces :

Options

A.
B. `C_(6)H_(5)underset(Br)underset(|)CHCH_(2)CH_(3)`
C. `C_(6)H_(5)CH_(2)underset(Br)underset(|)CHCH_(3)`
D. `C_(6)H_(5)CH_(2)CH_(2)CH_(2)Br`

Answer

To solve the problem of the reaction of C6H5CH=CHCH3 with HBr, we will follow these steps:

### Step 1: Identify the Structure
The compound C6H5CH=CHCH3 can be broken down as follows:
- C6H5 represents a phenyl group (benzene ring).
- CH=CH represents a double bond between two carbon atoms.
- CH3 represents a methyl group.

So the full structure can be visualized as:
```
C6H5
|
C
||
C
|
CH3
```

### Step 2: Apply Markovnikov's Rule
When HBr reacts with an alkene, Markovnikov's rule states that the hydrogen atom (H) will attach to the carbon atom that has more hydrogen atoms already attached. In this case, we have two possible sites for the HBr to add across the double bond.

### Step 3: Determine Possible Carbocation Intermediates
1. **First Possible Carbocation (Intermediate A)**:
- If H adds to the carbon of the double bond that is part of the phenyl group (C6H5), the structure will be:
```
C6H5
|
CH
|
CH3
```
This leads to a carbocation at the other carbon (CH) which is less stable.

2. **Second Possible Carbocation (Intermediate B)**:
- If H adds to the other carbon (the one attached to CH3), the structure will be:
```
C6H5
|
CH2+
|
CH3
```
This carbocation is more stable due to resonance stabilization from the phenyl group.

### Step 4: Nucleophilic Attack by Bromide Ion
Once the more stable carbocation (Intermediate B) is formed, the bromide ion (Br-) will attack the positively charged carbon:
- The Br will attach to the carbon that has the positive charge, leading to the final product.

### Step 5: Write the Final Product
The final product can be represented as:
```
C6H5
|
CH
|
CH2Br
```
In condensed form, this is C6H5CH(CH2Br)CH3.

### Conclusion
The reaction of C6H5CH=CHCH3 with HBr produces C6H5CH(CH2Br)CH3.