The reaction of `C_6H_5CH=CHCH_3` with HBr produces

Options

A. `C_6H_5CH_2CH_2CH_2Br`
B.
C. `C_6H_5undersetunderset"Br"|CHCH_2CH_3`
D. `C_6H_5CH_2undersetunderset"Br"|CH_2CH_3`

Explanation

To solve the question regarding the reaction of `C_6H_5CH=CHCH_3` (which is styrylpropane) with HBr, we can follow these steps:

### Step 1: Identify the Structure
The compound `C_6H_5CH=CHCH_3` can be represented as:
- A phenyl group (C6H5) attached to a double bond (CH=CH) followed by a methyl group (CH3).

### Step 2: Understand the Reaction with HBr
When HBr reacts with alkenes, it adds across the double bond. The addition occurs through a mechanism that involves the formation of a carbocation intermediate.

### Step 3: Determine the Site of Protonation
In this case, the double bond is between the second and third carbon atoms (C2 and C3). HBr will first protonate one of the carbons in the double bond. The more stable carbocation will form, which is determined by the stability of the resulting carbocation.

### Step 4: Analyze Carbocation Stability
The possible carbocations that can form are:
1. Protonation at C2 (leading to a secondary carbocation).
2. Protonation at C3 (leading to a primary carbocation).

Since secondary carbocations are more stable than primary carbocations due to greater hyperconjugation and inductive effects, the proton will add to C2.

### Step 5: Form the Carbocation
After protonation, we will have the following structure:
- The carbocation will be at C2, which is attached to the phenyl group (C6H5) and has a positive charge.

### Step 6: Nucleophilic Attack by Bromide Ion
The bromide ion (Br-) will then attack the carbocation at C2, leading to the formation of the final product.

### Step 7: Write the Final Product
The final product of the reaction will be:
- `C6H5CHBrCH2CH3` (where Br is attached to C2).

### Final Answer
The reaction of `C_6H_5CH=CHCH_3` with HBr produces `C6H5CHBrCH2CH3`.

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