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

Options

A. `C_(6)H_(5)-CH_(2)CH_(2)-CH_(2) -Br`
B.
C. `C_(6)H_(5)- overset(Br)overset(|)(CH)-CH_(2)-CH_(3)`
D. `C_(6)H_(5)-CH_(2)-overset(Br)overset(|)(CH_(2)) - CH_(3)`

Answer

To determine the product of the reaction between `C6H5-CH=CH-CH3` (which is styrylpropene) and HBr, we can follow these steps:

### Step 1: Identify the Reactants
The reactant is `C6H5-CH=CH-CH3`, which consists of a phenyl group (C6H5) attached to a double bond (C=CH) and a propyl group (CH3). HBr is a strong acid that dissociates into H+ and Br- ions.

### Step 2: Protonation of the Alkene
When HBr reacts with the alkene, the double bond (C=CH) can undergo protonation. The H+ from HBr can add to one of the carbon atoms in the double bond. Due to Markovnikov's rule, the hydrogen will add to the carbon that is less substituted.

In this case, the double bond is between the second and third carbon atoms:
```
C6H5 - CH = CH - CH3
```
The proton will add to the carbon atom that is bonded to the CH3 group (the less substituted carbon), leading to the formation of a more stable carbocation.

### Step 3: Formation of the Carbocation
After the protonation, the structure becomes:
```
C6H5 - CH2 - CH+ - CH3
```
This results in a carbocation at the second carbon (C2) of the double bond.

### Step 4: Nucleophilic Attack by Bromide Ion
The Br- ion, which is the nucleophile, will then attack the positively charged carbon (C2) of the carbocation. This will lead to the formation of the final product.

### Step 5: Final Product
The final product after the nucleophilic attack will be:
```
C6H5 - CH2 - CHBr - CH3
```
This compound is 1-bromo-3-phenylpropane.

### Summary of the Reaction
The reaction of `C6H5-CH=CH-CH3` with HBr produces 1-bromo-3-phenylpropane.

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