Propene on reaction with `Br_(2) "in" C Cl_(4)` yields

Answer

To solve the question regarding the reaction of propene with bromine in carbon tetrachloride (CCl4), we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are propene (C3H6) and bromine (Br2). Propene has the structure CH3-CH=CH2.

### Step 2: Understand the Reaction Mechanism
When bromine (Br2) is added to propene, the bromine molecule undergoes a homolytic cleavage. This means that one of the bromine atoms will become positively charged (Br+) and the other will become negatively charged (Br-).

### Step 3: Formation of the Bromonium Ion
The positively charged bromine (Br+) will attack the double bond of propene. This results in the formation of a bromonium ion intermediate. The double bond between the two carbon atoms in propene breaks, and one of the bromine atoms forms a bond with one of the carbon atoms, creating a three-membered ring structure with the bromonium ion.

### Step 4: Nucleophilic Attack
The negatively charged bromine ion (Br-) acts as a nucleophile and attacks the more substituted carbon atom of the bromonium ion. This leads to the opening of the bromonium ion and the formation of a dibrominated product.

### Step 5: Final Product Formation
The final product of this reaction is 1,2-dibromopropane. The structure of 1,2-dibromopropane can be represented as:
```
Br
|
CH3-CH-CH2
|
Br
```
This shows that bromine atoms are added to both carbon atoms that were involved in the double bond of propene.

### Conclusion
The reaction of propene with bromine in CCl4 yields 1,2-dibromopropane.