Ethylene reacts with `Br_(2)` to give 1,2-dibromoethane.The anti addition takes place due to the formation of which intermediate?

Options

A. `CH_(2)Br-overset(+)CH_(2)`
B.
C. `BrCH-CH^(+)`
D.

Explanation

To solve the question of how ethylene reacts with bromine to form 1,2-dibromoethane through anti-addition, we can break down the process into several steps:

### Step 1: Understanding the Reactants
Ethylene (C2H4) has a double bond between the two carbon atoms. The structure can be represented as:
\[ \text{CH}_2= \text{CH}_2 \]

### Step 2: Reaction with Bromine
When ethylene reacts with bromine (Br2), the double bond acts as a nucleophile and attacks one of the bromine atoms. This leads to the formation of a cyclic bromonium ion intermediate.

### Step 3: Formation of the Bromonium Ion
As the double bond attacks one of the bromine atoms, the bond between the bromine atoms breaks. This results in the formation of a bromonium ion, which is a three-membered cyclic structure where one bromine atom is positively charged and is bonded to both carbon atoms of the ethylene:
\[ \text{Br}^+ \text{C}_2\text{H}_4 \]

### Step 4: Nucleophilic Attack
The negatively charged bromide ion (Br-) that was released during the formation of the bromonium ion can now attack the more substituted carbon atom of the bromonium ion. This attack occurs from the opposite side of the bromonium ion, leading to anti-addition.

### Step 5: Final Product Formation
This results in the formation of 1,2-dibromoethane (CH2Br-CH2Br), where the two bromine atoms are added to opposite sides of the original double bond:
\[ \text{CH}_2\text{Br}-\text{CH}_2\text{Br} \]

### Conclusion
The anti-addition occurs due to the formation of the bromonium ion intermediate, which allows the bromide ion to attack from the opposite side, leading to the final product.

### Answer
The anti-addition takes place due to the formation of the bromonium ion intermediate.

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