How many total organic products are obtained in following reaction.
`CH_(2)=CH_(2)+Br_(2)overset(aq.NaCl)rarr?`

Answer

To determine the total organic products obtained from the reaction of ethene (CH₂=CH₂) with bromine (Br₂) in the presence of aqueous sodium chloride (NaCl), we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- Ethene (CH₂=CH₂)
- Bromine (Br₂)
- Aqueous sodium chloride (NaCl)

### Step 2: Understand the Reaction Mechanism
When ethene reacts with bromine, a bromonium ion is formed as an intermediate. This is due to the electrophilic addition mechanism where Br₂ adds across the double bond of ethene.

### Step 3: Formation of Bromonium Ion
The bromine molecule (Br₂) approaches the double bond of ethene, and one of the bromine atoms forms a bond with one of the carbon atoms, creating a cyclic bromonium ion. The other bromine atom becomes a bromide ion (Br⁻).

### Step 4: Nucleophilic Attack
In the presence of aqueous NaCl, there are chloride ions (Cl⁻) available. The bromide ion (Br⁻) can act as a nucleophile, and the chloride ion (Cl⁻) can also attack the bromonium ion. This leads to two possible pathways for nucleophilic attack:
1. **Br⁻ attacks**: This results in the formation of 1,2-dibromoethane (Ethylene dibromide).
2. **Cl⁻ attacks**: This results in the formation of 1-bromo-2-chloroethane.

### Step 5: Identify the Products
From the above pathways, we can identify the products:
1. **1,2-Dibromoethane** (Ethylene dibromide)
2. **1-Bromo-2-chloroethane**

### Step 6: Count the Total Organic Products
Thus, the total organic products obtained from the reaction are:
- 1,2-Dibromoethane
- 1-Bromo-2-chloroethane

### Conclusion
The total number of organic products obtained from the reaction is **2**.

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