Consider the following reaction `CH_3-CH=CH_2overset(Br_2//NaCl)rarr` Product of the reaction will be

Options

A. Only 1,2 - dibromopropane
B. Only 1- bromo -2- chloropropane
C. Only -2- bromo -1- chloropropane
D. Mixture of 1,2 dibromopropane and 1 - bromo -2- chloropropane

Answer

To solve the question regarding the reaction of propene (CH3-CH=CH2) with bromine (Br2) in the presence of sodium chloride (NaCl), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactants are propene (CH3-CH=CH2) and bromine (Br2) in the presence of sodium chloride (NaCl).

2. **Understand the Reaction Mechanism**: The double bond in propene is electron-rich, making it susceptible to electrophilic attack. The bromine molecule (Br2) will undergo homolytic cleavage to form two bromine radicals (Br•).

3. **Formation of Bromonium Ion**: When bromine approaches the double bond, one of the bromine atoms will form a bond with one of the carbon atoms in the double bond, resulting in the formation of a bromonium ion. The other bromine atom will carry a positive charge.

- Intermediate: CH3-CH(Br+)-CH2

4. **Nucleophilic Attack**: The bromide ion (Br-) that is generated from the cleavage of Br2 will now act as a nucleophile. It will attack the more substituted carbon of the bromonium ion from the opposite side (anti-addition), leading to the formation of a vicinal dibromide.

5. **Product Formation**: The major product formed from this reaction will be 1,2-dibromopropane (CH3-CHBr-CH2Br) due to the anti-addition of bromine across the double bond.

6. **Alternative Product Formation**: In the presence of NaCl, the chloride ion (Cl-) can also act as a nucleophile. If Cl- attacks instead of Br-, we can form 1-bromo-2-chloropropane (CH3-CHCl-CH2Br) through a similar anti-addition mechanism.

### Summary of Products:
- **Major Product**: 1,2-Dibromopropane (CH3-CHBr-CH2Br)
- **Minor Product**: 1-Bromo-2-chloropropane (CH3-CHCl-CH2Br)