Silver Acetate + `Br_(2) overset(CS_(2)) to` Main Product . The major product of this reaction is

Options

A. `CH_(3)Br`
B. `CH_(3)COI`
C. `CH_(3)COOH`
D. None of these

Answer

To determine the major product of the reaction between silver acetate and bromine in the presence of carbon disulfide (CS2), we can follow these steps:

### Step-by-Step Solution

1. **Identify the Reactants**:
- The reactants are silver acetate (CH3COOAg) and bromine (Br2) in the solvent carbon disulfide (CS2).

2. **Structure of Silver Acetate**:
- Silver acetate can be represented as CH3COOAg, where CH3 is the methyl group, COO is the acetate part, and Ag is the silver ion.

3. **Mechanism Initiation**:
- In the presence of Br2, the acetate ion (OOCCH3) acts as a nucleophile due to the presence of the negatively charged oxygen. This nucleophile will attack one of the bromine atoms in Br2.

4. **Formation of Brominated Product**:
- The attack of the acetate ion on Br2 leads to the cleavage of the Br-Br bond. One bromine atom attaches to the acetate, forming a brominated acetate intermediate (CH3COOBr) and releasing AgBr as a byproduct.

5. **Decarboxylation**:
- The brominated acetate can undergo decarboxylation, where the acetate group (COO) loses carbon dioxide (CO2). This results in the formation of a methyl radical (CH3•).

6. **Formation of the Final Product**:
- The methyl radical (CH3•) can then combine with the bromine radical (Br•) formed during the initial reaction, leading to the formation of methyl bromide (CH3Br).

7. **Conclusion**:
- The major product of the reaction between silver acetate and bromine in the presence of carbon disulfide is methyl bromide (CH3Br).

### Final Answer:
The major product of the reaction is **Methyl Bromide (CH3Br)**.

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