The product obtained by the reaction of silver propanoate with bromine in carbon tetrachloride solution is:

Options

A. Bromoethane
B. Silver-3-bromopropanoate
C. Ethanoyl bromide
D. None of the above

Answer

To determine the product obtained by the reaction of silver propanoate with bromine in carbon tetrachloride (CCl4), we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- Silver propanoate (CH3CH2COOAg)
- Bromine (Br2)
- Solvent: Carbon tetrachloride (CCl4)

### Step 2: Understand the Reaction Type
This reaction is a type of halogenation involving a silver salt of a carboxylic acid. Specifically, it follows the Hunsdiecker reaction mechanism, which involves the formation of an alkyl halide from the silver salt of a carboxylic acid.

### Step 3: Analyze the Structure of Silver Propanoate
The structure of silver propanoate can be represented as:
- CH3-CH2-C(=O)O^- Ag^+
This indicates that it has a three-carbon chain (propanoate) with a silver ion.

### Step 4: Reaction Mechanism
In the presence of bromine and CCl4, the following steps occur:
1. The Br2 molecule undergoes homolytic cleavage to generate two bromine radicals (Br·).
2. The silver propanoate reacts with Br2, leading to the formation of a radical intermediate and the release of carbon dioxide (CO2).
3. The alkyl radical (CH3CH2·) formed from the propanoate will then react with a bromine radical to form bromoethane (CH3CH2Br).

### Step 5: Determine the Product
From the above steps, the product formed is:
- Bromoethane (CH3CH2Br)

### Step 6: Byproducts
In addition to bromoethane, the reaction also produces:
- Silver bromide (AgBr)
- Carbon dioxide (CO2)

### Conclusion
The final product obtained from the reaction of silver propanoate with bromine in CCl4 is bromoethane (CH3CH2Br).

### Summary of the Final Product
- **Product**: Bromoethane (CH3CH2Br)
- **Byproducts**: Silver bromide (AgBr) and carbon dioxide (CO2)

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