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 solve the problem of determining the product obtained by the reaction of silver propanoate with bromine in carbon tetrachloride solution, we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- Silver propanoate (C3H5O2Ag), which can be represented as CH3CH2COOAg.
- Bromine (Br2) in carbon tetrachloride (CCl4) as the solvent.

### Step 2: Recognize the Type of Reaction
This reaction is known as the **Hunsdiecker reaction**, which involves the reaction of a silver salt of a carboxylic acid with halogens. It typically results in the formation of a haloalkane and the release of carbon dioxide (CO2).

### Step 3: Mechanism of the Reaction
1. **Formation of Radicals**: The bromine (Br2) will react with the silver propanoate, leading to the formation of radicals. The bromine will attack the carbon adjacent to the carboxylate group (COOAg).
2. **Cleavage of the Silver Salt**: The lone pair of electrons on bromine will facilitate the cleavage of the AgBr bond, resulting in the formation of a radical species and the release of AgBr.

### Step 4: Loss of Carbon Dioxide
In the Hunsdiecker reaction, the carboxylate group (COO) will lose a carbon dioxide molecule (CO2) during the reaction. This results in a decrease in the number of carbon atoms in the product.

### Step 5: Determine the Product
After the loss of CO2, we will have a two-carbon radical (CH3CH2•) and a bromine radical (Br•). The two radicals will combine to form the final product:
- The two-carbon radical (ethyl radical) combines with bromine to form **bromoethane (C2H5Br)**.

### Final Answer
The product obtained by the reaction of silver propanoate with bromine in carbon tetrachloride solution is **bromoethane (C2H5Br)**.

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