`CH_(3)-CH_(2)-overset(O)overset(||)C-O Ag overset(Br_(2))underset("CCl"_(4))rarrX overset("(1) Li")underset("(2) CuI")rarr Y overset(CH_(3)-CH_(2)-Br)underset("dry ether")rarr Z`

Options

A. `CH_(3)-CH_(3)`
B. `CH_(3)-CH_(2)-CH_(3)`
C. `CH_(3)-cH_(2)-CH_(2)-CH_(3)`
D. `CH_(3)-overset(CH_(3))overset("| ")"CH"-CH_(3)`.

Answer

To solve the given question step by step, we will analyze each reaction and its corresponding product.

### Step 1: Initial Reaction
We start with silver propionate, which is represented as:
\[ \text{CH}_3\text{CH}_2\text{C}(=O)\text{Ag} \]

This compound is treated with bromine (\( \text{Br}_2 \)) in the presence of carbon tetrachloride (\( \text{CCl}_4 \)). This reaction is known as the Hans-Dekker reaction.

**Mechanism:**
1. The oxygen atom in the silver propionate acts as a nucleophile and attacks the carbon atom of the carbonyl group, leading to the formation of a brominated product.
2. The silver ion (\( \text{Ag}^+ \)) is released as \( \text{AgBr} \).

The product after this reaction can be represented as:
\[ \text{CH}_3\text{CH}_2\text{C}(=O)\text{OBr} \]

### Step 2: Formation of Free Radicals
The bromine atom can then dissociate, leading to the formation of free radicals:
1. The \( \text{OBr} \) can break down to form a free radical \( \text{O}^\cdot \) and \( \text{Br}^\cdot \).
2. The carbonyl group can lose carbon dioxide (\( \text{CO}_2 \)), leading to the formation of a radical.

This results in:
\[ \text{CH}_3\text{CH}_2\cdot + \text{Br}^\cdot \]

### Step 3: Product X
The combination of the ethyl radical and bromine radical gives us product X:
\[ \text{Product X: CH}_3\text{CH}_2\text{Br} \]

### Step 4: Treatment with Lithium and Copper Iodide
Product X is then treated with lithium (\( \text{Li} \)) and copper iodide (\( \text{CuI} \)). This reaction is known as the Corey-House synthesis, which is used to prepare Gilman's reagent.

**Mechanism:**
1. Lithium replaces the bromine in the ethyl bromide, forming ethyl lithium:
\[ \text{CH}_3\text{CH}_2\text{Li} \]
2. The copper iodide facilitates the formation of the Gilman's reagent:
\[ \text{CH}_3\text{CH}_2\text{CuLi} \]

### Step 5: Reaction with Alkyl Halide
Now, the Gilman's reagent is treated with another alkyl halide, which is \( \text{CH}_3\text{CH}_2\text{Br} \).

**Mechanism:**
1. The nucleophilic ethyl group from the Gilman's reagent attacks the alkyl halide, leading to the formation of a new carbon-carbon bond.
2. The bromine is displaced.

The final product Y can be represented as:
\[ \text{Product Y: CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 \]

### Final Product Z
Thus, the final product Z is:
\[ \text{Z: CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 \]
This corresponds to butane.

### Summary of Products
- **Product X:** \( \text{CH}_3\text{CH}_2\text{Br} \)
- **Product Y:** \( \text{CH}_3\text{CH}_2\text{CuLi} \)
- **Final Product Z:** \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 \) (Butane)