`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 reaction sequence step by step, let's break down each part of the process:

### Step 1: Reaction of Silver Salt of Carboxylic Acid with Bromine
The starting compound is a silver salt of a carboxylic acid, which can be represented as CH₃-CH₂-C(=O)OAg. When this compound reacts with Br₂ in the presence of CCl₄, it undergoes a reaction known as the **Hunsdiecker reaction**.

**Reaction:**
\[ \text{CH}_3\text{CH}_2\text{C(=O)OAg} + \text{Br}_2 \xrightarrow{\text{CCl}_4} \text{CH}_3\text{CH}_2\text{C(=O)Br} + \text{CO}_2 + \text{AgBr} \]

**Product:** The product formed here is CH₃-CH₂-C(=O)Br, and carbon dioxide (CO₂) is released.

### Step 2: Free Radical Mechanism
In the presence of bromine, the reaction proceeds via a free radical mechanism. The bromine attacks the carbonyl carbon, leading to the formation of a radical:

1. The bromine adds to the carbonyl carbon, forming a brominated intermediate.
2. The bond between the carbon and the bromine breaks, leading to the formation of a radical and the release of CO₂.

### Step 3: Formation of Alkyl Bromide (X)
The radical formed (CH₃-CH₂•) can then combine with Br• to form the alkyl bromide:

**Product:**
\[ \text{CH}_3\text{CH}_2\text{Br} \]
This compound is denoted as X.

### Step 4: Reaction with Lithium
Next, we treat the alkyl bromide (X) with lithium (Li):

**Reaction:**
\[ \text{CH}_3\text{CH}_2\text{Br} + \text{Li} \rightarrow \text{CH}_3\text{CH}_2\text{Li} + \text{LiBr} \]

**Product:** The product formed is CH₃-CH₂-Li, which is denoted as Y.

### Step 5: Reaction with Copper(I) Iodide (CuI)
Now, we react the lithium alkyl (Y) with copper(I) iodide:

**Reaction:**
\[ \text{CH}_3\text{CH}_2\text{Li} + \text{CuI} \rightarrow \text{CH}_3\text{CH}_2\text{Cu} + \text{LiI} \]

**Product:** The product formed is CH₃-CH₂-Cu, which is still denoted as Y.

### Step 6: Reaction with Alkyl Bromide
Finally, we react Y with another molecule of alkyl bromide (CH₃-CH₂-Br) in dry ether:

**Reaction:**
\[ \text{CH}_3\text{CH}_2\text{Cu} + \text{CH}_3\text{CH}_2\text{Br} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_3 + \text{CuBr} \]

**Product:** The product formed is butane (CH₃-CH₂-CH₂-CH₃), which is denoted as Z.

### Final Summary
- The final product Z is butane (C₄H₁₀).