`CH_3-underset(Br)underset(|)overset(Br)overset(|)C-underset(Br)underset(|)overset(Br)overset(|)C-CH_3overset(xNaNH_(2))to overset(yCH_(3)I)to CH_(3)-C-=C-C-=C-CH_(3)`
value of x+y =

Answer

To solve the problem step by step, we will analyze the reaction of tetrabromobutane with sodium amide (NaNH₂) and methyl iodide (CH₃I) to determine the values of x and y.

### Step 1: Identify the starting compound
The starting compound is tetrabromobutane, which can be represented as:
\[ \text{Br-CH}_2\text{-C(Br)(Br)-C(Br)(Br)-CH}_3 \]

### Step 2: Determine the reaction with sodium amide (NaNH₂)
Sodium amide is a strong base and acts as a dehydrohalogenating agent. In this case, it will remove the bromine atoms along with hydrogen atoms from adjacent carbon atoms to form a triple bond.

Since there are four bromine atoms in the tetrabromobutane, we will need:
- **4 moles of NaNH₂** to remove all four bromine atoms and form two triple bonds.

### Step 3: Form the intermediate product
After the reaction with 4 moles of NaNH₂, the compound will convert to:
\[ \text{CH}_3-\text{C} \equiv \text{C}-\text{C} \equiv \text{C}-\text{CH}_3 \]
This is a diacetylene compound.

### Step 4: Reaction with methyl iodide (CH₃I)
Next, we will react the intermediate product with methyl iodide. Each terminal alkyne can react with methyl iodide to form a new carbon-carbon bond.

To convert the diacetylene to the desired product, we need:
- **2 moles of CH₃I** to add a methyl group to each terminal carbon of the diacetylene.

### Step 5: Calculate x and y
From the above analysis:
- \( x = 4 \) (moles of NaNH₂)
- \( y = 2 \) (moles of CH₃I)

### Step 6: Find the sum of x and y
Now, we can find the sum:
\[ x + y = 4 + 2 = 6 \]

### Final Answer
The value of \( x + y \) is **6**.