In the reaction , `C_2H_5Br underset((ii)CuI)overset((i)Li)to "(X)" overset(CH_3Br)to (Y) `, the products X and Y are ________ and ____________ respectively.

Answer

To solve the problem, we need to analyze the given reaction step by step.

### Step 1: Identify the starting compound
The starting compound is `C2H5Br` (ethyl bromide).

### Step 2: Reaction with Lithium
When `C2H5Br` reacts with lithium (Li), it forms an organolithium compound. The reaction can be represented as:
\[ C2H5Br + Li \rightarrow C2H5Li + LiBr \]
Here, `C2H5Li` is the organolithium compound formed, and `LiBr` is a byproduct.

### Step 3: Reaction with Copper(I) Iodide
Next, the organolithium compound `C2H5Li` reacts with copper(I) iodide (`CuI`). The reaction can be represented as:
\[ C2H5Li + CuI \rightarrow C2H5Cu + LiI \]
In this step, we form an organocopper compound, `C2H5Cu`, and lithium iodide (`LiI`) is formed as a byproduct.

### Step 4: Reaction with Methyl Bromide
Now, we take the organocopper compound `C2H5Cu` and react it with methyl bromide (`CH3Br`). The reaction can be represented as:
\[ C2H5Cu + CH3Br \rightarrow C2H5CH3 + CuBr \]
In this step, we form the final product `C2H5CH3` (which is butane) and copper(I) bromide (`CuBr`) as a byproduct.

### Conclusion
From the above steps, we can conclude that:
- The product \( X \) is `C2H5Cu` (the organocopper compound).
- The product \( Y \) is `C2H5CH3` (butane).

Thus, the products \( X \) and \( Y \) are **C2H5Cu** and **C2H5CH3** respectively.

### Final Answer
The products \( X \) and \( Y \) are **C2H5Cu** and **C2H5CH3** respectively.

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