`(X)overset(KOH+CHCl_(3))rarr(Y)overset(LiAlH_(4))rarrCH_(3)CH_(2)NHCH_(3)`
Identify compound X

Options

A. `CH_(3)-CH_(2)-overset(O)overset(||)C-OH`
B. `CH_(3)-CH_(2)-overset(O)overset(||)C-NH_(2)`
C. `CH_(3)-CH_(2)-overset(O)overset(||)C-Cl`
D. `CH_(3)-CH_(2)-NH_(2)`

Answer

To identify compound X in the given reaction sequence, we can follow these steps:

### Step 1: Analyze the Final Product
The final product is given as CH3CH2NHCH3. This indicates that the product is a secondary amine, specifically N-ethylmethylamine.

### Step 2: Identify the Reducing Agent
The reaction involves LiAlH4, which is a strong reducing agent. It typically reduces carbonyl compounds, imines, or similar functional groups to amines.

### Step 3: Determine the Preceding Compound (Y)
Since LiAlH4 is reducing Y to the final product, Y must be a compound that can be reduced to form the secondary amine. A likely candidate is an imine or a similar nitrogen-containing compound. Given the structure of the final product, Y could be an isocyanate or a nitrile.

### Step 4: Identify the Precursor Compound (X)
The reaction of X with KOH and CHCl3 suggests that X is likely a primary amine or an amide that can react to form an isocyanate. The presence of KOH and chloroform indicates a reaction that can lead to the formation of a carbylamine (isocyanate).

### Step 5: Construct the Structure of X
If we assume that X is a primary amine, it should have the structure CH3CH2NH2 (ethylamine). When ethylamine reacts with KOH and CHCl3, it can form ethyl isocyanate (Y), which can then be reduced by LiAlH4 to give the final product CH3CH2NHCH3.

### Conclusion
Thus, the compound X is identified as ethylamine (CH3CH2NH2).

### Summary of Steps:
1. Analyze the final product to determine its structure.
2. Identify the role of LiAlH4 as a reducing agent.
3. Determine the structure of compound Y that can be reduced to the final product.
4. Identify compound X that can react with KOH and CHCl3 to form Y.
5. Conclude that compound X is ethylamine.