`(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 reaction sequence provided, we need to analyze the transformations that occur step by step.

### Step 1: Understanding the Final Product
The final product after the reaction with Lithium Aluminum Hydride (LiAlH4) is given as CH3CH2NHCH3. This compound is a secondary amine, specifically N-ethylmethylamine.

### Step 2: Identifying Compound Y
Before the reduction step, we denote the compound formed after the reaction with KOH and chloroform as Y. The reaction of KOH with chloroform is known as the "base-induced reaction with chloroform," which typically involves the formation of an intermediate that can lead to the formation of an amine.

### Step 3: Determining the Precursor of Y
Since Y is converted to the final product (CH3CH2NHCH3) by reduction, we can deduce that Y must be an intermediate that contains a nitrogen atom and is likely an imine or an amine derivative. Given that we are starting with a secondary amine, we can assume that Y is an N-substituted derivative of the amine.

### Step 4: Identifying Compound X
To form Y from X using KOH and chloroform, we need to consider the reaction of a primary amine with chloroform in the presence of a strong base. The general reaction of a primary amine (RNH2) with chloroform and KOH leads to the formation of an isocyanate (R-N=C=O) or a carbamate, which can then be reduced to the corresponding amine.

Thus, if we assume that X is a primary amine, specifically ethylamine (CH3CH2NH2), the reaction with KOH and chloroform would yield an intermediate that can be reduced to yield the final product.

### Conclusion
Therefore, the compound X that leads to the formation of Y, which upon reduction gives CH3CH2NHCH3, is:

**X = Ethylamine (CH3CH2NH2)**