In the given reaction sequence `C_(6)H_(5)-CH_(2)-NH_(2) overset(CHCl_(3)//Alc.KOH)underset(Delta)rarr [X] overset(H_(2)O //NaOH)rarr [Y],[Y]` will be:

Options

A. `C_(6)H_(5)-CN`
B. `C_(6)H_(5)NC`
C. `C_(6)H_(5)-CH_(2)-NH_(2)`
D. `C_(6)H_(5)-CH_(2)OH`

Answer

To solve the given reaction sequence step by step, we will analyze the reaction of benzylamine (C6H5-CH2-NH2) with chloroform (CHCl3) and alcoholic KOH, followed by hydrolysis with water and sodium hydroxide.

### Step 1: Reaction of Benzylamine with Chloroform and Alcoholic KOH
1. **Starting Compound**: The starting compound is benzylamine (C6H5-CH2-NH2).
2. **Reagents**: We are using chloroform (CHCl3) and alcoholic KOH.
3. **Mechanism**: In the presence of alcoholic KOH, chloroform acts as a source of the isocyanide (or carbylamine) formation. The basic medium facilitates the deprotonation of the amine, allowing it to react with chloroform.
4. **Formation of Isocyanide**: The reaction leads to the formation of benzyl isocyanide (C6H5-CH2-NC), which can be represented as:
\[ C6H5-CH2-NH2 + CHCl3 + KOH \rightarrow C6H5-CH2-NC + KCl + H2O \]
This compound is the product [X].

### Step 2: Hydrolysis of Isocyanide
1. **Next Step**: The product [X] (benzyl isocyanide) is then treated with water and sodium hydroxide (NaOH).
2. **Hydrolysis Reaction**: The isocyanide undergoes hydrolysis in the presence of water and NaOH to yield benzylamine and formic acid (HCOOH):
\[ C6H5-CH2-NC + H2O \rightarrow C6H5-CH2-NH2 + HCOOH \]
Here, the isocyanide is converted back to the original amine (benzylamine) and formic acid.

### Final Product
- The final product [Y] after hydrolysis is benzylamine (C6H5-CH2-NH2).

### Conclusion
Thus, the correct answer for [Y] is:
\[ \text{C6H5-CH2-NH2 (benzylamine)} \]

### Answer
The correct option is C: C6H5-CH2-NH2.

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