`CH_(3)CH_(2)Cl overset("NaCN")(rarr) X overset("Ni"//H_(2))(rarr)`?
Y in the above reacting sequence is

Options

A. `CH_(3)CH_(2)CH_(2)"NHCOCH"_(3)`
B. `CH_(3)CH_(2)CH_(2)NH_(2)`
C. `CH_(3)CH_(2)"CONHCH"_(3)`
D. `CH_(3) CH_(2) CH_(2) "CONHCOCH"_(3)`

Answer

To solve the question, we need to analyze the reaction step by step:

1. **Identify the starting compound**: The starting compound is CH₃CH₂Cl, which is ethyl chloride.

2. **Reaction with NaCN**: When ethyl chloride (CH₃CH₂Cl) reacts with sodium cyanide (NaCN), the cyanide ion (CN⁻) acts as a nucleophile. The nucleophile attacks the carbon atom bonded to the chlorine atom because chlorine is a good leaving group.

- **Mechanism**: The CN⁻ ion attacks the carbon atom in CH₃CH₂Cl, leading to the displacement of the Cl⁻ ion.
- **Product formed**: The product of this reaction is CH₃CH₂CN, which is ethyl cyanide (or propionitrile).

3. **Reduction of the nitrile**: The next step involves the reduction of the nitrile (CH₃CH₂CN) in the presence of nickel (Ni) and hydrogen (H₂).

- **Reduction process**: Nitriles can be reduced to primary amines. In this case, the nitrile group (–C≡N) will be reduced to an amine group (–NH₂).
- **Product formed**: The product of this reduction is CH₃CH₂NH₂, which is ethylamine.

4. **Final product**: Therefore, the compound Y in the given reaction sequence is ethylamine (CH₃CH₂NH₂).

### Summary of the Reaction Sequence:
- **Step 1**: CH₃CH₂Cl + NaCN → CH₃CH₂CN (ethyl cyanide)
- **Step 2**: CH₃CH₂CN + H₂ (Ni) → CH₃CH₂NH₂ (ethylamine)

### Final Answer:
Y in the above reacting sequence is **ethylamine (CH₃CH₂NH₂)**.

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