The compound obtained by reducing `CH_3 CN` with `Na//C_2 H_5 OH` is

Options

A. methyl alcohol
B. acetic acid
C. ethylamine
D. methane

Explanation

To solve the problem of identifying the compound obtained by reducing methyl cyanide (CH₃CN) with sodium in ethanol (Na/C₂H₅OH), we can follow these steps:

### Step 1: Understand the Reactants
The reactant is methyl cyanide (CH₃CN), which contains a carbon-nitrogen triple bond (C≡N).

### Step 2: Identify the Reduction Process
When methyl cyanide is treated with sodium in ethanol, a reduction reaction occurs. In this context, reduction refers to the addition of hydrogen atoms to the molecule.

### Step 3: Analyze the Structure of Methyl Cyanide
Methyl cyanide has the following structure:
- CH₃ (methyl group) is attached to a carbon that is triple-bonded to nitrogen (C≡N).

### Step 4: Perform the Reduction
During the reduction:
- The triple bond (C≡N) will be converted to a single bond (C-N).
- Two hydrogen atoms will be added to the carbon atom and one hydrogen atom will be added to the nitrogen atom.

The reaction can be summarized as follows:
- CH₃CN + 2H → CH₃CH₂NH₂

### Step 5: Identify the Product
The product of the reduction is CH₃CH₂NH₂, which is ethylamine.

### Step 6: Conclusion
Thus, the compound obtained by reducing methyl cyanide (CH₃CN) with sodium in ethanol is ethylamine.

### Final Answer
**The compound obtained is Ethylamine (C).**

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