In the sequence of the following reactions
`CH_3OH overset(HI)to CH_3I overset(KCN)`
`CH_3CN overset("reduction")to X overset(HNO_2)to Y`
X and Y are respectively

Options

A. `CH_3CH_2NH_2 and CH_3CH_2OH`
B. `CH_3CH_2NH_2 and CH_3COOH`
C. `CH_3CH_2OH and CH_3CHO`
D. `CH_3OCH_3and CH_3CHO`

Explanation

To solve the given sequence of reactions, we will go through each step carefully.

### Step 1: Reaction of Methanol with HI
The first reaction is:
\[ \text{CH}_3\text{OH} \overset{\text{HI}}{\rightarrow} \text{CH}_3\text{I} \]

In this reaction, methanol (CH₃OH) reacts with hydroiodic acid (HI) to form methyl iodide (CH₃I). This is a substitution reaction where the hydroxyl group (-OH) of methanol is replaced by an iodide ion (I⁻) from HI.

### Step 2: Reaction of Methyl Iodide with KCN
Next, we have:
\[ \text{CH}_3\text{I} \overset{\text{KCN}}{\rightarrow} \text{CH}_3\text{CN} \]

In this step, methyl iodide (CH₃I) reacts with potassium cyanide (KCN). This is a nucleophilic substitution reaction where the iodide ion is replaced by a cyanide ion (CN⁻), resulting in the formation of acetonitrile (CH₃CN).

### Step 3: Reduction of Acetonitrile
Now we proceed to the reduction:
\[ \text{CH}_3\text{CN} \overset{\text{reduction}}{\rightarrow} X \]

During the reduction of acetonitrile (CH₃CN), it is converted to ethylamine (C₂H₅NH₂). This can be achieved using reducing agents like lithium aluminum hydride (LiAlH₄) or other suitable reducing agents.

### Step 4: Reaction of Ethylamine with HNO₂
Next, we have:
\[ X \overset{\text{HNO}_2}{\rightarrow} Y \]

Here, ethylamine (C₂H₅NH₂) reacts with nitrous acid (HNO₂). In this reaction, ethylamine is converted into an unstable diazonium salt (C₂H₅N₂⁺Cl⁻). This diazonium salt decomposes to form ethanol (C₂H₅OH) and nitrogen gas (N₂).

### Conclusion
Thus, the final compounds X and Y are:
- **X**: Ethylamine (C₂H₅NH₂)
- **Y**: Ethanol (C₂H₅OH)

### Summary of the Solution
- **X** is Ethylamine (C₂H₅NH₂).
- **Y** is Ethanol (C₂H₅OH).

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