In the sequence of following reactions : `CH_(3) OH overset(HI)to CH_(3) I overset(KCN)to CH_(3) CNoverset("Reduction")toXoverset(HNO_(2))to Y, X ` and Y are respectively

Options

A. `CH_(3) CH_(2) NH_(2) and CH_(3) CH_(2) OH`
B. `CH_(3) CH_(2) NH_(2) and CH_(3) COOH`
C. `CH_(3) CH_(2) OH and CH_(3) CHO`
D. `CH_(3) OCH_(3) and CH_(3) CHO`

Answer

To solve the question step by step, let’s break down the reactions involved:

### Step 1: Reaction of Methanol with HI
- **Reaction**: CH₃OH + HI → CH₃I + H₂O
- **Explanation**: Methanol (CH₃OH) reacts with hydroiodic acid (HI) to form methyl iodide (CH₃I) and water. This reaction occurs via an SN2 mechanism because methanol is a primary alcohol.

### Step 2: Reaction of Methyl Iodide with KCN
- **Reaction**: CH₃I + KCN → CH₃CN + KI
- **Explanation**: Methyl iodide (CH₃I) reacts with potassium cyanide (KCN) to form methyl cyanide (CH₃CN) and potassium iodide (KI). This is also an SN2 reaction, where the cyanide ion (CN⁻) acts as a nucleophile.

### Step 3: Reduction of Methyl Cyanide
- **Reaction**: CH₃CN + H₂ (or reducing agent) → CH₃CH₂NH₂
- **Explanation**: Methyl cyanide (CH₃CN) is reduced to ethylamine (CH₃CH₂NH₂) using a reducing agent such as hydrogen gas in the presence of a catalyst (e.g., nickel). Ethylamine is a primary amine.

### Step 4: Reaction of Ethylamine with HNO₂
- **Reaction**: CH₃CH₂NH₂ + HNO₂ → CH₃CH₂OH + N₂ + H₂O
- **Explanation**: Ethylamine (CH₃CH₂NH₂) reacts with nitrous acid (HNO₂) to form ethyl alcohol (CH₃CH₂OH), nitrogen gas (N₂), and water (H₂O). This reaction involves the formation of a diazonium salt intermediate, which decomposes to release nitrogen gas and form the corresponding alcohol.

### Conclusion
- **X** is Ethylamine (CH₃CH₂NH₂).
- **Y** is Ethyl alcohol (CH₃CH₂OH).

### Final Answer
- X = Ethylamine (CH₃CH₂NH₂)
- Y = Ethyl alcohol (CH₃CH₂OH)

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