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 given problem step by step, we will analyze each reaction in the sequence provided:

1. **Starting Material: Methanol (CH₃OH)**
- The first reaction involves methanol reacting with hydroiodic acid (HI).

2. **Reaction 1: CH₃OH + HI → CH₃I**
- Methanol reacts with HI to form methyl iodide (CH₃I). This is a substitution reaction where the hydroxyl group (-OH) of methanol is replaced by the iodide ion (I⁻).

3. **Reaction 2: CH₃I + KCN → CH₃CN**
- Methyl iodide then reacts with potassium cyanide (KCN) to produce methyl cyanide (also known as acetonitrile, CH₃CN). This is another substitution reaction where the iodide ion is replaced by the cyanide ion (CN⁻).

4. **Reaction 3: CH₃CN → X (Reduction)**
- The next step involves the reduction of methyl cyanide (CH₃CN). This can be done using reducing agents like lithium aluminum hydride (LiAlH₄) or sodium borohydride (NaBH₄).
- During the reduction, the triple bond (C≡N) in the nitrile is converted to a primary amine. The product of this reduction is ethylamine (CH₃CH₂NH₂).

5. **Reaction 4: X (Ethylamine) + HNO₂ → Y**
- Ethylamine (X) is then treated with nitrous acid (HNO₂).
- Since ethylamine is a primary amine, it reacts with nitrous acid to form a diazonium salt, which is unstable and decomposes to form ethanol (CH₃CH₂OH) as the final product (Y).

### Summary of Products:
- **X (after reduction of CH₃CN)**: Ethylamine (CH₃CH₂NH₂)
- **Y (after reaction with HNO₂)**: Ethanol (CH₃CH₂OH)

### Final Answer:
- X = Ethylamine (CH₃CH₂NH₂)
- Y = Ethanol (CH₃CH₂OH)