Two compounds [A] and [B] have molecular formula `C_(2)H_(6)O.` On reacting with HI [A] gives alkyl iodide and water while [B] gives alkyl iodide and alcohol . What are the compounds [A] and [B] ? Write the reactions involved.

Options

A. `CH_3OC_2H_5 and CH_OC_H_5`
B. `CH_3OCH_3 and C_2H_5OCH_3 `
C. `C_2H_5OH and CH_3OCH_3 `
D. `CH_3OH and CH_3OCH_3 `

Explanation

To solve the problem, we need to identify the two compounds [A] and [B] that have the molecular formula C₂H₆O and their reactions with HI.

### Step 1: Identify the possible compounds with the formula C₂H₆O
The molecular formula C₂H₆O can represent two types of compounds: an alcohol and an ether. The two functional isomers are:
- **Compound A (Alcohol)**: Ethanol (C₂H₅OH)
- **Compound B (Ether)**: Dimethyl ether (CH₃OCH₃)

### Step 2: Analyze the reaction of compound A with HI
When ethanol (C₂H₅OH) reacts with hydroiodic acid (HI), it undergoes substitution to form an alkyl iodide and water:
\[ \text{C₂H₅OH} + \text{HI} \rightarrow \text{C₂H₅I} + \text{H₂O} \]
Here, C₂H₅I is the alkyl iodide (ethyl iodide), and water is produced.

### Step 3: Analyze the reaction of compound B with HI
When dimethyl ether (CH₃OCH₃) reacts with hydroiodic acid (HI), it undergoes cleavage to form an alkyl iodide and an alcohol:
\[ \text{CH₃OCH₃} + \text{HI} \rightarrow \text{CH₃I} + \text{CH₃OH} \]
In this case, CH₃I is the alkyl iodide (methyl iodide), and CH₃OH is the alcohol (methanol).

### Conclusion
Based on the reactions:
- Compound [A] is Ethanol (C₂H₅OH).
- Compound [B] is Dimethyl ether (CH₃OCH₃).

### Summary of Reactions
1. **For Compound A (Ethanol)**:
\[ \text{C₂H₅OH} + \text{HI} \rightarrow \text{C₂H₅I} + \text{H₂O} \]

2. **For Compound B (Dimethyl ether)**:
\[ \text{CH₃OCH₃} + \text{HI} \rightarrow \text{CH₃I} + \text{CH₃OH} \]