The reaction of `CH_(3) OC_(2) H_(5)` with HI gives :

Options

A. `CH_(3) I`
B. `C_(2) H_(5) OH`
C. `CH_(3) I + C_(2) H_(5) OH`
D. `C_(2) H_(5) + CH_(3) OH`

Answer

To solve the problem of what happens when `CH3OC2H5` (ethyl methyl ether) reacts with HI (hydroiodic acid), we can follow these steps:

### Step 1: Identify the Reactants
The reactant is ethyl methyl ether, which has the structure:
\[ CH_3OCH_2CH_5 \]

### Step 2: Understand the Reaction with HI
When ethyl methyl ether reacts with hydroiodic acid (HI), the ether bond (C-O) is cleaved. Hydroiodic acid provides H⁺ and I⁻ ions.

### Step 3: Protonation of the Ether
The oxygen atom in the ether has lone pairs and can donate one of its lone pairs to bond with H⁺ from HI. This results in the formation of a protonated ether:
\[ CH_3O^+HCH_2CH_5 \]

### Step 4: Nucleophilic Attack
The protonated ether is now more reactive. The iodide ion (I⁻) will attack one of the alkyl groups. Due to steric hindrance, I⁻ will preferentially attack the less hindered methyl group (CH₃), leading to the formation of methyl iodide (CH₃I).

### Step 5: Formation of Products
The reaction will yield two products:
1. Methyl iodide (CH₃I)
2. Ethanol (C₂H₅OH) from the remaining part of the ether.

### Final Products
Thus, the products of the reaction of `CH3OC2H5` with HI are:
- Methyl iodide (CH₃I)
- Ethanol (C₂H₅OH)

### Summary of the Reaction
The overall reaction can be summarized as:
\[ CH_3OC_2H_5 + HI \rightarrow CH_3I + C_2H_5OH \]