In the reaction
`CH_(3)overset(CH_(3))overset(|)(C )H-CH_(2)-O-CH_(2)CH_(3)+HIoverset(Heated)(rarr)`
Which of the following compounds will be formed?

Options

A. `CH_(3)-underset(CH_(3))underset(|)(CH)-CH_(3)+CH_(3)CH_(2)OH`
B. `CH_(3)-underset(CH_(3))underset(|)(CH)-CH_(2)OH+CH_(3)CH_(3)`
C. `CH_(3)-underset(CH_(3))underset(|)(CH)-CH_(2)OH+CH_(3)CH_(2)I`
D. `CH_(3)-underset(CH_(3))underset(|)(CH)-CH_(2)I+CH_(3)CH_(2)OH.`

Explanation

To solve the problem, we need to analyze the reaction of the given ether with HI when heated. The ether in question is:

\[ \text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{OCH}_2\text{CH}_3 \]

### Step 1: Identify the Ether Structure
The structure of the ether can be broken down as follows:
- The ether consists of two alkyl groups: one is a propyl group (from the left side, CH3CH(CH3)CH2) and the other is an ethyl group (from the right side, CH2CH3).

### Step 2: Reaction with HI
When the ether is heated with HI, it undergoes cleavage. The oxygen atom in the ether has a lone pair of electrons that can attack the hydrogen ion (H+) from HI, leading to the formation of an alcohol and an alkyl iodide.

### Step 3: Protonation of Ether
The lone pair on the oxygen atom attacks the H+ from HI, resulting in the protonation of the ether:
\[ \text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{OH}^+ \text{CH}_2\text{CH}_3 \]

### Step 4: Nucleophilic Attack by Iodide Ion
After protonation, the iodide ion (I-) from HI acts as a nucleophile. It will attack one of the alkyl groups. The iodide ion will preferentially attack the less hindered alkyl group due to steric factors.

In this case, the less hindered group is the ethyl group (CH2CH3), while the more hindered group is the propyl group (CH3CH(CH3)CH2).

### Step 5: Formation of Products
The final products of the reaction will be:
1. **Alcohol**: From the protonated ether, we will get:
\[ \text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{OH} \]
2. **Alkyl Iodide**: The iodide will attach to the ethyl group:
\[ \text{I-CH}_2\text{CH}_3 \]

Thus, the products formed are:
- **Alcohol**: Propyl alcohol (or isopropyl alcohol)
- **Alkyl Iodide**: Ethyl iodide

### Final Answer
The compounds formed from the reaction are:
- Propyl alcohol (or isopropyl alcohol)
- Ethyl iodide