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

Options

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

Explanation

To solve the problem, we need to analyze the reaction of the given ether with hydroiodic acid (HI). The ether in question is:

\[ CH_3 - C(CH_3)(CH_2) - O - CH_2 - CH_3 \]

### Step 1: Identify the Ether
The structure given is an ether, which consists of an alkoxy group (–O–) connecting two alkyl groups. In this case, we have a branched ether.

**Hint:** Recognize the structure of ethers and how they are formed from alcohols.

### Step 2: Protonation of Ether
When an ether reacts with a strong acid like HI, the first step is the protonation of the ether oxygen. This forms an oxonium ion:

\[ CH_3 - C(CH_3)(CH_2) - O^+H - CH_2 - CH_3 \]

**Hint:** Remember that protonation increases the electrophilicity of the ether, making it more reactive.

### Step 3: Cleavage of the Ether
The oxonium ion can undergo cleavage. The bond between the oxygen and the carbon atoms can break. The cleavage will occur preferentially at the less sterically hindered carbon atom. In this case, we have two possible cleavage sites:

1. The bond between the oxygen and the ethyl group (–O–CH₂–CH₃).
2. The bond between the oxygen and the branched carbon (–O–C(CH₃)–).

Due to steric hindrance, the bond to the ethyl group will break preferentially.

**Hint:** Consider steric effects when determining which bond will break in ether cleavage.

### Step 4: Formation of Products
Upon cleavage, we will form ethyl iodide (from the ethyl group) and isobutanol (from the branched carbon):

1. Ethyl iodide: \( CH_3 - CH_2 - I \)
2. Isobutanol: \( CH_3 - C(CH_3)(CH_2) - OH \)

The final products are:

\[ CH_3 - CH_2 - I + CH_3 - C(CH_3)(CH_2) - OH \]

**Hint:** Write down the structures of the products to visualize the reaction outcome.

### Conclusion
The compounds formed from the reaction of the given ether with HI are ethyl iodide and isobutanol.

### Final Answer
The products formed are:
1. Ethyl iodide: \( CH_3 - CH_2 - I \)
2. Isobutanol: \( CH_3 - C(CH_3)(CH_2) - OH \)

### Summary of Steps
1. Identify the ether structure.
2. Protonate the ether with HI.
3. Determine the site of cleavage based on sterics.
4. Write down the resulting products.