`CH_(3)CH_(2)CH_(2)OC_(2)H_(5)` is treated with hydroiodic acid. The fragments after reaction obtained are

Explanation

To solve the problem of what fragments are obtained when ethyl propyl ether (CH₃CH₂CH₂OC₂H₅) is treated with hydroiodic acid (HI), we can follow these steps:

### Step 1: Understand the Structure of Ethyl Propyl Ether
The given compound is ethyl propyl ether, which can be represented as:
- Propyl group: CH₃CH₂CH₂-
- Ether linkage: -O-
- Ethyl group: -C₂H₅

So, the complete structure is CH₃CH₂CH₂-O-C₂H₅.

### Step 2: Reaction with Hydroiodic Acid
When ethyl propyl ether is treated with hydroiodic acid (HI), the ether undergoes protonation. The oxygen atom in the ether has lone pairs that can attract the hydrogen ion (H⁺) from HI, resulting in the formation of an oxonium ion (protonated ether).

### Step 3: Formation of the Oxonium Ion
The reaction can be represented as:
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{-O-C}_2\text{H}_5 + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{-O}^+\text{H-C}_2\text{H}_5 \]
This results in the formation of an oxonium ion where the oxygen carries a positive charge.

### Step 4: Nucleophilic Attack
In the next step, the iodide ion (I⁻) from hydroiodic acid will attack the oxonium ion. This reaction proceeds via an SN2 mechanism, where the iodide ion attacks the less sterically hindered (smaller) alkyl group.

### Step 5: Identify the Alkyl Groups
In this case, the propyl group (CH₃CH₂CH₂-) is larger than the ethyl group (C₂H₅-). Therefore, the iodide ion will attack the ethyl group.

### Step 6: Fragment Formation
The attack leads to the cleavage of the ether bond, resulting in the formation of two fragments:
1. Propanol (from the remaining propyl group)
2. Ethyl iodide (from the ethyl group that was attacked)

### Final Answer
The fragments obtained after the reaction are:
- Propanol (CH₃CH₂CH₂OH)
- Ethyl iodide (C₂H₅I)