In the given reaction `CH_3-CH_2-CH_2-O-CH_2-CH_3 overset("Conc.Hl)rarr(X)+(Y)`

Options

A. `CH_3-CH_3-CH_2OH and CH_3-CH_2-I`
B. `CH_3-CH_2-CH_2-I andCH_2-CH_2-OH`
C. `CH_3-CH_2-CH_2-I and CH_2=CH_2`
D. `CH_3-CH=CH_2and CH_2=CH_2`

Answer

To solve the given reaction `CH3-CH2-CH2-O-CH2-CH3` when treated with concentrated HI, we will follow these steps:

### Step 1: Identify the Reactants
The reactant is an ether, specifically diethyl ether, which has the structure `CH3-CH2-O-CH2-CH3`.

### Step 2: Understand the Reaction with HI
When the ether is treated with concentrated hydrogen iodide (HI), the ether undergoes cleavage. The reaction involves the protonation of the ether oxygen followed by nucleophilic attack by iodide ions.

### Step 3: Protonation of the Ether
The oxygen atom in the ether has two lone pairs and can donate one pair to bond with the hydrogen ion (H+) from HI. This results in the formation of a positively charged oxonium ion:
\[ CH3-CH2-O^+H-CH2-CH3 \]

### Step 4: Nucleophilic Attack by Iodide Ion
The iodide ion (I-) will then attack one of the carbon atoms bonded to the oxygen. In this case, the attack will occur at the less hindered carbon. The structure can be visualized as:
- The left side has a methyl group (less hindered),
- The right side has an ethyl group (more hindered).

Since the iodide ion prefers to attack the less hindered carbon, it will attack the carbon attached to the methyl group.

### Step 5: Formation of Products
1. The product formed from the attack on the less hindered carbon will be ethyl iodide (CH3-CH2-I).
2. The remaining part of the ether will be propanol (CH3-CH2-CH2-OH).

Thus, the products are:
- Product X: Propanol (C3H8O)
- Product Y: Ethyl iodide (C2H5I)

### Step 6: Finalize the Answer
Therefore, the correct products formed from the reaction are:
- X = Propanol
- Y = Ethyl iodide

### Conclusion
The correct option from the given choices is the first option, which states that the products are propanol and ethyl iodide.

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