In the given reaction:
`CH_(3)-CH_(2)-CH_(2)-O-CH_(2)-CH_(3) overset(HCl//Delta)rarr [X] +[Y]`
` [X]` and `[Y]` wil respectively be:

Options

A. `CH_(3)CH_(2)CH_(2)OH` and `CH_(3)CH_(2)Cl`
B. `CH_(3)CH_(2)CH_(2)Cl` and `CH_(3)CH_(2)OH`
C. `CH_(3)CH_(2)CH_(2)Cl` and `CH_(2)=CH_(2)`
D. `CH_(3)-CH=CH_(2)` and `CH_(2)=CH_(2)`

Answer

To solve the problem, we need to analyze the reaction of the ether with HCl under heating conditions. The ether given is:

\[ CH_3-CH_2-CH_2-O-CH_2-CH_3 \]

### Step 1: Identify the Structure of the Ether
The ether consists of two alkyl groups:
- On one side of the oxygen (O), we have a propyl group (CH3-CH2-CH2-).
- On the other side, we have an ethyl group (CH2-CH3).

### Step 2: Understand the Reaction Mechanism
When ethers react with strong acids like HCl, they undergo cleavage. The mechanism of cleavage depends on the structure of the ether. In this case, we have an unsymmetrical ether, which means we have two different alkyl groups.

### Step 3: Determine the Type of Alkyl Groups
In our ether:
- The propyl group is a primary alkyl group.
- The ethyl group is also a primary alkyl group.

Since both groups are primary, the reaction will proceed via an SN2 mechanism. This mechanism is favored for primary alkyl groups because there is less steric hindrance.

### Step 4: Reaction with HCl
When HCl is added, the oxygen atom of the ether will get protonated, making it a better leaving group. The chloride ion (Cl-) will then attack the less sterically hindered carbon atom. In this case, both groups are primary, but the reaction will preferentially cleave at the ethyl side due to less steric hindrance compared to the propyl side.

### Step 5: Products Formation
The cleavage will yield:
1. An alcohol from the propyl side:
\[ CH_3-CH_2-CH_2-OH \] (1-butanol)
2. An alkyl halide from the ethyl side:
\[ CH_3-CH_2-Cl \] (ethyl chloride)

### Conclusion
Thus, the products of the reaction are:
- [X] = 1-butanol (CH3-CH2-CH2-OH)
- [Y] = ethyl chloride (CH3-CH2-Cl)

### Final Answer
- [X] = 1-butanol
- [Y] = ethyl chloride

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