`CH_(3)CH_(2)-underset(CH_(3))underset(|)(CH)-CH_(2)-O-CH_(2)CH_(3)+HIto `?
The product is

Options

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

Answer

To solve the problem of the reaction of the ether \( CH_3CH_2-CH(CH_3)-CH_2-O-CH_2CH_3 \) with HI, we can follow these steps:

### Step 1: Identify the Structure of the Ether
The given ether can be represented as:
\[ CH_3CH_2-CH(CH_3)-CH_2-O-CH_2CH_3 \]
This is an unsymmetrical ether because the two alkyl groups attached to the oxygen atom are different.

### Step 2: Understand the Reaction with HI
When ethers react with hydrogen iodide (HI), they undergo cleavage. The reaction typically results in the formation of an alcohol and an alkyl halide. The cleavage occurs at the ether bond, breaking it into two parts.

### Step 3: Determine the Mechanism
The reaction proceeds via an SN2 mechanism, especially for unsymmetrical ethers. In this mechanism, the nucleophile (I⁻ from HI) attacks the less sterically hindered carbon atom adjacent to the ether oxygen.

### Step 4: Identify the Alkyl Groups
In our ether:
- The left side has the alkyl group \( CH_3CH_2-CH(CH_3)-CH_2 \) (which is more sterically hindered).
- The right side has the alkyl group \( CH_2CH_3 \) (which is less sterically hindered).

### Step 5: Cleavage of the Ether
The I⁻ ion will preferentially attack the less hindered side (the ethyl group \( CH_2CH_3 \)). Therefore, the ether bond will break, leading to:
- Formation of an alcohol from the more hindered side.
- Formation of an alkyl halide from the less hindered side.

### Step 6: Write the Products
After the cleavage:
1. The alcohol formed will be:
\[ CH_3CH_2-CH(CH_3)-CH_2-OH \] (which is butanol)
2. The alkyl halide formed will be:
\[ CH_3CH_2I \] (ethyl iodide)

### Final Products
The final products of the reaction are:
- Alcohol: \( CH_3CH_2-CH(CH_3)-CH_2OH \) (butanol)
- Alkyl Halide: \( CH_3CH_2I \) (ethyl iodide)

### Summary
The products of the reaction of the given ether with HI are:
\[ CH_3CH_2-CH(CH_3)-CH_2OH + CH_3CH_2I \]

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