major product of the following reaction is : `CH_(3)CH=CHCO_(2)CH_(3) overset("LiAIH"_(4))rarr`

Options

A. `CH_(3)CH_(2)CH_(2)CHO`
B. `CH_(3)CH_(2)CH_(2)CO_(2)CH_(3)`
C. `CH_(3)CH_(2)CH_(2)OH`
D. `CH_(3)CH=CHCH_(2)OH`

Explanation

To determine the major product of the reaction between `CH₃CH=CHCO₂CH₃` (an ester) and lithium aluminium hydride (LiAlH₄), we can follow these steps:

### Step 1: Identify the Functional Groups
The compound `CH₃CH=CHCO₂CH₃` contains an alkene (the `CH=CH` part) and an ester functional group (`CO₂CH₃`).

### Step 2: Understand the Reactivity of LiAlH₄
Lithium aluminium hydride is a strong reducing agent that can reduce esters to primary alcohols. In this case, the ester will be reduced.

### Step 3: Mechanism of Reduction
1. The hydride ion (H⁻) from LiAlH₄ attacks the carbonyl carbon of the ester.
2. This forms a tetrahedral intermediate.
3. The alkoxy group (the `OCH₃` part) leaves, resulting in the formation of an aldehyde.
4. The aldehyde is then further reduced by another hydride from LiAlH₄ to form a primary alcohol.

### Step 4: Determine the Structure of the Product
After the reduction:
- The initial ester `CH₃CH=CHCO₂CH₃` will first be converted to an aldehyde.
- The aldehyde will then be reduced to a primary alcohol.

The final product will be `CH₃CH=CHCH₂OH`, which is a primary alcohol.

### Step 5: Conclusion
The major product of the reaction between `CH₃CH=CHCO₂CH₃` and LiAlH₄ is `CH₃CH=CHCH₂OH`.

### Final Answer
The major product is `CH₃CH=CHCH₂OH`.

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