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

Options

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

Answer

To determine the major product of the reaction `CH₃CH=CHCO₂CH₃` with lithium aluminium hydride (LiAlH₄), we will follow these steps:

### Step 1: Identify the functional groups
The compound `CH₃CH=CHCO₂CH₃` contains:
- A double bond (alkene) between the second and third carbon atoms.
- An ester functional group (CO₂CH₃) at the end of the molecule.

### Step 2: Understand the reducing agent
Lithium aluminium hydride (LiAlH₄) is a strong reducing agent that can reduce:
- Carbonyl groups (like those in esters and aldehydes) to alcohols.
- Alkenes to alkanes.

### Step 3: Determine the order of reduction
In this reaction, we must determine which functional group will be reduced first. Generally, esters are more reactive towards reduction than alkenes unless the alkene is in conjugation with another reactive group. Since the alkene in this case is not conjugated with the carbonyl group, the ester will be reduced first.

### Step 4: Perform the reduction of the ester
When the ester (CO₂CH₃) is reduced by LiAlH₄, it will convert to an alcohol. The reaction can be represented as follows:
- The ester `CO₂CH₃` is reduced to `CH₂OH` (an alcohol).

### Step 5: Write the structure of the product
After the reduction of the ester, we still have the alkene part of the molecule intact. The structure of the product will be:
- The alkene remains as `CH₃CH=CH` (the double bond is unchanged).
- The ester is converted to an alcohol, resulting in `CH₃CH=CHCH₂OH`.

### Final Product
Thus, the major product of the reaction is:
- `CH₃CH=CHCH₂OH`

### Summary
The major product of the reaction `CH₃CH=CHCO₂CH₃` with LiAlH₄ is `CH₃CH=CHCH₂OH`.

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