The major product of the following reaction is
`CH_(3)-CH=CH-overset(O)overset(||)C-O-CH_(3)overset(LiAlH_(4))rarr`

Options

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

Explanation

To solve the problem, we need to determine the major product of the reaction involving the compound `CH3-CH=CH-C(=O)-O-CH3` when treated with `LiAlH4`.

### Step-by-Step Solution:

1. **Identify the Reactant and Reagent**:
The reactant is `CH3-CH=CH-C(=O)-O-CH3`, which is an alkene with an ester functional group. The reagent is `LiAlH4`, a strong reducing agent.

**Hint**: Recognize the functional groups present in the reactant.

2. **Understand the Function of LiAlH4**:
`LiAlH4` is a powerful reducing agent that can reduce esters, aldehydes, and ketones to their corresponding alcohols. However, it does not reduce alkenes.

**Hint**: Recall the reactivity of `LiAlH4` with different functional groups.

3. **Analyze the Reaction**:
In the given reactant, the ester part `C(=O)-O-CH3` will be reduced by `LiAlH4`. The alkene part `CH3-CH=CH-` will remain unchanged.

**Hint**: Focus on the ester functional group for reduction.

4. **Write the Product of the Reduction**:
The ester `C(=O)-O-CH3` will be reduced to a primary alcohol. The carbonyl carbon (C=O) will be converted to a methylene group (CH2), and the oxygen will become an alcohol (OH). The overall structure will be:

- The alkene part remains: `CH3-CH=CH-`
- The ester part is reduced to: `-CH2-OH`

Therefore, the product will be `CH3-CH=CH-CH2-OH`.

**Hint**: Ensure you account for the unchanged alkene and the reduced ester.

5. **Final Product Structure**:
Combine both parts to write the final product:
\[ \text{Product: } CH3-CH=CH-CH2-OH \]

**Hint**: Double-check the connectivity of the atoms in the final product.

### Conclusion:
The major product of the reaction is `CH3-CH=CH-CH2-OH`.