The compound `CH_(3)COCH_(2)COOC_(2)H_(5)` is treated with `LiAlH_(4)` in dry ether. The product formed is

Options

A. `CH_(3)CH_(2)CH_(2)COOC_(2)H_(5)`
B. `CH_(3)CHOHCH_(2)CO_(2)C_(2)H_(5)`
C. `CH_(3)CHOHCH_(2)OH`
D. `CH_(3)COCH_(2)CH_(2)OH`

Explanation

To solve the question, we need to analyze the compound `CH3COCH2COOC2H5` and determine what happens when it is treated with `LiAlH4` in dry ether.

### Step-by-Step Solution:

1. **Identify the Functional Groups**:
The compound `CH3COCH2COOC2H5` contains:
- A ketone group (`CH3CO-`).
- An ester group (`-COOC2H5`).

2. **Understand the Reducing Agent**:
`LiAlH4` (Lithium Aluminium Hydride) is a strong reducing agent that can reduce:
- Aldehydes to primary alcohols.
- Ketones to secondary alcohols.
- Esters to alcohols.

3. **Reduction of the Ketone**:
The ketone part of the compound is `CH3COCH2-`. When treated with `LiAlH4`, it will be reduced to a secondary alcohol:
- `CH3COCH2-` becomes `CH3C(OH)CH2-`.

4. **Reduction of the Ester**:
The ester part of the compound is `-COOC2H5`. When treated with `LiAlH4`, it will be reduced to an alcohol:
- `-COOC2H5` becomes `-C(OH)C2H5`.

5. **Combine the Results**:
After the reduction of both functional groups, we combine the results:
- From the ketone: `CH3C(OH)CH2-`.
- From the ester: `-C(OH)C2H5`.

Thus, the final product is:
- `CH3C(OH)CH2C(OH)C2H5`.

6. **Final Structure**:
The final structure can be represented as:
- `CH3C(OH)CH2C(OH)C2H5`, which is a compound with two hydroxyl (-OH) groups.

### Final Answer:
The product formed is `CH3C(OH)CH2C(OH)C2H5`.