Propene is allowed to react with m-chloroperoxybenzoic acid. The product (A) is then reduced with `LiAlH_(4)` in dry ether to give (B).
`CH_(3)CH=CH_(2)overset(MCPBA)rarrAunderset(2. H_(3)O^(+))overset(1. LiAlH_(4))rarrB`
The structure of the product (B) is :

Options

A. `CH_(3)CHOHCH_(2)OH`
B. `CH_(3)CH_(2)CH_(2)OH`
C. `CH_(3)CHOHCH_(3)`
D.

Answer

To solve the problem, we will follow the steps of the reaction of propene with m-chloroperoxybenzoic acid (MCPBA) and then the reduction with lithium aluminium hydride (LiAlH₄).

### Step 1: Reaction of Propene with MCPBA
1. **Identify the Reactants**: The reactants are propene (CH₃CH=CH₂) and m-chloroperoxybenzoic acid (MCPBA).
2. **Mechanism of Reaction**: MCPBA is a peracid that can convert alkenes into epoxides. When propene reacts with MCPBA, the double bond of propene attacks the electrophilic oxygen of the peracid, leading to the formation of an epoxide.
3. **Product Formation**: The product (A) is an epoxide, specifically propylene oxide. The structure of product (A) can be represented as:
- Structure of A:
```
O
/ \
CH3 CH
|
CH2
```

### Step 2: Reduction of the Epoxide with LiAlH₄
1. **Identify the Reducing Agent**: Lithium aluminium hydride (LiAlH₄) is a strong reducing agent that can reduce epoxides.
2. **Mechanism of Reduction**: In the presence of LiAlH₄, the epoxide ring opens. The nucleophile (hydride ion from LiAlH₄) attacks the less hindered carbon atom of the epoxide, leading to the formation of an alcohol.
3. **Product Formation**: The product (B) after the reduction of the epoxide will be propane-1-ol. The structure of product (B) can be represented as:
- Structure of B:
```
CH3-CH2-CH2-OH
```

### Final Answer
The structure of product (B) is propane-1-ol.