When `C_6 H_5 COCOC_6 H_5` is reduced with `LiAlH_4`, the product formed has ________stereoisomers.

Options

A. 2
B. 3
C. 4
D. 6

Explanation

To determine the number of stereoisomers formed when \( C_6H_5COCOC_6H_5 \) is reduced with \( LiAlH_4 \), we can follow these steps:

### Step 1: Identify the Structure
The compound \( C_6H_5COCOC_6H_5 \) has the following structure:

- It consists of two phenyl groups (\( C_6H_5 \)) connected by a carbon chain that contains two carbonyl groups (C=O).
- The structure can be represented as \( C_6H_5-CO-CO-C_6H_5 \).

### Step 2: Reduction with \( LiAlH_4 \)
When this compound is reduced with \( LiAlH_4 \), a strong reducing agent, both carbonyl groups (C=O) will be reduced to hydroxyl groups (–OH), resulting in the formation of a diol:

- The product will be \( C_6H_5-CHOH-CHOH-C_6H_5 \).

### Step 3: Analyze the Product for Chirality
Next, we need to analyze the product for chiral centers:

- The two hydroxyl groups are attached to the central carbon atoms, which can potentially create chiral centers.
- Each of the two central carbon atoms (where the –OH groups are attached) can have four different substituents, making them chiral centers.

### Step 4: Count the Chiral Centers
In the product \( C_6H_5-CHOH-CHOH-C_6H_5 \):

- There are **two chiral centers** (the two carbon atoms bonded to the –OH groups).

### Step 5: Calculate the Number of Stereoisomers
To find the number of stereoisomers, we can use the formula:

\[ \text{Number of stereoisomers} = 2^n \]

where \( n \) is the number of chiral centers.

- Here, \( n = 2 \), so:

\[ \text{Number of stereoisomers} = 2^2 = 4 \]

### Conclusion
Thus, the product formed when \( C_6H_5COCOC_6H_5 \) is reduced with \( LiAlH_4 \) has **4 stereoisomers**.

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