When a mixture of benzaldehyde and acetophenone is treated with dilute `NaOH` at `293 K`, it form

Options

A. 2, 3-diphenylpropanal
B. 1, 1-diphenylpropan-2-one-1-ol
C. 1, 3-diphenylprop-2-en-1-one
D. 1, 2-diphenylprop-2-en-1-one

Explanation

To solve the question regarding the reaction of a mixture of benzaldehyde and acetophenone when treated with dilute NaOH at 293 K, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- Benzaldehyde (C6H5CHO) is an aromatic aldehyde.
- Acetophenone (C6H5C(O)CH3) is an aromatic ketone.

2. **Understand the Reaction Conditions**:
- The reaction occurs in the presence of dilute NaOH at a temperature of 293 K. This suggests that a condensation reaction may take place.

3. **Recognize the Type of Reaction**:
- The reaction between benzaldehyde and acetophenone in the presence of a base like NaOH is likely to be an aldol condensation. In this case, the aldehyde and ketone can react to form a β-hydroxy ketone or an α,β-unsaturated carbonyl compound.

4. **Write the Reaction**:
- The reaction can be represented as:
\[ \text{C6H5CHO} + \text{C6H5C(O)CH3} \xrightarrow{\text{NaOH}} \text{C6H5CH=CHC(O)C6H5} \]
- This results in the formation of a compound with the structure of 1,3-diphenyl-1,3-prop-2-en-1-one.

5. **Name the Product**:
- The product formed is 1,3-diphenylprop-2-en-1-one, which is also known as chalcone.

6. **Conclusion**:
- Therefore, the final product of the reaction is 1,3-diphenylprop-2-en-1-one.

### Final Answer:
The mixture of benzaldehyde and acetophenone treated with dilute NaOH at 293 K forms 1,3-diphenylprop-2-en-1-one.