Acetaldehyde when treated with dilute `NaOH` gives.

Options

A. `CH_(3)CH_(2)OH`
B. `CH_(3)COOH`
C. `beta-`hydroxybutanal
D. Ethane

Answer

To solve the question regarding the reaction of acetaldehyde with dilute NaOH, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactant**:
The reactant in this case is acetaldehyde, which has the molecular formula CH3CHO.

2. **Understand the Reaction Conditions**:
The reaction is conducted in the presence of dilute NaOH. This indicates that we are likely dealing with an aldol condensation reaction.

3. **Recognize the Aldol Reaction**:
In an aldol reaction, aldehydes or ketones with at least one alpha hydrogen can react with a base to form a beta-hydroxy aldehyde or ketone. Acetaldehyde has alpha hydrogens (the hydrogens attached to the carbon adjacent to the carbonyl group).

4. **Formation of Enolate Ion**:
The hydroxide ion (OH-) from NaOH will abstract one of the alpha hydrogens from acetaldehyde, forming an enolate ion. The structure of the enolate ion can be represented as:
\[ CH_2=C(OH)CHO \]

5. **Nucleophilic Attack**:
The enolate ion then acts as a nucleophile and attacks the carbonyl carbon of another molecule of acetaldehyde. This leads to the formation of a beta-hydroxy aldehyde.

6. **Final Product Formation**:
After the nucleophilic attack, the resulting compound is a beta-hydroxybutanal (also known as 3-hydroxybutanal). The structure can be represented as:
\[ CH_3-CH(OH)-CH_2-CHO \]

7. **Conclusion**:
Therefore, the product formed when acetaldehyde is treated with dilute NaOH is beta-hydroxybutanal.

### Final Answer:
Acetaldehyde when treated with dilute NaOH gives **beta-hydroxybutanal**.