Acetaldehyde when treated with dilute `NaOH` gives.

Options

A. `CH_(3) CH_(2) OH`
B. `CH_(3) COOH`
C. `CH_(3)-underset(OH)underset(|)(CH)-CH_(2)-CHO`
D. `CH_(3) - CH_(3)`

Explanation

To solve the question "Acetaldehyde when treated with dilute NaOH gives," we will follow these steps:

### Step 1: Identify the Reactants
Acetaldehyde (CH₃CHO) is the reactant in this case. It is an aldehyde with the structure:
```
H
|
H--C--C=O
|
H
```

### Step 2: Understand the Reaction Conditions
The reaction is conducted in the presence of dilute sodium hydroxide (NaOH), which acts as a base. This indicates that we are likely dealing with an aldol condensation reaction.

### Step 3: Mechanism of the Reaction
1. **Enolate Ion Formation**: Under basic conditions, acetaldehyde can undergo deprotonation at the alpha-carbon (the carbon adjacent to the carbonyl carbon) to form an enolate ion:
```
CH₃CHO + NaOH → CH₂=CHO⁻ + H₂O
```

2. **Nucleophilic Addition**: The enolate ion then acts as a nucleophile and attacks the carbonyl carbon of another acetaldehyde molecule:
```
CH₂=CHO⁻ + CH₃CHO → CH₃-CH(OH)-CH₂-CHO
```

### Step 4: Product Formation
The product formed is a β-hydroxy aldehyde, commonly referred to as "aldol." The structure of the aldol product is:
```
H
|
H--C--C(OH)--C=O
| |
H H
```
This compound is known as 3-hydroxybutanal or simply aldol.

### Step 5: Conclusion
Thus, when acetaldehyde is treated with dilute NaOH, the product formed is 3-hydroxybutanal (aldol).

### Final Answer
Acetaldehyde when treated with dilute NaOH gives 3-hydroxybutanal (aldol).

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