Butanal with dilute NaOH gives

Options

A. `CH_(3)CH_(2)CH_(2)underset(H)underset(|)overset(OH)overset(|)"C"CH_(2)CH_(2)CH_(2)CH_(2)OH`
B. `CH_(3)CH_(2)CH_(2)overset(O)overset(||)C"CHO"CH_(2)CH_(2)CHO`
C. `OH"C"CH_(2)CH_(2)CH_(2)CH_(2)CH_(2)CH_(2)CHO`
D. `CH_(3)CH_(2)CH_(2)underset(H)underset(|)overset(OH)overset(|)C-underset(CH_(3))underset(|)underset(CH_(2))underset(|)CHCHO`

Explanation

To solve the question "Butanal with dilute NaOH gives," we need to understand the reaction that occurs when butanal (an aldehyde) is treated with dilute sodium hydroxide (NaOH). This reaction is an example of an aldol condensation.

### Step-by-Step Solution:

1. **Identify the Reactant**:
- The reactant is butanal, which has the chemical formula CH₃CH₂CH₂CHO.

2. **Understand the Reaction Type**:
- When butanal reacts with dilute NaOH, it undergoes an aldol condensation reaction. In this reaction, the aldehyde can form an enolate ion in the presence of a base.

3. **Formation of Enolate Ion**:
- The dilute NaOH acts as a base and abstracts a proton from the alpha carbon (the carbon adjacent to the carbonyl group) of butanal, forming an enolate ion:
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{C}^-\text{H}=\text{C}(\text{O})\text{H} \]

4. **Nucleophilic Attack**:
- This enolate ion then attacks the carbonyl carbon of another molecule of butanal, leading to the formation of a β-hydroxy aldehyde (aldol):
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{C}^-\text{H} + \text{CH}_3\text{CH}_2\text{CH}_2\text{CHO} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{C(OH)(CH}_2\text{CH}_3)\text{CHO} \]

5. **Final Product**:
- The product formed is 3-hydroxybutanal (an aldol), which is a β-hydroxy carbonyl compound. Since there is no heating involved, the aldol does not undergo dehydration to form an α,β-unsaturated carbonyl compound.

### Conclusion:
The product of the reaction of butanal with dilute NaOH is 3-hydroxybutanal.