Aldol condensation between which of the following two compounds followed by dehydration gives methyl vinyl ketone?

Options

A. Formaldehyde and acetone
B. Formaldehyde and acetaldehyde
C. Two molecules of acetaldehyde
D. Two molecules of acetone

Explanation

To determine which two compounds undergo aldol condensation followed by dehydration to yield methyl vinyl ketone, we will analyze the process step by step.

### Step-by-Step Solution:

1. **Identify Methyl Vinyl Ketone**:
- Methyl vinyl ketone has the structure CH3C(=O)CH=CH2. This means it contains a ketone group (C=O) and a vinyl group (C=CH2).

2. **Consider Possible Reactants**:
- The question provides options, but we will focus on the common reactants for aldol condensation: aldehydes and ketones. In this case, we will consider formaldehyde (HCHO) and acetone (CH3COCH3).

3. **Aldol Condensation Mechanism**:
- **Formation of Enolate Ion**: Acetone has alpha-hydrogens that can be deprotonated to form an enolate ion. The enolate ion is formed from acetone:
\[ \text{CH}_3\text{COCH}_3 \xrightarrow{\text{Base}} \text{CH}_3\text{C}^- \text{(O)}\text{CH}_3 \]
- **Nucleophilic Attack**: The enolate ion then attacks the carbonyl carbon of formaldehyde:
\[ \text{CH}_3\text{C}^- \text{(O)}\text{CH}_3 + \text{HCHO} \rightarrow \text{CH}_3\text{C(OH)}\text{(O)}\text{CH}_2\text{(OH)} \]

4. **Formation of β-Hydroxy Ketone**:
- The product of the nucleophilic attack is a β-hydroxy ketone:
\[ \text{CH}_3\text{C(OH)(O)}\text{CH}_2\text{(OH)} \]

5. **Dehydration Step**:
- Upon heating, the β-hydroxy ketone undergoes dehydration (loss of water) to form methyl vinyl ketone:
\[ \text{CH}_3\text{C(OH)(O)}\text{CH}_2\text{(OH)} \xrightarrow{\text{Heat}} \text{CH}_3\text{C(=O)}\text{CH=CH}_2 + \text{H}_2\text{O} \]

6. **Final Product**:
- The final product after dehydration is methyl vinyl ketone (CH3C(=O)CH=CH2).

### Conclusion:
The two compounds that undergo aldol condensation followed by dehydration to yield methyl vinyl ketone are **formaldehyde and acetone**.