Reaction of a carbonyl compound with dilute `NaOH` gives 4-methylpent-3-en-2-one. The carbonly compound is:

Options

A. Acetaldehyde
B. Acetone
C. Formaldehyde
D. Propanal

Answer

To determine the carbonyl compound that reacts with dilute NaOH to yield 4-methylpent-3-en-2-one, we can follow these steps:

### Step 1: Understand the Reaction Type
The reaction of a carbonyl compound with dilute NaOH suggests that we are dealing with an aldol condensation reaction. This is because the product is a single compound, indicating that it is not a case of the Cannizzaro reaction, which yields two products.

**Hint:** Identify the type of reaction based on the number of products formed.

### Step 2: Identify the Product Structure
The product given is 4-methylpent-3-en-2-one. We can break down the name:
- "pent" indicates a five-carbon chain.
- "3-en" indicates a double bond starting at the third carbon.
- "2-one" indicates a ketone functional group at the second carbon.

We can draw the structure of 4-methylpent-3-en-2-one:
```
CH3
|
CH3-CH-CH=CH-CO-CH3
```
This structure has a total of five carbons.

**Hint:** Draw the structure of the product based on its IUPAC name.

### Step 3: Determine the Carbonyl Compound
Since we are performing an aldol condensation, we need a carbonyl compound that can provide the necessary alpha hydrogen for the reaction. The general form of a ketone is RCO-R', where R and R' can be alkyl groups.

Given that we need to form a product with five carbons, we can consider possible ketones. The simplest ketone that fits this description and has alpha hydrogens is acetone (CH3COCH3).

**Hint:** Consider the structure of possible carbonyl compounds that can yield the desired product.

### Step 4: Verify the Reaction Mechanism
1. **Deprotonation**: The alpha hydrogen of acetone is removed by hydroxide ion (OH-) to form an enolate ion.
2. **Nucleophilic Attack**: The enolate ion then attacks the carbonyl carbon of another acetone molecule.
3. **Formation of β-hydroxy ketone**: This intermediate undergoes dehydration (loss of water) to form the final product, 4-methylpent-3-en-2-one.

**Hint:** Understand the mechanism of aldol condensation to confirm the formation of the product.

### Conclusion
The carbonyl compound that reacts with dilute NaOH to yield 4-methylpent-3-en-2-one is **acetone (CH3COCH3)**.

**Final Answer:** Acetone (option B).