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: Identify the product
The product given is 4-methylpent-3-en-2-one. This compound has a specific structure that we need to analyze to deduce the possible carbonyl compound.

### Step 2: Analyze the structure of the product
The structure of 4-methylpent-3-en-2-one can be broken down as follows:
- It has a pentane backbone (5 carbon atoms).
- There is a double bond between the 3rd and 4th carbon atoms (indicating it is an enone).
- The ketone functional group is located at the 2nd carbon.

### Step 3: Consider the reaction conditions
The reaction involves a carbonyl compound and dilute NaOH. This suggests that we might be dealing with an aldol condensation or a similar reaction where a carbonyl compound with alpha hydrogens reacts under basic conditions.

### Step 4: Identify the carbonyl compound
Given that the product is a ketone and considering the structure of 4-methylpent-3-en-2-one, we can deduce that the starting carbonyl compound is likely acetone (2-propanone). Acetone has the structure:
- CH3-CO-CH3

### Step 5: Confirm the reaction
When acetone reacts with dilute NaOH, it can undergo an aldol condensation. The reaction proceeds as follows:
1. The alpha hydrogen of acetone is deprotonated by NaOH, forming an enolate ion.
2. This enolate ion can then attack another molecule of acetone, leading to the formation of a β-hydroxy ketone.
3. Upon heating, this β-hydroxy ketone can dehydrate to form the final product, which is 4-methylpent-3-en-2-one.

### Conclusion
Thus, the carbonyl compound that reacts with dilute NaOH to give 4-methylpent-3-en-2-one is **acetone**.