An olefinic compound on reductive ozonolysis produces formaldehyde, acetaldehyde and 1, 3-propanedial. The IUPAC name of the alkene is

Options

A. 1, 4-Hexadiene
B. 2-Methyl-1, 3-pentadiene
C. 1, 2-Hexadiene
D. None of these

Answer

To solve the problem, we need to identify the olefinic compound (alkene) that, upon reductive ozonolysis, produces formaldehyde, acetaldehyde, and 1,3-propanediol. Let's break down the steps to find the IUPAC name of the alkene.

### Step 1: Understand the products of ozonolysis
- The products given are:
1. Formaldehyde (HCHO)
2. Acetaldehyde (CH3CHO)
3. 1,3-Propanediol (CHO-CH2-CHO)

### Step 2: Analyze the structure of the products
- **Formaldehyde** has 1 carbon atom.
- **Acetaldehyde** has 2 carbon atoms.
- **1,3-Propanediol** has 3 carbon atoms and contains two aldehyde groups.

### Step 3: Determine the total number of carbon atoms
- Adding the carbon atoms from the products:
- Formaldehyde: 1 C
- Acetaldehyde: 2 C
- 1,3-Propanediol: 3 C
- Total = 1 + 2 + 3 = 6 carbon atoms.

### Step 4: Identify the possible alkene structure
- Since the alkene must have 6 carbon atoms and produce the given aldehydes, we can deduce that the alkene must have a double bond that allows for the formation of these products upon ozonolysis.
- The structure must allow for the cleavage that leads to the formation of the aldehydes.

### Step 5: Write the possible alkene
- A suitable structure that fits this requirement is **1,4-hexadiene**. This structure has:
- A double bond between carbon 1 and carbon 2.
- A double bond between carbon 4 and carbon 5.
- This structure allows for the cleavage to yield the specified products.

### Step 6: Confirm the IUPAC name
- The IUPAC name of the alkene is **1,4-hexadiene**.

### Final Answer
The IUPAC name of the alkene is **1,4-hexadiene**.

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