Question

One mole of an alkene on ozonolysis gives a mixture of one mole pentan-3-one and one mole methanal. The alkene is (2024)

• A. 3-ethylbut-1-ene
• B. 2-methylpent-1-ene
• C. 2-ethylbut-1-ene
• D. 4-methylpent-2-ene
• E. 4-methylpent-1-ene

Hint:

To solve ozonolysis problems quickly, just look at the names of the products. "Methanal" always indicates a terminal double bond ($=CH_2$). "Pentan-3-one" shows the branching occurs at the third carbon of a 5-carbon sequence.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Ozonolysis is a powerful analytical reaction used to determine the position of double bonds in alkenes. During ozonolysis, ozone (\(\text{O}_3\)) completely cleaves the carbon-carbon double bond. Each of the carbon atoms that originally formed the double bond becomes double-bonded to an oxygen atom, resulting in the formation of smaller carbonyl compounds like aldehydes and ketones.
Step 2: Key Formula or Approach:
To deduce the structure of the original alkene from its given ozonolysis products, we use a simple reverse-engineering strategy:
1. Draw the chemical structures of the given product molecules.
2. Orient the molecules so that their carbonyl oxygen atoms (\(\text{=O}\)) are facing each other.
3. "Remove" these oxygen atoms and connect the two bare carbon atoms with a new carbon-carbon double bond.
Step 3: Detailed Explanation:
The problem states the products are pentan-3-one and methanal.
- First, let's draw pentan-3-one. It's a five-carbon chain ketone with the carbonyl group on the third carbon:
\[ \text{CH}_3\text{-CH}_2\text{-C(=O)-CH}_2\text{-CH}_3 \]
- Next, let's draw methanal (also known as formaldehyde). It's the simplest aldehyde:
\[ \text{H}_2\text{C=O} \]
Now, applying our retro-synthetic "cut and paste" strategy:
Align the products:
\[ (\text{CH}_3\text{CH}_2)_2\text{C=O} \quad + \quad \text{O=CH}_2 \]
Remove the oxygen atoms and join the central carbons:
\[ (\text{CH}_3\text{CH}_2)_2\text{C=CH}_2 \]
Finally, we need to determine the proper IUPAC name for this reconstructed alkene:
1. Identify the longest parent chain: The longest continuous carbon chain that includes both carbons of the double bond has 4 carbon atoms. Thus, the parent name is based on 'butene'.
2. Number the chain: Number from the end that gives the lowest numbers to the double bond carbons.
\[ \text{CH}_2=\text{C}(\text{CH}_2\text{CH}_3)-\text{CH}_2-\text{CH}_3 \]
\(\quad 1\quad \quad \quad 2\quad \quad \quad \quad \quad \quad 3\quad \quad \quad 4\)
The double bond starts at C-1, so it is a 'but-1-ene'.
3. Identify substituents: There is an ethyl group (\(-\text{CH}_2\text{CH}_3\)) attached to carbon number 2.
4. Assemble the name: Putting it all together, the IUPAC name is 2-ethylbut-1-ene.
This matches option (c).
Step 4: Final Answer:
The original alkene is 2-ethylbut-1-ene.