Question

Ozonolysis of 2-Methylbut-2-ene followed by reaction with Zn/H\(_2\)O gives:

• A. Propanone and Ethanal
• B. Propanal and Ethanal
• C. Two moles of Propanone
• D. Butanone and Methanal

Hint:

A quick shortcut for reductive ozonolysis: mentally erase the double bond \( \text{=} \) and write \( \text{=O} \) facing \( \text{O=} \) in its place.
This instantly gives you the structures of the final carbonyl compounds.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Ozonolysis is an organic reaction used to break double or triple bonds using ozone (\( \text{O}_3 \)).
When followed by \( \text{Zn/H}_2\text{O} \), it undergoes reductive cleavage, converting the alkene into respective aldehydes or ketones without further oxidizing them to carboxylic acids.

Step 2: Key Formula or Approach:
The simplest approach to predict products of reductive ozonolysis is to formally break the carbon-carbon double bond.
Then, cap both ends of the broken bond with a doubly-bonded oxygen atom (\( \text{=O} \)).

Step 3: Detailed Explanation:
First, we must write down the structural formula for the starting material, 2-Methylbut-2-ene.
The structure is \( \text{CH}_3\text{-C(CH}_3\text{)=CH-CH}_3 \).
Next, we identify the exact location of the \( \text{C=C} \) double bond to perform the cleavage.
We split the molecule into two fragments at the double bond.
The left fragment is \( \text{CH}_3\text{-C(CH}_3\text{)=} \).
The right fragment is \( \text{=CH-CH}_3 \).
Now, we attach an oxygen atom to the double bond of each fragment.
The left fragment becomes \( \text{CH}_3\text{-C(=O)-CH}_3 \), which is a ketone known as Propanone (or acetone).
The right fragment becomes \( \text{O=CH-CH}_3 \), which is an aldehyde known as Ethanal (or acetaldehyde).
Therefore, the final mixture of products consists of Propanone and Ethanal.

Step 4: Final Answer:
The reaction gives Propanone and Ethanal.