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:
Reductive ozonolysis of an alkene is a chemical reaction that cleaves the carbon-carbon double bond (\( \text{C=C} \)).
The reagents used are ozone (\( \text{O}_3 \)) followed by a reducing agent like zinc dust and water (\( \text{Zn/H}_2\text{O} \)).
Step 2: Key Formula or Approach:
In reductive ozonolysis, the \( \text{C=C} \) double bond is broken completely, and an oxygen atom is added to each of the two carbon atoms to form carbonyl compounds (aldehydes or ketones).
The general approach is to "cut" the double bond and attach an \( \text{=O} \) to each side.
Step 3: Detailed Explanation:
The given reactant is 2-Methylbut-2-ene.
Its chemical structure is: \( \text{CH}_3-\text{C}(\text{CH}_3)=\text{CH}-\text{CH}_3 \).
When we cleave the \( \text{C=C} \) double bond:
Left fragment: \( \text{CH}_3-\text{C}(\text{CH}_3)= \) becomes \( \text{CH}_3-\text{C}(=\text{O})-\text{CH}_3 \), which is Propanone (Acetone).
Right fragment: \( =\text{CH}-\text{CH}_3 \) becomes \( \text{O=CH}-\text{CH}_3 \), which is Ethanal (Acetaldehyde).
Therefore, the products formed are Propanone and Ethanal.
Step 4: Final Answer:
The reaction yields Propanone and Ethanal.