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Ozonolysis &
Oxidative Cleavage

Learn how ozone cleaves carbon-carbon double bonds to produce aldehydes, ketones, and carboxylic acids: powerful tools for structure determination.

Organic Chemistry 1

What is Oxidative Cleavage?

Oxidative cleavage is a reaction in which a carbon-carbon double bond (\( \mathrm{C=C} \)) is completely broken. Both the \( \sigma \) bond and the \( \pi \) bond are severed, and the two carbon atoms that were part of the double bond each become bonded to oxygen instead.

Meet Ozone (\( \mathrm{O_3} \))

\( \mathrm{O} \)
\( \mathrm{O} \)
+
\( \mathrm{O} \)
-
Bent structure, \( 116.8^\circ \) bond angle
Ozone is a powerful electrophile

Ozone (\( \mathrm{O_3} \)) is an unstable, highly reactive allotrope of oxygen. Its central oxygen bears a formal positive charge, making it an excellent electrophile that attacks \( \pi \) bonds readily.

Why Does Ozonolysis Matter?

Structure Determination
By analyzing the carbonyl fragments, chemists can work backwards to deduce the position of a double bond in an unknown molecule.
Synthetic Utility
Ozonolysis provides a clean route to aldehydes and ketones from alkenes: products that are useful building blocks in organic synthesis.
Selectivity
The reductive vs. oxidative workup lets you choose your products: stop at aldehydes or push all the way to carboxylic acids.

The Ozonolysis Mechanism

Alkene
\( \mathrm{R_2C=CR_2} \)
\( \mathrm{O_3} \)
Molozonide
1,2,3-trioxolane
rearranges
Ozonide
1,2,4-trioxolane
workup
Reductive Workup
\( \mathrm{Zn/AcOH} \) or \( \mathrm{Me_2S} \)
Oxidative Workup
\( \mathrm{H_2O_2} \)
Aldehydes + Ketones
Carboxylic Acids + Ketones

Reductive Workup: \( \mathrm{Zn / AcOH} \) or \( \mathrm{Me_2S} \)

In reductive ozonolysis, the ozonide is treated with a mild reducing agent. This prevents further oxidation and preserves any aldehydes that form.

Disubstituted C
\( \mathrm{R_2C=} \) gives a ketone: \( \mathrm{R_2C=O} \)
Monosubstituted C
\( \mathrm{RHC=} \) gives an aldehyde: \( \mathrm{RCHO} \)
Unsubstituted C
\( \mathrm{H_2C=} \) gives formaldehyde: \( \mathrm{HCHO} \)

Oxidative Workup: \( \mathrm{H_2O_2} \)

In oxidative ozonolysis, the ozonide is treated with hydrogen peroxide (\( \mathrm{H_2O_2} \)). This oxidizes any aldehyde products further to carboxylic acids.

Disubstituted C
\( \mathrm{R_2C=} \) still gives a ketone: \( \mathrm{R_2C=O} \) (cannot be oxidized further)
Monosubstituted C
\( \mathrm{RHC=} \) gives a carboxylic acid: \( \mathrm{RCOOH} \)
Unsubstituted C
\( \mathrm{H_2C=} \) gives \( \mathrm{CO_2 + H_2O} \)

Reductive vs. Oxidative Workup

Reductive Oxidative
Reagent \( \mathrm{Zn/AcOH} \), \( \mathrm{Me_2S} \), or \( \mathrm{PPh_3} \) \( \mathrm{H_2O_2} \)
\( \mathrm{RHC=} \) gives Aldehyde (RCHO) Carboxylic acid (RCOOH)
\( \mathrm{R_2C=} \) gives Ketone Ketone
\( \mathrm{H_2C=} \) gives HCHO (formaldehyde) \( \mathrm{CO_2 + H_2O} \)
Key Idea Preserves aldehydes Oxidizes aldehydes further

How to Solve Ozonolysis Problems

1
Locate the \( \mathrm{C=C} \) double bond. Draw the alkene and identify the substituents on each carbon of the double bond.
2
Break the double bond. Erase both the \( \sigma \) and \( \pi \) bonds. You now have two separate carbon fragments.
3
Cap each carbon with \( =O \). Replace the broken bond on each carbon with a carbonyl (\( \mathrm{C=O} \)).
4
Check the workup. Reductive: keep aldehydes. Oxidative: convert aldehydes to carboxylic acids.
5
Write the products. Draw each carbonyl fragment as its final form (aldehyde, ketone, or carboxylic acid).

Example 1: 2-Butene + Reductive Ozonolysis

\( \mathrm{CH_3} \)
\( \mathrm{CH} \)
\( \mathrm{CH} \)
\( \mathrm{CH_3} \)
1) \( \mathrm{O_3} \)
2) \( \mathrm{Zn/AcOH} \)
\( \mathrm{CH_3} \)
\( \mathrm{CHO} \)
acetaldehyde
+
\( \mathrm{CH_3} \)
\( \mathrm{CHO} \)
acetaldehyde

Example 2: 2-Methylpropene + Reductive

\( \mathrm{CH_3} \)
\( \mathrm{C} \)
\( \mathrm{CH_3} \)
\( \mathrm{CH_2} \)
1) \( \mathrm{O_3} \)
2) \( \mathrm{Me_2S} \)
\( \mathrm{CH_3} \)
\( \mathrm{C=O} \)
\( \mathrm{CH_3} \)
acetone (ketone)
+
\( \mathrm{H} \)
\( \mathrm{CHO} \)
formaldehyde

Example 3: 1-Methylcyclohexene + Oxidative

\( \mathrm{CH_3} \)
1) \( \mathrm{O_3} \)
2) \( \mathrm{H_2O_2} \)
\( \mathrm{HOOC(CH_2)_4COCH_3} \)
6-oxoheptanoic acid
(single open-chain product)

Cyclic Alkenes: The Ring-Opening Rule

When a cyclic alkene undergoes ozonolysis, the ring opens. Instead of producing two separate fragments, you get one single molecule with two carbonyl groups at the positions where the double bond used to be.

cyclopentene
\( \mathrm{O_3} \), then workup
\( \mathrm{O=CH} \)
\( \mathrm{(CH_2)_3} \)
\( \mathrm{CH=O} \)
pentanedial (reductive)

Working Backwards: Finding the Alkene

On exams you may be given the ozonolysis products and asked to determine the original alkene. Reverse the process:

1
Remove the \( \mathrm{=O} \) from each fragment. If oxidative workup was used, first convert any \( \mathrm{-COOH} \) back to \( \mathrm{-CHO} \).
2
Join the two carbons that lost their oxygens with a double bond (\( \mathrm{C=C} \)).
3
Check your answer: re-run the forward ozonolysis mentally to confirm the products match.

Common Mistakes to Avoid

✕ Confusing the Workup
Writing aldehydes when \( \mathrm{H_2O_2} \) is used (oxidative), or carboxylic acids when \( \mathrm{Zn/AcOH} \) is used (reductive). Always check the second reagent.
✕ Forgetting Formaldehyde
Terminal alkenes (\( \mathrm{R-CH=CH_2} \)) give formaldehyde (\( \mathrm{HCHO} \)) as one product. Students often forget to write it.
✕ Splitting a Ring
Cyclic alkenes do not give two separate fragments. The ring opens to give a single open-chain dicarbonyl product.

Summary Table

C=C Substitution Reductive Products Oxidative Products
\( \mathrm{R_2C=CR_2} \) 2 Ketones 2 Ketones
\( \mathrm{R_2C=CHR} \) Ketone + Aldehyde Ketone + Carboxylic acid
\( \mathrm{RHC=CHR} \) 2 Aldehydes 2 Carboxylic acids
\( \mathrm{R_2C=CH_2} \) Ketone + HCHO Ketone + \( \mathrm{CO_2} \)
\( \mathrm{RHC=CH_2} \) Aldehyde + HCHO Carboxylic acid + \( \mathrm{CO_2} \)
Cyclic alkene 1 Dicarbonyl 1 Dicarbonyl (acids/ketones)

Question 1 of 5

What are the products of ozonolysis of 1-butene (\( \mathrm{CH_3CH_2CH=CH_2} \)) with reductive workup (\( \mathrm{Zn/AcOH} \))?

Question 2 of 5

Which reagent is used in the second step of oxidative ozonolysis to convert aldehydes into carboxylic acids?

Question 3 of 5

Ozonolysis of cyclohexene followed by reductive workup (\( \mathrm{Me_2S} \)) produces:

Question 4 of 5

Oxidative ozonolysis of 2-methylpropene (\( \mathrm{(CH_3)_2C=CH_2} \)) yields:

Question 5 of 5

An unknown alkene undergoes reductive ozonolysis to give acetone and acetaldehyde as the only products. What is the original alkene?