Alkene Reactions: Ozonolysis using O3 and H2O2

Ozone (O3) and hydrogen peroxide (H2O2) react with alkenes to form ketones and/or carboxylic acids (Oxidative Workup):

To visualize this reaction, cut the double bond in half and add an oxygen to each side of the double bond. Next, determine how many allylic carbons are attached to each carbon on the alkene bond: if there are 2 carbons, a ketone will be formed and if there is 1 carbon, a carboxylic acid will be formed. If there are no allylic carbons, CO2 will be formed. If both alkene carbons have 2 allylic carbons, 2 ketones will be formed. Finally, if the alkene bond is part of a ring structure, the ring will unwind and there will be an alkane chain with ketone/carboxylic acid functional groups on either side :

Double Ketone

Double Carboxylic Acid

CO2 formation

Ring Unwinding

The reaction mechanism is presented below:

In the first step, ozone (O3) reacts with the alkene bond to form a ring, known as cycloaddition.

In the second step, the ring breaks due to free lone pair electrons attacking more favorable positions.

In the third step, the negatively charged oxygen atom attacks the carbon atom that is part of the ketone bond, forming a single molecule.

In the fourth step, the newly formed negative oxygen atom attacks the positively charged oxygen atom, once again forming a ring.

In the fifth step, the “electron dance” takes place again, this time breaking the ring to form one aldehyde molecule and one carbonyl oxide molecule.

In the sixth step, hydrogen peroxide (H2O2) reacts with the aldehyde molecule while water (H2O) reacts with the carbonyl oxide molecule.

In the seventh step, both molecules undergo proton transfer steps to add protons (H+) to negatively charged oxygen atoms and remove protons from positively charged oxygen atoms.

In the eighth step, chaos ensues. Water removes a proton on the first molecule (left), leading to the formation of a carbon-oxygen double bond, which kicks the -OH group off, forming the carboxylic acid molecule. A similar sequence occurs on the other molecule (right) which results in the formation of a ketone molecule. A water molecule removes a proton from an oxygen (forming H3O+), which then uses those electrons to form a double bond with carbon. This results in the removal of the OOH- group by reacting with H3O+ to regenerate H2O2 (not pictured).

As was stated above, the number of allylic carbons determines whether a ketone, carboxylic acid, or CO2 will be formed. The mechanism shown above results in 1 ketone and 1 carboxylic acid functional group. For other conditions (2 ketone, 2 carboxy acid, etc…) an identical mechanism would be true until step 8 where the deprotonation would depend on which protons are available to be removed.

This is the oxidative workup; different products would be obtained if performed under a reductive workup (O3 and DMS).

Practice this reaction using our Reaction Solver!