Alkene Reactions: Ozonolysis using O3 and DMS

Ozone (O3) and Dimethyl Sulfate (DMS or (CH3)2S) react with alkenes to form ketones and/or aldehydes (Reductive Workup):

Alkene Reactions: Ozonolysis using O3 and DMS - alkene ozonolysis o3 dms reaction

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, an aldehyde will be formed. If there are no allylic carbons, formaldehyde (CH2O) 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/aldehyde functional groups on either side :

Double Ketone

Alkene Reactions: Ozonolysis using O3 and DMS - alkene ozonolysis o3 dms reaction double ketone

Double Aldehyde

Alkene Reactions: Ozonolysis using O3 and DMS - alkene ozonolysis o3 dms reaction double aldehyde

Formaldehyde formation

Alkene Reactions: Ozonolysis using O3 and DMS - alkene ozonolysis o3 dms reaction formaldehyde

Ring Unwinding

Alkene Reactions: Ozonolysis using O3 and DMS - alkene ozonolysis o3 dms reaction ring unwinding

The reaction mechanism is presented below:

Alkene Reactions: Ozonolysis using O3 and DMS - alkene ozonolysis o3 dms reaction mechanism

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 DMS attacks an oxygen atom which then breaks the oxygen-oxygen bond with the other oxygen atom.

In the sixth step, the negatively charged oxygen sends its lone pair electrons to the carbon atom and breaks the other oxygen-carbon bond, forming the ketone product.

In the seventh step, free electrons from the negatively charged oxygen atom form a double bond with the carbon atom, causing the oxygen atom attached to DMS to release from the molecule, thereby forming the aldehyde product.

As was stated above, the number of allylic carbons determines whether a ketone, aldehyde, or formaldehyde will be formed. The mechanism shown above results in 1 ketone and 1 aldehyde functional group. For other conditions (2 ketone, 2 aldehyde, etc…) an identical mechanism would be followed.

This is the reductive workup; different products would be obtained if performed under a oxidative workup (O3 and H2O2).

Practice this reaction using our Reaction Solver!