Imagine you have a chain with a weak link in the middle — a carbon-carbon double bond. You want to cut the chain at exactly that point, and you want to know what was on each side of the cut. Ozonolysis is the chemical scissors that does this.
The double bond is electron-rich. Ozone (O3) is a bent, reactive molecule that loves to attack electron-rich spots. When ozone hits the double bond, it forms a temporary, unstable ring called a molozonide. This quickly rearranges into an ozonide — a five-membered ring with three oxygen atoms. The ozonide is explosive if isolated, so we never keep it; we immediately break it down with a reductive work-up (typically using zinc dust and water, or dimethyl sulfide).
The reductive work-up is the key. It cleaves the ozonide in a way that each carbon of the original double bond ends up with a double bond to oxygen — an aldehyde (−CHO) or a ketone (>C=O). Which one you get depends on what was attached to that carbon originally.
R2C=CR2′1.O3,2.Zn/H2OR2C=O+O=CR2′
The carbon that had two hydrogens on it becomes formaldehyde (HCHO).
The carbon that had one hydrogen and one alkyl group becomes an aldehyde (RCHO).
The carbon that had two alkyl groups becomes a ketone (R2C=O).
Tip
To predict the products, just mentally "cut" the double bond and put an oxygen atom on each cut end. The number of alkyl groups on each carbon tells you if it's an aldehyde or a ketone.
Example: Ozonolysis of 2-methyl-2-butene.
The structure is CH3−C(CH3)=CH−CH3. Cut the double bond:
Left carbon: attached to two methyl groups → ketone: acetone, (CH3)2C=O.
Right carbon: attached to one H and one methyl → aldehyde: acetaldehyde, CH3CHO.
Step 1. Interpret the product. Getting 2 moles of the SAME carbonyl compound, ethanal (CH3-CHO), from ozonolysis of one alkene means both halves of the cleaved double bond were identical to each other.
Step 2. Reconstruct each carbonyl carbon's substituents. Ethanal's carbonyl carbon carries one CH3 and one H; reading this back onto an alkene carbon means that carbon carried a CH3 and an H (i.e. =CH-CH3). …
Proposing but-1-ene or an unsymmetrical alkene, which would give TWO DIFFERENT carbonyl products on ozonolysis (e.g. formaldehyde + propanal), not 2 moles of the SAME ethanal. …