Q.Identify X, Y and Z in the following sequence of reactions : CH3-CH2-CH2Br --(Alcoholic KOH)--> X --(O3)--> Y --(Zn/H2O)--> Z + HCHO

Ozonolysis – Breaking a Double Bond Cleanly

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 ($O_3$) 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.

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 ($R_2C=O$).

Example: Ozonolysis of 2-methyl-2-butene.

The structure is $CH_3-C(CH_3)=CH-CH_3$. Cut the double bond:

• Left carbon: attached to two methyl groups → ketone: acetone, $(CH_3)_2C=O$.
• Right carbon: attached to one H and one methyl → aldehyde: acetaldehyde, $CH_3CHO$.

So the products are acetone and acetaldehyde. …