91Ó°ÊÓ

When an alkene is treated with ozone and subjected to reductive workup with zinc then the carbon-carbon double bond is cleaved to give ketones or aldehydes which depend on the structure of the alkene.

Oxidative cleavage reaction can be performed by ${\mathbf{O}}_{\mathbf{3}}$

To distinguish these two structures we have given the number to the structure. Then we will study how we distinguish these two reactions with the help of reagents

Structure 1

Structure 2

Reagent ${\mathrm{O}}_3$ with Zn lead to the breaking of the double bond present in the structure 1 and lead to the formation of unsymmetrical aldehydes.

The reaction can be shown as:

${\mathrm{O}}_3$and Zn reaction

Reagent ${\mathrm{O}}_3$with Zn lead to the breaking of the double bond present in the structure 2 and lead to the formation of symmetrical aldehydes.

The reaction can be shown as:

${\mathrm{O}}_3$and Zn reaction

• To distinduish between the two structures we can perform oxidative cleavage by use of ${\mathrm{O}}_3$ and Zn.
• In this reaction, double bonds are broken between alkenes.
• As here double bonds are present at different locations so we will get different products.
• If on the first structure we perform ozonolysis then we will get Butane-1,4-dial and Ethane-1,4-dial.
• If on the second structure we perform ozonolysis then we will get2 Propane -1,3-dial.