Animated Solution for Chemistry - Hydrocarbons: In the following sequence of reactions, the alkene affords the compound 'B' $C{H}_{3}CH=CHC{H}_3\stackrel{O_3}{\to }A\underset{Zn}{\overset{H_{2}O}{\to }}B$ The compound $B$ is

Welcome to one of the most elegant and visually satisfying reactions in all of organic chemistry: Ozonolysis! Imagine you have a long, complex carbon chain, and you want to cleanly snap it into two pieces exactly at the double bond. Ozonolysis is the molecular pair of scissors that lets you do exactly that. In this problem, we are going to walk through this fascinating mechanism step by step.

Meet But-2-ene

Look closely at the starting material provided in the question. We have a symmetrical alkene, specifically But-2-ene ($C{H}_3-CH=CH-C{H}_3$).

Our goal is to subject this alkene to a two-step reaction sequence. The first reagent is ozone ($O_3$), and the second reagent is a mixture of zinc dust and water ($Zn/H_{2}O$). This specific combination of reagents is the classic recipe for reductive ozonolysis.

The Ozone Attack and the Ozonide Hut

The first step is the reaction with ozone. Ozone is a highly reactive, electrophilic molecule. When it encounters the electron-rich pi bond of the alkene, it doesn't just break the pi bond; it breaks the sigma bond as well!

Visualize this: The double bond breaks completely, and the three oxygen atoms from the ozone molecule insert themselves between the two carbon atoms. They form a five-membered ring structure that looks remarkably like a little hut or a house. This intermediate is called an ozonide (specifically, a secondary ozonide).

At this stage, the original carbon-carbon double bond is entirely gone, replaced by carbon-oxygen single bonds.

The Reductive Cleavage with Zinc

Now comes the crucial second step. In the presence of zinc and water, this ozonide hut undergoes cleavage.

This is where mistakes happen if you aren't careful. How does the cleavage occur? Imagine taking a sword and slicing right down the middle of the hut in a 'Y' shape. You cut it such that both of the original alkene carbons get to keep one oxygen atom each, forming a new carbon-oxygen double bond ($C=O$).

But what happens to the third oxygen atom at the top of the hut? That oxygen is picked up by the zinc, forming zinc oxide ($ZnO$). Zinc acts as a reducing agent here. Its primary job is to prevent the formation of hydrogen peroxide ($H_2{O}_2$), which would otherwise oxidize our delicate products.

Compound B

So, what do we get after the dust settles? Let's look at the two halves of our cleaved molecule.

From the left side of the original But-2-ene, we get $C{H}_3-CH=O$. From the right side, we also get $O=CH-C{H}_3$. Because our starting alkene was symmetrical, we end up with two identical molecules of acetaldehyde (also known by its IUPAC name, ethanal).

Therefore, our final product, Compound B, is $C{H}_{3}CHO$.

Reductive vs. Oxidative

Always remember one critical distinction in organic chemistry. If the zinc wasn't there—if the second step was just $H_{2}O$ or $H_2{O}_2$—the reaction would be oxidative ozonolysis.

In an oxidative environment, the aldehydes we just formed would be further oxidized into carboxylic acids. In that alternate universe, our final product would have been acetic acid ($C{H}_{3}COOH$). But since zinc is present to keep things reductive, the reaction safely stops at the aldehyde stage. This is a favorite concept for JEE, so always keep an eye out for that zinc!