Animated Solution for Chemistry - Organic Chemistry: Natural rubber on complete ozonolysis (O${}_3$/Zn-H${}_2$O) gives compound \textbf{X} as the major product. \textbf{X} gives positive iodoform and Tollen's tests. \textbf{X} on heating with aqueous NaOH gives \textbf{Y} as the major product. \textbf{Y} is

Natural Rubber

Every great organic chemistry problem begins with a solid foundation. Here, our starting material is natural rubber. If you recall your polymer chemistry, natural rubber is a naturally occurring polymer of isoprene, specifically known as cis-1,4-polyisoprene.

The repeating unit of this polymer chain is $[-C{H}_2-C(C{H}_3)=CH-C{H}_2-{]}_n$. This long, continuous chain of carbon-carbon double bonds is the perfect canvas for our first major chemical transformation.

Reductive Ozonolysis

The first step in our reaction sequence is complete ozonolysis using ozone ($O_3$) followed by a reductive workup with zinc and water ($Zn/H_{2}O$). Ozonolysis is a powerful tool that acts like molecular scissors, cleanly cleaving every single carbon-carbon double bond it encounters.

When we apply these scissors to the repeating unit of natural rubber, the double bond breaks, and oxygen atoms cap the newly formed ends. The resulting molecule is Compound X, which has the structure $C{H}_3-C(=O)-C{H}_2-C{H}_2-CHO$.

This molecule is formally named 4-oxopentanal. It is a fascinating bifunctional molecule, possessing both a ketone group on one end and an aldehyde group on the other.

Iodoform and Tollen's Tests

The problem provides two massive hints to confirm that our structure for Compound X is absolutely correct.

First, Compound X gives a positive iodoform test. The iodoform test specifically identifies the presence of a methyl ketone group ($-C(=O)C{H}_3$). Looking at the left side of our 4-oxopentanal structure, we see exactly that!

Second, Compound X gives a positive Tollen's test. Tollen's reagent is a mild oxidizing agent that specifically reacts with aldehydes, reducing silver ions to form a beautiful silver mirror. Looking at the right side of our structure, we have an aldehyde group ($-CHO$). These two tests perfectly validate our structural deduction.

Intramolecular Aldol Condensation

Now we reach the climax of the problem. We treat Compound X with aqueous sodium hydroxide ($NaOH$) and heat ($Delta$). Whenever you see a molecule with two carbonyl groups placed in a basic medium, your mind should immediately jump to an intramolecular aldol condensation.

The base will abstract an alpha-hydrogen to form an enolate. But which alpha-hydrogen? We have a choice between the terminal methyl group and the internal methylene group next to the ketone.

If the enolate forms at the internal methylene group and attacks the aldehyde, it would create a highly strained 3-membered ring. However, if the enolate forms at the terminal methyl group, it can swing around and attack the aldehyde carbon to form a highly stable 5-membered ring. Thermodynamics always wins, so the 5-membered ring forms.

Cyclopent-2-en-1-one

Upon the enolate's attack on the aldehyde, a 5-membered ring is formed containing a newly created hydroxyl ($-OH$) group.

Because the reaction is heated, this intermediate beta-hydroxy ketone immediately undergoes dehydration. A water molecule is eliminated, creating a double bond that is conjugated with the ketone.

The final product, Compound Y, is an alpha,beta-unsaturated ketone known as cyclopent-2-en-1-one. Comparing this elegant cyclic structure to our given options, it perfectly matches Option (A).