JEE Challenger
More from Polymers
ChemMedAdv 2026 • Paper 2

Ozonolysis of Natural Rubber and Subsequent Aldol Condensation

Natural rubber on complete ozonolysis (O3/Zn-H2O\text{O}_3/\text{Zn-H}_2\text{O}) gives compound X\mathbf{X} as the major product. X\mathbf{X} gives positive iodoform and Tollen's tests. X\mathbf{X} on heating with aqueous NaOH\text{NaOH} gives Y\mathbf{Y} as the major product. Y\mathbf{Y} is

Options

A
Option A
Correct
B
Option B
C
Option C
D
Option D

Step-by-Step Solution

1. Structure of Natural Rubber: Natural rubber is cis-1,4-polyisoprene, which consists of the repeating monomeric unit: [−CH2−C(CH3)=CH−CH2−]n\left[ -\text{CH}_2-\text{C}(\text{CH}_3)=\text{CH}-\text{CH}_2- \right]_n

2. Reductive Ozonolysis: Complete reductive ozonolysis (O3/Zn-H2O\text{O}_3/\text{Zn-H}_2\text{O}) of natural rubber cleaves the double bonds in the polymer chain to yield 4-oxopentanal (levulinic aldehyde) as the major monomeric product X\mathbf{X}: [−CH2−C(CH3)=CH−CH2−]n→O3/Zn-H2On  CH3−C(=O)−CH2−CH2−CHO(X)\left[ -\text{CH}_2-\text{C}(\text{CH}_3)=\text{CH}-\text{CH}_2- \right]_n \xrightarrow{\text{O}_3/\text{Zn-H}_2\text{O}} n \,\, \text{CH}_3-\text{C}(=\text{O})-\text{CH}_2-\text{CH}_2-\text{CHO} \quad (\mathbf{X})

3. Confirmatory Tests for Compound X\mathbf{X}:

  • Tollen's Test: Compound X\mathbf{X} contains a terminal aldehyde group (−CHO-\text{CHO}), giving a positive Tollen's test (silver mirror).
  • Iodoform Test: Compound X\mathbf{X} contains a methyl ketone group (CH3−C(=O)−\text{CH}_3-\text{C}(=\text{O})-), giving a positive iodoform test (yellow precipitate of CHI3\text{CHI}_3).

4. Intramolecular Aldol Condensation: Numbering the carbon chain of 4-oxopentanal (X\mathbf{X}) as follows: C5H3−C4(=O)−C3H2−C2H2−C1HO\overset{5}{\text{C}}\text{H}_3-\overset{4}{\text{C}}(=\text{O})-\overset{3}{\text{C}}\text{H}_2-\overset{2}{\text{C}}\text{H}_2-\overset{1}{\text{C}}\text{HO}

When X\mathbf{X} is treated with aqueous NaOH\text{NaOH} and heated:

  1. Enolate Formation: Base deprotonates the α\alpha-carbon of the methyl group (C5\text{C}_5) to generate an enolate nucleophile. Deprotonation at C5\text{C}_5 is preferred over C3\text{C}_3 because attack from C5\text{C}_5 onto the aldehyde carbon (C1\text{C}_1) forms a thermodynamically stable 5-membered ring. CH3−C(=O)−CH2−CH2−CHO→OH−−CH2−C(=O)−CH2−CH2−CHO\text{CH}_3-\text{C}(=\text{O})-\text{CH}_2-\text{CH}_2-\text{CHO} \xrightarrow{\text{OH}^-} {}^-\text{CH}_2-\text{C}(=\text{O})-\text{CH}_2-\text{CH}_2-\text{CHO}

  2. Ring Closure: The enolate carbon (C5\text{C}_5) attacks the more reactive electrophilic aldehyde carbon (C1\text{C}_1), yielding a 5-membered β\beta-hydroxy ketone intermediate (3-hydroxycyclopentanone).

  3. Dehydration: Upon heating, dehydration (elimination of H2O\text{H}_2\text{O}) occurs to yield an α,β\alpha,\beta-unsaturated ketone, cyclopent-2-en-1-one, as the major product Y\mathbf{Y}.

X→aq. NaOH, Δcyclopent-2-en-1-one(Y)\mathbf{X} \xrightarrow{\text{aq. NaOH, } \Delta} \text{cyclopent-2-en-1-one} \quad (\mathbf{Y})

This structure matches Option (A).