Natural rubber on complete ozonolysis (O$_3$/Zn-H$_2$O) gives compound X as the major product. X gives positive iodoform and Tollen's tests. X on heating with aqueous NaOH gives Y as the major product. Y is
Show Hint
For intramolecular Aldol condensations, always look for the formation of stable 5- or 6-membered rings. The aldehyde group is usually the electrophile, and the ketone's \(\alpha\)-carbon is often the nucleophile. Peroxide-catalyzed addition gives anti-Markovnikov products.
Step 1: Understanding the Question:
The question describes a reaction sequence starting with natural rubber, followed by ozonolysis to form compound X. Compound X undergoes specific chemical tests (iodoform and Tollen's) and then reacts with aqueous NaOH to form product Y. We need to identify the structure of Y. Step 2: Key Formula or Approach:
1. Natural rubber structure: Natural rubber is a polymer of isoprene (2-methylbuta-1,3-diene) units, with *cis*-1,4-polymerization. It contains C=C double bonds.
2. Ozonolysis (O$_3$/Zn-H$_2$O): This reaction cleaves C=C double bonds and forms aldehydes or ketones at the sites of cleavage.
3. Iodoform test: Positive for methyl ketones (R-CO-CH$_3$) or methyl carbinols (R-CH(OH)-CH$_3$).
4. Tollen's test: Positive for aldehydes (R-CHO).
5. Reaction with aqueous NaOH (Aldol Condensation/Cannizzaro): Aldehydes with \(\alpha\)-hydrogens undergo Aldol condensation. Aldehydes without \(\alpha\)-hydrogens undergo Cannizzaro reaction. Ketones with \(\alpha\)-hydrogens can undergo Aldol condensation. Step 3: Detailed Explanation:
1. Natural rubber ozonolysis to form X:
Natural rubber is poly(cis-1,4-isoprene). The repeating unit is:
\[ -\text{CH}_2 - \text{C}(\text{CH}_3)=\text{CH} - \text{CH}_2- \]
Ozonolysis ($O_3/Zn-H_2O$) cleaves all C=C double bonds. Each cleavage site yields a carbonyl group. The structure of the repeating unit means that after ozonolysis, the main product will be a dialdehyde-diketone mixture or fragments. Since it's a polymer, breaking every double bond will give a repeating unit like:
\[ \text{O=CH}-\text{CH}_2-\text{C}(=\text{O})-\text{CH}_3 + \text{O=CH}-\text{CH}_2-\text{C}(=\text{O})-\text{CH}_3 + \dots \]
A major product will be levulinaldehyde (4-oxopentanal), if it's the monomer. But natural rubber cleaves to give a single unit. It would give a structure that corresponds to CH$_3$-CO-CH$_2$-CH$_2$-CHO. This compound has both aldehyde and methyl ketone groups.
So, X = $\text{CH_3\text{COCH}_2\text{CH}_2\text{CHO}$} (Levulinaldehyde).
2. Tests for X: Positive Iodoform test: Levulinaldehyde has a methyl ketone group ($\text{CH}_3\text{CO}- $), so it will give a positive iodoform test. Positive Tollen's test: Levulinaldehyde has an aldehyde group ($\text{CHO}$), so it will give a positive Tollen's test.
These facts are consistent with X being Levulinaldehyde.
3. Reaction of X with aqueous NaOH (to form Y):
Compound X ($\text{CH}_3\text{COCH}_2\text{CH}_2\text{CHO}$) has both an aldehyde group and a methyl ketone group, and it possesses \(\alpha\)-hydrogens on the $\text{CH}_2$ adjacent to the ketone and aldehyde.
Under heating with aqueous NaOH, an intramolecular Aldol condensation is highly favored, especially if it can form a stable 5- or 6-membered ring.
Levulinaldehyde has:
- \(\alpha\)-hydrogens on $\text{CH}_3$ (adjacent to ketone).
- \(\alpha\)-hydrogens on $\text{CH}_2$ (adjacent to ketone).
- \(\alpha\)-hydrogens on $\text{CH}_2$ (adjacent to aldehyde).
The aldehyde group is generally more reactive as an electrophile. The \(\alpha\)-hydrogens adjacent to the ketone are often more acidic than those next to an aldehyde.
However, to form a 5-membered ring, the enolate formed from the $\alpha$-methylene group of the ketone ($\text{CH}_3\text{CO}\underline{\text{CH}_2}\text{CH}_2\text{CHO}$) would attack the aldehyde carbonyl.
Let's number the carbons starting from the ketone carbonyl:
$\overset{1}{\text{CH}_3}\overset{2}{\text{CO}}\overset{3}{\text{CH}_2}\overset{4}{\text{CH}_2}\overset{5}{\text{CHO}}$
Deprotonation at C3 ($\alpha$-carbon to ketone) gives enolate. Attack on C5 (aldehyde carbonyl). This would form a 5-membered ring.
The mechanism would be:
- Deprotonation of C3 to form enolate.
- Intramolecular nucleophilic attack of C3 on C5.
- Formation of a cyclic \(\beta\)-hydroxy ketone.
- Dehydration (loss of water) to form an \(\alpha,\beta\)-unsaturated ketone in a 5-membered ring.
This intramolecular aldol condensation (followed by dehydration) would yield cyclopent-2-enone.
\[ \text{CH}_3\text{COCH}_2\text{CH}_2\text{CHO} \xrightarrow{\text{NaOH, } \Delta} \text{cyclopent-2-enone} \]
Structure of Y (cyclopent-2-enone):
This matches option (A). Step 4: Final Answer:
Compound Y is cyclopent-2-enone.
