Click an option to check it. Structures redrawn from the printed paper.
Natural rubber is cis-1,4-polyisoprene: the repeat unit –CH2–C(CH3)=CH–CH2– carries exactly one C=C. Reductive ozonolysis (O3 then Zn/H2O) cuts every one of them and caps each cut carbon with =O — the trisubstituted end becomes a ketone, the monosubstituted end an aldehyde. Drag the slider to run the reaction and watch what a single fragment turns out to be.
Notice the fragment straddles two repeat units: the aldehyde carbon comes from one unit's =CH–, the methyl ketone from the next unit's =C(CH3)–.
The question hands you two diagnostic tests. Iodoform (I2/NaOH) needs a CH3–CO– group (or a CH3–CH(OH)– that oxidises to one); Tollens' reagent (Ag(NH3)2+) needs an aldehyde. Pick a candidate and run both.
Warm 4-oxopentanal with aqueous NaOH and an enolate forms, then attacks a carbonyl in the same molecule. There are three ways to do that. Pick the α-carbon that gets deprotonated and the carbonyl it attacks — the ring size is then fixed by the carbon skeleton, and only one choice survives.
Total strain energies of the cycloalkanes, from heats of combustion. A three-membered ring costs about 27 kcal mol−1; a five-membered ring costs 6. That gap is the whole reason the C5→C1 closure wins, and it is why cyclopentenones are such common aldol products.
| Ring size | Cycloalkane | Total strain (kcal mol⁻¹) | Reachable here? | Closure |
|---|
Natural rubber is cis-1,4-polyisoprene. Its repeat unit is –CH2–C(CH3)=CH–CH2–, C5H8, with one trisubstituted C=C per unit.
O3 followed by Zn/H2O breaks each C=C and puts =O on both carbons, stopping at the carbonyl stage (Zn destroys H2O2, so the aldehyde is not oxidised further). The =C(CH3)– end becomes a methyl ketone, the =CH– end an aldehyde.
Each fragment straddles two repeat units, so one repeat unit gives exactly one molecule of X.
Both tests fit, which confirms X rather than levulinic acid (an oxidative work-up product, no CHO) or pentane-2,4-dione (no CHO).
Number X as C1 (CHO), C2, C3, C4 (C=O), C5 (CH3). An enolate can form at C2 (α to the aldehyde), or at C3 and C5 (α to the ketone). Ring size = |α − carbonyl| + 1:
Three-membered rings carry ~27 kcal mol−1 of strain and do not form here; the five-membered closure is both unstrained and kinetically fast. The ketone enolate attacking the more electrophilic aldehyde is also the favoured direction.
The OH sits β to the ketone, so E1cb elimination of water gives the conjugated enone — the thermodynamic sink that makes aldol condensation irreversible.
Answer: (A).
[CH3]C(=O)[#6] and [CX3H1](=O)[#6]): X matches both; every
other C5H8O2 candidate fails at least one.| Simulation ID | ADV-2026-P2-CHE-Q03 |
| Exam / Year / Paper | JEE Advanced · 2026 · Paper 2 |
| Subject | Chemistry (Organic) |
| Chapter | Aldehydes, Ketones and Carboxylic Acids |
| Topic | Aldol condensation |
| Subtopics | Reductive ozonolysis · Structure of natural rubber (cis-1,4-polyisoprene) · Iodoform test · Tollens' test · Intramolecular aldol and ring size · Dehydration to an α,β-unsaturated ketone |
| Question type | Single correct MCQ (+3 / 0 / −1) |
| Difficulty | Moderate–Hard |
| Answer | (A) cyclopent-2-en-1-one |