Ozonolysis–Aldol Ring Lab

QMission brief — the question, verbatim

JEE (Advanced) 2026, Paper 1, Chemistry, Section 4 — matching list set, one correct option: +4 if it is chosen, 0 if unanswered, −1 otherwise.

Q.15 The List-II contains products obtained from the reaction of compounds in List-I with O₃/Zn-H₂O followed by cyclization (via more stable enolate) in the presence of aqueous NaOH. Match each entry in List-I with appropriate entry in List-II and choose the correct option.

List-I

(P)

(Q)

(R)

(S)

List-II

(1)

(2)

(3)

(4)

(5)

(A) P ⟶ 2; Q ⟶ 4; R ⟶ 1; S ⟶ 3
(B) P ⟶ 3; Q ⟶ 4; R ⟶ 5; S ⟶ 2
(C) P ⟶ 2; Q ⟶ 1; R ⟶ 5; S ⟶ 3
(D) P ⟶ 3; Q ⟶ 5; R ⟶ 4; S ⟶ 2

The structures are redrawn from the printed ones; ring sizes and the positions of CH₃, OH and C=O are exactly as in the paper.

P →

Q →

R →

S →

OPTION

The mystery. Each alkene has one double bond — the bond both rings share. Ozone cuts it and opens both rings into one ten-membered ring with a C=O at each end. Then base pulls an α-H off and the enolate bites the other C=O, closing two new rings. Which rings, and where do the CH₃, OH and C=O end up? The lab pushes every electron pair to find out.

01Interactive experiment

One flask and the molecules, watched at two levels at once. Press START EXPERIMENT: cool → ozone adds → molozonide splits → ozonide → zinc → dione → enolate → ring closure → proton → Q, R, S → List-II → options → answer.

Ready

01 Cool02 +O₃03 Split04 Ozonide 05 Zn06 Dione07 Enolate08 Aldol 09 H⁺10 Q R S11 List-II12 Options13 Answer

Set your values or use the question defaults.

Answer——

step 1 of 7

Four fused alkenes, one sequence: O₃, then Zn/H₂O, then aqueous NaOH. The question's conditions are loaded — press START EXPERIMENT.

✓ QUESTION VALUES LOADED

Customize experiment — optional

Instrument panel

step

—

temperature

—

progress

0 %

alkene

—

C=O now

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arrows now

—

new rings

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P · Q · R · S

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option

—

Pink curved arrows move one electron pair each; red dashes are bonds breaking, green are bonds forming; gold dots are the lone pairs that move. The flask on the left changes at the same moment.

Nothing needs to be chosen. Press START EXPERIMENT.

C=O and C=C in P after every step

read from the molecule graph — the C=C becomes two C=O, then one C=O is used by the aldol

Every possible enolate and the ring it would close

one bar per α-C–H in each dione; rings smaller than five cannot close

Option analysis — claim → evidence → verdict

(A)P → 2; Q → 4; R → 1; S → 3

not yet examined

UNEVALUATED

(B)P → 3; Q → 4; R → 5; S → 2

not yet examined

UNEVALUATED

(C)P → 2; Q → 1; R → 5; S → 3

not yet examined

UNEVALUATED

(D)P → 3; Q → 5; R → 4; S → 2

not yet examined

UNEVALUATED

💨Criegee Master

⚙️Zinc Reducer

🎯Enolate Picker

⭕Ring Builder

🧩Structure Matcher

🏆Chemistry Master

Reasoning log — input → action → change → observation → conclusion

Nothing investigated yet.

Evidence table (0)

02Mechanism explorer

Pick an alkene and an enolate, then step through the eight arrow-pushing steps — or drag through one step and watch each electron pair move.

Alkene

Enolate

Step: 1 / 8

Through the step

03Detailed solution

The complete worked solution — every structure, step and calculation.

Understand the concept

Ozonolysis opens both rings

The only C=C is the ring-fusion bond. Cutting it turns a bicyclic alkene into one ten-membered ring with two C=O groups.

6-6 → cyclodecane-1,6-dione · 7-5 → cyclodecane-1,5-dione

Zn / H₂O keeps ketones

Reductive work-up cuts the ozonide's O–O bond without oxidising further: two ketones, and ZnO.

ozonide + Zn → 2 C=O + ZnO

More stable enolate

The more substituted enolate (from the CH that carries CH₃) is the more stable one — but it must be able to close a ring of five or more atoms.

C=C with more alkyl groups = more stable

Intramolecular aldol

The enolate carbon attacks the other C=O across the ring: one new C–C bond makes two rings, a β-hydroxy ketone with OH on a ring-fusion carbon.

6-6 → 5-7 · 5-7 → 6-6

P, step by step — ozonolysis (Criegee mechanism)
P — reductive work-up with Zn / H₂O
xmlns:xlink='http://www.w3.org/1999/xlink' xml:space='preserve'
P — enolate, aldol, protonation
width='360px' height='250px' viewBox='0 0 360 250'>
Step 1 — 1,3-dipolar cycloaddition: ozone's O⁻ lone pair and the C=C π pair form two C–O bonds while the O=O⁺ π pair falls back onto the central oxygen — the molozonide (1,2,3-trioxolane).

Step 2 — retro-cycloaddition: a ring-O lone pair makes C=O⁺, the ring-fusion C–C bond breaks into C=O and the O–O bond breaks: a ketone and a carbonyl oxide on one ten-membered ring.

Step 3 — recombination: the carbonyl oxide O⁻ adds to the ketone carbon and the ketone O bonds to the other carbon — the secondary ozonide (1,2,4-trioxolane) bridging the ring.

Step 4 — zinc gives an electron pair to a peroxide oxygen; the weak O–O bond breaks.

Step 5 — the O⁻ lone pair forms C=O, the bridging O takes the C–O pair and forms the second C=O, releasing ZnO: cyclodecane-1,6-dione (from a 6-6 alkene).

Step 6 — OH⁻ removes the α-H on the more substituted carbon (the CH bearing CH₃): the more stable, more substituted enolate.

Step 7 — intramolecular aldol: the enolate carbon attacks the other C=O; the new C–C bond closes a five-membered and a seven-membered ring.

Step 8 — the alkoxide takes a proton from water and OH⁻ is regenerated: the β-hydroxy ketone.

P → the 5-7 fused β-hydroxy ketone (new C–C bond highlighted): CH₃ and C=O at one ring-fusion carbon side, OH at the other with the second CH₃ beside it in the seven-membered ring — List-II (2).

P: dione ring a–C1(Me)–C2–C3–C4–b–C5(Me)–C6–C7–C8 · enolate at C1 (CH–CH₃, more substituted) attacks C=O at b
new rings: C1–C2–C3–C4–b (5) and C1–a–C8–C7–C6–C5–b (7) ⇒ 5-7 product, CH₃ + C=O beside one fusion carbon, OH + CH₃ at the other ⇒ (2)

Q, R and S — the same eight steps; only the ring closure differs

Q: ozonolysis gives cyclodecane-1,5-dione; the more stable enolate closes rings of 6 and 6 atoms → List-II (1). The other CH(CH₃) enolate would have to close a 4-membered ring, so it cannot react.

R: ozonolysis gives cyclodecane-1,5-dione; the more stable enolate closes rings of 6 and 6 atoms → List-II (5).

S: ozonolysis gives cyclodecane-1,6-dione; the more stable enolate closes rings of 5 and 7 atoms → List-II (3).

Alkene	Dione	Enolate (more substituted, ring ≥ 5)	New rings	Product
P (6-6)	1,6-dione	CH(CH₃) next to C=O	5 + 7	(2)
Q (7-5)	1,5-dione	CH(CH₃) from the 7-ring side (the 5-ring CH(CH₃) would need a 4-ring)	6 + 6	(1)
R (7-5)	1,5-dione	CH(CH₃) from the 7-ring side	6 + 6	(5)
S (6-6)	1,6-dione	CH(CH₃) next to C=O	5 + 7	(3)

Match and choose

P → 2, Q → 1, R → 5, S → 3 ⇒ correct option = —

Option	P	Q	R	S	Verdict
(A)	2 ✓	4 ✗	1 ✗	3 ✓	rejected
(B)	3 ✗	4 ✗	5 ✓	2 ✗	rejected
(C)	2 ✓	1 ✓	5 ✓	3 ✓	correct
(D)	3 ✗	5 ✗	4 ✗	2 ✗	rejected

Traps: using the less substituted (kinetic) enolate gives rings without the quaternary CH₃-bearing carbon — none of those is in List-II; and (4) has the CH₃ and OH arranged as no enolate of these diones can give.

Independent verification

Arrow-pushing, twice. The page and a separate Python script push the same eight steps; every intermediate has valid valences and the total charge is kept.
RDKit. Reaction SMARTS for ozonolysis and aldol, applied independently, give the same four products; canonical SMILES match List-II (2), (1), (5), (3).
Enolate audit. For each dione every α-C–H enolate is listed with the ring it would close; only the chosen ones are both most substituted and able to close rings of 5–7.
Solution websites. MathonGo's published solution (their Q47) gives (C).
This page. Products are made by pushing the arrows; the match is found by comparing graphs with the printed structures; nothing is written in.
Answer key. Checked afterwards against the official final key printed with the paper: Q15 → C.

04How to use this simulation

For a student opening it for the first time. The whole investigation runs by itself after one press.

STEP 1

Read the mission brief

Four bicyclic alkenes, five products. Ozone, zinc and water, then aqueous NaOH. Which product comes from which alkene?

STEP 2

Check the values

The question's conditions are loaded. You do not need to change anything. The answer is shown above the experiment.

STEP 3

Optional: customize

Use the less substituted enolate, or leave out the base, and see what changes. ↺ RESTORE QUESTION CONDITIONS puts everything back.

STEP 4

Start

Press ▶ START EXPERIMENT under the experiment screen. Nothing else to click (about 1½ minutes; 2× speeds it up).

STEP 5

Watch both levels

Left: the flask — ozone bubbles, the blue end-point, zinc, NaOH and heat. Right, at the same moment: every electron pair moving as a curved arrow.

STEP 6

Q, R, S and the match

The other three alkenes run through the same steps quickly; every product is compared with List-II and the options are tested.

STEP 7

Explore

In the Mechanism explorer step through any alkene and drag through any step. Then read the Detailed solution.

STEP 8

Experiment again

Press ▶ REPLAY EXPERIMENT, or change a condition and press ⚙ RUN WITH CUSTOM VALUES. CLASSROOM MODE enlarges everything for a projector.

Classification

Exam	JEE Advanced 2026, Paper 1, Chemistry, Section 4, Q.15
Chapter	Aldehydes, Ketones and Carboxylic Acids
Topic	Reductive ozonolysis of cyclic alkenes and intramolecular aldol cyclization via the more stable enolate
Difficulty	Difficult
Answer	— run the experiment

Ozonolysis–Aldol Ring Lab — cut the ring-fusion C=C, close a new ring

QMission brief — the question, verbatim

List-I

List-II

01Interactive experiment