Synthesis Pathway Lab

QMission brief — the question, verbatim

JEE (Advanced) 2026, Paper 1, Chemistry, Section 2 — one or more options correct: +4 only if all the correct options are chosen, partial marks +3 / +2 / +1 for a correct subset, 0 if unanswered, −1 otherwise.

Q.8 In the following reaction sequence, Q, R, S and T are the major products.

CH₃CH₂CH₂COONa —(1. Kolbe's electrolysis; 2. V₂O₅, 500 °C, 10-20 atm)→ Q —(phthalic anhydride, anhyd. AlCl₃)→ R —(1. PCl₅; 2. H₂-Pd/BaSO₄)→ S —(NH₂NH₂, heat)→ T

The correct statement(s) about Q, R, S and T is(are)

(A)S on warming with ammoniacal AgNO₃ results in the formation of silver mirror.
(B)Q on treatment with Cl₂(excess)/UV gives gammaxane.
(C)T is a heterocyclic compound.
(D)R on acid catalyzed intramolecular cyclization followed by treatment with Zn-Hg/HCl gives 9,10-dihydroxyanthracene.

PRODUCT Q

PRODUCT R

PRODUCT S

PRODUCT T

The mystery. A four-carbon salt goes in; five named reactions later a nitrogen-containing ring comes out. Each unknown is made in the lab, drawn from its bonds, and then put through the test each option describes — Tollens' reagent, chlorine under UV, a ring-atom audit and a Clemmensen reduction.

01Interactive experiment

Five reactions, four tests and an analysis board. Press START EXPERIMENT — the whole synthesis runs by itself: electrolysis → aromatisation → chlorination under UV → Friedel–Crafts → cyclisation and Clemmensen → Rosenmund → Tollens → ring closure → verdict on each option.

Ready

01 Kolbe02 Aromatise → Q03 Test B 04 Friedel–Crafts → R05 Test D06 Rosenmund → S 07 Test A08 Ring closure → T09 Test options10 Final answer

Set your values or use the question defaults.

step 1 of 5

One salt, five reactions, four unknowns. The question's conditions are loaded — press START EXPERIMENT.

✓ QUESTION VALUES LOADED

Customize experiment — optional

Instrument panel

step

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reaction status

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conditions

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progress

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product

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formula

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key group

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test result

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supported options

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Every structure is drawn from its bonds and every formula is counted from those bonds; the temperatures and pressures are the question's printed conditions.

Option analysis lab

(A)S on warming with ammoniacal AgNO₃ gives a silver mirror.

not yet examined

UNEVALUATED

(B)Q with Cl₂(excess)/UV gives gammaxane.

not yet examined

UNEVALUATED

(C)T is a heterocyclic compound.

not yet examined

UNEVALUATED

(D)R → acid cyclization → Zn-Hg/HCl gives 9,10-dihydroxyanthracene.

not yet examined

UNEVALUATED

⚡Electrolysis Expert

☀️Photochemist

🧪Friedel–Crafts Pro

🔍Reduction Detective

🪞Silver Mirror

💍Ring Builder

🏆Chemistry Master

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

Nothing investigated yet.

Evidence table (0)

Nothing needs to be chosen. Press START EXPERIMENT.

02Structure explorer

Every compound of the sequence, drawn from its bonds. Pick one, highlight its functional groups and ring atoms, and read its formula, degree of unsaturation and what each test would do to it. Compare puts the two (D) candidates side by side.

03Detailed solution

Run the experiment — the solution opens with the answer.

The solution is still locked

The experiment has not finished yet. Press START and let it run to the end — or use Reveal anyway in the header if you are revising.

Understand the concept

Q — Kolbe, then aromatisation

Kolbe's electrolysis couples two propyl radicals: n-hexane. V₂O₅ at 500 °C closes the chain and strips 4 H₂.

2 C₃H₇COO⁻ → C₆H₁₄ → C₆H₆

R — Friedel–Crafts acylation

AlCl₃ opens phthalic anhydride to an acylium ion, which acylates benzene.

R = 2-benzoylbenzoic acid, C₁₄H₁₀O₃

S — Rosenmund reduction

PCl₅ makes the acid chloride; H₂ over poisoned Pd/BaSO₄ stops at the aldehyde.

S = 2-benzoylbenzaldehyde, C₁₄H₁₀O₂

T — hydrazine closes a ring

Both carbonyls of S condense with the two NH₂ of hydrazine (−2H₂O): a six-membered ring with N–N.

T = 1-phenylphthalazine, C₁₄H₁₀N₂

Step 1 — Q

2 CH₃CH₂CH₂COO⁻ → CH₃(CH₂)₄CH₃ + 2CO₂ + 2e⁻ ; C₆H₁₄ —V₂O₅, 500 °C→ C₆H₆ + 4H₂

Q = benzene. With excess Cl₂ under UV, benzene adds three Cl₂ (a free-radical addition, not substitution) to give benzene hexachloride C₆H₆Cl₆; its γ-isomer is gammaxane (lindane), an insecticide. (B) ✓

On the bench: CO₂ bubbled at the anode, the hexane chain folded into a ring in the V₂O₅ furnace, and six Cl atoms added under the UV lamp.
Step 2 — R, and the (D) trap

C₆H₆ + phthalic anhydride —AlCl₃→ 2-benzoylbenzoic acid (R)

Acid (conc. H₂SO₄) closes R into anthraquinone (−H₂O). Zn-Hg/HCl (Clemmensen) turns each C=O into CH₂: the product is 9,10-dihydroanthracene, C₁₄H₁₂ — no oxygen at all. 9,10-Dihydroxyanthracene (C₁₄H₁₀O₂) is not formed. (D) ✗
Step 3 — S

R —PCl₅→ R–COCl —H₂, Pd/BaSO₄→ 2-benzoylbenzaldehyde (S)

S contains a –CHO group, so warming it with Tollens' reagent (ammoniacal AgNO₃) reduces Ag⁺ to metallic silver: a silver mirror. (A) ✓
Step 4 — T

S + NH₂NH₂ —heat→ 1-phenylphthalazine (T) + 2H₂O

The aldehyde and the ketone are 1,4-related on the ring, so hydrazine bridges them into a six-membered ring containing N=N–C=N… two nitrogen atoms in the ring: T is heterocyclic. (C) ✓

Answer: —

Independent verification

Structures and formulas. Every compound was built with RDKit and its formula counted from its bonds: Q C₆H₆, R C₁₄H₁₀O₃, S C₁₄H₁₀O₂, T C₁₄H₁₀N₂; each step balances.
Reactions as rules. The Friedel–Crafts, Rosenmund, hydrazine and Clemmensen steps were applied as reaction templates to the structures, and the products compared with the named compounds.
Functional-group tests. S has an aldehyde C–H (Tollens positive); T has ring nitrogen; the Clemmensen product has no oxygen.
Solution websites. MathonGo's published solution (their Q40) gives (A), (B), (C) with the same pathway.
This page. Each option is decided from the structures the lab builds; the answer letters are not written into the page.
Answer key. Checked afterwards against the official final key printed with the paper: Q8 → (A), (B), (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

Sodium butanoate is carried through five reactions to Q, R, S and T. Which of the four statements are correct? More than one can be.

STEP 2

Check the conditions

The question's reagents are loaded — sodium butanoate and the poisoned Pd/BaSO₄ catalyst. You do not need to change anything.

STEP 3

Optional: customize

Start from sodium propanoate, or hydrogenate over ordinary Pd/C. The lab tells you honestly when the question's pathway no longer applies. ↺ RESTORE QUESTION CONDITIONS puts everything back.

STEP 4

Start

Press ▶ START EXPERIMENT. There is nothing else to click (about a minute; 2× makes it faster).

STEP 5

Watch

CO₂ bubbles at the anode, a chain folds into a ring, chlorine adds under UV, an acylium ion attacks benzene, silver coats the test tube and hydrazine stitches a new ring.

STEP 6

Answer reveal

When all four statements are decided the lab says ✓ EXPERIMENT COMPLETE and reveals the answer. Badges unlock on the way.

STEP 7

Explore and understand

In the Structure explorer open each compound, highlight its groups, and compare the two (D) products. 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 2, Q.8
Chapter	Aldehydes, Ketones and Carboxylic Acids
Topic	Reaction sequence - Kolbe electrolysis, Friedel-Crafts acylation, Rosenmund reduction, Tollens test and hydrazine ring closure
Difficulty	Moderate
Answer	— run the experiment

QMission brief — the question, verbatim

01Interactive experiment