ADV-2026-P2-CHE-Q08 · Diazonium and Azo Dye Virtual Lab

01Original question — verbatim

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

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

(A)

Q on treatment with ethanol generates an aromatic aldehyde.
(B)

S gives positive phthalein dye test.
(C)

P is a dinitro compound.
(D)

T is a coloured compound.

Multiple options may be correct (+4 / partial / −1). Walk both routes first — three of the four statements are decided the moment you know what P, Q, S and T actually are.

02Route 1 — aniline to the diazonium salt

Four steps, each one there for a reason. Watch what the acetyl group does to the nitration: it is the whole reason P comes out with one nitro group and not two.

Flask

compound

aniline

in the flask

formula

—

molecular

nitro groups

−NO₂ on the ring

ring activation

—

towards EAS

Steps

Why acetylate at all?

Try the nitration with and without the acetyl protecting group and count what comes out.

03Route 2 — cumene to salicylic acid

The cumene process and then Kolbe–Schmitt. The product S is the one the phthalein test is asked about, so look carefully at where its ring positions are free.

Flask

compound

cumene

in the flask

formula

—

molecular

phenolic OH

present?

para to OH

—

position 5

Steps

Bench tests on S

Run the phthalein dye test on whatever is in the flask.

no colour yet

04The coupling — and what ethanol does instead

A diazonium salt has two very different fates. Give it a activated ring and it couples to a dye; give it ethanol and it is reduced away. Statement (A) hinges on knowing which is which.

Offer Q a partner

Pick what Q meets. Only one of these makes T.

Why T is coloured

Extend the conjugation and watch the absorption move. A dye is a molecule whose absorption has walked into the visible.

Conjugated π system 2 rings + N=N

λ absorbed

—

colour absorbed

—

at that λ

colour you see

—

complementary

05Verdict on each statement

Each statement is marked from the bench that tests it — the identity you established, not a stored key.

Lab notebook

Written as you work. Only things you actually demonstrated appear here.

Nothing recorded yet — walk both routes, then run the benches.

Predict, then check

06Step-by-step solution

The solution is still locked

Test all four statements on the bench above and the derivation opens by itself.
Statements tested: 0 of 4 — finish both routes, run the phthalein test, and offer Q both ethanol and S

Route 1, steps 1–2: acetylate, then nitrate — this gives P

Aniline itself cannot be nitrated cleanly. The −NH₂ group is protonated by the nitrating mixture to −NH₃⁺, which is strongly deactivating and meta-directing, and aniline is oxidised by nitric acid as well. Acetic anhydride converts it to acetanilide first.

PhNH₂ + (CH₃CO)₂O / pyridine ⟶ PhNHCOCH₃

−NHCOCH₃ is still activating and ortho, para-directing, but much more mildly, and it is bulky. Nitration at 288 K therefore stops at one nitro group, overwhelmingly at the para position because ortho is blocked sterically.

P = p-nitroacetanilide — C₈H₈N₂O₃, a mononitro compound
Route 1, steps 3–4: hydrolyse, then diazotise — this gives Q

P + H₃O⁺ ⟶ p-nitroaniline · + NaNO₂/HCl, 273–278 K ⟶ Q = p-nitrobenzenediazonium chloride

The low temperature matters: above about 278 K the diazonium salt decomposes to the phenol and nitrogen.
Route 2: cumene process, then Kolbe–Schmitt — this gives S

cumene —O₂, then H₃O⁺—> phenol + acetone

phenol —NaOH, CO₂—> sodium salicylate —H₃O⁺—> S = salicylic acid

Kolbe–Schmitt carboxylates the phenoxide ortho to the oxygen, so S is 2-hydroxybenzoic acid: OH on C2, COOH on C1.
(A) Q with ethanol — the aldehyde is not aromatic

Ethanol reduces a diazonium salt: the diazonium group is replaced by hydrogen and the ethanol is oxidised.

ArN₂⁺Cl⁻ + C₂H₅OH ⟶ ArH + CH₃CHO + N₂ + HCl

With Ar = p-nitrophenyl, the aromatic product is nitrobenzene. The only aldehyde formed is acetaldehyde — two carbons, no ring. The statement asks for an aromatic aldehyde, and there isn't one. (A) is incorrect.
(B) S in the phthalein dye test

The test condenses a phenol with phthalic anhydride under conc. H₂SO₄; two phenol rings attack the anhydride carbonyl at the position para to their −OH. The requirement is therefore a free para position, which is why p-cresol fails the test.

salicylic acid: OH at C2 · COOH at C1 · para to the OH is C5, and C5 carries a hydrogen

S is a phenol with its para position open, so it condenses and gives the coloured phthalein in alkali. (B) is correct.
(C) Is P a dinitro compound?

No — and this is exactly what the acetylation was for. Count the nitrogens in P: one from the amide, one from a single nitro group.

P = CH₃CONH–C₆H₄–NO₂ : one −NO₂

Only an unprotected, strongly activated ring under forcing conditions would take a second nitro group. (C) is incorrect.
(D) T, the azo dye

In aqueous NaOH salicylic acid is deprotonated to the phenoxide — a very strongly activating group. The diazonium ion is a weak electrophile, so it needs exactly that activation, and it couples para to the phenoxide oxygen, at C5.

S + Q / aq. NaOH ⟶ T, an −N=N− bridged azo compound, C₁₃H₉N₃O₅

The −N=N− links the two rings into one long conjugated system, with an electron-donating −O⁻/−OH at one end and an electron-withdrawing −NO₂ at the other. That push–pull arrangement drops the HOMO–LUMO gap far enough that the absorption lands in the visible. Azo compounds of this kind are the classic synthetic dyes. (D) is correct.
Answer

(B) and (D)

Marking: +4 for choosing both; +2 for either one alone; −1 for including (A) or (C).

✓ Independent verification

Structures checked computationally: every intermediate was written as a SMILES string and parsed, and the counts quoted here come from substructure matching rather than from reading the drawing — P returns exactly one nitro match, T returns exactly one aryl–N=N–aryl match.
The para position of S was located programmatically: starting from the phenolic oxygen, walking three atoms round the ring lands on C5, and that atom carries one hydrogen. That is the structural requirement for the phthalein test, established without assuming the answer.
Formula bookkeeping: aniline C₆H₇N → acetanilide C₈H₉NO → P C₈H₈N₂O₃ → p-nitroaniline C₆H₆N₂O₂ → Q C₆H₄N₃O₂⁺; cumene C₉H₁₂ → phenol C₆H₆O → S C₇H₆O₃; T C₁₃H₉N₃O₅.
Why (A) is a good trap: a diazonium salt does generate an aldehyde with ethanol, so the statement is half true. The word that breaks it is "aromatic" — the aldehyde is acetaldehyde, and the aromatic fragment leaves as nitrobenzene.
Why (C) is a good trap: "nitrating mixture" suggests forcing conditions, and aniline is highly activated. The acetyl group and the 288 K are precisely what hold it to one substitution.
Cross-checked against a published answer key for this paper after the derivation was complete — the key agrees on (B) and (D), and on the coupling occurring at position 5.

07Concepts and classification

Relations used

−NH₂ + H⁺ → −NH₃⁺ : activating becomes deactivating

acetylation moderates the ring and holds nitration to mono, para

ArNH₂ + NaNO₂/HCl, 273–278 K → ArN₂⁺Cl⁻

ArN₂⁺ + C₂H₅OH → ArH + CH₃CHO + N₂

cumene + O₂, then H₃O⁺ → phenol + acetone

phenoxide + CO₂ (Kolbe–Schmitt) → salicylate, ortho

phthalein dye test needs a free para position to the phenolic −OH

azo coupling: weak electrophile needs a phenoxide or amine, couples para

Classification

Simulation ID	ADV-2026-P2-CHE-Q08
Exam / Year / Paper	JEE Advanced · 2026 · Paper 2
Subject	Chemistry (Organic)
Chapter	Amines and Diazonium Salts
Topic	Diazotisation and azo coupling
Subtopics	Acetylation as a protecting group · Nitration of acetanilide · Diazotisation at 273–278 K · Reduction of diazonium salts by ethanol · Cumene process · Kolbe–Schmitt reaction · Phthalein dye test · Azo dyes and colour
Question type	Multiple correct MCQ (+4 / partial / −1)
Difficulty	Moderate–Hard
Answer	— run the experiment

01Original question — verbatim

02Route 1 — aniline to the diazonium salt

Flask

Steps

Why acetylate at all?

03Route 2 — cumene to salicylic acid

Flask

Steps

Bench tests on S

04The coupling — and what ethanol does instead

Offer Q a partner

Why T is coloured

05Verdict on each statement

Lab notebook

Predict, then check

06Step-by-step solution

The solution is still locked

Route 1, steps 1–2: acetylate, then nitrate — this gives P

Route 1, steps 3–4: hydrolyse, then diazotise — this gives Q

Route 2: cumene process, then Kolbe–Schmitt — this gives S

(A) Q with ethanol — the aldehyde is not aromatic

(B) S in the phthalein dye test

(C) Is P a dinitro compound?

(D) T, the azo dye

Answer

✓ Independent verification

07Concepts and classification

Relations used

Classification