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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Aromatic Synthesis Route
Aromatic Synthesis Route – First Principles
Imagine you are a chef who has been given a plain wooden board and told to carve a specific shape out of it. You can cut away wood, but you cannot add wood back. That is exactly the problem in aromatic synthesis: you start with a simple, cheap aromatic ring (like benzene) and you need to attach specific groups at specific positions. The ring itself is already there — you cannot rearrange its carbon skeleton. So the entire challenge is where to put the next group, and how to get it there.
The "route" is the sequence of reactions you choose. The order matters enormously because the groups already on the ring control where the next group will go. A wrong order can give you the wrong isomer, or force you into a dead end.
The Core Idea: The Ring Directs
Every substituent already on a benzene ring has a directing effect — it tells the next incoming group to go to certain positions. There are two families:
- Ortho/para directors (e.g., –OH, –NH₂, –CH₃, –Cl) — they push the next group to positions 2 and 4 (ortho and para).
- Meta directors (e.g., –NO₂, –CN, –CHO, –SO₃H) — they push the next group to position 3 (meta).
A common mistake is to think you can just "add any group in any order." The ring is not passive — it has a memory of what is already attached. If you ignore directing effects, you will get a mixture of products, often with the wrong isomer as the major one.
The Precise Statement
An aromatic synthesis route is a planned sequence of electrophilic aromatic substitution (EAS) reactions, chosen so that each new substituent is introduced at the correct position relative to the existing ones. The route must account for:
- Directing effects of all current substituents.
- Activation/deactivation — some groups make the ring more reactive (activators), some make it less reactive (deactivators). You cannot do a reaction on a strongly deactivated ring without special conditions.
- Order of introduction — sometimes you must introduce a meta director first, then an ortho/para director, or vice versa, to get the desired final pattern.
A Concrete Example: Making 4-Nitrobenzoic Acid
You want a benzene ring with –COOH at position 1 and –NO₂ at position 4 (para to each other).
–COOH is a meta director. –NO₂ is also a meta director. If you put –COOH first and then nitrate, the –COOH will send the –NO₂ to the meta position (3), not para (4). That gives the wrong isomer.
Correct route:
- Nitrate benzene first → nitrobenzene ( –NO₂ is meta directing).
- Then oxidise the methyl group (if you started with toluene) or use a different method to introduce –COOH. But wait — –NO₂ deactivates the ring strongly. So you cannot easily do Friedel-Crafts acylation on nitrobenzene.
So the actual correct route is different:
- Start with toluene (methylbenzene). The –CH₃ is an ortho/para director and an activator.
- Nitrate toluene → you get a mixture of ortho and para nitrotoluene. Separate the para isomer.
- Oxidise the –CH₃ to –COOH using KMnO₄. The –NO₂ survives this oxidation.
The order here is: introduce the ortho/para director first ( –CH₃ ), then nitrate to get para, then convert the –CH₃ to –COOH. If you had tried to put –COOH first, you would have a meta director that would send –NO₂ to the wrong place.
The General Strategy
When planning a route, ask yourself in order:
- What is the final substitution pattern? (1,2- ; 1,3- ; 1,4- ; etc.)
- Which groups are ortho/para directors and which are meta directors?
- Can I introduce the meta director first, then the ortho/para director? (Often yes, because meta directors deactivate the ring, making further substitution harder — so you want to do the deactivating step last if possible.)
- If I need a 1,3 pattern, I usually put a meta director first, then an ortho/para director. If I need a 1,4 pattern, I usually put an ortho/para director first, then a meta director (because the ortho/para director will send the next group to para, and the meta director will then be at the correct position). …
Why this formula?
Aromatic Synthesis Route: Understanding the Why Behind the Key Principles
In organic chemistry, an aromatic synthesis route refers to a sequence of reactions designed to construct or modify an aromatic ring (typically benzene or its derivatives). The "key formulas" here are not single equations but rather rules and principles that govern reactivity and orientation. Let's break down the reasoning behind the most critical ones.
1. The 4n+2 Hückel Rule — Why Aromaticity Exists
Formula: A planar, cyclic, conjugated molecule is aromatic if it has (4n+2) π electrons, where n=0,1,2,…
Why this holds (the derivation):
- In a cyclic conjugated system, the π electrons occupy molecular orbitals (MOs) that form a ring.
- The energy levels of these MOs are given by the Frost circle (or polygon rule):
- For a regular polygon with N vertices (atoms), inscribe it in a circle with one vertex at the bottom.
- The energy of each MO corresponds to the vertical coordinate of each vertex.
- For benzene (N=6), the MOs split into:
- 1 low-energy bonding orbital
- 2 degenerate bonding orbitals
- 2 degenerate antibonding orbitals
- 1 high-energy antibonding orbital
- Key insight: The 6 π electrons fill the 3 bonding MOs completely. This gives a closed-shell, highly stable configuration — the aromatic stabilization energy (~150 kJ/mol for benzene).
- For N=4 (cyclobutadiene), the MO pattern gives 2 degenerate non-bonding orbitals — filling with 4 electrons creates an open-shell, antiaromatic (unstable) system.
Takeaway: The (4n+2) rule is not arbitrary — it emerges from the symmetry of cyclic π systems and the filling of bonding MOs.
2. Electrophilic Aromatic Substitution (EAS) — The Reactivity Formula
General reaction:
Ar-H+E+catalystAr-E+H+
Why this is the only viable route for aromatic rings:
- Aromatic rings are electron-rich (due to the π cloud) but resistant to addition — addition would break aromaticity.
- Mechanism reasoning:
- The electrophile E+ attacks the ring, forming a σ-complex (arenium ion) — this step is slow (rate-determining).
- The σ-complex is non-aromatic (4 π electrons in the ring) — it is high-energy and unstable.
- To regain aromaticity, the complex loses a proton (H+) — this step is fast and thermodynamically driven.
- Why substitution, not addition: Addition would permanently destroy aromaticity; substitution restores it.
Key formula: The rate law is Rate=k[Ar-H][E+] — first order in both, because the slow step involves both reactants.
3. Orientation Rules — Why Substituents Direct Where the Next Group Goes
Rule:
- Activating groups (e.g., −OH,−NH2,−OCH3) direct to ortho/para positions.
- Deactivating groups (e.g., −NO2,−CN,−CHO) direct to meta positions.
Why this happens (resonance + inductive reasoning):
For ortho/para directors:
- The substituent has a lone pair or π bond that can donate electrons into the ring via resonance.
- Draw the resonance structures of the σ-complex for attack at ortho, meta, and para:
- Ortho attack: The positive charge can be delocalized onto the substituent (e.g., −OH becomes =OH+). This stabilizes the intermediate.
- Para attack: Similar stabilization — charge delocalized to the substituent.
- Meta attack: The positive charge cannot reach the substituent — less stable.
- Result: Ortho/para intermediates are lower in energy → faster reaction.
For meta directors:
- The substituent is electron-withdrawing (by induction or resonance, e.g., −NO2).
- Draw resonance for ortho attack: The positive charge is placed directly on the carbon bearing the withdrawing group — this is highly destabilizing (like putting a + charge next to a + pole).
- For meta attack: The positive charge is never on the carbon with the withdrawing group — relatively more stable.
- Result: Meta attack is the least destabilized → preferred. …
Concept: Aromatic Synthesis via Diazonium Salts — the diazonium group can either be swapped for another substituent at the very same ring position, or deleted (replaced by H), depending on the reagent chosen.
- 3-Methylaniline → 3-Nitrotoluene In 3-methylaniline the −NH2 (C-1) and −CH3 (C-3) are already meta to each other — exactly the relationship the product needs between −NO2 and −CH3. So the amino group only needs to be replaced in place, not removed: diazotise with NaNO2/HCl at 273–278 K, convert the diazonium chloride into the stable diazonium fluoroborate with HBF4, then heat the fluoroborate with aqueous NaNO2 in the presence of copper (Δ). The −N2+ group is replaced directly by −NO2 at the same carbon, giving 3-nitrotoluene.
- Aniline → 1,3,5-Tribromobenzene …
Use diazonium chemistry to swap or delete the amino group without disturbing anything else on the ring. For (i), the amino and methyl groups of 3-methylaniline are already meta to each other — diazotise, form the diazonium fluoroborate with HBF4, then heat with NaNO2/Cu to replace −N2+ directly with −NO2 at the same position, giving 3-nitrotoluene. For (ii), brominate free aniline directly (excess bromine water → 2,4,6-tribromoaniline), then delete the amino group via diazotisation and H3PO2, giving 1,3,5-tribromobenzene.
The Core Concept: The Diazonium Route
The amino group (−NH2) is a powerful ortho/para director and a strong activator. This makes it excellent for placing substituents at specific positions — but it also means aniline cannot be nitrated or mono-brominated cleanly. Diazonium chemistry resolves this: the amino group does its directing work (or simply marks a position), and is then either converted into the substituent needed at that very carbon, or removed entirely.
(i) 3-Methylaniline → 3-Nitrotoluene
In 3-methylaniline, the methyl group sits at C-3 relative to the amino group at C-1 — they are already meta to each other, which is exactly the −NO2/−CH3 relationship the target needs. So the cleanest route is not to nitrate the ring at all: it is to convert the existing amino group directly into a nitro group at the position it already occupies.
1. Diazotise the amine.
Treat 3-methylaniline with NaNO2 and dilute HCl at 273–278 K:
C6H4(CH3)(NH2)+NaNO2+2HCl273–278 KC6H4(CH3)(N2+Cl−)+NaCl+2H2O
2. Convert the diazonium chloride into the fluoroborate.
Treat the diazonium salt with fluoroboric acid; the stable, sparingly soluble diazonium fluoroborate separates:
Ar-N2+Cl−+HBF4→Ar-N2+BF4−+HCl
3. Replace −N2+ with −NO2.
Heat the diazonium fluoroborate with aqueous NaNO2 in the presence of copper — nitrite displaces the diazonium group:
Ar-N2+BF4−+NaNO2Cu, ΔAr-NO2+N2+NaBF4
Because the substitution happens at the same carbon the amino group occupied, the methyl group never moves and the new nitro group inherits the meta relationship. The product is 3-nitrotoluene.
Do not reach for H3PO2 here — that reagent replaces −N2+ with plain −H (deamination) and would simply regenerate toluene, deleting the nitrogen instead of converting it into the −NO2 group the target needs. H3PO2 is the right reagent for part (ii) below, where the amino group must disappear; it is the wrong reagent here.
Diazonium → nitro: Ar-N2+Cl−HBF4Ar-N2+BF4−NaNO2/Cu, ΔAr-NO2 — one of the standard diazonium substitutions, alongside Ar-N2+CuClAr-Cl, Ar-N2+CuBrAr-Br, Ar-N2+KIAr-I and Ar-N2+H3PO2Ar-H.
(ii) Aniline → 1,3,5-Tribromobenzene
The target is a benzene ring with three bromines in a 1,3,5 pattern and no nitrogen. The free amino group is the perfect tool: it activates the ring so strongly that bromine water substitutes all three ortho/para positions almost instantly.
1. Brominate free aniline directly.
Add excess bromine water at room temperature — the reaction is immediate and gives a white precipitate:
C6H5NH2+3Br2H2O2,4,6-Br3C6H2NH2+3HBr …
Aromatic Synthesis Route — Two Clear Methods
(i) 3-Methylaniline → 3-Nitrotoluene
Method: Diazotisation → diazonium fluoroborate (HBF4) → replacement of −N2+ by −NO2 (NaNO2/Cu, Δ)
Why this works:
In 3-methylaniline the amino group (C-1) and the methyl group (C-3) are already meta to each other — the exact geometry the product needs between −NO2 and −CH3. So no new substitution on the ring is required at all: the amino group is converted in place into a nitro group. No nitration step, no protection step, no isomer problem.
Steps:
- Diazotise Treat 3-methylaniline with NaNO2 + dilute HCl at 273–278 K to form the diazonium salt:
C6H4(CH3)(NH2)NaNO2/HCl273–278 KC6H4(CH3)(N2+Cl−)
- Form the diazonium fluoroborate Treat the diazonium chloride with HBF4; the stable fluoroborate separates:
C6H4(CH3)(N2+Cl−)HBF4C6H4(CH3)(N2+BF4−)+HCl
- Replace −N2+ by −NO2 Heat the fluoroborate with aqueous NaNO2 in the presence of copper. Nitrite displaces the diazonium group at the same carbon:
C6H4(CH3)(N2+BF4−)NaNO2/Cu, ΔC6H4(CH3)(NO2)+N2+NaBF4
The methyl group never moves, so the product keeps the meta relationship.
Final product: 3-nitrotoluene
(ii) Aniline → 1,3,5-Tribromobenzene
Method: Direct tribromination → Diazotisation → Reduction (replacement of −N2+ by −H)
Why this works:
The free −NH2 group is strongly activating and ortho/para-directing, so bromine water substitutes all three ortho/para positions of aniline at once — giving 2,4,6-tribromoaniline directly. Deleting the amino group afterwards leaves the three bromines in the 1,3,5 pattern.
Steps:
- Brominate aniline directly Add excess bromine water at room temperature; the reaction is instantaneous:
C6H5NH2+3Br2H2O2,4,6-Br3C6H2NH2+3HBr
Product: 2,4,6-tribromoaniline (white precipitate). No acetylation/protection step is used — protection would moderate the ring and stop bromination at the mono stage.
- Diazotise NaNO2 + dilute HCl at 273–278 K → the diazonium salt. …
These two conversions test whether you can tell when to swap the amino group for another substituent and when to delete it — and when a protecting group helps versus when it ruins the synthesis.
(i) 3-Methylaniline → 3-Nitrotoluene
✗ Common Mistake 1: Direct nitration of 3-methylaniline
Why it's wrong:
The free −NH2 group is a strongly activating ortho/para director, so nitration would go ortho/para to the amino group — not where the target needs it — and the amine itself is prone to oxidation in the nitrating mixture. More fundamentally, no nitration is needed at all: the product's −NO2 belongs at the very carbon the −NH2 already occupies.
✗ Common Mistake 2: The protect–nitrate–deaminate detour
Why it's wrong — check the geometry:
Acetylating to 3-methylacetanilide and then nitrating does not save this route, because −NHCOCH3 is an ortho/para director (the textbook's own bromination of acetanilide gives the para product as the major one — the amide group does not direct meta). Nitration would therefore land ortho/para to the acetamido group at C-1 — e.g. at C-4, which is ortho to the methyl at C-3. After hydrolysis and deamination the nitro group would sit ortho to the methyl group (a 2-nitrotoluene-type product), not meta. The detour is longer and delivers the wrong isomer.
✗ Common Mistake 3: Using H3PO2 on the diazonium salt
Why it's wrong:
H3PO2 replaces −N2+ with plain −H — that deletes the nitrogen entirely and just gives toluene. In this part the nitrogen position must become a nitro group, not vanish.
✓ Correct Route
- Diazotise 3-methylaniline (NaNO2/HCl, 273–278 K).
- Form the diazonium fluoroborate with HBF4.
- Heat with NaNO2/Cu, Δ — the diazonium group is replaced by −NO2 at the same carbon → 3-nitrotoluene. The original meta relationship between −CH3 and the nitrogen position is preserved automatically.
(ii) Aniline → 1,3,5-Tribromobenzene
✗ Common Mistake 1: Acetylating (protecting) the amine before bromination
Why it's wrong:
Protection is the tool for mono-bromination: the acetamido group moderates the ring precisely so that bromination stops at one position (mainly para) — that is why acetanilide, not aniline, is brominated when p-bromoaniline is the target. Here the target needs three bromines, so protection defeats the whole plan. The free amine's full activation is exactly what delivers 2,4,6-tribromoaniline in one step.
✗ Common Mistake 2: Stopping at 2,4,6-tribromoaniline
Why it's wrong:
The target contains no nitrogen. The amino group must still be removed after it has done its directing work.
✗ Common Mistake 3: Using NaNO2/Cu on the diazonium salt
Why it's wrong: …
Showing the 12 most recent of 96 on this concept.
- AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQQ.An aromatic compound (C7H8) on reaction with Br2/Fe in dark gave X. An alkene (C3H6) on reaction with HBr/(C6H5CO)2O2 gave Y as major product. Reaction of X and Y in the presence of Na/dry ether gave Z. What is Z? (A) 4-Isopropyl-1-methylbenzene (p-cymene) — a benzene ring bearing a methyl group and an isopropyl group in the para positions (B) n-Butylbenzene — a benzene ring bearing a straight-chain butyl group (C) 1-Methyl-4-propylbenzene — a benzene ring bearing a methyl group and a n-propyl group in the para positions (D) 1-Methyl-3-propylbenzene — a benzene ring bearing a methyl group and a n-propyl group in the meta positions
›Reveal solutionSolution
This tests ring bromination direction, the peroxide (anti-Markovnikov) effect, and the Wurtz–Fittig coupling. The final product is p-methylpropylbenzene.
Concept and Intuition
Three separate organic ideas are chained together here. First, Br2/Fe on an arene (dark, ionic mechanism, Fe acting as Lewis-acid catalyst by forming FeBr3 in situ) is classic electrophilic aromatic substitution, not the radical benzylic bromination you'd get with light. Toluene's methyl group is an activating ortho/para director, and because ortho substitution is more sterically hindered, the para isomer is the major product. Second, HBr addition to an unsymmetrical alkene normally follows Markovnikov's rule, but in the presence of a peroxide (here (C6H5CO)2O2, benzoyl peroxide) it proceeds by a radical chain mechanism, and the bromine ends up on the less substituted carbon — the anti-Markovnikov or 'peroxide effect'. Third, the Wurtz–Fittig reaction couples an aryl halide with an alkyl halide using sodium in dry ether, forming a new C–C bond between the ring and the alkyl chain.
Step-by-Step Solution
- Toluene (C6H5CH3) + Br2/Fe (dark) → ring bromination, methyl directs ortho/para, para is major: X = p-bromotoluene (4-BrC6H4CH3).
- Propene (CH3CH=CH2) + HBr in presence of peroxide → anti-Markovnikov addition: Br adds to the terminal (less substituted) carbon. Y = 1-bromopropane (CH3CH2CH2Br), i.e. n-propyl bromide. …
- AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQQ.What are X and Y respectively in the following set of reactions? C6H5FYC6H5N2ClXC6H5Cl (A) KCl ; KF (B) HCl ; KF (C) HCl ;(i) HBF4(ii) Δ (D) Cu | HCl ;(i) HBF4(ii) Δ
›Reveal solutionSolution
Diazonium salt → chlorobenzene needs the Sandmeyer route (Cu/HCl); diazonium salt → fluorobenzene needs the Balz–Schiemann route (HBF₄ then heat). So X = Cu|HCl, Y = (i) HBF₄ (ii) Δ.
Concept and Intuition
Aryl diazonium salts are versatile intermediates because the −N2+ group can be replaced by a wide range of other groups. Simple treatment with aqueous HCl alone does not replace −N2+ by Cl (it just hydrolyses to phenol on warming with water); a Cu(I) catalyst is essential to mediate the radical-type substitution — this is the Sandmeyer reaction. Fluorine, on the other hand, cannot be introduced by a Sandmeyer-type Cu-catalysed exchange at all; instead the diazonium salt is first converted to the stable, isolable diazonium tetrafluoroborate salt using HBF4, which on gentle heating undergoes clean thermal decomposition to expel N2 and BF3, leaving the aryl fluoride — the Balz–Schiemann reaction.
Step-by-Step Solution
- Reaction to X: C6H5N2ClXC6H5Cl — this is a diazonium-to-chloride exchange, which requires the Sandmeyer conditions: Cu / HCl (cuprous chloride generated in situ, catalysing −N2+→−Cl). …
- AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQQ.What are X and Y in the following set of reactions? Y(i) Br2∣Fe (ii) KMnO4∣OH−(iii) H3O+ C6H5CH3 (i) KMnO4∣OH− (ii) H3O+(iii) Br2∣Fe X (A) 4-bromobenzoic acid ; 4-bromobenzoic acid (B) 3-bromobenzoic acid ; 4-bromobenzoic acid (C) 4-bromobenzoic acid ; 3-bromobenzoic acid (D) 3-bromobenzoic acid ; 3-bromobenzoic acid
›Reveal solutionSolution
The order of oxidation vs bromination flips the directing group in play. Oxidise-then-brominate (toluene → benzoic acid → Br, meta-directed) gives 3-bromobenzoic acid (X); brominate-then-oxidise (toluene → p-bromotoluene → oxidise) gives 4-bromobenzoic acid (Y).
Concept and Intuition
This question tests how the order of reagent addition changes the regiochemistry of electrophilic aromatic substitution on a disubstituted benzene ring, because the directing effect of a substituent depends on what's already on the ring at the time of substitution:
- The methyl group (−CH3) in toluene is an activating, ortho/para-director.
- The carboxylic acid group (−COOH) is a deactivating, meta-director.
So bromination is directed to the para position when methyl is still present, but to the meta position once the methyl has already been oxidised to −COOH.
Step-by-Step Solution
- Path to X (given order: KMnO₄/OH⁻ first, then H₃O⁺, then Br₂/Fe):
- Step (i): C6H5CH3KMnO4/OH− oxidises the benzylic −CH3 to a carboxylate.
- Step (ii): H3O+ protonates to give benzoic acid, C6H5COOH.
- Step (iii): Br2/Fe brominates the ring — but now −COOH is the only ring substituent, and it is a meta-director, so bromination occurs at the meta position, giving 3-bromobenzoic acid. This is X.
- Path to Y (given order: Br₂/Fe first, then KMnO₄/OH⁻, then H₃O⁺):
- Step (i): C6H5CH3Br2/Fe brominates the ring while −CH3 is still present; methyl is an ortho/para-director, and the major (less hindered) product is para-bromotoluene. …
- AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQQ.Identify the major product 'z' in the given sequence of reactions (anhy = anhydrous) Benzenediazonium chloride (C6H5N2Cl, i.e. benzene ring with −N2Cl substituent) H2O283K X (i) NaOH(ii) C2H5Cl Y (i) CH3COCl/anhy. AlCl3(ii) Zn-Hg/HCl Z (A) para-substituted benzene: OC2H5 and COCl (i.e. 4−(C2H5O)−C6H4−COCl) (B) para-substituted benzene: OCOCH3 and OC2H5 (i.e. 4−(C2H5O)−C6H4−OCOCH3) (C) para-substituted benzene: OC2H5 and COCH3 (i.e. 4−(C2H5O)−C6H4−COCH3) (D) para-substituted benzene: OC2H5 and CH2CH3 (i.e. 4−(C2H5O)−C6H4−CH2CH3)
›Reveal solutionSolution
This is a 3-step synthesis test: diazonium hydrolysis → Williamson ether synthesis → Friedel–Crafts acylation followed by Clemmensen reduction. The final answer is 4-ethoxyethylbenzene.
Concept and Intuition
A diazonium salt hydrolyzed at low temperature (283K, i.e. ice-cold water, warmed slightly) replaces −N2+ with −OH, giving a phenol — this is the classic way to make phenols where direct ring hydroxylation is impossible. Once you have a phenol, treating it with base and then an alkyl halide is the Williamson ether synthesis: the phenoxide ion (a good nucleophile) displaces chloride from C2H5Cl to give an aryl alkyl ether. An alkoxy group like −OC2H5 is a powerful electron donor by resonance, so it strongly activates the ring towards electrophilic substitution and directs ortho/para — with the bulky acylium electrophile, the para product dominates. The last step, Zn(Hg)/HCl, is the Clemmensen reduction, a classic way to fully deoxygenate a ketone C=O down to CH2 under acidic conditions (as opposed to Wolff–Kishner, which is basic).
Step-by-Step Solution
- C6H5N2ClH2O283K hydrolysis of diazonium salt → X = C6H5OH (phenol), releasing N2 and HCl.
- X (i) NaOH sodium phenoxide C6H5O−Na+; (ii) C2H5Cl Williamson ether synthesis → Y = C6H5OC2H5 (phenetole / ethoxybenzene).
- Y CH3COCl, anhy. AlCl3 Friedel–Crafts acylation; the −OC2H5 group directs the acetyl group para (steric preference over ortho) → 4-ethoxyacetophenone, 4−(C2H5O)C6H4COCH3. …
- AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQQ.What are A and B in the following set of reactions respectively? Y (does not give iodoform test)Banhy. AlCl3BenzeneAanhy. AlCl3X (gets oxidised with Tollens′ reagent) (A) CO,HCl ; CH3CH2COCl (B) CO,HCl ; CH3COCl (C) CH3COCl ; CH3CH2COCl (D) CH3COCl ; CO,HCl
›Reveal solutionSolution
A = CO/HCl (Gatterman–Koch formylation of benzene to benzaldehyde, which is Tollens'-positive); B = propionyl chloride (Friedel–Crafts acylation to propiophenone, which is iodoform-negative because it lacks a methyl ketone group).
Concept and Intuition
Tollens' reagent is a test for aldehydes (and other easily oxidised groups) — a positive test on X tells us X must carry a −CHO group, i.e. X is benzaldehyde. Direct formylation of an aromatic ring is done by the Gatterman–Koch reaction: C6H6+CO+HClanhyd. AlCl3/CuClC6H5CHO.
The iodoform test is positive only for compounds containing a CH3−CO− (methyl ketone) or CH3−CH(OH)− group. Friedel–Crafts acylation of benzene with acetyl chloride (CH3COCl) gives acetophenone C6H5−CO−CH3, which is iodoform-positive. But the question specifies Y does not give the iodoform test, so the acylating agent must not leave a methyl ketone — using propionyl chloride (CH3CH2COCl) gives propiophenone C6H5−CO−CH2CH3, whose carbonyl is flanked by an ethyl group, not methyl, so it is iodoform-negative.
Step-by-Step Solution
- X gives positive Tollens' test ⟹ X has −CHO ⟹ X = benzaldehyde.
- Benzene → benzaldehyde requires the Gatterman–Koch formylation: reagent A = CO,HCl (with anhydrous AlCl3). …
- AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQQ.What are X and Y in the following set of reactions? Cumene (major product) Yanhy. AlCl3 C6H6 X C6H6Cl6 (A) Cl2/anhy. AlCl3, dark ; CH3CH2CH2Cl (B) Cl2/dark ; (CH3)2CHCl (C) Cl2/UV light ; CH3CH2CH2Cl (D) Cl2/anhy. AlCl3 ; (CH3)2CHCl
›Reveal solutionSolution
C6H6Cl6 forms by photochemical addition of Cl2 to benzene (UV light, no catalyst); cumene forms as the major (rearranged) product when benzene is Friedel–Crafts alkylated with n-propyl chloride, since the initial 1° carbocation rearranges to the more stable 2° isopropyl cation.
Concept and Intuition
The formula C6H6Cl6 keeps all 6 hydrogens of benzene while adding 6 chlorines — this can only happen by addition across all three double bonds (giving 1,2,3,4,5,6-hexachlorocyclohexane), not substitution. This addition is initiated by UV light (free-radical mechanism), with no Lewis-acid catalyst — AlCl3 would instead catalyze electrophilic aromatic substitution (chlorobenzene, losing HCl), which cannot give this formula.
The word "major product" attached to cumene is the giveaway for the second blank: if the reagent were straightforwardly isopropyl chloride, there'd be nothing to call "major" about — the alkylation would go directly to cumene. Calling it the major product signals a carbocation rearrangement is involved: n-propyl chloride with AlCl3 generates a 1° carbocation that undergoes a hydride shift to the more stable 2° carbocation, which then alkylates benzene to give cumene (isopropylbenzene) as the major product (with some n-propylbenzene as minor product).
Step-by-Step Solution
- Identify X (benzene → C6H6Cl6): must be an addition reaction, achieved by Cl2 under UV light — X = Cl2/UV light. …
- AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQQ.Which of the following is not an example of Sandmeyer reaction? (A) C6H5N2+Cl−CuCl ∣ HClC6H5Cl (B) C6H5N2+Cl−KI ∣ WarmC6H5I (C) C6H5N2+Cl−CuBr ∣ HBrC6H5Br (D) C6H5N2+Cl−CuCN ∣ KCNC6H5CN
›Reveal solutionSolution
This tests the precise scope of the "Sandmeyer reaction" name — it specifically refers to Cu(I)-catalysed replacement of a diazonium group, not just any halogen-substitution reaction of a diazonium salt.
Concept and Intuition
Aryl diazonium salts are versatile intermediates: the −N2+ group can be replaced by many different groups. When the replacement is carried out using a cuprous salt as catalyst (CuCl, CuBr, or CuCN), the reaction is specifically named the Sandmeyer reaction. When iodide replaces the diazonium group, however, iodide ion is nucleophilic and reducing enough on its own that no copper catalyst is needed — simply warming the diazonium salt with KI solution gives the aryl iodide directly. Because this reaction proceeds without the defining Cu(I) catalyst, it is conventionally not called a Sandmeyer reaction (even though it accomplishes a similar net transformation).
Step-by-Step Solution
- (A) C6H5N2+Cl−CuCl/HClC6H5Cl: diazonium + CuCl/HCl → chlorobenzene. Uses Cu(I) catalyst → is a Sandmeyer reaction.
- (B) C6H5N2+Cl−KI/WarmC6H5I: diazonium + KI, simply warmed, no copper salt used → not a Sandmeyer reaction (it is just direct nucleophilic substitution by iodide). …
- AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQQ.What is the end product (Y) in the given sequence of reactions? C6H5N2+Cl−(i) Cu2Cl2/HCl(ii) C2H5Cl/Na, dry etherX(i) KMnO4/OH−(ii) H3O+Y (A) Phenol, C6H5OH (benzene ring with −OH) (B) Benzoic acid, C6H5COOH (benzene ring with −COOH) (C) Phenylacetic acid, C6H5CH2COOH (benzene ring with −CH2COOH) (D) 2-Ethylphenol (benzene ring with −OH and, ortho to it, −C2H5)
›Reveal solutionSolution
This tests a two-step named-reaction sequence (Sandmeyer, then Wurtz–Fittig) followed by vigorous side-chain oxidation of an alkylbenzene. The final product Y is benzoic acid.
Concept and Intuition
Aryl diazonium salts are versatile synthetic handles. Converting −N2+ to −Cl via Cu(I) (Sandmeyer) gives an aryl halide, which can then be coupled to an alkyl halide via Na/dry ether (Wurtz–Fittig) to build a longer alkyl side chain on the ring. Once you have an alkylbenzene, hot alkaline KMnO4 is a powerful oxidant that attacks the benzylic C–H bonds and oxidizes the entire side chain down to a single carboxylic acid carbon attached to the ring — the specific length or branching of the original side chain doesn't matter (as long as there's at least one benzylic hydrogen), the product is always the aromatic carboxylic acid, i.e. benzoic acid for a monosubstituted benzene.
Step-by-Step Solution
- Step (i): C6H5N2+Cl−Cu2Cl2/HCl — this is the Sandmeyer reaction, replacing −N2+ with −Cl: product is chlorobenzene, C6H5Cl.
- Step (ii): Chlorobenzene C2H5Cl, Na, dry ether — this is the Wurtz–Fittig reaction: sodium metal couples the aryl halide with the alkyl halide, forming a new C–C bond between the ring and the ethyl group. Product X = ethylbenzene, C6H5CH2CH3. …
- AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQQ.The incorrect statement about Z formed in the sequence of reactions given below is C6H14V2O5, 773 K10-20 atmXCH3ClAnhy. AlCl3Yi) CrO2Cl2, CS2ii) H3O+Z (A) It does not give test with Fehling's solution (B) It can also be obtained by Gatterman-Koch reaction (C) It undergoes Cannizaro reaction in the presence of Conc. NaOH solution (D) It gives p-nitro derivative as major product in nitration reaction
›Reveal solutionSolution
This tests a multi-step named-reaction sequence (aromatization → Friedel–Crafts alkylation → Étard reaction) ending at benzaldehyde, then checking four textbook facts about benzaldehyde. The false one is the claim about para-nitration, since −CHO is a meta-director.
Concept and Intuition
Building the product chain first tells you exactly what Z is; then each answer choice is simply a fact-check against known benzaldehyde chemistry. The trap is statement (D), which sounds plausible if you forget that an electron-withdrawing carbonyl group directs meta, not para, during electrophilic aromatic substitution.
Step-by-Step Solution
- C6H14 (n-hexane) V2O5,773K,10-20atm X: catalytic reforming/dehydrocyclization converts the straight-chain hexane into an aromatic ring: X = benzene.
- X CH3Cl, Anhy. AlCl3 Y: Friedel–Crafts alkylation installs a methyl group on the ring: Y = toluene.
- Y (i)CrO2Cl2,CS2 (ii)H3O+ Z: this is the Étard reaction, oxidising the aromatic methyl group directly to an aldehyde: Z = benzaldehyde, C6H5CHO.
- Check (A): benzaldehyde is an aromatic aldehyde and famously does not reduce Fehling's solution — statement is true.
- Check (B): benzaldehyde can indeed be made directly from benzene by the Gattermann–Koch reaction (CO/HCl, anhydrous AlCl3+CuCl) — statement is true.
- Check (C): benzaldehyde has no α-hydrogen (the carbon adjacent to −CHO is the aromatic ring carbon, with no abstractable H for the usual aldol pathway), so with concentrated NaOH it undergoes Cannizzaro reaction (disproportionation to benzyl alcohol and benzoate) — statement is true. …
- AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQQ.What is Y in the following reaction sequence? (X = major product) C7H8Br2/FedarkX2Nadry etherY (A) [Structure: C6H5−CH2−CH2−C6H5 (1,2-diphenylethane / bibenzyl)] (B) [Structure: 4,4'-dimethylbiphenyl, H3C−C6H4−C6H4−CH3 (methyl groups para on each ring, rings joined at the 1,1'-positions)] (C) [Structure: 3,3'-dimethylbiphenyl — two benzene rings joined at the 1,1'-positions, each bearing a methyl group at its 3-position] (D) [Structure: C6H5−CH2−C6H4−CH3 (4-methyldiphenylmethane, methyl para to the CH2 link)]
›Reveal solutionSolution
Toluene + Br2/Fe (dark) gives p-bromotoluene; Wurtz-Fittig coupling of that with Na/dry ether joins the two rings at the C–Br carbons, giving 4,4'-dimethylbiphenyl.
Concept and Intuition
"Dark" with Fe as catalyst signals ionic (electrophilic aromatic substitution), not free-radical, bromination — so substitution happens on the ring, directed by the methyl group's activating, ortho/para-directing effect. The major product (less steric hindrance) is the para isomer. The Wurtz-Fittig reaction then couples an aryl halide with sodium in dry ether, forming a new C–C bond between two aryl rings at the position the halogen occupied.
Step-by-Step Solution
- Toluene + Br2, catalytic Fe, in the dark ⇒ electrophilic aromatic substitution on the ring (methyl is o,p-director); major product X = p-bromotoluene (CH3−C6H4−Br, Br para to CH3).
- X + 2Na / dry ether ⇒ Wurtz-Fittig reaction: two molecules of p-bromotoluene couple at their C–Br carbons, eliminating 2NaBr, joining the two rings directly (biphenyl linkage) at the carbons that bore Br. …
- AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQQ.Identify the end product Z in the given sequence of reactions (Anhy = anhydrous) C3H6HBr(C6H5CO)2O2XC6H6Anhy. AlCl3Y(i) KMnO4/OH−(ii) H3O+(iii) NaOH/CaO,ΔZ (A) Toluene (B) Cumene (C) Benzene (D) Xylene
›Reveal solutionSolution
Propene gives 1-bromopropane (anti-Markovnikov, peroxide effect); Friedel–Crafts alkylation with AlCl3 rearranges it to cumene; oxidation to benzoic acid followed by soda-lime decarboxylation regenerates benzene as Z.
Concept and Intuition
Three separate named reactions are chained here: (1) the peroxide (Kharasch) effect reverses the regiochemistry of HBr addition to an alkene via a radical mechanism; (2) Friedel–Crafts alkylations with primary alkyl halides are notorious for carbocation rearrangement to the most stable carbocation before the ring attacks it; (3) vigorous oxidation of any alkylbenzene with KMnO4 always converts the entire side chain to a single −COOH directly on the ring (regardless of chain length), and heating the sodium salt of a carboxylic acid with soda lime removes the −COOH group entirely, replacing it with −H.
Step-by-Step Solution
- C3H6 (propene) + HBr / (C6H5CO)2O2 (peroxide) → anti-Markovnikov addition (Br adds to the terminal, less-substituted carbon via a radical chain mechanism) → X = CH3CH2CH2Br (1-bromopropane, n-propyl bromide).
- X + C6H6 / anhydrous AlCl3 → Friedel–Crafts alkylation. The primary carbocation generated from n-propyl bromide is unstable and undergoes a 1,2-hydride shift to the more stable secondary (isopropyl) carbocation before attacking benzene → Y = isopropylbenzene (cumene), not n-propylbenzene. …
- AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQQ.Identify the product 'Z' formed in the given sequence of reactions (Anhy = anhydrous) C6H5N2+Cl−H2O283 KXZn/ΔYCO+HClAnhy. AlCl3Z (A) Chlorobenzene (C6H5Cl) (B) Benzaldehyde (C6H5CHO) (C) Benzoyl chloride (C6H5COCl) (D) Benzyl chloride (C6H5CH2Cl)
›Reveal solutionSolution
This chains three named reactions — diazonium hydrolysis, Zn-dust reduction, and Gattermann–Koch formylation — to arrive at benzaldehyde as Z.
Concept and Intuition
Diazonium salts are versatile synthetic hubs: warming them in water hydrolyses −N2+Cl− to −OH, releasing N2. Phenol's −OH can then be completely removed (deoxygenated) by distillation with zinc dust, regenerating the parent arene. From benzene, the Gattermann–Koch reaction is the classic method to install an aldehyde group directly onto the ring using CO and HCl with anhydrous AlCl3/CuCl as catalyst (a Friedel–Crafts-type formylation, effectively delivering a formyl cation electrophile).
Step-by-Step Solution
- C6H5N2+Cl−H2O,283KX: warming the diazonium salt in water hydrolyses it, replacing −N2+ with −OH and releasing N2 gas ⇒ X = phenol (C6H5OH).
- XZn/ΔY: distillation of phenol with zinc dust reduces/removes the phenolic −OH group entirely, giving the parent hydrocarbon ⇒ Y = benzene (C6H6). …
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