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MHT-CET Chemistry · Aromatic Compounds

Side-Chain Oxidation, Addition to Benzene, and Diazonium Chemistry

The reactions that do not substitute the ring: strong oxidants cut any alkyl side chain down to –COOH, chromyl chloride stops at –CHO, benzene adds Cl₂ or ozone under forcing conditions, and a diazonium salt or a Grignard reagent carries the ring into a new compound.

Why this matters

15 PYQs; the chapter's one HARD question is here. Eight are side-chain oxidation — KMnO₄, dilute HNO₃, chromyl chloride, CrO₃ on phenol; three are addition — ozonolysis to glyoxal and BHC; four are the other transformations — diazotisation, removing the diazo group, Wurtz–Fittig, a Grignard on a nitrile. Three cards.

Concept 1 of 3: Oxidising the Side Chain

The ring survives oxidation; the side chain does not. Hot alkaline KMnO₄ (or dilute HNO₃) attacks the benzylic carbon and cuts off everything beyond it, so ethylbenzene, propylbenzene and cumene all end at benzoic acid. Chromyl chloride (the Étard reaction) is gentler and stops at the aldehyde.

Definition

  • Alk. KMnO₄, then H₃O⁺: any side chain with a benzylic H → C₆H₅COOH (ethylbenzene, cumene).
  • Dilute HNO₃ also oxidises the side chain: ethylbenzene → benzoic acid (the paper's key).
  • Étard reaction: toluene + CrO₂Cl₂ in CS₂, then H₃O⁺ → benzaldehyde.
  • Phenol + CrO₃ → p-benzoquinone.
StartReagentProduct
Ethylbenzenei) alk. KMnO₄ ii) H₃O⁺Benzoic acidQ
CumeneKMnO₄, KOH, Δ; then H₃O⁺Benzoic acidQ
EthylbenzeneDilute HNO₃Benzoic acidQ
TolueneCrO₂Cl₂ / CS₂; then H₃O⁺BenzaldehydeQ
PhenolCrO₃p-BenzoquinoneQ
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 13th May Shift 2 · Q62Moderate

Example 1 · Aromatic Compounds · Side-Chain Reactions, Oxidation and Other Transformations

Which from following compounds is obtained when toluene is treated with CrO2Cl2\text{CrO}_2\text{Cl}_2 in presence of CS2\text{CS}_2 followed by acid hydrolysis?

Keeping the side-chain carbons

C₆H₅CH₂COOH and longer acids are offered because they keep the chain length. KMnO₄ keeps only the benzylic carbon: the product is always C₆H₅COOH.

Benzal chloride from chromyl chloride

The Étard complex is hydrolysed to the aldehyde. Benzal chloride, C₆H₅CHCl₂, comes from chlorinating toluene's side chain in light — a different reaction.

Concept 2 of 3: Addition to Benzene: Ozonolysis and BHC

Benzene resists addition, but under forcing conditions it adds three molecules at once: three O₃ give a triozonide, and three Cl₂ in UV light give benzene hexachloride. Each C=C of the Kekulé structure is cut or saturated.

Definition

  • Ozonolysis: benzene + excess O₃ → benzene triozonide; Zn/H₂O → 3 glyoxal (OHC–CHO) + H₂O₂. Zn is there to destroy the H₂O₂ so the aldehyde is not oxidised.
  • BHC: benzene + 3Cl₂ in UV light → C₆H₆Cl₆ (benzene hexachloride). Its γ-isomer is gammexane (lindane), an insecticide.
ReactionProduct
Benzene triozonide + Zn/H₂OGlyoxalQ
Reagent that converts the triozonide to glyoxalZn + H₂OQ
Benzene + Cl₂, UVBHC; γ-isomer = gammexaneQ
Practice this conceptself-check

The same idea in a real exam question:

MHT-CET · 2024 · 10th May Shift 1 · Q81Moderate

Example 2 · Aromatic Compounds · Side-Chain Reactions, Oxidation and Other Transformations

Benzene + ozone (excess) →CCl4\xrightarrow{\text{CCl}_4} benzene triozonide →Zn/H2O\xrightarrow{\text{Zn/H}_2\text{O}} P + H2_2O2_2. Product P is

Gammexane as a herbicide

Gammexane kills insects, not weeds — it is an insecticide.

Concept 3 of 3: Diazonium Salts, Wurtz–Fittig and a Grignard on a Nitrile

These carry the ring into a new compound. Aniline becomes a diazonium salt at 0–5 °C, and the –N₂⁺ can then be replaced — by H, using a mild reducing agent. Sodium couples an aryl halide with an alkyl halide. A Grignard reagent adds to a C≡N to give an imine that hydrolyses to a ketone.

Definition

  • Diazotisation: aniline + NaNO₂ + HCl, 0–5 °C → benzene diazonium chloride.
  • Removing –N₂⁺: C₆H₅N₂Cl + CH₃CH₂OH (or H₃PO₂) → benzene.
  • Wurtz–Fittig: aryl halide + alkyl halide + Na (dry ether) → alkylbenzene. Two aryl halides = Fittig; two alkyl halides = Wurtz.
  • Nitrile + Grignard: C₆H₅CN + C₆H₅MgBr → imine salt; H₃O⁺ → benzophenone, C₆H₅COC₆H₅.
ReactionProduct or reagent
Aniline + NaNO₂/HCl, coldBenzene diazonium chlorideQ
C₆H₅N₂Cl + R → benzeneR = ethanol (paper's key)Q
H₃PO₂/H₂O does the same job; the paper keys ethanol.
Wurtz–FittigAryl + alkyl halide with NaQ
Benzonitrile + C₆H₅MgBr, then H₃O⁺BenzophenoneQ
Practice this conceptself-check

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 2 · Q98Hard

Example 3 · Aromatic Compounds · Side-Chain Reactions, Oxidation and Other Transformations

Identify product 'B' in following reaction. Benzonitrile →C6H5MgBr/dry etherA→H3O+B\xrightarrow{\text{C}_6\text{H}_5\text{MgBr/dry ether}} A \xrightarrow{\text{H}_3\text{O}^+} B

Stopping at the aldehyde

The Grignard adds a SECOND carbon group to the nitrile carbon, so the product is a ketone. Benzaldehyde would need a hydride, not C₆H₅MgBr.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Reference tables (3)

Oxidising the Side Chain5 rows
StartReagentProduct
Ethylbenzenei) alk. KMnO₄ ii) H₃O⁺Benzoic acidQ
CumeneKMnO₄, KOH, Δ; then H₃O⁺Benzoic acidQ
EthylbenzeneDilute HNO₃Benzoic acidQ
TolueneCrO₂Cl₂ / CS₂; then H₃O⁺BenzaldehydeQ
PhenolCrO₃p-BenzoquinoneQ
Addition to Benzene: Ozonolysis and BHC3 rows
ReactionProduct
Benzene triozonide + Zn/H₂OGlyoxalQ
Reagent that converts the triozonide to glyoxalZn + H₂OQ
Benzene + Cl₂, UVBHC; γ-isomer = gammexaneQ
Diazonium Salts, Wurtz–Fittig and a Grignard on a Nitrile4 rows
ReactionProduct or reagent
Aniline + NaNO₂/HCl, coldBenzene diazonium chlorideQ
C₆H₅N₂Cl + R → benzeneR = ethanol (paper's key)Q
H₃PO₂/H₂O does the same job; the paper keys ethanol.
Wurtz–FittigAryl + alkyl halide with NaQ
Benzonitrile + C₆H₅MgBr, then H₃O⁺BenzophenoneQ

Watch out for (4)

Test yourself on Aromatic Compounds

15 past MHT-CET questions from this chapter, timed at 14 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.