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Exercises · 7.12

Q.How do you count for the following observations ?

(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why ? Write a balanced redox equation for the reaction.
(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why ?
Andaman Nicobar CbseNCERTSubjective· 3mImportance★★★★★est
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The key idea is that the choice of oxidising agent depends on the reaction medium and the desired product. For (a), alcoholic KMnO₄ is used to avoid over-oxidation of toluene to CO₂, giving benzoic acid selectively. For (b), concentrated H₂SO₄ oxidises Br⁻ to Br₂ (red vapour) but not Cl⁻ to Cl₂ because Br⁻ is a stronger reducing agent than Cl⁻.

(a) Why alcoholic KMnO₄ for benzoic acid from toluene?

Concept & Intuition:

Potassium permanganate (KMnO₄) is a powerful oxidising agent. Its oxidising strength depends heavily on the pH of the medium. In acidic medium, MnO₄⁻ is reduced to Mn²⁺ (strongest oxidising power). In alkaline medium, it reduces to MnO₂ (moderate oxidising power). In neutral or alcoholic medium, the reaction is milder and more controlled.

When toluene (methylbenzene) is oxidised, the methyl group (-CH₃) is first converted to -CH₂OH, then -CHO, and finally -COOH (benzoic acid). If the oxidising agent is too strong (like acidic KMnO₄), the benzene ring itself may get attacked, leading to ring cleavage and CO₂ formation — over-oxidation. Alcoholic KMnO₄ provides a mild, controlled oxidation that stops at the carboxylic acid stage without destroying the aromatic ring. Just as importantly, alcohol dissolves BOTH reactants — toluene (an organic liquid, immiscible with water) and KMnO₄ — so the two meet in a single phase, with far better contact than a water-only medium could give.

Watch out

A common mistake is to think that "alkaline" and "alcoholic" KMnO₄ are the same. They are not. Alkaline KMnO₄ (aqueous KOH + KMnO₄) is still quite strong and can over-oxidise. Alcoholic KMnO₄ (KMnO₄ used in an alcohol medium) is milder, and — crucially — the alcohol acts as a common solvent that brings the water-loving oxidant and the water-repelling hydrocarbon into one phase.

Balanced redox equation:

The oxidation of toluene to benzoic acid by KMnO₄ in alcoholic medium:

C6H5CH3+2KMnO4→C6H5COOK+2MnO2+KOH+H2O\text{C}_6\text{H}_5\text{CH}_3 + 2\text{KMnO}_4 \rightarrow \text{C}_6\text{H}_5\text{COOK} + 2\text{MnO}_2 + \text{KOH} + \text{H}_2\text{O}

Then acidify to get benzoic acid:

C6H5COOK+HCl→C6H5COOH+KCl\text{C}_6\text{H}_5\text{COOK} + \text{HCl} \rightarrow \text{C}_6\text{H}_5\text{COOH} + \text{KCl}

The half-reactions are:

  • Oxidation: C6H5CH3+2H2O→C6H5COOH+6H++6e−\text{C}_6\text{H}_5\text{CH}_3 + 2\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_5\text{COOH} + 6\text{H}^+ + 6e^-
  • Reduction: MnO4−+2H2O+3e−→MnO2+4OH−\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^-

Multiplying reduction by 2 to balance electrons (6e⁻ each side) gives the net ionic equation.

Tip

In exam problems, you don't need to write the full ionic equation every time. The molecular equation with KMnO₄ → MnO₂ + benzoate salt is sufficient for most board-level questions.

(b) Why HCl gas from chloride but Br₂ vapour from bromide with conc. H₂SO₄?

Concept & Intuition:

Concentrated sulphuric acid (H₂SO₄) is both a strong acid and an oxidising agent. When added to halide salts, it first displaces the hydrogen halide (HX). But for Br⁻ and I⁻, the H₂SO₄ further oxidises the HX to the elemental halogen.

The key lies in the reducing power of the halide ions:

Reducing power: I−>Br−>Cl−>F−\text{Reducing power: } \text{I}^- > \text{Br}^- > \text{Cl}^- > \text{F}^-

  • Chloride (Cl⁻): Cl⁻ is a weak reducing agent. Concentrated H₂SO₄ cannot oxidise Cl⁻ to Cl₂. So only the acid displacement occurs:

NaCl+H2SO4→NaHSO4+HCl↑\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl} \uparrow

HCl is a colourless, pungent gas. …

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