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MHT-CET Chemistry · Aldehydes, Ketones and Carboxylic Acids

Oxidation, Reduction and Identification Tests

A carbonyl is reduced to CH₂ by Clemmensen (Zn–Hg/conc. HCl) or Wolff–Kishner (hydrazine, then KOH in ethylene glycol), to the alcohol by LiAlH₄; aldehydes alone are oxidised by Tollens and Fehling and turn Schiff's reagent pink, and methyl ketones and ethanal give the haloform reaction.

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Why this matters

21 PYQs, 1 HARD. Twelve are the two carbonyl-to-methylene reductions — name from reagent, reagent from name, the propiophenone → n-propylbenzene product — plus LiAlH₄ leaving a C=C alone; nine are the tests — Tollens' silver mirror with ethanal, Schiff's magenta, why aldehydes oxidise and ketones do not, which compound lacks the CH₃CO group for the haloform reaction. Two cards.

Concept 1 of 2

Clemmensen, Wolff–Kishner and Hydride Reductions

Intuition

Two reactions take C=O all the way to CH₂: Clemmensen in ACID (zinc amalgam, concentrated HCl) and Wolff–Kishner in BASE (hydrazine first, then KOH in hot ethylene glycol). Choose by what else the molecule tolerates. Hydride reagents stop at the alcohol: LiAlH₄ and NaBH₄ reduce C=O but leave an isolated C=C untouched.

Definition

  • Clemmensen: RCHO→Zn-Hg / conc. HClRCH3\text{RCHO} \xrightarrow{\text{Zn-Hg / conc. HCl}} \text{RCH}_3; RCOR’→RCH2R’\text{RCOR'} \to \text{RCH}_2\text{R'}. Acidic.
  • Wolff–Kishner: RCOR’→NH2NH2hydrazone→KOH, (HOCH2)2, ΔRCH2R’+N2\text{RCOR'} \xrightarrow{\text{NH}_2\text{NH}_2} \text{hydrazone} \xrightarrow{\text{KOH},\ (\text{HOCH}_2)_2,\ \Delta} \text{RCH}_2\text{R'} + \text{N}_2. Basic. Ethyl phenyl ketone → n-propylbenzene (all three side-chain carbons kept).
  • LiAlH₄ / NaBH₄: C=O → CH–OH only. CH3CH=CHCH2CHO→CH3CH=CHCH2CH2OH\text{CH}_3\text{CH=CHCH}_2\text{CHO} \to \text{CH}_3\text{CH=CHCH}_2\text{CH}_2\text{OH}; the C=C stays.
  • Reagent ↔ name: Stephen SnCl₂/HCl · Etard CrO₂Cl₂ · Rosenmund H₂/Pd–BaSO₄ · Gattermann–Koch CO/HCl/AlCl₃ (NOT CrO₃/Ac₂O).

C=O to CH₂

Clemmensen: Zn-Hg / conc. HCl;Wolff–Kishner: NH2NH2 then KOH / ethylene glycol\text{Clemmensen: Zn-Hg / conc. HCl};\qquad \text{Wolff–Kishner: NH}_2\text{NH}_2 \text{ then KOH / ethylene glycol}

Worked example

Convert acetophenone into ethylbenzene by two different named reductions, and say what NaBH₄ would give instead.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Aldehydes, Ketones and Carboxylic AcidsMODERATE
Which of the following is Wolf-Kishner reduction?

[Q78 · 13th May Shift 2 · 2024]

Losing a carbon in Wolff–Kishner

The reduction replaces O by two H and changes nothing else. Propiophenone (three side-chain carbons) gives n-PROPYLbenzene; 'ethylbenzene' is the planted option.
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Concept 2 of 2

Tollens, Fehling, Schiff and the Haloform Test

Intuition

An aldehyde carries a hydrogen on its carbonyl carbon that an oxidant can take, so mild oxidants convert it to the acid; a ketone has no such hydrogen and resists. Tollens' reagent (ammoniacal AgNO₃) is reduced to a silver mirror, Fehling's to red Cu₂O, Schiff's reagent turns magenta — all by aldehydes only. The haloform (iodoform) reaction is different: it needs a CH₃CO– group, so ethanal and methyl ketones give it, propanal does not.

Definition

  • Tollens' test: ammoniacal AgNO₃ + aldehyde, boiled → silver mirror. Ethanal yes; ethanol, ethoxyethane, ethanoic acid no.
  • Fehling's test: aldehydes (aliphatic) → red Cu₂O. Schiff's test: aldehydes restore the magenta/pink colour of decolourised fuchsin; ketones do not.
  • Why: aldehydes have the abstractable C–H on the carbonyl carbon; ketones lack it.
  • Haloform: needs CH3CO-\text{CH}_3\text{CO-} (or CH₃CH(OH)–): ethanal, propanone, butanone give CHI₃; propanal does not.
  • Strong oxidants: alkaline KMnO₄ takes ethylbenzene (any side chain) to benzoic acid; cyclohexene with acidic KMnO₄ opens to adipic acid.

Tollens' reaction

RCHO+2[Ag(NH3)2]++3OH−→RCOO−+2Ag↓+4NH3+2H2O\text{RCHO} + 2[\text{Ag(NH}_3)_2]^+ + 3\text{OH}^- \to \text{RCOO}^- + 2\text{Ag}\downarrow + 4\text{NH}_3 + 2\text{H}_2\text{O}

Worked example

Three liquids are propanal, propanone and propan-1-ol. Assign a test that identifies each.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Aldehydes, Ketones and Carboxylic AcidsEASY
Which of the following compounds when treated with ammoniacal silver nitrate exhibits silver mirror test?

[Q83 · 4th May Shift 1 · 2023]

Expecting the haloform reaction from every carbonyl

It is a test for the CH₃CO– fragment, not for carbonyls in general. Ethanal passes; propanal, one carbon longer on the wrong side, fails.
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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.

Formulas (2)

  • Clemmensen, Wolff–Kishner and Hydride Reductions

    C=O to CH₂

    Clemmensen: Zn-Hg / conc. HCl;Wolff–Kishner: NH2NH2 then KOH / ethylene glycol\text{Clemmensen: Zn-Hg / conc. HCl};\qquad \text{Wolff–Kishner: NH}_2\text{NH}_2 \text{ then KOH / ethylene glycol}
  • Tollens, Fehling, Schiff and the Haloform Test

    Tollens' reaction

    RCHO+2[Ag(NH3)2]++3OH−→RCOO−+2Ag↓+4NH3+2H2O\text{RCHO} + 2[\text{Ag(NH}_3)_2]^+ + 3\text{OH}^- \to \text{RCOO}^- + 2\text{Ag}\downarrow + 4\text{NH}_3 + 2\text{H}_2\text{O}

Watch out for (2)

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