Why Aldehydes Oxidise While Ketones Resist
Aldehydes and ketones share the same polar carbonyl group, yet they behave entirely differently towards oxidation. The reason is structural and absolute: the carbonyl carbon of an aldehyde still carries a hydrogen atom, while in a ketone that carbon is flanked by two carbon substituents. Oxidation, in essence, is the insertion of oxygen into an existing bond. For an aldehyde, oxygen slips into the weak C–H bond, lifting it to the carboxylic acid level without disturbing the carbon skeleton.
A ketone offers no such hydrogen. To oxidise it, an oxidant must rupture a strong carbon–carbon bond — a process that demands vigorous conditions and yields a mixture of smaller carboxylic acids. NCERT states the distinction plainly: aldehydes are easily oxidised by even mild oxidising agents, whereas "ketones are generally oxidised under vigorous conditions, i.e., strong oxidising agents and at elevated temperatures."
The presence or absence of a hydrogen on the carbonyl carbon decides whether mild oxidation is even possible. This one feature underpins every distinguishing test in this subtopic.
Oxidation of Aldehydes to Carboxylic Acids
Aldehydes are oxidised to carboxylic acids by common reagents such as nitric acid, acidified potassium permanganate and potassium dichromate. The general transformation simply caps the carbonyl carbon with an –OH:
$$\ce{R-CHO ->[\text{[O]}] R-COOH}$$
What makes aldehydes pedagogically special — and what NEET exploits relentlessly — is that they fall even to mild oxidising agents that leave ketones untouched. NCERT names two such reagents: Tollens' reagent and Fehling's reagent. Because these reagents oxidise aldehydes but not ketones, they double as diagnostic tests that tell the two carbonyl classes apart in a test tube. In every case the aldehyde is oxidised to the corresponding carboxylate anion, since the reactions run in alkaline medium.
Distinguishing Tests: Tollens', Fehling's and Benedict's
Three closely related mild-oxidant tests dominate this topic. In each, the metal ion of the reagent is the oxidant; as the aldehyde is oxidised, the metal ion is reduced to a visually obvious product — a silver mirror or a red precipitate.
Tollens' test — the silver mirror
Tollens' reagent is freshly prepared ammoniacal silver nitrate, containing the diamminesilver(I) ion $\ce{[Ag(NH3)2]+}$. On gentle warming with an aldehyde, a bright silver mirror deposits on the tube wall as silver(I) is reduced to silver metal:
$$\ce{R-CHO + 2[Ag(NH3)2]^+ + 3OH^- -> R-COO^- + 2Ag(v) + 4NH3 + 2H2O}$$
Tollens' test is the most general of the three: both aliphatic and aromatic aldehydes give a positive silver mirror. Benzaldehyde, the standard aromatic aldehyde, reduces Tollens' reagent.
Fehling's test — the red precipitate
Fehling's reagent comes as two solutions mixed in equal amounts just before use. Fehling solution A is aqueous copper sulphate; Fehling solution B is alkaline sodium-potassium tartrate (Rochelle salt), which keeps Cu²⁺ in solution as a complex. On heating with an aliphatic aldehyde, a red-brown precipitate of cuprous oxide forms:
$$\ce{R-CHO + 2Cu^{2+} + 5OH^- -> R-COO^- + Cu2O(v) + 3H2O}$$
The deep blue cupric solution turns to a brick-red Cu₂O precipitate. Crucially, NCERT records that aromatic aldehydes do not respond to this test — a fact that makes Fehling's reagent the standard way to separate benzaldehyde from an aliphatic aldehyde.
Benedict's test
Benedict's reagent is a variant of Fehling's in which citrate replaces tartrate as the complexing agent; it behaves identically, giving a red Cu₂O precipitate with reducing (aliphatic) aldehydes and forming the basis of the classical test for reducing sugars.
| Test | Reagent / active species | Positive result | Aromatic aldehyde? |
|---|---|---|---|
| Tollens' | Ammoniacal AgNO₃, [Ag(NH3)2]+ | Silver mirror (Ag metal) | Yes — gives positive test |
| Fehling's | Alkaline Cu²⁺ + Rochelle salt (tartrate) | Red-brown Cu₂O precipitate | No — does not respond |
| Benedict's | Alkaline Cu²⁺ + citrate | Red Cu₂O precipitate | No — does not respond |
A ketone leaves both reagents unchanged. The silver mirror and the red Cu₂O precipitate are the two visual signatures NEET asks you to recognise.
Benzaldehyde: yes to Tollens', no to Fehling's
A frequent error is assuming every aldehyde responds to every mild oxidant. Aromatic aldehydes such as benzaldehyde do reduce Tollens' reagent (silver mirror) but do not respond to Fehling's or Benedict's test. So to tell benzaldehyde apart from an aliphatic aldehyde, Fehling's is the discriminating test, not Tollens'.
If both compounds give a silver mirror but only one gives a red precipitate, the Fehling-positive one is the aliphatic aldehyde.
These oxidations follow the carbonyl chemistry set up in Nucleophilic Addition to the Carbonyl Group. Revise that mechanism for the full picture.
Oxidation of Ketones and Popoff's Rule
Ketones do oxidise, but only under forcing conditions — strong oxidants at elevated temperature. Because there is no C–H on the carbonyl carbon, oxidation proceeds by cleaving a carbon–carbon bond adjacent to the carbonyl. NCERT notes this gives "a mixture of carboxylic acids having lesser number of carbon atoms than the parent ketone." For an unsymmetrical ketone, cleavage can occur on either side of the carbonyl, so a mixture of acids results.
The empirical guideline governing where cleavage prefers to occur is Popoff's rule: the carbonyl group tends to stay with the smaller alkyl group during oxidative cleavage. A symmetrical ketone such as pentan-3-one cleaves cleanly to two molecules of the same acid:
$$\ce{CH3CH2-CO-CH2CH3 ->[\text{strong [O]}] 2\,CH3CH2COOH}$$
For an unsymmetrical ketone, the product spread reflects competing cleavage on both sides, which is exactly why ketone oxidation is preparatively poor and why mild oxidants simply leave ketones alone.
Reduction to Alcohols
Turning to reduction, the gentlest outcome is conversion of the carbonyl group to a hydroxyl group. Aldehydes and ketones are reduced to primary and secondary alcohols respectively by sodium borohydride ($\ce{NaBH4}$), lithium aluminium hydride ($\ce{LiAlH4}$), or by catalytic hydrogenation over a metal catalyst:
$$\ce{R-CHO ->[NaBH4] R-CH2-OH} \qquad \ce{R-CO-R' ->[NaBH4] R-CH(OH)-R'}$$
The product level — alcohol, not hydrocarbon — is the defining feature here. The number of carbon atoms is preserved, and the oxygen stays in the molecule as an –OH group. $\ce{NaBH4}$ is the milder, more selective hydride; $\ce{LiAlH4}$ is the more powerful reagent and must be used in anhydrous conditions.
| Substrate | Reagent | Product | Alcohol class |
|---|---|---|---|
Aldehyde, RCHO | NaBH₄ / LiAlH₄ / H₂–catalyst | RCH2OH | Primary (1°) |
Ketone, RCOR' | NaBH₄ / LiAlH₄ / H₂–catalyst | RCH(OH)R' | Secondary (2°) |
Reduction to Hydrocarbons: Clemmensen and Wolff-Kishner
A deeper reduction strips the oxygen out entirely, converting the carbonyl group all the way to a methylene ($\ce{CH2}$) group. Two named reactions accomplish this, and NEET distinguishes them chiefly by the medium they operate in.
Clemmensen reduction — acidic medium
The carbonyl group is reduced to $\ce{CH2}$ using zinc amalgam (Zn-Hg) and concentrated hydrochloric acid:
$$\ce{R-CO-R' ->[\text{Zn-Hg, conc. HCl}] R-CH2-R'}$$
Because the conditions are strongly acidic, Clemmensen reduction suits substrates that are stable to acid but would be destroyed by strong base.
Wolff-Kishner reduction — basic medium
The same C=O → CH₂ transformation is achieved by first forming the hydrazone with hydrazine ($\ce{NH2NH2}$), then heating it with a strong base — sodium or potassium hydroxide — in a high-boiling solvent such as ethylene glycol:
$$\ce{R-CO-R' ->[NH2NH2] R-C(=N-NH2)-R' ->[\text{KOH, ethylene glycol, }\Delta] R-CH2-R' + N2 ^}$$
The conditions are strongly basic, so Wolff-Kishner is the complement to Clemmensen: choose it when the molecule is base-stable but acid-sensitive.
Two depths of reduction: hydride reagents stop at the alcohol, while Clemmensen (acid) and Wolff-Kishner (base) drive all the way to the hydrocarbon.
NaBH₄ and LiAlH₄ stop at the alcohol
Do not expect $\ce{NaBH4}$ or $\ce{LiAlH4}$ to give a hydrocarbon. These hydrides reduce the carbonyl only to the alcohol stage. Reaching $\ce{CH2}$ requires Clemmensen or Wolff-Kishner. Conversely, do not write an alcohol as the product of Clemmensen or Wolff-Kishner — those go straight to the methylene group.
Hydride / H₂ → alcohol. Zn-Hg/HCl or NH₂NH₂/KOH → CH₂ hydrocarbon.
The Haloform Reaction and the Iodoform Test
The haloform reaction is a special oxidative cleavage that sits at the intersection of oxidation and α-substitution. Aldehydes and ketones bearing a methyl group directly attached to the carbonyl carbon (methyl ketones) are oxidised by sodium hypohalite (X₂ in NaOH) to the sodium salt of a carboxylic acid having one carbon less than the carbonyl compound; the methyl group departs as the haloform $\ce{CHX3}$:
$$\ce{R-CO-CH3 + 3X2 + 4NaOH -> R-COONa + CHX3 + 3NaX + 3H2O}$$
NCERT adds an important qualifier: this oxidation does not affect a carbon–carbon double bond if one is present in the molecule, so an α,β-unsaturated methyl ketone keeps its alkene.
The iodoform test
When the halogen is iodine, the haloform produced is iodoform, $\ce{CHI3}$ — a bright yellow solid with a characteristic melting point. This makes the reaction a sensitive qualitative test. A positive iodoform test (yellow precipitate with I₂ and NaOH) signals the presence of the $\ce{CH3CO-}$ group or a $\ce{CH3CH(OH)-}$ group, the latter because it is first oxidised to the methyl-carbonyl unit:
$$\ce{CH3-CO-R + 3I2 + 4NaOH -> R-COONa + CHI3(v) + 3NaI + 3H2O}$$
Among aldehydes, only ethanal (acetaldehyde, $\ce{CH3CHO}$) responds, since it alone carries the $\ce{CH3CO-}$ unit. Higher aldehydes and non-methyl ketones give a negative result, which is what makes the test diagnostically powerful.
A compound C₈H₈O gives an orange-red precipitate with 2,4-DNP and a yellow precipitate with I₂/NaOH, but does not reduce Tollens' or Fehling's reagent. On drastic oxidation it gives benzoic acid (C₇H₆O₂). Identify it.
The 2,4-DNP test confirms a carbonyl. The negative Tollens'/Fehling's result rules out an aldehyde, so it is a ketone. The positive iodoform test marks it as a methyl ketone. Drastic oxidation to benzoic acid fixes the aromatic ring with one carbon bearing the carbonyl. The compound is acetophenone, $\ce{C6H5COCH3}$: a methyl ketone (positive iodoform) that does not reduce mild oxidants (no carbonyl C–H) yet is cleaved to benzoic acid under forcing oxidation.
Oxidation & reduction in one screen
- Aldehydes oxidise easily (even mild oxidants) because the carbonyl carbon has a C–H bond; ketones resist and need strong oxidants that cleave C–C (Popoff's rule).
- Tollens' (ammoniacal AgNO₃) → silver mirror; works for aromatic aldehydes too. Fehling's / Benedict's → red Cu₂O; aromatic aldehydes do not respond.
- NaBH₄ / LiAlH₄ / H₂-catalyst reduce to alcohols: aldehyde → 1°, ketone → 2°.
- Clemmensen (Zn-Hg, conc. HCl — acidic) and Wolff-Kishner (NH₂NH₂ then KOH in glycol — basic) reduce C=O all the way to CH₂.
- Haloform reaction: methyl ketones (and CH₃CHO) with X₂/NaOH give haloform + carboxylate with one carbon less; iodoform test (yellow CHI₃) detects CH₃CO– or CH₃CH(OH)– groups and does not touch C=C.