Why aldehydes and ketones diverge
Both aldehydes ($\ce{RCHO}$) and ketones ($\ce{R2CO}$) contain a carbonyl group, and most addition reactions of the carbonyl carbon are common to both. The decisive difference appears in oxidation. In an aldehyde, the carbonyl carbon carries a hydrogen atom; this aldehydic $\ce{C-H}$ bond can be replaced by a $\ce{C-OH}$ on oxidation, converting the aldehyde to a carboxylic acid (or its salt in alkaline medium) without breaking any carbon–carbon bond.
A ketone has no such hydrogen — the carbonyl carbon is flanked by two carbon groups. To oxidise a ketone you must rupture a strong $\ce{C-C}$ bond, which only harsh oxidants achieve and which always degrades the molecule. The mild, alkaline reagents used in distinguishing tests cannot do this. Consequently:
Aldehydes are oxidised by mild oxidising agents and therefore behave as reducing agents — they reduce $\ce{Ag+}$ to silver and $\ce{Cu^2+}$ to $\ce{Cu+}$. Ketones do not, so they leave these reagents unchanged. This single asymmetry is the engine behind Tollens', Fehling's and Benedict's tests.
"Both have a carbonyl, so both reduce Tollens'."
A common error is to treat aldehydes and ketones as interchangeable because they share $\ce{>C=O}$. The carbonyl governs addition chemistry, not oxidation. Only the aldehyde's $\ce{C-H}$ makes it a reducing agent.
Rule: mild oxidation distinguishes them — aldehyde reduces the reagent, ketone does not.
Tollens' test: the silver mirror
Tollens' reagent is ammoniacal silver nitrate — the colourless diamminesilver(I) complex $\ce{[Ag(NH3)2]+}$. It must be prepared fresh just before use. Adding a little dilute $\ce{NaOH}$ to silver nitrate first throws down a brown precipitate of silver oxide; aqueous ammonia is then added dropwise until that precipitate just dissolves to a clear solution:
$$\ce{2AgNO3 + 2NaOH -> Ag2O v + 2NaNO3 + H2O}$$
$$\ce{Ag2O + 4NH3 + H2O -> 2[Ag(NH3)2]OH}$$
When an aldehyde is warmed gently with this reagent, the silver(I) is reduced to metallic silver, which deposits as a bright, mirror-like film on the clean inner wall of the tube — the famous silver mirror. The aldehyde is oxidised to the carboxylate ion (the medium is basic):
$$\ce{RCHO + 2[Ag(NH3)2]+ + 3OH- -> RCOO- + 2Ag v + 4NH3 + 2H2O}$$
Ketones give no reaction — the solution stays clear and no silver deposits. The test is therefore positive for aldehydes and negative for ketones. Importantly, both aliphatic and aromatic aldehydes respond: benzaldehyde gives a positive Tollens' test even though it fails Fehling's.
Fehling's test: the brick-red precipitate
Fehling's solution comes in two bottles that are mixed in equal volumes immediately before use:
| Component | Contents | Role |
|---|---|---|
| Fehling A | Aqueous copper(II) sulphate, CuSO₄ | Source of the oxidising Cu(II) |
| Fehling B | Alkaline sodium potassium tartrate (Rochelle salt) in NaOH | Keeps Cu(II) in solution; provides alkali |
On mixing, the tartrate chelates copper(II) to give a deep-blue complex; this stops $\ce{Cu(OH)2}$ from precipitating prematurely in the alkaline medium. When an aldehyde is warmed with this blue solution, the copper(II) is reduced to copper(I), which separates as a brick-red precipitate of cuprous oxide, $\ce{Cu2O}$. The disappearance of blue and appearance of red is the positive signal:
$$\ce{RCHO + 2Cu^2+ + 5OH- -> RCOO- + Cu2O v + 3H2O}$$
The selectivity of Fehling's is sharper than Tollens'. Aliphatic aldehydes give a positive test; aromatic aldehydes (e.g. benzaldehyde) generally do not reduce Fehling's solution, and ketones are negative. This makes the pair of tests jointly diagnostic, as the comparison table below shows.
Methyl ketones and ethanal also answer to a colour-and-precipitate test of their own — see the Iodoform test for the yellow $\ce{CHI3}$ that separates $\ce{CH3CO-}$ compounds.
Benedict's test (the sugar variant)
Benedict's reagent is a single, more stable solution containing copper(II) sulphate, sodium citrate and sodium carbonate. The citrate plays the same stabilising role as the tartrate in Fehling's, but the one-bottle formulation keeps better, which is why Benedict's is preferred for routine detection of reducing sugars. The chemistry mirrors Fehling's: an aldehyde (or a reducing sugar) reduces the blue $\ce{Cu^2+}$ to a brick-red $\ce{Cu2O}$ precipitate. For NEET it is enough to know that Benedict's is the milder, citrate-based analogue of Fehling's and gives the same brick-red result with reducing sugars.
The redox basis of all three
All three tests are the same redox event seen through different metal ions. The aldehyde is the reducing agent: it is oxidised from the aldehyde to the carboxylate, surrendering electrons. The metal ion is the oxidising agent: it is reduced and reports the result as a visible product — a silver mirror or a red oxide.
| Test | Oxidising species (reduced) | Visible product | Aldehyde fate |
|---|---|---|---|
| Tollens' | $\ce{Ag+}$ → $\ce{Ag^0}$ | Silver mirror | $\ce{RCHO -> RCOO-}$ |
| Fehling's | $\ce{Cu^2+}$ → $\ce{Cu+}$ | Brick-red $\ce{Cu2O}$ | $\ce{RCHO -> RCOO-}$ |
| Benedict's | $\ce{Cu^2+}$ → $\ce{Cu+}$ | Brick-red $\ce{Cu2O}$ | $\ce{RCHO -> RCOO-}$ |
The half-reaction for the aldehyde is the same in every case; in alkaline medium it is written as the oxidation of the aldehyde to the carboxylate ion, $\ce{RCHO + 3OH- -> RCOO- + 2H2O + 2e-}$. The two electrons released reduce two $\ce{Ag+}$ to silver, or reduce $\ce{Cu^2+}$ to $\ce{Cu+}$ which then appears as $\ce{Cu2O}$.
What tells aldehyde from ketone
The two tests, read together, classify almost any carbonyl an examiner will offer. The table below is the one to memorise.
| Compound class | Tollens' (silver mirror) | Fehling's (brick-red Cu₂O) |
|---|---|---|
| Aliphatic aldehyde (e.g. ethanal) | Positive | Positive |
| Aromatic aldehyde (e.g. benzaldehyde) | Positive | Negative (generally) |
| Ketone (e.g. propanone) | Negative | Negative |
| Formic acid / formate | Positive | Positive |
| Reducing sugar (glucose, fructose) | Positive | Positive |
| Sucrose (non-reducing) | Negative | Negative |
The decision logic follows directly: a silver mirror confirms an aldehyde; if the same sample also gives brick-red with Fehling's it is aliphatic, and if it fails Fehling's it is aromatic. A compound that is negative to both Tollens' and Fehling's is a ketone (or a non-reducing species).
Worked distinguishing examples
Distinguish ethanal ($\ce{CH3CHO}$) from propanone ($\ce{CH3COCH3}$) chemically.
Warm each separately with Tollens' reagent. Ethanal, an aldehyde, reduces $\ce{[Ag(NH3)2]+}$ and deposits a silver mirror; propanone, a ketone, gives no change. The reaction with ethanal:
$$\ce{CH3CHO + 2[Ag(NH3)2]+ + 3OH- -> CH3COO- + 2Ag v + 4NH3 + 2H2O}$$
A confirmatory Fehling's test would give a brick-red $\ce{Cu2O}$ with ethanal and nothing with propanone. (Note: both ethanal and propanone also give a positive iodoform test, so iodoform alone cannot tell them apart — only an oxidation test can.)
Benzaldehyde and acetophenone ($\ce{C6H5COCH3}$) are both colourless liquids. How do you distinguish them?
Use Tollens' reagent. Benzaldehyde, being an aldehyde, gives a silver mirror; acetophenone, a ketone, gives no mirror. Fehling's would not help separate them here, because benzaldehyde (aromatic aldehyde) generally fails Fehling's just as the ketone does. So Tollens' is the decisive test.
$$\ce{C6H5CHO + 2[Ag(NH3)2]+ + 3OH- -> C6H5COO- + 2Ag v + 4NH3 + 2H2O}$$
A sample gives a silver mirror with Tollens' but no brick-red precipitate with Fehling's. Identify the class.
Positive Tollens' rules out a ketone and confirms an aldehyde. A negative Fehling's points to an aromatic aldehyde such as benzaldehyde, since aliphatic aldehydes would have given the brick-red $\ce{Cu2O}$ as well.
Tollens' reagent must be freshly prepared.
On standing, ammoniacal silver nitrate can form explosive silver nitride ("fulminating silver"). Examiners often test the procedural fact that the reagent is made fresh and not stored, and that the test tube must be scrupulously clean for the mirror to form.
Remember: Tollens' = ammoniacal AgNO₃, prepared fresh; warm gently, do not boil hard.
Special cases & reducing sugars
A few species behave as aldehydes even though they are not simple $\ce{RCHO}$ molecules, and they recur in NEET options:
| Species | Behaviour | Reason |
|---|---|---|
| Formic acid, $\ce{HCOOH}$ | Positive Tollens' and Fehling's | Retains an aldehydic $\ce{-CHO}$-like H on the carbon; acts as a reducing agent |
| Glucose, fructose | Positive (reducing sugars) | Free or potential carbonyl that can open to an aldehyde/α-hydroxy ketone form |
| Maltose, lactose | Positive (reducing sugars) | Have a free anomeric –OH that opens to a reducing aldehyde |
| Sucrose | Negative (non-reducing) | Both anomeric carbons are locked in the glycosidic bond |
Note that fructose is a ketose yet still reduces Tollens' and Fehling's: under the basic conditions of the test it isomerises to glucose/mannose (an aldose), which then reduces the reagent. This is why "ketone → always negative" must be read carefully — it applies to ordinary ketones, not to α-hydroxy ketones such as reducing sugars in alkaline medium.
Tollens' & Fehling's in one screen
- Why: aldehydes have a $\ce{C-H}$ on the carbonyl and act as reducing agents; ketones lack it and do not react with mild oxidants.
- Tollens': ammoniacal $\ce{AgNO3}$, $\ce{[Ag(NH3)2]+}$, prepared fresh → silver mirror; positive for all aldehydes (aliphatic and aromatic), negative for ketones.
- Fehling's: A = $\ce{CuSO4}$, B = Rochelle salt + $\ce{NaOH}$ → brick-red $\ce{Cu2O}$; positive for aliphatic aldehydes, generally negative for aromatic aldehydes and ketones.
- Benedict's: citrate-based one-bottle analogue of Fehling's; same brick-red result, preferred for reducing sugars.
- Redox: aldehyde is oxidised to the carboxylate; the metal ion ($\ce{Ag+}$ or $\ce{Cu^2+}$) is reduced and reports the result.
- Watch: formic acid and reducing sugars are positive; sucrose is negative; benzaldehyde is positive to Tollens' but not Fehling's.