JEE Mains Chemistry · Aldehydes, Ketones and Carboxylic Acids

Oxidation and Identification Tests

Tollens' and Fehling's reagents detect aldehydes by oxidising them, 2,4-DNP detects any aldehyde or ketone, and the iodoform test detects a CH₃CO or CH₃CH(OH) group.

All Aldehydes, Ketones and Carboxylic Acids notes JEE Mains Chemistry strategy

Why this matters

Twenty-three PYQs, four numerical, seven from 2026, more recent questions than any other page. Twelve use Tollens', Fehling's or 2,4-DNP to tell compounds apart; eleven ask which compounds give the iodoform test, or what it produces.

Concept 1 of 2: Tollens', Fehling's and the 2,4-DNP test

An aldehyde has an H on its carbonyl carbon, so mild oxidants turn it into an acid; a ketone has no such H and resists. Tollens' reagent is strong enough for every aldehyde. Fehling's is weaker and misses aromatic aldehydes. 2,4-DNP does not oxidise anything: it condenses with any C=O of an aldehyde or ketone.

Definition

Tollens' reagent, ammoniacal silver nitrate $\mathrm{[Ag(NH_3)_2]^+}$ : the aldehyde is oxidised to the carboxylate and silver is deposited as a mirror.
Fehling's solution ( $\mathrm{Cu^{2+}}$ with tartrate in alkali) and Benedict's solution ( $\mathrm{Cu^{2+}}$ with citrate): aliphatic aldehydes reduce $\mathrm{Cu^{2+}}$ to a red-brown precipitate of $\mathrm{Cu_2O}$ . Aromatic aldehydes do not.
An α-hydroxy ketone ( $\mathrm{RCOCH_2OH}$ , as in fructose) passes both tests, although it is a ketone. Methanoic acid, which contains an aldehyde-like C–H, passes Tollens'.
2,4-DNP gives a yellow, orange or red precipitate with any aldehyde or ketone. It cannot tell the two apart. The C=O of acids, esters and amides does not react.
To tell an aldehyde from a ketone of the same formula (propanal from propanone), use Tollens' or Fehling's, never 2,4-DNP.

Test and reagent	Positive sign	Positive for	Negative for
Tollens': $\mathrm{[Ag(NH_3)_2]^+}$ , $\mathrm{OH^-}$	Silver mirror	All aldehydes, aliphatic and aromatic; methanoic acid; α-hydroxy ketones; reducing sugars	Simple ketones; carboxylic acids other than methanoic acid
Fehling's: $\mathrm{Cu^{2+}}$ , tartrate, NaOH	Red-brown precipitate of $\mathrm{Cu_2O}$	Aliphatic aldehydes; α-hydroxy ketones such as fructose	Aromatic aldehydes; simple ketones
Benedict's: $\mathrm{Cu^{2+}}$ , citrate, $\mathrm{Na_2CO_3}$	Red-brown precipitate of $\mathrm{Cu_2O}$	Aliphatic aldehydes; α-hydroxy ketones such as fructose	Aromatic aldehydes; simple ketones
2,4-DNP	Yellow, orange or red precipitate	Any aldehyde or ketone	Carboxylic acids, esters, amides, alcohols, ethers
Iodoform: $\mathrm{I_2}$ , NaOH	Yellow precipitate of $\mathrm{CHI_3}$	$\mathrm{CH_3CO{-}}$ on C or H; $\mathrm{CH_3CH(OH){-}}$	Ketones and alcohols without these groups; acetic acid and its esters
$\mathrm{NaHCO_3}$ solution	Effervescence of $\mathrm{CO_2}$	Carboxylic acids; picric acid	Aldehydes, ketones, alcohols, most phenols

Tollens' catches every aldehyde; Fehling's catches only aliphatic ones.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 29 January 2024 · Q59Moderate

Example 1 · Aldehydes, Ketones and Carboxylic Acids · Oxidation and Identification Tests

From the compounds given below, number of compounds which give positive Fehling's test is ____ . Benzaldehyde, Acetaldehyde, Acetone, Acetophenone, Methanal, 4-nitrobenzaldehyde, cyclohexane carbaldehyde.

Aromatic aldehydes fail Fehling's but pass Tollens'

Benzaldehyde and its ring-substituted relatives give a silver mirror but no red precipitate. Counting 'aldehydes' is not enough for a Fehling's count.

2,4-DNP does not separate aldehydes from ketones

Both give the orange precipitate. The test only shows that a C=O of an aldehyde or ketone is present.

Drill 11 more on tollens', fehling's and the 2,4-dnp test

Concept 2 of 2: The iodoform test and the haloform reaction

Hypoiodite in base replaces all three hydrogens of a methyl group next to a C=O by iodine. The

\mathrm{CI_3}

group then leaves as iodoform,

\mathrm{CHI_3}

, a yellow solid, and the rest becomes a carboxylate with one carbon fewer. Hypoiodite is also an oxidant, so an alcohol that it can oxidise to such a methyl ketone passes too.

Definition

Positive: $\mathrm{CH_3CO{-}}$ joined to H or C, so ethanal and every methyl ketone, including aryl methyl ketones and α,β-unsaturated methyl ketones.
Also positive: $\mathrm{CH_3CH(OH){-}}$ joined to H or C, so ethanol and every 2° alcohol with a methyl on the carbinol carbon. They are first oxidised to the methyl ketone.
Negative: methanol; methanoic acid; acetic acid, its esters and amides (the $\mathrm{CH_3CO}$ is joined to O or N); 3° alcohols; ketones and alcohols with no methyl next to the C=O or CHOH (pentan-3-one, pentan-3-ol).
The reagent is $\mathrm{I_2}$ with NaOH, or KI with NaOCl, which also gives hypoiodite $\mathrm{OI^-}$ .

Haloform reaction of a methyl ketone

\mathrm{RCOCH_3 + 3I_2 + 4NaOH \to RCOONa + CHI_3\downarrow + 3NaI + 3H_2O}

Worked example

How many of these give the iodoform test: methanol, propan-1-ol, 1-phenylethanol, 3-methylbutan-2-one, propanoic acid, 2-methylpropan-2-ol, hexan-2-ol, ethyl ethanoate?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 3 Apr 2025 · Q36Moderate

Example 2 · Aldehydes, Ketones and Carboxylic Acids · Oxidation and Identification Tests

Number of molecules from below which cannot give iodoform reaction is :Ethanol, Isopropyl alcohol, Bromoacetone, 2-Butanol, 2-Butanone, Butanal, 2-Pentanone, 3-Pentanone, Pentanal and 3-Pentanol

Acetic acid and its esters are negative

They contain

\mathrm{CH_3CO}

, but it is joined to oxygen. The test needs

\mathrm{CH_3CO}

joined to carbon or hydrogen.

Ethanol and ethanal are the only positives in their classes

Ethanol is the only 1° alcohol that gives the test and ethanal the only aldehyde. Methanol and methanal are both negative.

Drill 10 more on the iodoform test and the haloform reaction

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 (1)

The iodoform test and the haloform reaction
Haloform reaction of a methyl ketone
$\mathrm{RCOCH_3 + 3I_2 + 4NaOH \to RCOONa + CHI_3\downarrow + 3NaI + 3H_2O}$

Reference tables (1)

Tollens', Fehling's and the 2,4-DNP test6 rows

Test and reagent	Positive sign	Positive for	Negative for
Tollens': $\mathrm{[Ag(NH_3)_2]^+}$ , $\mathrm{OH^-}$	Silver mirror	All aldehydes, aliphatic and aromatic; methanoic acid; α-hydroxy ketones; reducing sugars	Simple ketones; carboxylic acids other than methanoic acid
Fehling's: $\mathrm{Cu^{2+}}$ , tartrate, NaOH	Red-brown precipitate of $\mathrm{Cu_2O}$	Aliphatic aldehydes; α-hydroxy ketones such as fructose	Aromatic aldehydes; simple ketones
Benedict's: $\mathrm{Cu^{2+}}$ , citrate, $\mathrm{Na_2CO_3}$	Red-brown precipitate of $\mathrm{Cu_2O}$	Aliphatic aldehydes; α-hydroxy ketones such as fructose	Aromatic aldehydes; simple ketones
2,4-DNP	Yellow, orange or red precipitate	Any aldehyde or ketone	Carboxylic acids, esters, amides, alcohols, ethers
Iodoform: $\mathrm{I_2}$ , NaOH	Yellow precipitate of $\mathrm{CHI_3}$	$\mathrm{CH_3CO{-}}$ on C or H; $\mathrm{CH_3CH(OH){-}}$	Ketones and alcohols without these groups; acetic acid and its esters
$\mathrm{NaHCO_3}$ solution	Effervescence of $\mathrm{CO_2}$	Carboxylic acids; picric acid	Aldehydes, ketones, alcohols, most phenols

Tollens' catches every aldehyde; Fehling's catches only aliphatic ones.

Watch out for (4)

Aromatic aldehydes fail Fehling's but pass Tollens'
→ Tollens', Fehling's and the 2,4-DNP test
2,4-DNP does not separate aldehydes from ketones
→ Tollens', Fehling's and the 2,4-DNP test
Acetic acid and its esters are negative
→ The iodoform test and the haloform reaction
Ethanol and ethanal are the only positives in their classes
→ The iodoform test and the haloform reaction

Test yourself on Aldehydes, Ketones and Carboxylic Acids

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