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Aldehydes and ketones carry more named reactions than almost any other chapter, and that is why students find it heavy. But nearly all of those reactions come from two features of the carbonyl group: an electrophilic carbon, and acidic hydrogens on the carbon next to it. This post builds the chapter on those two ideas, works through the reactions that appear most often, and gives you a decision list for identification questions.
What do the syllabi include?
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus list the nature of the carbonyl group, nucleophilic addition, relative reactivity of aldehydes and ketones, addition of HCN, NH₃ and its derivatives, Grignard reagents, oxidation, Clemmensen and Wolff-Kishner reductions, acidity of the α-hydrogen, aldol condensation, the Cannizzaro and haloform reactions, and chemical tests to tell aldehydes from ketones. The JEE (Advanced) 2026 brochure adds preparations from acid chlorides, nitriles and esters and reactions with NaHSO₃, alcohols and amines. The NCERT Class 12 chapter on aldehydes, ketones and carboxylic acids is the base text. Check the current year's syllabus.
What is nucleophilic addition?
Oxygen is more electronegative than carbon, so the C=O bond is polar: the carbon carries a partial positive charge. A nucleophile attacks that carbon, the π electrons move onto oxygen, and the flat sp² carbon becomes a tetrahedral sp³ carbon. A proton then adds to the oxygen.
With HCN, the product is a cyanohydrin.
With a Grignard reagent RMgX, followed by water, the product is an alcohol: methanal gives a primary alcohol, other aldehydes give secondary alcohols, and ketones give tertiary alcohols.
With ammonia derivatives (hydroxylamine, hydrazine, 2,4-DNP), addition is followed by loss of water to give C=N compounds such as oximes and hydrazones.
Why aldehydes react faster. NCERT gives two reasons. Ketones have two relatively large groups that block the nucleophile, and two alkyl groups reduce the positive charge on the carbonyl carbon more than one does. The order is methanal > ethanal > propanone. Benzaldehyde is less reactive than propanal, because the ring's resonance reduces the polarity of its carbonyl group.
Why does the α-hydrogen matter?
A hydrogen on the carbon next to C=O is weakly acidic, because the anion left behind (the enolate) spreads its charge onto the oxygen. That anion is itself a nucleophile, and it drives the aldol reaction.
Worked example 1: aldol condensation of ethanal
NCERT defines the aldol reaction as the reaction of aldehydes or ketones with at least one α-hydrogen in dilute alkali, giving a β-hydroxy aldehyde or ketone, which loses water to give an α,β-unsaturated compound.
Hydroxide removes an α-hydrogen from one ethanal molecule to give an enolate.
The enolate attacks the carbonyl carbon of a second ethanal molecule.
Protonation gives 3-hydroxybutanal, CH₃CH(OH)CH₂CHO, the "aldol".
On heating, it loses water to give but-2-enal, CH₃CH=CHCHO.
Cross aldol. With two different aldehydes that both have α-hydrogens, NCERT notes that you get four products. The count is simple: either compound can form the enolate, and either can be attacked, so 2 × 2 = 4.
Worked example 2: when there is no α-hydrogen
Methanal, benzaldehyde and 2,2-dimethylpropanal have no α-hydrogen, so they cannot form an enolate. With concentrated alkali they undergo the Cannizzaro reaction instead: one molecule is reduced to an alcohol and another is oxidised to a carboxylate salt.
2HCHO + KOH (concentrated) → CH₃OH + HCOOK
Benzaldehyde gives benzyl alcohol and a benzoate salt in the same way.
The common mistake: writing an aldol product for benzaldehyde or methanal alone. Check for an α-hydrogen first, every time.
How do the tests tell aldehydes from ketones?
NCERT lists three:
Tollens' test: warming an aldehyde with ammoniacal silver nitrate gives a silver mirror.
Fehling's test: heating an aldehyde with Fehling's reagent gives a reddish-brown precipitate. Aromatic aldehydes such as benzaldehyde do not respond.
Iodoform test: compounds with a CH₃CO group, or a CH₃CH(OH) group that oxidises to one, give a yellow precipitate of iodoform with sodium hypoiodite. This picks out methyl ketones, ethanal and ethanol.
Worked example 3: identifying C₃H₆O
A compound C₃H₆O gives a yellow precipitate with iodine and sodium hydroxide but does not reduce Tollens' reagent. What is it?
C₃H₆O with a carbonyl group is either propanal (CH₃CH₂CHO) or propanone (CH₃COCH₃).
No reaction with Tollens' reagent rules out the aldehyde.
A positive iodoform test needs CH₃CO, which propanone has and propanal does not.
The compound is propanone.
Which reductions should you know?
NCERT gives two ways to turn C=O into CH₂: Clemmensen reduction with zinc amalgam and concentrated HCl, and Wolff-Kishner reduction with hydrazine followed by heating with KOH or NaOH in ethylene glycol. Sodium borohydride or lithium aluminium hydride stop at the alcohol. Choose Clemmensen for acid-stable molecules and Wolff-Kishner for base-stable ones.
Why are nucleophiles and electrophiles hard to spot?
Anzovino and Bretz (2015) studied second-semester organic chemistry students and found they relied mainly on charges to identify nucleophiles and electrophiles, and often needed to know the mechanism before they could tell which was which. The carbonyl carbon is the test case: it has no formal charge, but it is an electrophile because of the bond's polarity. Practise naming both partners before drawing arrows. Our post on learning organic mechanisms instead of lists has more on this.
If you get stuck partway through a mechanism, Fermi's AI tutor notices where you are stuck and decides whether to explain, ask or show, so you work through the step instead of reading the answer.
A practice routine for carbonyl chemistry
Write the nucleophilic addition mechanism with HCN and with a Grignard reagent from memory.
For ten compounds, mark α-hydrogens and predict aldol, Cannizzaro or neither.
Do five identification questions using Tollens', Fehling's and iodoform tests.
Make a one-line card for each named reaction.
Mix in five conversions that also use alcohols; see our guide to alcohols, phenols and ethers.
Today, write the aldol product of propanal and say whether it gives a positive iodoform test.
