Aldehydes and ketones both contain the carbonyl group (C=O). CBSE Class 12 and NEET cover them in detail, with preparation, reactions and distinguishing tests. Expect two questions a year on this topic.
Core Concepts
Structure
Aldehyde has C=O with at least one H attached to the carbonyl carbon (R-CHO). Ketone has C=O with two alkyl groups (R-CO-R’). Both are planar at the carbonyl carbon (sp2).
The carbonyl group is polar: oxygen is more electronegative, creating a partial negative charge on O and partial positive charge on C. This polarity makes the carbonyl carbon electrophilic — the site of nucleophilic attack in most reactions.
| Property | Explanation |
|---|---|
| Boiling point | Higher than alkanes (dipole-dipole), lower than alcohols (no H-bonding between carbonyl molecules) |
| Solubility | Lower members soluble in water (H-bond with water), solubility decreases with chain length |
| Smell | Lower aldehydes have pungent smell; higher ones and ketones have pleasant/fruity smell |
Preparation of aldehydes
Oxidation of primary alcohols (controlled, to prevent further oxidation to acid). Ozonolysis of alkenes. Rosenmund reduction of acyl chlorides. Stephen reduction of nitriles.
| Method | Reagent | Equation |
|---|---|---|
| Oxidation of 1° alcohol | PCC in CH2Cl2 | |
| Rosenmund reduction | H2/Pd on BaSO4 (poisoned) | |
| Stephen reduction | SnCl2/HCl, then H2O | |
| Ozonolysis | O3 then Zn/H2O | Alkene → two carbonyl fragments |
The Rosenmund reduction uses a “poisoned” catalyst — Pd on BaSO4 with added sulphur or quinoline. The poison prevents the aldehyde from being reduced further to the alcohol. This controlled reduction is a classic NEET concept.
Preparation of ketones
Oxidation of secondary alcohols. From acyl chlorides and organocadmium compounds. Friedel-Crafts acylation for aromatic ketones.
Friedel-Crafts acylation:
This is preferred over Friedel-Crafts alkylation because: (1) no carbocation rearrangement, (2) the ketone product deactivates the ring, preventing polysubstitution.
Reactions — Nucleophilic addition
The carbonyl carbon is electrophilic. Nucleophiles attack the carbon, breaking the pi bond and forming a tetrahedral intermediate.
| Nucleophile | Reagent | Product | Name |
|---|---|---|---|
| CN- | HCN | Cyanohydrin (R-CH(OH)CN) | Cyanohydrin formation |
| HSO3- | NaHSO3 | Bisulphite adduct | Bisulphite addition |
| NH2OH | Hydroxylamine | Oxime (R-CH=NOH) | Oximation |
| NH2-NH2 | Hydrazine | Hydrazone (R-CH=NNH2) | Hydrazone formation |
| 2,4-DNP | 2,4-Dinitrophenylhydrazine | 2,4-DNP derivative (yellow/orange ppt) | Brady’s test |
| NaBH4 | Sodium borohydride | Alcohol (R-CH2OH) | Reduction |
| RMgX | Grignard reagent | Alcohol (after hydrolysis) | Grignard addition |
Aldol condensation — the most important C-C bond forming reaction:
The first product (aldol) is a beta-hydroxy aldehyde. On heating, it dehydrates to form an alpha-beta unsaturated aldehyde (crotonaldehyde). This reaction requires at least one alpha-hydrogen.
Cannizzaro reaction — for aldehydes WITHOUT alpha-hydrogen:
One molecule is oxidised to the acid; the other is reduced to the alcohol. This is a disproportionation reaction.
The aldol vs Cannizzaro distinction is a favourite NEET/JEE question: “Which reaction will this aldehyde undergo — aldol or Cannizzaro?” Check for alpha-hydrogens. If present → aldol. If absent (formaldehyde, benzaldehyde) → Cannizzaro.
Distinguishing tests
Tollen’s test — aldehydes give silver mirror; ketones do not. Fehling’s test — aldehydes give red precipitate of Cu2O; ketones do not. Iodoform test — methyl ketones and ethanol give yellow precipitate.
2,4-DNP test (Brady’s reagent): Both aldehydes AND ketones give a yellow/orange precipitate with 2,4-dinitrophenylhydrazine. This test confirms the presence of a carbonyl group but does not distinguish aldehyde from ketone.
Oxidation comparison
| Compound | Tollen’s | Fehling’s | KMnO4 |
|---|---|---|---|
| Aliphatic aldehyde | Positive (silver mirror) | Positive (red ppt) | Oxidised to acid |
| Aromatic aldehyde (benzaldehyde) | Positive | Negative | Oxidised to acid |
| Ketone | Negative | Negative | No reaction (mild) |
| Formic acid | Positive | Positive | Oxidised to CO2 |
Worked Examples
In ketones, two alkyl groups donate electrons via +I effect, reducing the positive charge on carbonyl carbon. Aldehydes have only one alkyl group (or none), so the carbon is more electrophilic and more reactive.
Two molecules of an aldehyde or ketone combine in the presence of dilute alkali to give a beta-hydroxy aldehyde or ketone, which may dehydrate to an alpha-beta unsaturated product. Classic C-C bond formation reaction.
If acetaldehyde and formaldehyde are mixed with dilute NaOH, formaldehyde (no alpha-H) cannot form an enolate. Only acetaldehyde forms an enolate, which attacks formaldehyde. The product is 3-hydroxypropanal. In practice, crossed aldols with two different enolisable aldehydes give mixtures.
Grignard reagent adds to formaldehyde → primary alcohol. To aldehyde → secondary alcohol. To ketone → tertiary alcohol. The product class depends on the carbonyl starting material.
Both convert C=O to CH2 (complete removal of oxygen). Clemmensen uses Zn-Hg/conc. HCl (acidic conditions). Wolff-Kishner uses NH2NH2/KOH (basic conditions). Use Clemmensen when the molecule is base-sensitive, Wolff-Kishner when it is acid-sensitive.
Common Mistakes
Writing that ketones give Tollen’s test. They do not — only aldehydes.
Confusing aldol with Cannizzaro. Aldol needs alpha hydrogens; Cannizzaro is for aldehydes without alpha hydrogens.
Saying aldehydes are harder to oxidise than ketones. Aldehydes oxidise easily; ketones do not oxidise under mild conditions.
Forgetting that benzaldehyde fails Fehling’s test. It gives Tollen’s (positive) but not Fehling’s (negative). The bulky tartrate complex in Fehling’s solution has steric issues with aromatic aldehydes.
Writing that 2,4-DNP test distinguishes aldehyde from ketone. It does not — both give a precipitate. It only confirms the presence of C=O. Use Tollen’s or Fehling’s to distinguish.
Exam Weightage and Revision
Aldehydes and Ketones carry 2-3 NEET questions per year. The most tested: distinguishing tests (Tollen’s, Fehling’s, iodoform), aldol vs Cannizzaro, and nucleophilic addition reactions. JEE also tests crossed aldol, Grignard additions, and selectivity of reducing agents.
| Question Type | NEET Frequency | JEE Frequency |
|---|---|---|
| Distinguishing tests | Every year | Most years |
| Aldol vs Cannizzaro | Most years | Every year |
| Nucleophilic addition product | Most years | Every year |
| Grignard with carbonyl | Every 2 years | Every year |
| Oxidation comparison | Most years | Most years |
The two most reliable NEET questions: (1) “Which gives a positive Tollen’s/Fehling’s test?” and (2) “Which undergoes aldol condensation?” For (1), check if it is an aldehyde. For (2), check for alpha-hydrogens.
Practice Questions
Q1. Compound A (C3H6O) gives a silver mirror with Tollen’s test and does not give iodoform test. Identify A.
Positive Tollen’s → aldehyde. No iodoform → does not have CH3CO- group. C3H6O aldehyde without methyl adjacent to C=O: propanal (CH3CH2CHO). Propanal is an aldehyde (Tollen’s positive) but does not have a CH3CO- group (iodoform negative). Acetone would give iodoform but not Tollen’s.
Q2. Why does formaldehyde undergo Cannizzaro reaction but acetaldehyde does not?
Cannizzaro reaction requires an aldehyde with no alpha-hydrogens. Formaldehyde (HCHO) has no carbon adjacent to the carbonyl, so no alpha-H. Acetaldehyde (CH3CHO) has three alpha-hydrogens on the methyl group, so it undergoes aldol condensation instead.
Q3. What happens when acetone is treated with CH3MgBr followed by hydrolysis?
Grignard reagent (CH3MgBr) attacks the electrophilic carbonyl carbon of acetone, forming an alkoxide intermediate. Hydrolysis gives 2-methylpropan-2-ol (tert-butyl alcohol). Grignard + ketone → tertiary alcohol.
FAQs
Why are aldehydes easier to oxidise than ketones?
Aldehydes have an H atom on the carbonyl carbon that can be removed as a hydrogen atom during oxidation. Ketones have only C-C bonds on the carbonyl carbon — these are much harder to break. Mild oxidising agents (Tollen’s, Fehling’s) can attack the C-H bond of aldehydes but cannot break the C-C bonds of ketones.
What is the difference between reduction and Clemmensen/Wolff-Kishner reduction?
Ordinary reduction (NaBH4, LiAlH4) converts C=O to C-OH (alcohol). Clemmensen and Wolff-Kishner convert C=O to CH2 (complete removal of oxygen). These are deoxygenation reactions, useful when you want to reduce a carbonyl all the way to a methylene group.
Can ketones undergo Cannizzaro reaction?
No. Cannizzaro requires an aldehyde hydrogen (the H on the CHO group) to be transferred. Ketones have no such hydrogen and cannot undergo this reaction.
Memorise three distinguishing tests — Tollen’s, Fehling’s, Iodoform. NEET asks one of these every year.
Reactivity Trends in Carbonyl Chemistry
Understanding why certain carbonyl compounds are more reactive helps you predict outcomes without memorising every reaction individually.
Why formaldehyde is the most reactive aldehyde
Formaldehyde (HCHO) has no alkyl groups on the carbonyl carbon — no +I effect to reduce the positive charge. It is the most electrophilic, smallest (least steric hindrance), and most reactive carbonyl compound. It undergoes Cannizzaro, crossed aldol, and polymerisation reactions that other aldehydes do not.
Why aromatic aldehydes are less reactive than aliphatic aldehydes
In benzaldehyde, the phenyl group delocalises electron density into the carbonyl carbon via resonance, reducing its positive character. Additionally, the bulky phenyl group creates steric hindrance for nucleophilic attack. This is why benzaldehyde fails Fehling’s test while aliphatic aldehydes pass.
Carbonyl reactivity order
Each step to the right adds either electron donation (+I effect of alkyl groups) or steric hindrance (bigger groups), or both, making the carbonyl less electrophilic.
Cross-aldol condensation — controlling the product
When two different carbonyl compounds are mixed with base, multiple products can form (self-aldol of each + crossed aldol). To get a specific crossed aldol product:
- Use one compound with no alpha-H (formaldehyde, benzaldehyde) — it cannot self-condense
- The other compound forms the enolate and attacks the first
- This gives a single crossed product
Example: benzaldehyde + acetaldehyde → cinnamaldehyde (C6H5CH=CHCHO) — used in cinnamon flavouring and perfume industry.
Carbonyls are the most reactive functional group in a typical organic course. Track the electrophilic carbon and most reactions explain themselves.