Aldehydes and Ketones
Aldehydes (R–CHO) and ketones (R–CO–R′) share the carbonyl (>C=O) group, one of the most reactive functional groups in organic chemistry. The AAMC MCAT content outline expects you to name them, predict their nucleophilic addition products, distinguish them with Tollens and Fehling reagents, and reduce them to alcohols. This is one of the highest-yield organic chapters.
Nomenclature and Structure
The carbonyl carbon is sp2-hybridized, planar, and bears a strong dipole (δ+C=Oδ−). The C=O π bond is polarized: C is electrophilic, O is nucleophilic. This polarization drives every reaction in the chapter.
Aldehyde nomenclature
- Replace the –e of the parent alkane with –al.
- The –CHO carbon is always C-1.
- Examples: HCHO = methanal (formaldehyde), CH3CHO = ethanal (acetaldehyde), CH3CH2CHO = propanal, C6H5CHO = benzaldehyde.
Ketone nomenclature
- Replace the –e of the parent alkane with –one. Number the chain so that the carbonyl carbon has the lowest locant.
- Examples: CH3COCH3 = propan-2-one (acetone), CH3COCH2CH3 = butan-2-one, C6H5COCH3 = acetophenone (1-phenylethan-1-one).
Preparation
Two of the most reliable preparation routes appear repeatedly in MCQs.
1° alcohol + [O] (acidified K2Cr2O7 or PCC) → aldehyde → further oxidation → carboxylic acid.
2° alcohol + [O] → ketone (no further oxidation under mild conditions).
Example: CH3CH2OH ⟶{K2Cr2O7/H+} CH3CHO ⟶{[O]} CH3COOH.
R–CH=CH–R′ ⟶{O3; Zn/H2O} R–CHO + R′–CHO. The position of the double bond is revealed by the carbonyl fragments produced.
HC≡CH + H2O ⟶{HgSO4/H2SO4} CH3CHO (via an enol that tautomerizes to the carbonyl). Markovnikov hydration of terminal alkynes gives methyl ketones; hydroboration–oxidation of a terminal alkyne gives the aldehyde instead.
Nucleophilic Addition Reactions
The defining reactivity of carbonyl compounds. A nucleophile attacks the δ+ carbonyl carbon; the π electrons shift to oxygen, which is then protonated. Aldehydes are more reactive than ketones because (i) their carbonyl carbon is less hindered and (ii) only one electron-donating alkyl group is present (so C remains more δ+).
R–CHO + HCN ⟶{CN−} R–CH(OH)CN. The cyanohydrin extends the carbon chain by one and is hydrolyzed to an α-hydroxy acid.
One equivalent of alcohol adds to the carbonyl to give a hemiacetal (a C bearing both –OH and –OR). Under acid catalysis a second alcohol replaces the –OH to give an acetal (C bearing two –OR) plus water:
R–CHO + 2 R′OH ⟶{H+} R–CH(OR′)2 + H2O.
The whole sequence is reversible — acetals hydrolyze back to the carbonyl in aqueous acid, so acetals are widely used as protecting groups for aldehydes and ketones. Cyclic hemiacetals form intramolecularly in sugars (e.g. glucose).
R′MgX + HCHO → (after H3O+) R′CH2OH (1° alcohol).
R′MgX + R–CHO → 2° alcohol.
R′MgX + R–CO–R″ → 3° alcohol. The carbanion-like R′ is the nucleophile that forms a new C–C bond.
A primary amine (R–NH2) condenses with a carbonyl, losing water, to give an imine (Schiff base), C=N–R. A secondary amine (R2NH) cannot form a stable C=N, so it loses water toward the α-carbon instead, giving an enamine (C=C–NR2). Both form fastest near pH 4–5. Related derivatives: + NH2OH → oxime; + 2,4-dinitrophenylhydrazine → an orange/yellow 2,4-dinitrophenylhydrazone, a classic test for any >C=O.
Alpha-Carbon Chemistry
The hydrogens on the carbon adjacent to the carbonyl (the α-carbon) are weakly acidic (pKa ≈ 20) because the resulting enolate is resonance-stabilized — the negative charge is shared between the α-carbon and the carbonyl oxygen. This acidity underlies tautomerism and the aldol reaction.
A carbonyl compound with at least one α-H is in equilibrium with its enol form (C=C–OH), interconverting by moving an α-H to the carbonyl oxygen. The keto form is normally far more stable and predominates. Tautomers are constitutional isomers (a bond is broken and made), not resonance structures. Acid or base catalyzes the interconversion.
Under dilute base, the enolate of one carbonyl attacks the carbonyl carbon of another, forming a new C–C bond and a β-hydroxy aldehyde/ketone (the "aldol"). On heating, the aldol undergoes dehydration (loss of water) to give an α,β-unsaturated carbonyl — the condensation product. Example: 2 CH3CHO → CH3CH(OH)CH2CHO ⟶{Δ} CH3CH=CHCHO. A carbonyl with no α-H (HCHO, C6H5CHO) cannot form an enolate and instead undergoes the Cannizzaro (disproportionation) reaction with concentrated base.
Oxidation Reactions
The single sharpest distinction between aldehydes and ketones — aldehydes oxidize easily to carboxylic acids, ketones do not (under normal conditions).
Reagent: ammoniacal AgNO3 = [Ag(NH3)2]+.
R–CHO + 2[Ag(NH3)2]+ + 3 OH− → R–COO− + 2 Ag↓ + 4 NH3 + 2 H2O.
A bright silver mirror confirms an aldehyde. Ketones give no mirror.
Reagent: alkaline Cu2+ tartrate complex (Fehling A + B).
R–CHO + 2 Cu2+ + 5 OH− → R–COO− + Cu2O↓ (red) + 3 H2O.
Aliphatic aldehydes give a brick-red Cu2O precipitate; aromatic aldehydes (benzaldehyde) and ketones do not.
CH3CHO and methyl ketones (CH3CO–R) + I2/NaOH → CHI3↓ (yellow crystals) + RCOO−Na+. Diagnostic for the CH3CO– or CH3CH(OH)– fragment.
| Test | Reagent | Aliphatic CHO | Aromatic CHO | Ketones |
|---|---|---|---|---|
| Tollens' | [Ag(NH3)2]+ (ammoniacal AgNO3) | Silver mirror | Silver mirror (still positive) | No reaction |
| Fehling's | Alkaline Cu2+ tartrate | Brick-red Cu2O | No reaction (key distinguisher) | No reaction |
| Benedict's | Alkaline Cu2+ citrate | Brick-red Cu2O | No reaction | No reaction |
| Iodoform | I2 + NaOH | CH3CHO only → yellow CHI3 | No reaction | Only methyl ketones → yellow CHI3 |
| 2,4-DNP | 2,4-dinitrophenylhydrazine | Yellow/orange precipitate | Yellow/orange precipitate | Yellow/orange precipitate |
Use Tollens to distinguish aldehyde from ketone; use Fehling to distinguish aliphatic from aromatic aldehyde; use 2,4-DNP to confirm any carbonyl group.
Reduction to Alcohols
The reverse of oxidation: hydride is delivered to the carbonyl carbon, giving an alkoxide that is protonated on workup to an alcohol.
R–CHO + NaBH4 ⟶{MeOH/H2O} R–CH2OH (1° alcohol).
R–CO–R′ + NaBH4 → R–CH(OH)–R′ (2° alcohol). Does not reduce esters or carboxylic acids — useful when those groups must be preserved.
Reduces aldehydes, ketones, esters, amides, and carboxylic acids all the way to alcohols (or amines). Used in dry ether under inert atmosphere — reacts violently with water.
R–CHO + H2 ⟶{Ni or Pt, heat/pressure} R–CH2OH. Reduces C=C bonds simultaneously, so it's not selective for the carbonyl.
Reduces >C=O all the way to >CH2:
Clemmensen: Zn(Hg)/conc. HCl (acidic conditions).
Wolff–Kishner: NH2NH2, KOH, ethylene glycol, heat (basic conditions).
Reactivity and Comparison
Aldehydes and ketones share the carbonyl group but differ in steric and electronic environment, leading to four key practical differences.
- Reactivity in nucleophilic addition: Aldehyde > ketone (less steric hindrance; one fewer +I alkyl group).
- Oxidation by Tollens/Fehling/K2Cr2O7: Aldehyde yes; ketone no (under mild conditions).
- Iodoform test: Positive for ethanal and all methyl ketones; negative for other aldehydes/ketones.
- 2,4-DNP: Both give orange/yellow precipitates — confirms a carbonyl group but does not differentiate the two.
- Aldol condensation: Both undergo it if they have an α-H. Compounds without α-H (HCHO, C6H5CHO, (CH3)3CCHO) instead undergo Cannizzaro reaction with strong base.
Worked MCQs
Five MCQs that capture the high-yield testing patterns for this chapter. Read the explanation even when you get the answer right — that's where the deeper concept lives.
Q1. Which of the following gives a silver mirror with Tollens reagent?
Tollens reagent oxidizes aldehydes to carboxylates and reduces Ag+ to metallic silver. Propanal (CH3CH2CHO) is an aldehyde; acetone is a ketone, ether is non-reactive, and methanol is an alcohol.
Q2. Reduction of butan-2-one with NaBH4 in methanol gives:
NaBH4 delivers a hydride to the carbonyl carbon; the alkoxide is protonated by methanol on workup. Butan-2-one (CH3COCH2CH3) becomes the secondary alcohol butan-2-ol (CH3CH(OH)CH2CH3).
Q3. Which compound does NOT give a positive iodoform test?
The iodoform test is positive for compounds containing a CH3CO– group or a CH3CH(OH)– group. Propanal (CH3CH2CHO) has neither — its α-carbon is –CH2–, not –CH3.
Q4. Aldehydes are more reactive than ketones in nucleophilic addition because:
Two effects work together: an aldehyde has only one alkyl group donating +I electrons (vs two in a ketone), so its carbonyl C is more δ+; and the H atom takes up less space than a second alkyl group, so the nucleophile approaches more easily.
Q5. Treating an aldehyde with excess ethanol and a trace of acid produces:
The first alcohol adds to give a hemiacetal; under acid catalysis a second alcohol displaces the –OH (as water) to give the acetal R–CH(OEt)2. The reaction is reversible, so acetals are used to protect carbonyls and are hydrolyzed back with aqueous acid. An imine would require an amine, not an alcohol.
Quick Recap
- Carbonyl C is sp2, planar, δ+; nucleophiles attack here.
- Reactivity order: HCHO > RCHO > RCOR′ (steric + electronic).
- Tollens silver mirror → aldehyde; Fehling red Cu2O → aliphatic aldehyde; iodoform yellow → CH3CO group.
- NaBH4 = mild (carbonyls only); LiAlH4 = strong (all carbonyls including esters/acids).
- Clemmensen (Zn/Hg, HCl) and Wolff–Kishner (NH2NH2, KOH) reduce >C=O all the way to >CH2.
- Alcohols → hemiacetal then acetal (acid-catalyzed, reversible, a protecting group); 1° amine → imine, 2° amine → enamine; HCN → cyanohydrin; 2,4-DNP → orange ppt confirms any >C=O.
- α-H's are acidic (enolate is resonance-stabilized): keto–enol tautomerism and the aldol reaction (β-hydroxy carbonyl, then dehydration); no α-H → Cannizzaro.