Aldehydes and Ketones

Carbonyl chemistry sits at the heart of MCAT organic chemistry. Almost every biologically relevant transformation (glycolysis, the citric acid cycle, amino acid metabolism) passes through a carbonyl. Master the logic of the carbonyl and the rest follows.

Priority labels: Must know = cold; Know the logic = mechanism not names; Passage-level = recognize, don't memorize; Optional = skippable.

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Structure and Nomenclature

The Carbonyl Group

Must know

The carbonyl group ($\ce{C=O}$) is a carbon doubly bonded to oxygen. The carbon is $sp^2$-hybridized, trigonal planar (~120° bond angles), with the unhybridized $p$ orbital forming the $\pi$ bond to oxygen.

Oxygen is more electronegative, so the $\pi$ electrons are pulled toward it, creating a permanent dipole — the carbonyl carbon is electrophilic ($\delta+$) and the oxygen is nucleophilic/basic ($\delta-$). This polarization is the single most important fact in carbonyl chemistry; everything else follows from it.

$\ce{R-\overset{\delta+}{C}=\overset{\delta-}{O}}$

• An aldehyde has at least one hydrogen on the carbonyl carbon: $\ce{R-CHO}$ (formaldehyde is $\ce{HCHO}$).
• A ketone has two carbon substituents: $\ce{R-CO-R'}$.

IUPAC Nomenclature

Must know

Aldehydes: replace the alkane -e with -al; the carbonyl carbon is C-1. Ketones: replace -e with -one with a locant for the lowest-numbered carbonyl position.

Know the common names (these appear directly in passages): formaldehyde (methanal, $\ce{HCHO}$), acetaldehyde (ethanal, $\ce{CH3CHO}$), benzaldehyde ($\ce{C6H5CHO}$), acetone (propan-2-one, $\ce{CH3COCH3}$).

Optional

Propionaldehyde = propanal; methyl ethyl ketone = butan-2-one; acetophenone = $\ce{C6H5COCH3}$; cyclic ketones keep -one (cyclohexanone). As substituents: oxo- (ketone), formyl-/-carbaldehyde (aldehyde).

Quick check: Name $\ce{CH3CH2CH2CHO}$ and $\ce{CH3COCH2CH3}$.

Answer: Butanal and butan-2-one.

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Physical Properties

Boiling Point and Intermolecular Forces

Must know

The $\ce{C=O}$ dipole gives significant dipole–dipole interactions, placing carbonyl boiling points between alkanes (dispersion only) and alcohols (H-bonding).

Aldehydes and ketones cannot donate H-bonds (no $\ce{O-H}$ or $\ce{N-H}$) but can accept them from water. So:

• Boiling points (similar MW): alcohol > aldehyde/ketone > alkane.
• Small aldehydes/ketones (≲ $\ce{C4}$) are water-miscible; beyond ~$\ce{C5}$ the hydrocarbon chain dominates and solubility drops.

Spectroscopic Signatures

Passage-level

The carbonyl $\ce{C=O}$ IR stretch is strong near 1700–1750 cm⁻¹; aldehydes add a distinctive C–H stretch near 2720/2820 cm⁻¹. In NMR, the aldehyde proton appears far downfield at δ 9–10 ppm.

Quick check: Butanal or pentan-1-ol — higher boiling point?

Answer: Pentan-1-ol, because it can donate and accept H-bonds. Butanal only accepts, so its intermolecular attractions are weaker at similar MW.

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General Principles of Carbonyl Reactivity

Electrophilicity of the Carbonyl Carbon

Must know

A resonance form places positive charge on carbon, negative on oxygen, so the carbonyl carbon is electrophilic and attacked by nucleophiles:

$\ce{R2C=O <-> R2\overset{+}{C}-\overset{-}{O}}$

Nucleophilic addition: a nucleophile attacks the $\delta+$ carbonyl carbon, the $\pi$ bond breaks (electrons to O) forming an alkoxide intermediate, which is then protonated.

Effect of Substituents on Reactivity

Must know

Two factors set reactivity toward nucleophilic attack:

1. Electronic: electron-withdrawing groups raise electrophilicity (more reactive); alkyl groups donate electron density (less reactive).
2. Steric: bulky substituents block nucleophile approach.

Both favor aldehydes over ketones (one small H vs. two alkyl groups), giving the order:
$\text{Formaldehyde} > \text{Aldehydes} > \text{Ketones}$

Quick check: Acetaldehyde vs. acetone — which reacts faster with a nucleophile? Give a steric and an electronic reason.

Answer: Acetaldehyde. Electronically, one methyl (vs. two) donates less, leaving the carbon more electrophilic. Sterically, one methyl + one H is less hindered than two methyls.

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Acidity of $\alpha$-Hydrogens and Carbanions

Why α-Hydrogens Are Acidic

Know the logic

The carbon adjacent to a carbonyl is the α-carbon; its α-hydrogens are far more acidic than normal C–H (pKₐ ~20 for a ketone vs. ~50 for an alkane).

Removing an α-H gives an enolate whose negative charge is delocalized onto the electronegative oxygen by resonance. A more stabilized conjugate base means a more acidic proton:

$\ce{R-CH2-C(=O)-R' -> R-\overset{-}{C}H-C(=O)-R' <-> R-CH=C(-\overset{-}{O})-R'}$

Passage-level

Numbers: alkane ~50, ester ~25, ketone ~20, aldehyde ~17, β-diketone ~9. A β-diketone has its α-carbon flanked by two carbonyls, so the enolate delocalizes into both — dramatically more acidic.

Bases: weak bases ($\ce{OH^-}$, $\ce{NaOEt}$) deprotonate only partially (equilibrium; used for aldols); the strong, hindered base LDA deprotonates completely (used for kinetic enolates below).

Quick check: Why is diethyl malonate ($\ce{CH2(CO2Et)2}$, pKₐ ~13) so much more acidic than acetone?

Answer: Its methylene is flanked by two ester carbonyls; the anion delocalizes into both, dispersing charge far better than acetone's single carbonyl.

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Nucleophilic Addition Reactions at C=O

Must know

Know the general mechanism — every reaction below is a variation on it:

1. Nucleophile attacks the carbonyl carbon (π bond breaks, electrons to O).
2. A tetrahedral alkoxide intermediate forms.
3. Protonation gives the product.

Hemiacetal and Acetal Formation

Must know

Hemiacetals form when an alcohol adds to an aldehyde — one $\ce{OH}$ and one $\ce{OR'}$ on the same carbon (with a ketone, a hemiketal):

$\ce{R-CHO + R'OH <=> R-CH(OH)(OR')}$

Hemiacetals are usually unstable and in equilibrium with the carbonyl, except in cyclic cases. A second alcohol under acid catalysis gives an acetal — two $\ce{OR'}$ groups on the same carbon, losing water:

$\ce{R-CH(OH)(OR') + R'OH ->[\ce{H+}] R-CH(OR')2 + H2O}$

Acetals are stable to base/nucleophiles but hydrolyze under aqueous acid — which is why they serve as protecting groups for aldehydes. Biologically, sugar ring forms are cyclic hemiacetals (e.g., glucose pyranose) and glycosidic bonds are acetals.

Quick check: Why are acetals used as protecting groups for aldehydes?

Answer: They are stable to base and nucleophiles, so reactions can be done elsewhere; dilute aqueous acid later hydrolyzes the acetal back to the aldehyde.

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Imine and Enamine Formation

Must know

A primary amine ($\ce{RNH2}$) adds to a carbonyl, then loses water under mild acid to give an imine (Schiff base, $\ce{C=N}$):

$\ce{R-CHO + H2N-R' -> R-CH=N-R' + H2O}$

Know the logic

Acid (pH ~4–5) is optimal — it activates the carbonyl, but too much acid protonates the amine and kills the nucleophile.

Passage-level

Biology: Schiff bases are central to amino acid metabolism (pyridoxal phosphate / vitamin B₆ in transamination) and to vision (retinal–opsin linkage in rhodopsin).

A secondary amine ($\ce{R2NH}$) has no N–H left after addition to eliminate as $\ce{C=N}$, so dehydration goes toward the α-carbon instead, giving an enamine ($\ce{C=C-NR2}$):

$\ce{R-CH2-CO-R' + HNR''2 -> R-CH=C(NR''2)-R' + H2O}$

Key difference: primary amine → imine; secondary amine → enamine.

Quick check: Why does imine formation slow dramatically at pH 1?

Answer: At pH 1 the amine is protonated ($\ce{RNH3+}$) and no longer nucleophilic, so it cannot attack the carbonyl.

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Hydride Reagents (Reduction of Carbonyls)

Must know

Know two reducing agents (hydride, $\ce{H^-}$, is the nucleophile; alkoxide is protonated on workup):

Reagent	Scope	Solvent
$\ce{NaBH4}$	Mild — aldehydes & ketones only	Protic (MeOH, H₂O)
$\ce{LiAlH4}$	Strong — also esters, carboxylic acids, amides	Anhydrous ether/THF only

$\ce{NaBH4}$ is safe in protic solvent; $\ce{LiAlH4}$ reacts violently with water, so anhydrous conditions then a separate aqueous workup are required. Reduction gives a 1° alcohol from an aldehyde, 2° from a ketone.

Optional

If reduction creates a new stereocenter, the product is racemic (hydride attacks both faces of the planar carbonyl equally).

Quick check: 0.010 mol acetone + excess $\ce{NaBH4}$ in methanol, then water. Product and moles?

Answer: $\ce{CH3CH(OH)CH3}$ (2-propanol), 0.010 mol — a ketone reduces to a secondary alcohol, 1:1.

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Cyanohydrin Formation

Must know

Cyanide ($\ce{CN^-}$, from $\ce{HCN}$) adds to give a cyanohydrin — $\ce{OH}$ and $\ce{CN}$ on the same carbon. Reversible; aldehydes favor the product, bulky ketones favor reactants:

$\ce{R-CHO + HCN <=> R-CH(OH)-CN}$

Passage-level

The $\ce{CN}$ can be hydrolyzed to $\ce{COOH}$ or reduced to an amine, so this lengthens a carbon chain by one (e.g., Kiliani aldose extension). Cyanide is also a metabolic poison (inhibits cytochrome c oxidase).

Quick check: Why does cyclohexanone react with HCN more slowly and in poorer yield than acetaldehyde?

Answer: (1) Steric — two ring carbons hinder $\ce{CN^-}$ approach more than one H + one CH₃. (2) Electronic — two flanking carbons make the ketone carbonyl less electrophilic.

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Organometallic (Grignard) Addition — Forming C–C Bonds

Must know

A Grignard reagent ($\ce{RMgX}$) or organolithium acts as a carbanion ($\ce{R^-}$) that attacks the carbonyl carbon, forming a new C–C bond and, after workup, an alcohol:

$\ce{R-CHO + R'MgX -> R-CH(OMgX)-R' ->[\text{H3O+}] R-CH(OH)-R'}$

Know the product class: formaldehyde → 1°, aldehyde → 2°, ketone → 3° alcohol. Grignards are strong bases — anhydrous conditions are mandatory (they destroy any $\ce{-OH}$, $\ce{-NH}$, $\ce{-COOH}$).

Hydrate (gem-Diol) Formation

Must know

Water adds to a carbonyl to give a hydrate (geminal diol):

$\ce{R2C=O + H2O <=> R2C(OH)2}$

Equilibrium usually favors the carbonyl but shifts toward the hydrate for formaldehyde and electron-poor carbonyls (e.g., chloral). This parallels hemiacetal formation (water vs. alcohol).

Oxidation of Aldehydes

Must know

An aldehyde's carbonyl C–H lets it oxidize to a carboxylic acid; ketones have no such C–H and resist oxidation.

$\ce{R-CHO ->[\text{oxidation}] R-COOH}$

Oxidants: $\ce{KMnO4}$ and $\ce{H2CrO4}$ (Jones) are strong; Tollens' reagent ($\ce{Ag(NH3)2+}$) and Benedict's/Fehling's ($\ce{Cu^{2+}}$) are mild and selective.

Named tests (high-yield):

• Tollens': aldehydes reduce $\ce{Ag+}$ to a metallic silver mirror; ketones give a negative result.
  $\ce{R-CHO + 2[Ag(NH3)2]+ + 2OH- -> R-COO- + 2Ag(s) + 4NH3 + H2O}$
• Benedict's/Fehling's: reducing sugars reduce $\ce{Cu^{2+}}$ (blue) to brick-red $\ce{Cu2O}$ — historically used to detect urinary glucose.

Passage-level

All open-chain aldoses are reducing sugars (free or potential aldehyde). Sucrose is non-reducing — both anomeric carbons are locked in the glycosidic bond.

Quick check: Glyceraldehyde + Tollens' — result and why?

Answer: A silver mirror (positive): glyceraldehyde is an aldehyde, so it reduces $\ce{Ag+}$ to $\ce{Ag^0}$ and is oxidized to glycerate.

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Enolate Chemistry

Keto-Enol Tautomerism

Must know

A carbonyl and its enol are tautomers — constitutional isomers interconverting by moving a proton between the α-carbon and oxygen (acid- or base-catalyzed):

$\ce{R-CH2-C(=O)-R' <=>[\ce{H+} or \ce{OH-}] R-CH=C(-OH)-R'}$

This is NOT resonance — tautomers have different connectivity (an atom moves); resonance moves only electrons. For most carbonyls the keto form dominates (>99%). Exceptions: β-diketones (>80% enol, stabilized by conjugation + intramolecular H-bonding) and phenol (enol favored — restores aromaticity).

Passage-level

Biology: the keto–enol step from phosphoenolpyruvate to pyruvate drives the pyruvate kinase step because the keto product is far more stable.

α-Racemization: because the enol α-carbon is planar ($sp^2$), an α-stereocenter is erased and re-formed without facial preference — so a carbonyl with an α-stereocenter racemizes under acid or base. (Tested conceptually.)

Quick check: Will 2-methylcyclohexanone racemize in dilute aqueous acid?

Answer: Yes. The methyl-bearing α-carbon is a stereocenter; enolization makes it $sp^2$, and re-protonation from either face regenerates both enantiomers.

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Aldol Condensation and Retro-Aldol

Know the logic

The aldol reaction is a key C–C bond-forming reaction and appears throughout biochemistry.

Base-catalyzed:

1. Base removes an α-H, giving an enolate.
2. The enolate (nucleophile) attacks the carbonyl carbon of a second molecule.
3. Protonation gives a β-hydroxy carbonyl (the aldol product).

$\ce{2 R-CH2-CHO ->[\ce{OH-}] R-CH2-CH(OH)-CH(R)-CHO}$

Heating dehydrates the aldol product to a conjugated α,β-unsaturated carbonyl — the aldol condensation; conjugation drives the dehydration:

$\ce{Aldol product ->[\Delta, -H2O] \alpha,\beta\text{-unsaturated carbonyl}}$

Optional

In a crossed aldol, a partner with no α-H (e.g., benzaldehyde) can only be the electrophile, avoiding a product mixture. Intramolecular aldols build 5- and 6-membered rings.

Passage-level

Retro-Aldol: the reverse cleavage of a β-hydroxy carbonyl into two fragments — how aldolase splits fructose-1,6-bisphosphate into G3P and DHAP in glycolysis.

Worked example: Aldol and condensation products from base + two equivalents of acetaldehyde ($\ce{CH3CHO}$).

The enolate $\ce{^-CH2CHO}$ attacks a second acetaldehyde, then protonation gives 3-hydroxybutanal (the aldol product, a β-hydroxy aldehyde):
$\ce{CH3CH(OH)CH2CHO}$
Heating dehydrates it to but-2-enal (crotonaldehyde), an α,β-unsaturated aldehyde:
$\ce{CH3CH(OH)CH2CHO ->[\Delta, \ce{OH-}] CH3CH=CHCHO + H2O}$

Sanity check: 4 carbons (2+2), β-OH at C-3; dehydration removes the β-OH and an α-H to give the conjugated enal. ✓

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Conjugate (1,4-) Addition and the Michael Reaction

Passage-level

An α,β-unsaturated carbonyl has two electrophilic sites (conjugated $\ce{C=C}$ and $\ce{C=O}$):

• 1,2-addition — at the carbonyl carbon (favored by hard/reactive nucleophiles like $\ce{RLi}$, $\ce{LiAlH4}$).
• 1,4-addition (conjugate) — at the β-carbon; the carbonyl is restored after tautomerization (favored by soft/stabilized nucleophiles).

When the nucleophile is an enolate, 1,4-addition is the Michael reaction (route to 1,5-dicarbonyls; cuprates add 1,4 selectively).

Kinetic vs. Thermodynamic Enolate

Must know

Know the distinction. An unsymmetrical ketone (classic: 2-methylcyclohexanone) can deprotonate at two α-carbons:

• Kinetic enolate = less-substituted (less hindered) — formed fast by a strong, bulky base (LDA) at low T (−78°C), irreversibly.
• Thermodynamic enolate = more-substituted (more stable) — formed under equilibrium with a weaker base at higher T.

This parallels kinetic vs. thermodynamic control in eliminations.

Quick check: 2-pentanone + LDA at −78°C — which enolate forms and why?

Answer: The kinetic enolate at the terminal methyl (C-1). Bulky LDA removes the least hindered proton, and −78°C prevents equilibration.

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Common Confusions & Tricks

1. Tautomers vs. resonance. Resonance moves only electrons (same connectivity); keto/enol tautomers require moving an H. Move an arrow only → resonance; move an H → tautomerism.

2. Acetal vs. hemiacetal. Count oxygens on that carbon: hemiacetal = one OR + one OH; acetal = two OR. A sugar's anomeric carbon in ring form is a hemiacetal; a glycosidic bond is an acetal. This determines whether the sugar is reducing.

3. NaBH₄ vs. LiAlH₄ scope. NaBH₄ reduces only aldehydes/ketones (and imines); LiAlH₄ also reduces esters, carboxylic acids, amides. For selective reduction of an aldehyde beside an ester, use NaBH₄.

4. Tollens' vs. Benedict's. Both oxidize aldehydes; Tollens' = silver mirror, Benedict's = red/orange precipitate; ketones are negative. Trap: fructose (a ketose) is positive for Benedict's because it tautomerizes to an aldehyde in base.

5. Imine vs. enamine. Primary amine → imine ($\ce{C=N-R}$); secondary amine → enamine ($\ce{C=C-NR2}$). Secondary amines have no N–H to keep in a $\ce{C=N}$, so the double bond goes to carbon.

6. Retro-aldol in biochem. Aldolase or fructose-1,6-bisphosphate in a passage → think retro-aldol: break the α–β bond of the β-hydroxy carbonyl.

7. Nucleophiles attack carbon, not oxygen. Oxygen is the nucleophilic site of the carbonyl; nucleophiles attack the electrophilic ($\delta+$) carbon.

8. Kinetic enolate = less substituted; thermodynamic = more substituted. Less substituted = less hindered = proton removed faster = kinetic.

9. Aldehydes oxidize; ketones resist. The aldehyde's carbonyl C–H is the handle for oxidation to $\ce{COOH}$; ketones lack it.

10. pKₐ vs. acidity run opposite. Lower pKₐ = stronger acid. A β-diketone (~9) is far more acidic than acetone (~20).

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Key Equations

Equation / Expression	When to Use
Nucleophilic addition: $\ce{R2C=O + Nu- -> R2C(Nu)(O-) ->[\ce{H+}] R2C(Nu)(OH)}$	General mechanism for all carbonyl additions (Nu⁻ = H⁻, CN⁻, RNH₂, RO⁻)
Hemiacetal: $\ce{R-CHO + R'OH <=> R-CH(OH)(OR')}$	Equilibrium; unstable unless cyclic
Acetal: $\ce{R-CH(OH)(OR') + R'OH ->[\ce{H+}] R-CH(OR')2 + H2O}$	Acid-catalyzed, reversible; protecting group; glycosidic bonds
Imine: $\ce{R-CHO + H2NR' <=>[\ce{H+}] R-CH=NR' + H2O}$	Primary amine; Schiff base; pH ~4–5
Enamine: $\ce{R-CH2COR' + HNR''2 <=>[\ce{H+}] R-CH=C(NR''2)R' + H2O}$	Secondary amine + carbonyl
Cyanohydrin: $\ce{R-CHO + HCN <=> R-CH(OH)CN}$	Reversible; adds one C
Aldehyde oxidation: $\ce{R-CHO ->[\text{oxidant}] R-COOH}$	KMnO₄, H₂CrO₄, Tollens', Benedict's; ketones resist
Tollens': $\ce{R-CHO + 2[Ag(NH3)2]+ + 2OH- -> R-COO- + 2Ag(s) + 4NH3 + H2O}$	Silver mirror test for aldehydes/reducing sugars
Aldol: $\ce{2 R-CH2CHO ->[\ce{OH-}] R-CH2CH(OH)-CHR-CHO}$	β-Hydroxy carbonyl; two molecules couple
Aldol condensation: $\text{Aldol product} \xrightarrow{\Delta,\,-\ce{H2O}} \alpha,\beta\text{-unsaturated carbonyl}$	Dehydration to conjugated enone/enal
Enolate resonance: $\ce{R-\overset{-}{C}H-C(=O)-R' <-> R-CH=C(-\overset{-}{O})-R'}$	Charge delocalized onto O
α-H acidity: $\mathrm{p}K_a(\text{ketone}) \approx 20$; $\mathrm{p}K_a(\beta\text{-diketone}) \approx 9$	Lower pKₐ = more acidic
NaBH₄ reduction: $\ce{R2C=O ->[\ce{NaBH4}][\ce{H2O}] R2CHOH}$	Mild; aldehyde/ketone → alcohol
LiAlH₄ reduction: $\ce{R2C=O ->[\text{1. }\ce{LiAlH4}\text{, THF}][\text{2. }\ce{H2O}] R2CHOH}$	Strong; also esters/acids; anhydrous