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Aldehydes & Ketones

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Aldehydes & Ketones - Complete Interactive Lesson

Part 1: Carbonyl Chemistry Introduction

Aldehydes & Ketones — The Carbonyl Group

Part 1 of 7 — Carbonyl Structure, Nomenclature & Reactivity

The carbonyl group, a carbon double-bonded to oxygen ( $\text{C=O}$ ), is one of the most important functional groups in organic chemistry. It is the defining feature of aldehydes, ketones, carboxylic acids, esters, amides, and acyl chlorides. In this suite we focus on the two simplest carbonyl families, the aldehydes and ketones, where the carbonyl carbon bears only carbon or hydrogen substituents (no electronegative leaving group).

Aldehyde: the carbonyl carbon is bonded to at least one hydrogen. General formula $\text{RCHO}$ (for formaldehyde, $\text{R} = \text{H}$ , giving $\text{HCHO}$ ). The carbonyl always sits at the end of a carbon chain.
Ketone: the carbonyl carbon is bonded to two other carbons. General formula $\text{RCOR}'$ . The carbonyl is internal, flanked on both sides.

Both are planar at the carbonyl carbon, which is $sp^2$ -hybridized with bond angles near $120^\circ$ .

Why the Carbonyl Carbon Is Electrophilic

The $\text{C=O}$ bond consists of a $\sigma$ bond and a $\pi$ bond, just like an alkene. The crucial difference is electronegativity: oxygen ( $3.44$ on the Pauling scale) is far more electronegative than carbon ( $2.55$ ). The shared electrons, especially the loosely held $\pi$ electrons, are pulled toward oxygen, leaving the carbon and the oxygen .

Checkpoint — Structure & Polarity

IUPAC Nomenclature

Aldehydes take the suffix -al. Because the carbonyl must be terminal, its carbon is always C1 — no locant is needed for the $\text{CHO}$ group itself.

$\text{CH}_3\text{CHO}$ = ethanal (acetaldehyde)
$\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}$ = butan

Checkpoint — Nomenclature

Why Aldehydes Are More Electrophilic Than Ketones

A recurring theme in this suite is that aldehydes react faster than ketones in nucleophilic addition. Two effects work together, both favoring the aldehyde:

1. Electronic (inductive/hyperconjugative) effect. Alkyl groups are weakly electron-donating. A ketone has two alkyl groups pushing electron density toward the carbonyl carbon, partially neutralizing its $\delta+$ and stabilizing the ground state. An aldehyde has only one alkyl group (plus an H, which donates essentially nothing). The aldehyde carbon is therefore left more electron-poor and more electrophilic.

2. Steric effect. Nucleophilic addition converts a flat, $sp^2$ carbon into a crowded, $sp^3$ center. In a ketone, the incoming nucleophile must squeeze past bulky R groups; in an aldehyde, one position is occupied only by a small hydrogen. Less steric strain in the transition state means a faster reaction and a more stable (less crowded) product.

Checkpoint — Relative Reactivity

Part 1 Summary

The carbonyl group ( $\text{C=O}$ ) is planar and $sp^2$ ; the carbon is $\delta+$ (electrophilic) and the oxygen is $\delta-$ (basic), thanks to oxygen's high electronegativity.
Aldehydes () are terminal and take the suffix (carbonyl = C1). () are internal and take the suffix with the lowest locant.

Part 2: Nucleophilic Addition

Nucleophilic Addition to the Carbonyl

Part 2 of 7 — The Mechanism, Hydrates, and Hemiacetals

In Part 1 we established that the carbonyl carbon is electrophilic. The single most important reaction of aldehydes and ketones follows directly: a nucleophile adds to the carbon, the $\pi$ bond breaks, and the electrons collapse onto oxygen. Because the net result is that two new groups (Nu and, ultimately, H) add across the $\text{C=O}$ with no atom leaving, the reaction is called nucleophilic addition.

$\text{R}_2\text{C=O} + \text{Nu}-\text{H} \rightarrow \text{R}_2\text{C(Nu)(OH)}$

Part 3: Grignard & Organolithium Reactions

Carbon and Hydride Nucleophiles

Part 3 of 7 — Grignard, Organolithium & Hydride Addition

Parts 2 covered reversible additions of $\text{O}$ -nucleophiles. We now meet two classes of strong, anionic nucleophiles that add irreversibly and are workhorses of synthesis:

Organometallics ( $\text{RMgX}$ Grignard reagents, $\text{RLi}$ organolithiums) deliver a carbanion ( $\text{R}^-$ ) to the carbonyl, forging a new . This is one of the most powerful ways to build a carbon skeleton.

Part 4: Reduction & Oxidation

Oxidation & Reduction of Carbonyls

Part 4 of 7 — Redox Chemistry and Diagnostic Tests

Aldehydes and ketones sit in the middle of carbon's oxidation ladder. Reading the ladder for a single carbon:

$\text{alkane} \rightarrow \text{alcohol} \rightarrow \text{aldehyde/ketone} \rightarrow \text{carboxylic acid}$

Each arrow to the right is an oxidation (loss of $\text{H}$ / gain of bonds); each step left is a . In Part 3 we moved (carbonyl alcohol) with hydride. Here we move .

Part 5: Acetals & Hemiacetals

Acetals, Hemiacetals & Protecting Groups

Part 5 of 7 — From Hemiacetal to Acetal, and Why It Matters

In Part 2 one molecule of alcohol added to a carbonyl to give a hemiacetal (one $\text{OH}$ , one $\text{OR}$ on the same carbon). Continue with acid catalysis and a second equivalent of alcohol and that hemiacetal converts to a full acetal — a carbon bearing two $\text{OR}$ groups and no oxygen–hydrogen bond:

$R_{2} C=O + 2 R^{'} OH \underset{H^{+}}{⇌} R_{2}^{}$

Part 6: Problem-Solving Workshop

Nitrogen Nucleophiles & the Wittig Reaction

Part 6 of 7 — Imines, Enamines, and C=C Formation

So far our nucleophiles have been oxygen ( $\text{H}_2\text{O}$ , $\text{ROH}$ ), carbon ( $\text{RMgX}$ ), and hydride ( $\text{H}^-$ ). This part adds two more strategically important reaction families and folds them into the problem-solving toolkit:

Part 7: Synthesis & Review

$\alpha$ -Acidity, Tautomerism & Synthesis Review

Part 7 of 7 — Enols, Enolates, and Putting It All Together

Parts 1–6 treated the carbonyl as an electrophile attacked at the carbon. This final part reveals a second face of carbonyl chemistry: the hydrogens on the carbon next to the carbonyl — the $\alpha$ -hydrogens — are unusually acidic, and removing one creates a nucleophile. This single idea (enol/enolate chemistry) underlies a huge fraction of carbon–carbon bond construction in organic synthesis. We finish with a synthesis-level review that ties every reaction in the suite together.

Why the $\alpha$ -Carbon Is Acidic

The $\alpha$ -carbon is the carbon directly bonded to the carbonyl carbon; its hydrogens are . A typical alkane C–H has . An -hydrogen of a ketone or aldehyde has — about .

We represent this polarization with a permanent dipole and with two resonance structures: the neutral form $\text{C=O}$ in equilibrium with the dipolar form $\overset{+}{\text{C}}-\overset{-}{\text{O}}$ (written $\text{C=O} \leftrightarrow \overset{+}{\text{C}}-\overset{-}{\text{O}}$ ).

The carbonyl carbon therefore carries a partial positive charge ( $\delta+$ ) and behaves as an electrophile, while the oxygen carries a partial negative charge ( $\delta-$ ) and is mildly nucleophilic/basic. This single fact drives nearly all carbonyl chemistry:

Key idea: Nucleophiles attack the carbon; electrophiles (including $\text{H}^+$ ) attach to the oxygen. The $sp^2$ , planar geometry leaves the carbon open to attack from either face, perpendicular to the molecular plane.

A second consequence of the polarity is physical: aldehydes and ketones are polar aprotic molecules. They cannot donate hydrogen bonds (no $\text{O}-\text{H}$ or $\text{N}-\text{H}$ ), but the lone pairs on oxygen can accept them, so small aldehydes and ketones are reasonably water-soluble and boil higher than comparable alkanes, yet lower than the corresponding alcohols.

\text{OHC}-\text{CH}_2-\text{CHO}

= propanedial

Ketones take the suffix -one, and the chain is numbered to give the carbonyl carbon the lowest possible locant.

$\text{CH}_3\text{COCH}_3$ = propanone (acetone)
$\text{CH}_3\text{COCH}_2\text{CH}_3$ = butan-2-one
$\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3$ = pentan-3-

When a carbonyl must be named as a substituent (e.g., when a higher-priority group like a carboxylic acid is present), use oxo- for the $\text{C=O}$ and formyl- or -carbaldehyde for $\text{CHO}$ on a ring. Priority order for choosing the principal characteristic group (highest first): carboxylic acid > ester > amide > nitrile > aldehyde > ketone > alcohol > amine.

Worked example. Name $\text{CH}_3\text{COCH}_2\text{CH}_2\text{CHO}$ . Both an aldehyde and a ketone are present. The aldehyde outranks the ketone, so the chain is numbered from the $\text{CHO}$ end (C1). The ketone $\text{C=O}$ at C4 becomes an "oxo" substituent: 4-oxopentanal.

Worked comparison. Rank the rate of nucleophilic addition: formaldehyde ( $\text{H}_2\text{C=O}$ ) vs. acetaldehyde ( $\text{CH}_3\text{CHO}$ ) vs. acetone ( $\text{(CH}_3)_2\text{C=O}$ ). Formaldehyde has zero electron-donating alkyl groups and the least steric bulk, so it is the most reactive. Acetaldehyde (one $\text{CH}_3$ ) is intermediate. Acetone (two $\text{CH}_3$ ) is the slowest. Order: $\text{H}_2\text{C=O} > \text{CH}_3\text{CHO} > \text{(CH}_3)_2\text{C=O}$ .

This same logic explains why, for instance, the equilibrium amount of hydrate formed from formaldehyde is enormous, while acetone barely hydrates at all — a result we will quantify in Part 2.

When carbonyls compete, the aldehyde outranks the ketone; the ketone becomes an oxo substituent.

Aldehydes are more reactive than ketones toward nucleophilic addition because they have fewer electron-donating alkyl groups (more

\delta+

) and less steric hindrance at the carbon.

In Part 2 we use this electrophilic carbon as the launch point for the master reaction of the entire suite: nucleophilic addition, illustrated with water (hydrates) and alcohols (hemiacetals).

This contrasts sharply with carboxylic acid derivatives (esters, amides, acyl chlorides), where a leaving group departs and the carbonyl is regenerated — that is nucleophilic acyl substitution. Aldehydes and ketones have no good leaving group on the carbonyl carbon, so they add rather than substitute.

The Two Mechanistic Pathways

Base-mediated (nucleophile is anionic/strong). A strong nucleophile attacks directly.

The nucleophile $\text{Nu}^-$ attacks the $\delta+$ carbonyl carbon; the carbon rehybridizes from $sp^2$ to $sp^3$ .
The $\pi$ electrons shift entirely onto oxygen, generating a tetrahedral alkoxide intermediate, $\text{R}_2\text{C(Nu)}-\text{O}^-$ .
The alkoxide is protonated (by solvent or workup) to give the neutral addition product, an alcohol bearing the new $\text{Nu}$ group.

Acid-mediated (nucleophile is weak/neutral, e.g. $\text{H}_2\text{O}$ , $\text{ROH}$ ). Here we activate the carbonyl first.

The carbonyl oxygen is protonated by acid, giving an oxocarbenium-like cation, $\text{R}_2\text{C=}\overset{+}{\text{O}}\text{H} \leftrightarrow \overset{+}{\text{R}_2\text{C}}-\text{OH}$ . Protonation makes the carbon .

Key idea: Acid catalysis increases the electrophilicity of the carbon (protonate the oxygen). Base catalysis increases the nucleophilicity of the attacking species (deprotonate Nu–H to make $\text{Nu}^-$ ). Many carbonyl additions can be run either way; you pick the pathway that fits the nucleophile.

Checkpoint — The Mechanism

Addition of Water — Hydrates (gem-Diols)

When water adds across a carbonyl, the product is a hydrate, also called a geminal diol (two $\text{OH}$ groups on the same carbon):

$\text{R}_2\text{C=O} + \text{H}_2\text{O} \rightleftharpoons \text{R}_2\text{C(OH)}_2$

This is a genuine equilibrium, and its position is exquisitely sensitive to the carbonyl's electrophilicity — exactly the trend from Part 1.

Carbonyl	% hydrate at equilibrium (aqueous)	Why
Formaldehyde $\text{H}_2\text{C=O}$	~ $99.9\%$	no electron-donating R, minimal sterics
Acetaldehyde $\text{CH}_3\text{CHO}$

Electron-withdrawing groups push the equilibrium the other way. Chloral ( $\text{CCl}_3\text{CHO}$ ) is so electrophilic that its hydrate, $\text{CCl}_3\text{CH(OH)}_2$ (chloral hydrate), is a stable, isolable solid.

Worked example. Predict whether cyclohexanone or cyclohexanecarbaldehyde forms more hydrate in water. The aldehyde (one alkyl substituent, an H, less steric crowding) is more electrophilic, so its hydration equilibrium lies farther to the right — it forms more hydrate than the ketone.

Addition of Alcohols — Hemiacetals

Replace water with an alcohol ( $\text{ROH}$ ) and the same addition gives a hemiacetal: one $\text{OH}$ and one $\text{OR}$ on the same carbon.

$\text{R}_2\text{C=O} + \text{R}'\text{OH} \rightleftharpoons \text{R}_2\text{C(OH)(OR}')$

The mechanism is the acid-catalyzed pathway from above, with $\text{R}'\text{OH}$ as the neutral nucleophile: protonate the carbonyl, alcohol attacks the activated carbon, then deprotonate the resulting oxonium.

Like hydration, hemiacetal formation is usually unfavorable for open-chain ketones (the equilibrium lies toward starting material). But there is a major exception that dominates carbohydrate chemistry: intramolecular hemiacetal formation. When the $\text{OH}$ and $\text{C=O}$ live in the same molecule (as in a 4- or 5-carbon-removed hydroxy-aldehyde), cyclization to a 5- or 6-membered ring is entropically and geometrically favored, so the cyclic hemiacetal is the dominant form.

Worked example — glucose. Open-chain glucose is a six-carbon aldehyde (an aldohexose) bearing an $\text{OH}$ on C5. The C5 $\text{OH}$ adds intramolecularly to the C1 aldehyde, forming a six-membered (pyranose) ring that is a cyclic hemiacetal. In aqueous solution, well over $99\%$ of glucose exists as this ring rather than the open-chain aldehyde. The new stereocenter at C1 (the anomeric carbon) gives the $\alpha$ and $\beta$ anomers.

A hemiacetal is only a half-step. With more alcohol and continued acid catalysis, it loses water and adds a second $\text{OR}$ to become a fully substituted acetal — the subject of Part 5.

Checkpoint — Hydrates & Hemiacetals

Part 2 Summary

Nucleophilic addition is the master reaction of aldehydes/ketones: Nu attacks the $\delta+$ carbon, the $\pi$ electrons go to oxygen forming a tetrahedral alkoxide, then protonation gives the product. No leaving group departs (that distinguishes it from acyl substitution).
Strong/anionic nucleophiles add directly (base pathway); weak neutral nucleophiles need acid activation of the carbonyl (protonate oxygen first).
Water adds to give a hydrate (gem-diol); alcohols add to give a hemiacetal (one $\text{OH}$ + one $\text{OR}$ ). Both are equilibria whose position tracks carbonyl electrophilicity (formaldehyde $\gg$ aldehyde $\gg$ ketone).
Intramolecular hemiacetal formation (e.g., glucose $\rightarrow$ pyranose ring) is strongly favored and underlies sugar chemistry.

Parts 3 and 4 turn to nucleophiles that add irreversibly and forge new C–C or C–H bonds: organometallics and hydrides.

carbon–carbon bond

Hydride reagents (

\text{NaBH}_4

\text{LiAlH}_4

) deliver a hydride (

\text{H}^-

) to the carbonyl, forming a new carbon–hydrogen bond and reducing the carbonyl to an alcohol.

Both follow the base-mediated addition mechanism from Part 2: the nucleophile attacks the carbon, an alkoxide forms, and a separate protonation step (aqueous workup) delivers the final alcohol.

Grignard & Organolithium Reagents

A Grignard reagent forms when an alkyl or aryl halide reacts with magnesium metal in dry ether: $\text{R}-\text{X} + \text{Mg} \rightarrow \text{R}-\text{MgX}$ . Because carbon is more electronegative than magnesium, the C–Mg bond is strongly polarized $\text{R}^{\delta-}-\text{Mg}^{\delta+}$ , so the carbon behaves as a carbanion — an excellent nucleophile and a strong base.

The reaction with a carbonyl:

The carbanion carbon of $\text{R}-\text{MgX}$ attacks the electrophilic carbonyl carbon, forming a new C–C bond and a magnesium alkoxide.
Aqueous acid workup (a separate step, written as " $\text{H}_3\text{O}^+$ ") protonates the alkoxide to give the alcohol.

The class of alcohol produced depends on the carbonyl:

Electrophile	Product after $\text{H}_3\text{O}^+$	Class
Formaldehyde $\text{H}_2\text{C=O}$

Crucial caveat — incompatibility. Grignard and organolithium reagents are strong bases. They are instantly destroyed by any acidic proton: $\text{O}-\text{H}$ (alcohols, water, carboxylic acids), $\text{N}-\text{H}$ , or terminal alkyne C–H. You cannot have a free $\text{OH}$ in the same flask. That is precisely why the proton source is added only after the addition is complete, as a separate workup.

Worked Example — Predicting a Grignard Product

Problem. Predict the product of $\text{CH}_3\text{CH}_2\text{MgBr}$ reacting with acetone $\text{(CH}_3)_2\text{C=O}$ , followed by aqueous acid.

Step 1 — identify the pieces. The nucleophile is the ethyl carbanion, $\text{CH}_3\text{CH}_2^-$ . The electrophile is the carbonyl carbon of acetone, which bears two methyl groups.

Step 2 — form the C–C bond. The ethyl group bonds to the carbonyl carbon. That carbon now carries: two $\text{CH}_3$ (from acetone), one $\text{CH}_2\text{CH}_3$ (the new group), and an (the former carbonyl oxygen, now a magnesium alkoxide).

Step 3 — protonate. Workup with $\text{H}_3\text{O}^+$ converts $\text{O}^-$ to $\text{OH}$ .

Product: $\text{(CH}_3)_2\text{C(OH)CH}_2\text{CH}_3$ = , a alcohol (the carbinol carbon is bonded to three other carbons). A ketone always yields a tertiary alcohol with a Grignard — consistent with the table above.

Retrosynthetic view. Working backward, any C–OH carbon can be "cut" at one of its C–C bonds: one fragment becomes the carbonyl, the other becomes the $\text{RMgX}$ . 2-Methylbutan-2-ol could equally be made from butan-2-one + $\text{CH}_3\text{MgBr}$ , or from acetone + ethyl Grignard. This disconnection logic is the heart of the Part 6 synthesis workshop.

Checkpoint — Grignard Reactions

Hydride Reduction — $\text{NaBH}_4$ and $\text{LiAlH}_4$

To convert a carbonyl into an alcohol without adding carbon, deliver a hydride ( $\text{H}^-$ ). The B–H and Al–H bonds are polarized so that hydrogen carries the negative character and attacks the carbonyl carbon.

$\text{R}_2\text{C=O} \xrightarrow{\text{H}^-} \text{R}_2\text{CH}-\text{O}^- \xrightarrow{\text{H}_3\text{O}^+} \text{R}_2\text{CH}-\text{OH}$

Aldehydes are reduced to primary alcohols; ketones are reduced to secondary alcohols. (No new C–C bond, so a ketone gives secondary, not tertiary — contrast with the Grignard.)

Choosing between the two reagents — strength and selectivity:

Reagent	Strength	Reduces	Tolerates	Solvent / workup
$\text{NaBH}_4$	mild	aldehydes, ketones	esters, amides, acids survive; protic solvents OK	can use in $\text{CH}_3\text{OH}/\text{H}_2\text{O}$

Selectivity in practice. A molecule containing both a ketone and an ester can be reduced at the ketone only with $\text{NaBH}_4$ , leaving the ester intact. $\text{LiAlH}_4$ would reduce both. Choosing the milder reagent to hit the more reactive group is a classic chemoselectivity tactic.

Worked example. Reduce 4-oxopentanal selectively at the aldehyde. The aldehyde is more electrophilic than the ketone (Part 1), but $\text{NaBH}_4$ reduces both aldehydes and ketones, so it is not selective enough here. In practice one protects the ketone as an acetal (Part 5), reduces the aldehyde, then removes the protecting group — illustrating why protecting groups exist.

Checkpoint — Hydride Reduction

Part 3 Summary

Grignard ( $\text{RMgX}$ ) and organolithium ( $\text{RLi}$ ) reagents are carbanion nucleophiles. They add to carbonyls to build C–C bonds: formaldehyde $\rightarrow$ 1° alcohol, other aldehyde $\rightarrow$ 2° alcohol, ketone $\rightarrow$ 3° alcohol, $\text{CO}_2 \rightarrow$ carboxylic acid.
Organometallics are strong bases — incompatible with $\text{O}-\text{H}$ / $\text{N}-\text{H}$ . Add the acidic proton only at workup.
Hydride reagents deliver $\text{H}^-$ and reduce carbonyls: aldehyde $\rightarrow$ 1° alcohol, ketone $\rightarrow$ 2° alcohol (no carbon added).
$\text{NaBH}_4$ is mild and selective (ketones/aldehydes only, tolerates protic solvents and esters); $\text{LiAlH}_4$ is strong (also reduces esters, amides, acids, nitriles) but must be kept anhydrous.

Part 4 completes the redox picture by going the other direction — oxidation of aldehydes, and the diagnostic tests that distinguish aldehydes from ketones.

The pivotal asymmetry that governs this entire lesson: an aldehyde still has a C–H bond on the carbonyl carbon, but a ketone does not. That lone hydrogen is the handle an oxidant grabs. It is why aldehydes are easily oxidized to carboxylic acids, while ketones strongly resist oxidation (oxidizing a ketone would require breaking a C–C bond, which mild reagents cannot do).

Oxidation of Aldehydes

$\text{RCHO} \xrightarrow{\text{[O]}} \text{RCOOH}$

A wide range of oxidants accomplish this: chromium(VI) reagents (Jones reagent, $\text{CrO}_3/\text{H}_2\text{SO}_4$ ), permanganate ( $\text{KMnO}_4$ ), and even mild ones (Tollens', $\text{Ag}_2\text{O}$ ). Mechanistically, water adds to the aldehyde to form the hydrate (Part 2); the oxidant then removes the C–H and an O–H, installing the second C=O of the acid. Because a ketone cannot form that same C–H-bearing hydrate carbon, it has no pathway to oxidize under these mild conditions.

The PCC distinction (controlled vs. over-oxidation). When making an aldehyde from a primary alcohol, the danger is over-oxidation straight through to the carboxylic acid (because the aldehyde, once formed, hydrates in water and keeps oxidizing).

PCC (pyridinium chlorochromate) is an anhydrous Cr(VI) reagent. With no water present, the aldehyde cannot hydrate, so oxidation stops at the aldehyde: $\text{RCH}_2\text{OH} \xrightarrow{\text{PCC}} \text{RCHO}$ .

Worked example. You need to convert 1-butanol to butanal (not butanoic acid). Choose PCC in $\text{CH}_2\text{Cl}_2$ : the anhydrous conditions prevent hydrate formation and halt oxidation at the aldehyde. Using aqueous $\text{KMnO}_4$ or Jones reagent would overshoot to butanoic acid.

Checkpoint — Oxidation

Diagnostic Tests — Telling an Aldehyde from a Ketone

Because only aldehydes are easily oxidized, several classic "wet" tests use a mild oxidant that produces a visible change when it is reduced. A positive result means "aldehyde present"; a negative result (no change) is consistent with a ketone.

Tollens' test (the silver mirror). Tollens' reagent is the diamminesilver(I) complex, $\text{[Ag(NH}_3)_2]^+$ . An aldehyde reduces $\text{Ag}^+$ to metallic silver, which deposits as a shiny silver mirror on the glass while the aldehyde is oxidized to a carboxylate:

$\text{RCHO} + 2\,\text{[Ag(NH}_3)_2]^+ + 3\,\text{OH}^- \rightarrow \text{RCOO}^- + 2\,\text{Ag}\downarrow + 4\,\text{NH}_3 + 2\,\text{H}_2\text{O}$

Aldehyde: positive — silver mirror forms.
Ketone: negative — no reaction, no mirror.

Related tests. Fehling's and Benedict's solutions use $\text{Cu}^{2+}$ (deep blue); an aldehyde reduces it to a brick-red $\text{Cu}_2\text{O}$ precipitate. These are the basis of historical tests for "reducing sugars," since the open-chain form of an aldose presents a free aldehyde.

Worked example. You have two unlabeled bottles: one is propanal ( $\text{CH}_3\text{CH}_2\text{CHO}$ ), the other propanone/acetone ( $\text{CH}_3\text{COCH}_3$ ). Add Tollens' reagent to each. The bottle that produces a silver mirror is the (propanal); the bottle that stays clear is the (acetone). One simple test distinguishes the two isomeric compounds.

Checkpoint — Diagnostic Tests

Bringing Redox Together

Transformation	Reagent(s)	Note
1° alcohol $\rightarrow$ aldehyde	PCC (anhydrous)	stops at aldehyde
1° alcohol $\rightarrow$ carboxylic acid	Jones, $\text{KMnO}_4$ (aqueous)	over-oxidizes through aldehyde
2° alcohol $\rightarrow$ ketone	PCC, Jones, $\text{KMnO}_4$	stops at ketone
aldehyde $\rightarrow$ carboxylic acid	Tollens', Jones, $\text{KMnO}_4$	ketones unreactive
aldehyde $\rightarrow$ 1° alcohol	$\text{NaBH}_4$ , $\text{LiAlH}_4$	reduction (Part 3)
ketone $\rightarrow$ 2° alcohol	$\text{NaBH}_4$ , $\text{LiAlH}_4$	reduction (Part 3)

Part 4 Summary

Carbon's oxidation ladder: alkane $\rightarrow$ alcohol $\rightarrow$ aldehyde/ketone $\rightarrow$ carboxylic acid.
Aldehydes oxidize easily to carboxylic acids (they have a removable C–H on the carbonyl carbon); ketones resist oxidation.
PCC (anhydrous) stops a primary alcohol at the aldehyde; aqueous oxidants (Jones, $\text{KMnO}_4$ ) push on to the acid.
Only aldehydes give positive (silver mirror) and (brick-red ) tests — the basis for distinguishing aldehydes from ketones and detecting reducing sugars.

Part 5 returns to the alcohol nucleophiles of Part 2 and completes the hemiacetal story: full acetal formation and its use as a protecting group.

R_{2} C=O + 2 R^{'} OH H^{+} ⇌ R_{2} C(OR^{'})_{2} + H_{2} O

The naming captures the count of $\text{OR}$ groups:

Species	Groups on the carbon	Note
Hydrate (gem-diol)	two $\text{OH}$	from water
Hemiacetal	one $\text{OH}$ + one $\text{OR}$	half-way
Acetal	two $\text{OR}$	fully substituted

(Older texts call the ketone-derived versions "hemiketals" and "ketals"; modern IUPAC uses hemiacetal/acetal for both.)

The Acetal Mechanism (Acid-Catalyzed, Reversible Throughout)

Every step is an equilibrium, and acid (not base) is required because the key intermediate is a cation.

Hemiacetal $\rightarrow$ acetal:

Protonate the hemiacetal's $\text{OH}$ to make it a good leaving group ( $\overset{+}{\text{O}}\text{H}_2$ ).
Lose water to form a resonance-stabilized oxocarbenium ion, $\text{R}_2\text{C=}\overset{+}{\text{O}}\text{R}' \leftrightarrow \overset{+}{\text{R}_2\text{C}}-\text{OR}'$ . This is the pivotal intermediate — it is why base cannot be used (base would deprotonate, not generate a cation).
A second alcohol attacks the oxocarbenium carbon.
Deprotonate to give the neutral acetal.

Because water appears as a product, Le Chatelier governs the outcome:

To form the acetal: use excess alcohol and remove water (e.g., a Dean–Stark trap, or molecular sieves). Often a diol such as ethylene glycol is used so the acetal is a cyclic 1,3-dioxolane, which is entropically favored (one molecule consumed, not two).
To hydrolyze the acetal back to the carbonyl: use excess water and aqueous acid.

Worked example — equilibrium control. Cyclohexanone + ethylene glycol with catalytic $\text{TsOH}$ , refluxing in toluene with a Dean–Stark trap, gives the cyclic acetal of cyclohexanone in high yield. The trap continuously removes the water by-product, dragging the equilibrium to the right. Stir that same acetal in aqueous $\text{HCl}$ and it reverts cleanly to cyclohexanone — the reaction simply runs backward.

Checkpoint — Forming Acetals

Acetals as Protecting Groups — The Key Application

Two facts make acetals invaluable in synthesis:

Acetals are stable to base, nucleophiles, hydride, and organometallics — none of the things that attack a carbonyl. An acetal has no $\text{C=O}$ and no acidic proton to expose.
Acetals are easily removed by mild aqueous acid, regenerating the original carbonyl.

Together this means a carbonyl can be temporarily hidden (protected) as an acetal while you carry out chemistry elsewhere in the molecule that would otherwise destroy or be destroyed by the carbonyl — then unmasked at the end.

Worked example — the protecting-group strategy. Suppose you want to add a Grignard reagent to an ester group in a molecule that also contains a ketone. Problem: the Grignard would attack the more reactive ketone too. Solution:

Protect the ketone as a cyclic acetal (ethylene glycol, $\text{H}^+$ , $-\text{H}_2\text{O}$ ). The ketone is now inert.

React the Grignard with the ester (the only electrophilic carbonyl left).

Deprotect with aqueous acid ( $\text{H}_3\text{O}^+$ ) to hydrolyze the acetal and restore the ketone.

A frequent exam trap: forgetting that a Grignard cannot be done in the presence of a free ketone or alcohol. The acetal protecting group is precisely the tool that resolves this.

This is also why carbohydrates behave as they do. A monosaccharide is a cyclic hemiacetal; when two sugars join, the anomeric $\text{OH}$ of one reacts with an $\text{OH}$ of another to form a glycosidic bond, which is a full acetal. Because that acetal is stable to base, disaccharides like sucrose are not "reducing sugars" — the masked carbonyl can no longer open to a free aldehyde under the basic Tollens'/Fehling's conditions.

Checkpoint — Protecting Groups

Part 5 Summary

A hemiacetal (one $\text{OH}$ + one $\text{OR}$ ) reacts with a second alcohol under acid catalysis to form an acetal (two $\text{OR}$ , no $\text{O}-\text{H}$ ), releasing water.
The mechanism runs through a resonance-stabilized oxocarbenium ion, so acid is required and every step is reversible: excess alcohol + water removal makes the acetal; excess aqueous acid hydrolyzes it back.
Acetals are stable to base, nucleophiles, hydride, and Grignard reagents but cleaved by mild aqueous acid, which makes them ideal protecting groups: mask a carbonyl, do chemistry elsewhere, then unmask.
Carbohydrate chemistry is acetal chemistry: monosaccharides are cyclic hemiacetals; glycosidic bonds are full acetals, which is why non-reducing sugars like sucrose exist.

Part 6 introduces nitrogen nucleophiles (forming imines and enamines) and the carbon-delivering Wittig reaction, broadening the addition story beyond oxygen, carbon, and hydride.

Nitrogen nucleophiles (amines) condense with carbonyls. A primary amine gives an imine (Schiff base, $\text{C=N}$ ); a secondary amine gives an enamine ( $\text{C=C}-\text{N}$ ). Both proceed by addition followed by loss of water (condensations).
The Wittig reaction uses a phosphorus ylide to convert a $\text{C=O}$ directly into a $\text{C=C}$ (an alkene), with the carbonyl oxygen leaving as a phosphine oxide.

The unifying skill of this part is predicting products and planning short syntheses, applying everything from Parts 1–5.

Imine Formation (Primary Amine $\rightarrow$ $\text{C=N}$ )

$\text{R}_2\text{C=O} + \text{R}'\text{NH}_2 \rightleftharpoons \text{R}_2\text{C=NR}' + \text{H}_2\text{O}$

Mechanism (addition then elimination of water):

The amine nitrogen (a good nucleophile) adds to the carbonyl carbon; proton transfer gives a neutral carbinolamine (hemiaminal), $\text{R}_2\text{C(OH)(NHR}')$ .
The $\text{OH}$ is protonated and leaves as water, giving an iminium ion, $R_{2} C= \overset{}{N}$ .

The pH sweet spot (a classic exam point). Imine formation is fastest near pH 4–5.

Too basic (high pH): not enough acid to protonate and expel the $\text{OH}$ in step 2, so dehydration stalls.
Too acidic (low pH): the amine is fully protonated to $\text{R}'\text{NH}_3^+$ , which has no lone pair and cannot act as a nucleophile in step 1.

Mildly acidic conditions balance "enough acid to dehydrate" against "enough free amine to attack." Imines are central in biochemistry (e.g., pyridoxal phosphate forms a Schiff base with amino acids in transamination).

Enamine Formation (Secondary Amine $\rightarrow$ $\text{C=C}-\text{N}$ )

Run the same condensation with a secondary amine ( $\text{R}_2'\text{NH}$ ) and step 3 changes. The iminium nitrogen now bears no hydrogen to lose, so the molecule instead removes a proton from the $\alpha$ -carbon, giving an enamine:

$\text{R}-\text{CH}_2-\text{CO}-\text{R} + \text{R}_2'\text{NH} \rightleftharpoons \text{R}-\text{CH=C(NR}_2')-\text{R} + \text{H}_2\text{O}$

The deciding factor — count the N–H bonds.

Primary amine ( $\text{R}'\text{NH}_2$ ): after dehydration the nitrogen keeps one H to lose, so the double bond ends up C=N $\rightarrow$ imine.

Secondary amine ( $R_{2}^{'}$ ): the nitrogen has no H left, so the double bond forms toward the -carbon () .

Enamines are valuable because the $\alpha$ -carbon becomes nucleophilic (the nitrogen lone pair pushes electron density onto it), making enamines useful for forming new bonds at the $\alpha$ position — a preview of the enol/enolate chemistry in Part 7.

Checkpoint — Nitrogen Nucleophiles

The Wittig Reaction — Building a $\text{C=C}$ from a $\text{C=O}$

The Wittig reaction converts a carbonyl directly into an alkene, replacing the $\text{O}$ with a $\text{=CR}_2$ unit supplied by a phosphorus ylide:

$\text{R}_2\text{C=O} + \text{Ph}_3\text{P=CR}_2' \rightarrow \text{R}_2\text{C=CR}_2' + \text{Ph}_3\text{P=O}$

Making the ylide. A primary (or secondary) alkyl halide reacts with triphenylphosphine to give a phosphonium salt, $\text{Ph}_3\overset{+}{\text{P}}-\text{CH}_2\text{R}\;\text{X}^-$ . A strong base (e.g., , ) then removes a proton from the carbon next to phosphorus to give the ylide, (a species with adjacent and formal charges).

Mechanism (brief). The nucleophilic ylide carbon adds to the carbonyl carbon; the alkoxide oxygen and phosphorus close into a four-membered oxaphosphetane, which collapses to expel triphenylphosphine oxide ( $\text{Ph}_3\text{P=O}$ ) and form the $\text{C=C}$ . The driving force for the whole reaction is the formation of the very strong P=O bond.

Worked example — predict and disconnect. What alkene forms from cyclohexanone + $\text{Ph}_3\text{P=CH}_2$ ? The carbonyl oxygen of cyclohexanone is replaced by $\text{=CH}_2$ , giving (an exocyclic ). Note the power of the method: the double bond appears , with no ambiguity about position. To make a specific alkene, disconnect at the — one carbon comes from a carbonyl, the other from the ylide (and therefore from an alkyl halide).

Checkpoint — Wittig & Synthesis

Part 6 Summary

Primary amines condense with carbonyls (addition + loss of water) to give imines / Schiff bases ( $\text{C=N}$ ); the reaction is optimal at pH 4–5 (acid enough to dehydrate, but the amine must stay unprotonated).
Secondary amines give enamines ( $\text{C=C}-\text{N}$ ) because the iminium nitrogen has no $\text{H}$ to lose, so an $\alpha$ -proton is removed instead; enamines make the $\alpha$ -carbon nucleophilic.
The Wittig reaction ( $\text{R}_2\text{C=O} + \text{Ph}_3\text{P=CR}_2' \rightarrow \text{R}_2\text{C=CR}_2'$ ) converts a carbonyl into a , expelling ; the strong bond drives it.
Synthesis discipline: disconnect alkenes at the $\text{C=C}$ (carbonyl + ylide); remember organometallics are incompatible with $\text{O}-\text{H}$ / $\text{N}-\text{H}$ and use protecting groups (Part 5) to work around it.

Part 7 closes the suite with keto–enol tautomerism, $\alpha$ -carbon acidity, and a synthesis-level review tying every reaction together.

$\alpha$ -hydrogens

\text{p}K_a \approx 50

\text{p}K_a \approx 20

thirty orders of magnitude more acidic

The reason is resonance stabilization of the conjugate base (the enolate). When base removes an $\alpha$ -proton, the resulting carbanion is not localized: its lone pair is conjugated with the $\text{C=O}$ , so the negative charge is delocalized onto the electronegative oxygen:

$\overset{-}{\text{C}}-\text{C=O} \;\leftrightarrow\; \text{C=C}-\overset{-}{\text{O}}$

The oxygen-bearing resonance form carries the charge on the more electronegative atom, which is the major contributor and the source of the stabilization. (Only $\alpha$ -hydrogens qualify — a $\beta$ -hydrogen, one carbon further out, has no such conjugation and remains as unreactive as any alkane C–H.)

Comparison. A $\beta$ -diketone such as pentane-2,4-dione ( $\text{CH}_3\text{COCH}_2\text{COCH}_3$ ) is far more acidic still ( $\text{p}K_a \approx 9$ ) because its central $\alpha$ -hydrogen is flanked by two carbonyls, so the enolate charge can delocalize onto two oxygens. More carbonyls sharing the negative charge = greater stabilization = stronger acid.

Keto–Enol Tautomerism

Tautomers are constitutional isomers that interconvert rapidly by moving a proton and shifting a double bond. Every carbonyl with an $\alpha$ -hydrogen is in equilibrium with its enol form ("ene" + "ol" = a C=C bearing an $\text{OH}$ ):

$\underset{\text{keto}}{\text{R}-\text{CH}_2-\text{C(=O)}-\text{R}} \;\rightleftharpoons\; \underset{\text{enol}}{\text{R}-\text{CH=C(OH)}-\text{R}}$

The proton moves from the $\alpha$ -carbon to the oxygen; the $\text{C=O}$ becomes $\text{C}-\text{OH}$ and a new $\text{C=C}$ forms. Tautomers are different molecules (bonds move), not resonance structures (only electrons move) — a distinction frequently tested.

Position of equilibrium. For simple aldehydes and ketones the keto form overwhelmingly dominates (e.g., acetone is ~ $99.9999\%$ keto), because the $\text{C=O}$ bond is much stronger than a $\text{C=C}$ bond plus an $\text{O}-\text{H}$ . The enol becomes significant only when it is specially stabilized — by conjugation (phenols are essentially "all enol"), or by intramolecular hydrogen bonding in -dicarbonyls (pentane-2,4-dione is appreciably enol).

Catalysis. Tautomerization is slow when truly neutral but is catalyzed by both acid and base:

Acid: protonate the carbonyl oxygen, then lose the $\alpha$ -proton to give the enol.
Base: remove the $\alpha$ -proton to give the enolate, then protonate oxygen to give the enol.

Checkpoint — Acidity & Tautomerism

The Payoff — Enols/Enolates as Nucleophiles

The reason $\alpha$ -acidity matters so much: the enol (and especially its conjugate base, the enolate) is nucleophilic at the $\alpha$ -carbon. This flips the carbonyl's usual polarity. Normally the carbonyl carbon is the electrophile; in an enolate, the $\alpha$ -carbon becomes a nucleophile that can attack electrophiles, forming new C–C bonds at the $\alpha$ position.

Two consequences worth previewing:

$\alpha$ -Halogenation: under acid or base, an enol/enolate reacts with $\text{X}_2$ to put a halogen on the $\alpha$ -carbon.
Aldol-type reactions: an enolate of one carbonyl attacks the electrophilic carbonyl carbon of another — one molecule supplies the nucleophile ( $\alpha$ -carbon), the other the electrophile (carbonyl carbon). This builds larger carbon skeletons and is the basis of much of biosynthesis (e.g., fatty-acid assembly).

This dual reactivity — electrophilic at the carbonyl carbon (Parts 2–6), nucleophilic at the $\alpha$ -carbon (Part 7) — is the conceptual heart of carbonyl chemistry.

Synthesis Review — One Carbonyl, Many Destinations

The entire suite reduces to a small set of transformations you can now combine. Starting from a generic aldehyde or ketone:

Reagent	Nucleophile delivered	Product
$\text{H}_2\text{O}$ , (H+)	$\text{HO}-$	hydrate (gem-diol)
$\text{ROH}$ , H+ (excess, – $\text{H}_2\text{O}$ )	$\text{RO}-$ (x2)	acetal (protecting group)
$\text{R}'\text{NH}_2$ , pH 4–5	$\text{R}'\text{N}<$
$\text{R}_2'\text{NH}$	$\text{R}_2'\text{N}<$
$\text{RMgX}$ then $\text{H}_3\text{O}^+$	$\text{R}-$ (carbanion)	1°/2°/3° alcohol (new C–C)
$\text{NaBH}_4$ or $\text{LiAlH}_4$	$\text{H}-$ (hydride)	1°/2° alcohol (reduction)
$\text{Ph}_3\text{P=CR}_2$	ylide carbon	alkene (+ $\text{Ph}_3\text{P=O}$ )
Tollens', Jones (aldehyde only)	— (oxidation)	carboxylic acid

Carbon-skeleton building comes from exactly two reactions in this list: the Grignard (C–C bond to make an alcohol) and the Wittig (C=C to make an alkene) — plus enolate chemistry at the $\alpha$ -carbon from this part.

Worked multistep example. Convert cyclohexanone into 1-methylcyclohexan-1-ol while a co-existing aldehyde elsewhere in the molecule must survive a Grignard.

Protect the aldehyde as an acetal ( $\text{HOCH}_2\text{CH}_2\text{OH}$ , H+, – $\text{H}_2\text{O}$ ) so the strongly basic Grignard cannot touch it.

Final Checkpoint — Synthesis Mastery

Suite Summary — Aldehydes & Ketones

You now command the chemistry of the carbonyl group:

Structure (Part 1): polar, planar $\text{C=O}$ ; electrophilic carbon; aldehydes > ketones in reactivity.
Addition (Part 2): the master mechanism; water $\rightarrow$ hydrate, alcohol $\rightarrow$ hemiacetal.
C and H nucleophiles (Part 3): Grignards build C–C bonds (1°/2°/3° alcohols); hydrides ( $\text{NaBH}_4$ , $\text{LiAlH}_4$ ) reduce.
Redox (Part 4): aldehydes oxidize to acids; PCC vs. Jones; Tollens'/Fehling's distinguish aldehydes from ketones.
Acetals (Part 5): full acetals as base-stable, acid-removable protecting groups.
N-nucleophiles & Wittig (Part 6): imines (1° amine), enamines (2° amine), and $\text{C=O} \rightarrow \text{C=C}$ .
$\alpha$ -Chemistry (Part 7): $\alpha$ -acidity, keto–enol tautomerism, and enolates as nucleophiles.

The carbonyl is electrophilic at carbon and nucleophilic at the $\alpha$ -carbon — two faces that together make it the central functional group of synthetic organic chemistry.

far more electrophilic

The weak, neutral nucleophile attacks the activated carbon.

Deprotonation of the newly attached

\text{Nu}-\text{H}^+

restores neutrality and regenerates the acid catalyst.

2-methylbutan-2-ol

Jones reagent (

\text{CrO}_3

, aqueous

\text{H}_2\text{SO}_4

) contains water, so a primary alcohol is driven all the way:

\text{RCH}_2\text{OH} \xrightarrow{\text{Jones}} \text{RCOOH}

A secondary alcohol gives a ketone with either reagent — and stops there, since ketones resist further oxidation.

\text{C}_3\text{H}_6\text{O}

Fehling's/Benedict's

\text{Cu}_2\text{O}

R_{2} C= N + HR^{'}

Deprotonation of nitrogen yields the neutral imine,

\text{R}_2\text{C=NR}'

Tertiary amine (

\text{R}_3'\text{N}

): cannot form a stable neutral product (no N–H at all) — only a transient iminium/ammonium.

\text{Ph}_3\text{P=CHR}

methylenecyclohexane

exactly where the carbonyl carbon was

specifically placed alkene

$\text{Ph}_3\text{P=O}$

Add

\text{CH}_3\text{MgBr}

to the ketone, then

\text{H}_3\text{O}^+

, giving the tertiary alcohol 1-methylcyclohexan-1-ol.

Deprotect the acetal with aqueous acid to reveal the original aldehyde. This single problem exercises reactivity ordering (Part 1), Grignard rules (Part 3), and protecting-group logic (Part 5) — the synthesis mindset the whole suite was building toward.