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Aldehydes Ketones and Carboxylic Acids · TOPIC LESSON

Carbonyl Synthesis and Reactions

Track the carbonyl carbon through preparation, addition, reduction and selective protection.

JEE Main · JEE Advanced · NEET UG55 min study guide12 indexed sections

BY THE END, YOU CAN

  • Explain carbonyl polarity, geometry and relative aldehyde–ketone reactivity.
  • Choose preparation routes that preserve the intended carbon skeleton and oxidation level.
  • Draw electron-pair flow for nucleophilic addition and subsequent proton transfers.
  • Predict hydride, organometallic, nitrogen-nucleophile and Wittig products.
  • Plan acetal protection and deprotection around a competing reaction with explicit compatibility checks.

BEFORE YOU START

  • Lewis structures, curved-arrow electron-pair bookkeeping and acid–base reactions.
  • Functional groups, skeletal formulas and stereocenters.
  • Oxidation and reduction in organic molecules.

Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.

CONNECT THE CONCEPT TO THE EXAM

Selected exam questions & worked solutions

These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.

No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.

01 / UNDERSTAND & APPLY

1. The carbonyl carbon is an electrophilic, nearly planar center

An aldehyde has a carbonyl carbon bonded to at least one hydrogen, written R–C(=O)H; formaldehyde has two hydrogens. A ketone has two carbon substituents, R–C(=O)–R′. In this lesson R and R′ denote specified carbon groups, and condensed formulas preserve the carbonyl carbon explicitly.

The C=O bond contains a σ component and a π component. Oxygen attracts electron density more strongly, leaving the carbon electrophilic. A charge-separated resonance contributor C⁺–O⁻ helps explain this polarization, but the actual molecule is not a rapidly alternating mixture of a neutral structure and a fully ionic structure.

The carbonyl carbon is approximately trigonal planar. Addition converts it to a tetrahedral center by forming a new σ bond and moving the carbonyl π electron pair onto oxygen. If four different substituents result, a new stereocenter can be formed. Two faces of an achiral planar substrate may be attacked, often giving a racemic product with achiral reagents and no other stereochemical bias.

Aldehydes generally undergo nucleophilic addition more readily than comparable ketones because they have less steric crowding and fewer electron-donating carbon substituents at the carbonyl carbon. Formaldehyde is especially accessible. Conjugation, substituent electronics, solvent and the actual nucleophile can modify simple rankings, so compare like conditions rather than asserting a universal order for all carbonyl compounds.

02 / UNDERSTAND & APPLY

2. Preparation is a carbon-skeleton and oxidation-level problem

A primary alcohol RCH₂OH can be oxidized to an aldehyde RCHO with a suitable controlled oxidant and conditions that avoid further oxidation. A secondary alcohol R₂CHOH gives a ketone R₂C=O. A tertiary alcohol lacks the required hydrogen on its hydroxyl-bearing carbon for the analogous simple oxidation without carbon–carbon bond cleavage.

Primary-alcohol oxidation does not always stop at the aldehyde. Strong aqueous oxidizing conditions commonly carry it onward to the carboxylic acid. A reagent choice such as a suitable mild nonaqueous alcohol oxidant is therefore part of the answer when an aldehyde is the target; writing only [O] conceals the selectivity question.

Alkene cleavage provides another route: each alkene carbon becomes a carbonyl carbon. Under a reductive ozonolysis workup, an alkene carbon bearing a hydrogen gives an aldehyde fragment, while one bonded to two carbon groups gives a ketone. An oxidative workup can further oxidize aldehyde products, so the workup must be specified.

Partial reduction of an ester can stop at an aldehyde with a suitably controlled reagent such as DIBAL-H, whereas an ordinary excess strong hydride reduction generally proceeds to alcohol products. Acid chlorides can give ketones with appropriate organocuprate reagents, avoiding the repeated addition commonly associated with more reactive organometallic reagents.

Aromatic ketones can be formed by Friedel–Crafts acylation on a compatible aromatic ring. The method has substrate restrictions: a strongly deactivated ring or an amine that binds the Lewis acid can prevent a simple expected reaction. Preparation questions require checking the whole substrate, not only matching a functional-group arrow.

ORIGINAL WORKED EXAMPLE

Choose oxidation products without losing a carbon

Predict controlled oxidation products of butan-1-ol and butan-2-ol, and explain why the conditions matter for the first.

  1. Butan-1-ol CH₃CH₂CH₂CH₂OH has a primary alcohol carbon. Controlled oxidation gives CH₃CH₂CH₂CHO, butanal.
  2. Butan-2-ol CH₃CH(OH)CH₂CH₃ is secondary and gives CH₃C(=O)CH₂CH₃, butan-2-one.
  3. Both retain all four carbon atoms.
  4. The first can be overoxidized to butanoic acid under stronger aqueous conditions; a selective aldehyde-forming method must be chosen.

Answer Controlled primary oxidation: butanal. Secondary oxidation: butan-2-one.

ORIGINAL WORKED EXAMPLE

Read an alkene cleavage

Predict carbonyl fragments from CH₃CH=C(CH₃)₂ under reductive ozonolysis workup.

  1. Split the C=C bond and place an oxygen double bond on each former alkene carbon.
  2. The left alkene carbon is attached to CH₃ and H, giving CH₃CHO.
  3. The right alkene carbon is attached to two CH₃ groups, giving (CH₃)₂CO.
  4. The products contain two plus three carbons, matching the five-carbon starting alkene.

Answer Ethanal and propanone, assuming reductive workup.

03 / UNDERSTAND & APPLY

3. Electron pairs start the arrows; atoms do not emit arrows by themselves

For an anionic nucleophile :Nu⁻, draw an electron-pair arrow from its lone pair or reactive bond to the carbonyl carbon. At the same time, draw an arrow from the C=O π bond to oxygen. This second arrow is essential: without it, carbon would be assigned an impossible extra bond beyond its octet in the simple addition product.

The immediate product is a tetrahedral alkoxide, R₂C(O⁻)Nu. Protonation by an appropriate proton donor then gives R₂C(OH)Nu. In a proton-transfer step, the arrow begins at the oxygen lone pair and ends at the proton, while the donor’s H–A bond electrons move to A. Charge should balance at every stage.

With a weak neutral nucleophile such as water or an alcohol, acid catalysis can first protonate the carbonyl oxygen, making the carbon more electrophilic. Nucleophilic attack then forms a positively charged oxygen-containing intermediate, followed by deprotonation. Acid activation and nucleophile protonation compete, so “more acid always accelerates everything” is not a general rule.

For aldehydes and ketones, a carbon or hydrogen substituent is not an ordinary leaving group that can simply be expelled to restore the carbonyl. Their common reaction is addition. Carboxylic acid derivatives have suitable heteroatom leaving groups and often undergo addition followed by elimination, giving acyl substitution; that is a distinct mechanistic family.

The nucleophile approaches out of the carbonyl plane toward an antibonding acceptor orbital rather than sliding along the C=O bond axis. A two-dimensional mechanism is an electron-accounting representation. Unless stereochemistry is explicitly specified, a flat product drawing does not by itself prove which face was attacked.

R₂C=O + Nu⁻ → R₂C(O⁻)Nu →[protonation] R₂C(OH)Nu

Generic addition sequence; the actual proton source and reagent compatibility must be specified.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Electron-pair flow in nucleophilic additionA lone pair on Nu⁻ attacks the carbonyl carbon while the carbonyl pi electron pair moves onto oxygen. The resulting tetrahedral alkoxide has four carbon sigma bonds, to R, R′, Nu and O⁻. A separate proton-transfer step gives R₂C(OH)Nu. Curved arrows begin at the nucleophile electron pair and the pi bond, respectively.Nucleophilic carbonyl additionCurved arrows show movement of electron pairsOCRR′Nu⁻Nu lone pair → carbon; C=O π pair → oxygenO⁻CRNuR′Tetrahedral alkoxide intermediateBonds drawn flat for connectivity, not geometryproton transferR₂C(OH)NuA suitable proton donor converts O⁻ into OH.No C or H substituent is expelled as a leaving group.
Generic anionic-nucleophile model. Intermediate bonds are drawn flat to show connectivity, not a planar tetrahedral geometry. Actual reagent and proton-source compatibility must be specified.

04 / UNDERSTAND & APPLY

4. Hydride adds hydrogen without adding a carbon atom

A hydride donor reduces an aldehyde to a primary alcohol and a ketone to a secondary alcohol after protonation. In the simplified electron-flow picture, a reagent H–B or H–Al bond transfers hydride equivalent to the carbonyl carbon while the π pair moves to oxygen. The reagent is not a bottle of freely solvated H⁻ in ordinary protic solution.

NaBH₄ commonly reduces aldehydes and ketones under conditions where many ordinary esters and carboxylic acids are not reduced. LiAlH₄ is more reactive and reduces a wider range of carbonyl derivatives. These are useful chemoselectivity patterns, but unusual substrates, additives and conditions can change reactivity; state the intended standard conditions.

LiAlH₄ and strong organometallic reagents are incompatible with water or acidic O–H and N–H groups before the intended reaction because proton transfer consumes them. A reaction scheme therefore separates the anhydrous reagent stage from the subsequent protonating workup. Writing a strong hydride together with aqueous acid as one simultaneous medium hides this incompatibility.

Reduction of an unsymmetrical ketone can create a stereocenter. With an achiral substrate environment and achiral reagent, equal access to enantiotopic faces gives a racemic mixture in the simplest case. Pre-existing stereocenters, chiral catalysts or constrained geometry can make the two faces nonequivalent, so a racemic assumption needs its stated conditions.

ORIGINAL WORKED EXAMPLE

Reduce an aldehyde and an unsymmetrical ketone

Predict standard NaBH₄ reduction products after protonating workup for propanal and butan-2-one, with achiral conditions.

  1. Propanal CH₃CH₂CHO becomes CH₃CH₂CH₂OH: the carbonyl carbon gains hydrogen and oxygen becomes OH.
  2. Butan-2-one CH₃COCH₂CH₃ becomes CH₃CH(OH)CH₂CH₃.
  3. The second product has four different groups at the alcohol carbon, so it is chiral.
  4. With the stated achiral environment, the simple face-attack model gives racemic butan-2-ol.

Answer Propan-1-ol and racemic butan-2-ol.

05 / UNDERSTAND & APPLY

5. Carbon nucleophiles build a new carbon–carbon bond

An organomagnesium reagent R–MgX behaves as a strong carbon nucleophile and base in the usual carbonyl-addition model. Its carbon group bonds to the carbonyl carbon, generating an alkoxide that is protonated during a separate workup. Track the incoming carbon group explicitly; unlike hydride reduction, this increases the carbon skeleton.

Formaldehyde plus an organometallic carbon group gives a primary alcohol, an ordinary aldehyde gives a secondary alcohol, and a ketone gives a tertiary alcohol after workup. These classifications follow the number of carbon substituents attached to the former carbonyl carbon, not the total number of carbons in the entire molecule.

Acidic protons can defeat a proposed addition. A carboxylic acid, phenol, alcohol or suitable N–H group can consume a Grignard reagent by acid–base chemistry before the desired C–C bond forms. Protecting an interfering group or choosing a different route may therefore be necessary.

Cyanohydrin formation is another carbon-adding nucleophilic addition: the carbon end of CN⁻ adds to the carbonyl carbon, followed by proton transfer, giving R₂C(OH)C≡N. The new nitrile carbon is part of the product skeleton. The reaction is reversible and steric crowding can reduce favorable addition. Cyanide chemistry is presented here only as reaction theory because the reagents are acutely hazardous.

ORIGINAL WORKED EXAMPLE

Count carbons in a Grignard addition

Predict the product of propanal with a methyl organomagnesium reagent followed by separate protonating workup.

  1. Propanal is CH₃CH₂–C(=O)H.
  2. The incoming CH₃ group bonds to the carbonyl carbon, giving CH₃CH₂–C(O⁻)(H)–CH₃ before workup.
  3. Protonation yields CH₃CH₂CH(OH)CH₃, butan-2-ol.
  4. The product has four carbons, one more than propanal, and is secondary. In achiral conditions it is racemic.

Answer Butan-2-ol, racemic under the stated achiral model.

06 / UNDERSTAND & APPLY

6. Hydration is an equilibrium, not automatic complete conversion

Water can add to an aldehyde or ketone to form a geminal diol, R₂C(OH)₂. The equilibrium depends strongly on substituents. Formaldehyde is strongly hydrated in water, whereas many simple ketones have only a small equilibrium hydrate fraction. Electron-withdrawing groups can favor hydration by making the carbonyl more electrophilic.

Acid or base can catalyze equilibration through different sequences, but a catalyst does not change the equilibrium constant at fixed conditions. Acid activation uses carbonyl protonation followed by water attack and deprotonation; base catalysis begins with hydroxide attack and then proton transfer from water.

Bisulfite addition can form ionic carbonyl adducts for suitable aldehydes and less hindered ketones. Their formation and regeneration can support separation or identification in appropriate analytical settings, but a negative result is not proof that a carbonyl group is absent. Structure, equilibrium and solubility all affect whether a useful adduct appears.

For any reversible adduct, changing reactant activities or removing product can shift composition. Do not infer a complete synthetic conversion solely from drawing an equilibrium arrow. A reaction-prediction problem should distinguish the major isolated product under specified conditions from the collection of species present at equilibrium.

07 / UNDERSTAND & APPLY

7. Nitrogen nucleophiles often add, then eliminate water

A primary amine R′NH₂ can react with an aldehyde or ketone to form an imine, R₂C=NR′, plus water. The pathway includes nucleophilic addition to give a carbinolamine, proton transfers, loss of water after activating the hydroxyl group, and deprotonation of an iminium intermediate. The carbonyl oxygen leaves as part of water, not as an unexplained disappearing atom.

Mild acid catalysis can facilitate dehydration, but strongly acidic conditions protonate the amine and suppress its nucleophilicity. The useful pH range balances carbonyl activation, proton transfers and availability of unprotonated amine. Water removal favors condensation, whereas aqueous acid can hydrolyze an imine back to carbonyl and amine.

A secondary amine can form an enamine when the carbonyl substrate has an appropriate alpha hydrogen. After carbinolamine formation and dehydration to an iminium ion, removal of an alpha proton forms a C=C bond adjacent to nitrogen. Because the nitrogen has no remaining N–H at that stage for the primary-amine-style step, the product pattern differs.

Hydroxylamine gives an oxime, R₂C=NOH, and hydrazine derivatives give hydrazones, R₂C=NNHR′. These are useful carbonyl derivatives and may show geometrical isomerism when the substituents permit it. A carbonyl test based on derivative formation identifies a reaction class; it does not by itself uniquely identify an unknown structure.

ORIGINAL WORKED EXAMPLE

Write an imine with atom accounting

Predict the condensation product of ethanal and methylamine, ignoring E/Z detail and assuming conditions favor imine formation.

  1. Ethanal supplies CH₃–C(=O)H and methylamine supplies CH₃NH₂.
  2. Replace C=O by C=N–CH₃ after the addition–dehydration sequence.
  3. The imine is CH₃CH=NCH₃.
  4. The net equation also produces H₂O; oxygen and the removed hydrogens are accounted for.

Answer CH₃CH=NCH₃ + H₂O as the net condensation products.

08 / UNDERSTAND & APPLY

8. A hemiacetal is not yet a full acetal

Addition of one alcohol molecule to a carbonyl gives a hemiacetal carbon bearing both OH and OR. A full acetal has two OR groups on that carbon and no OH group there. Ketone-derived acetals are also often called ketals. The distinction is local to that carbon, not based on whether the molecule contains an unrelated hydroxyl group elsewhere.

Acid-catalyzed acetal formation begins with carbonyl activation, alcohol attack and deprotonation to a hemiacetal. Protonating the hemiacetal OH converts it into a good leaving group as water. Loss of water gives an oxocarbenium-type electrophilic intermediate, another alcohol attacks, and deprotonation produces the acetal while regenerating acid.

The net transformation is R₂C=O+2R′OH⇌R₂C(OR′)₂+H₂O. Excess alcohol or removal of water can favor acetal formation. Aqueous acid favors hydrolysis back to carbonyl and alcohol. A diol can form a cyclic acetal, often useful because both oxygen substituents belong to one tethered molecule.

Ordinary base does not convert a hemiacetal OH into a good leaving group by protonation, so the acid-catalyzed pathway to full acetal is not simply reproduced by adding base. Acetals are generally stable toward many bases and nucleophiles but acid-labile in water. This contrasting behavior underlies their use as carbonyl protecting groups.

R₂C=O + 2R′OH ⇌ R₂C(OR′)₂ + H₂O

Acid-catalyzed equilibrium; water activity helps control formation versus hydrolysis.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Distinguish carbonyl, hemiacetal and acetal connectivityThree local structures preserve R and R′ substituents. The carbonyl has a C=O double bond; the hemiacetal carbon has OH and OR″ groups; the full acetal carbon has two OR″ groups and no OH at that center. Acid-catalyzed alcohol addition under low water activity favors acetal formation; aqueous acid permits carbonyl regeneration.Carbonyl, hemiacetal and acetalFollow the oxygen bonds at the same carbonCRR′OCarbonyl: one C=O double bondCRR′OHOR″Hemiacetal: OH and OR on one carbonCRR′OR″OR″Acetal: two OR groups, no OH at this carbonAcid + alcohol, low water activity → formationAqueous acid → carbonyl regenerationConnectivity sketches; real tetrahedral bonds are 3D.
Local connectivity sketches, not full acid-catalyzed mechanisms. Ketone-derived acetals are often called ketals. R, R′ and R″ denote the stated substituent groups.

ORIGINAL WORKED EXAMPLE

Distinguish three oxygen arrangements

Classify CH₃CH(OH)(OCH₃), CH₃CH(OCH₃)₂ and CH₃CH₂OCH₃ at their indicated carbon frameworks.

  1. CH₃CH(OH)(OCH₃) has OH and OR on the same carbon, so it is a hemiacetal.
  2. CH₃CH(OCH₃)₂ has two OR groups on one carbon and no OH there, so it is an acetal.
  3. CH₃CH₂OCH₃ is an ordinary ether; no carbon bears the paired oxygen pattern of an acetal.
  4. Counting total oxygen atoms without checking connectivity would not make these distinctions.

Answer Hemiacetal; acetal; ether, respectively.

09 / UNDERSTAND & APPLY

9. Protection is a temporary change in reactivity, followed by restoration

When a molecule contains two groups that react with the intended reagent, a protecting group can temporarily mask one. For a ketone and an ester in the same molecule, LiAlH₄ would ordinarily reduce both. Converting the ketone into an acetal can allow the ester reduction to proceed while the masked ketone survives the basic hydride stage.

The sequence must include all three operations: install the protecting group under compatible conditions, perform the desired transformation elsewhere, then remove the protecting group to restore the carbonyl. Stopping at the acetal gives a different final functional group from the stated target.

For example, consider methyl 5-oxohexanoate, CH₃CO(CH₂)₃CO₂CH₃. Protect the ketone as a cyclic acetal with a suitable diol under acid-catalyzed water-removing conditions. Reduce the ester to CH₂OH with a suitable strong hydride stage and workup, then hydrolyze the acetal with aqueous acid. The displayed target CH₃CO(CH₂)₃CH₂OH specifies the open-chain connectivity. This hydroxyketone can also form a six-membered cyclic hemiketal by intramolecular addition of its terminal OH to the ketone; open-chain and cyclic forms may coexist, especially as acid catalyzes equilibration during deprotection. The route does not promise an exclusively open-chain isolated product.

The example illustrates compatibility, not a guarantee that every multifunctional substrate tolerates those operations. Acid-sensitive alkenes, other acetals, stereocenters prone to epimerization or groups reactive toward the hydride can create competing pathways. Protection can also introduce equilibrium and isolation costs; use it when it solves a real selectivity conflict.

Chemoselectivity can sometimes avoid protection entirely. Under suitable standard conditions, NaBH₄ can reduce a ketone in the presence of an ordinary ester, leaving the ester intact. That gives the opposite selective target from the protect-ketone/reduce-ester strategy. Decide which group must change before choosing the reagent.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Cyclic acetal protection enables selective ester reductionMethyl 5-oxohexanoate CH₃CO(CH₂)₃CO₂Me is shown with its ketone carbon protected in a five-membered ethylene acetal ring. That carbon retains CH₃ and (CH₂)₃CO₂Me substituents and bonds to both ring oxygens. A suitable strong hydride stage reduces the ester side chain to (CH₂)₃CH₂OH while the acetal remains. Aqueous acid then removes the acetal, yielding CH₃CO(CH₂)₃CH₂OH. The final formula specifies open-chain connectivity, not exclusive solution speciation: terminal OH addition to the restored ketone can give a six-membered cyclic hemiketal.Mask the ketone; transform the esterStarting skeleton: CH₃CO(CH₂)₃CO₂Me1. Cyclic acetal protection of the ketoneCCH₃(CH₂)₃CO₂MeOOCH₂CH₂Five-membered ethylene acetal ringThe ester remains outside the protecting ring.2. Suitable strong hydride: ester → CH₂OHKetone remains masked during this step.Protected center–(CH₂)₃CH₂OH3. Aqueous acid: remove the acetalCH₃CO(CH₂)₃CH₂OHKetone restored; ester carbon becomes CH₂OH.Me = CH₃. Check other functional-group compatibility.Final formula: open-chain connectivity.A six-membered cyclic hemiketal can also form.Open/cyclic equilibrium may accompany deprotection.
Conceptual protection–transformation–deprotection sequence. Reagent stages and workup are separate; compatibility with all other functional groups is required. This is not an experimental protocol. The final formula is the open-chain hydroxyketone; a six-membered cyclic hemiketal may coexist in equilibrium, particularly during acid-catalyzed deprotection.

ORIGINAL WORKED EXAMPLE

Choose between two selective targets

For CH₃CO(CH₂)₃CO₂CH₃, propose a conceptual route to reduce only the ketone, and another to reduce only the ester while restoring the ketone.

  1. To reduce only the ketone, use a suitable standard NaBH₄ reduction and workup, giving CH₃CH(OH)(CH₂)₃CO₂CH₃ while the ordinary ester remains.
  2. To reduce only the ester, first mask the ketone as an acetal.
  3. Reduce the ester with a suitable strong hydride stage, then hydrolyze the acetal to restore C=O.
  4. The second target CH₃CO(CH₂)₃CH₂OH is written in its open-chain form. Its terminal OH can add intramolecularly to give a six-membered cyclic hemiketal, so account for possible open/cyclic equilibrium under the acidic deprotection conditions. Check every intermediate for reagent compatibility.

Answer Ketone-only: selective mild hydride. Ester-only: ketone acetal protection → ester reduction → acetal hydrolysis. The ester-only target is displayed as open-chain connectivity; a six-membered cyclic hemiketal may also be present.

10 / UNDERSTAND & APPLY

10. Oxidation and complete deoxygenation change different bonds

Aldehydes are readily oxidized to carboxylic acids or carboxylates under suitable conditions. Ordinary ketones resist many mild oxidants, but stronger oxidative conditions can cleave carbon–carbon bonds. “Ketones cannot be oxidized” is therefore an overstatement; their oxidation often changes the carbon skeleton rather than simply adding oxygen without cleavage.

Tollens-type silver-ion oxidation commonly distinguishes aldehydes from ordinary ketones, but some other reducing substrates, including suitable alpha-hydroxy ketones, can also respond. A positive test is evidence for reducing behavior under the test conditions, not an infallible unique label for an aldehyde. Laboratory test preparation and disposal are outside this conceptual reaction note.

Reducing C=O to CH₂ is different from reducing C=O to CHOH. The Wolff–Kishner transformation proceeds through a hydrazone and base-promoted steps with nitrogen loss, giving a methylene group at the former carbonyl carbon. The overall carbon skeleton is retained when no competing reaction occurs.

Clemmensen reduction also converts suitable aldehydes or ketones to the corresponding hydrocarbon framework but uses strongly acidic metal-mediated conditions rather than the strongly basic hydrazone route. Acid-sensitive and base-sensitive groups therefore influence the choice. These named reactions are not interchangeable merely because their net carbonyl removal looks the same.

ORIGINAL WORKED EXAMPLE

Distinguish alcohol formation from oxygen removal

Compare the products of ordinary hydride reduction and complete carbonyl deoxygenation of CH₃COCH₂CH₃.

  1. Hydride reduction changes C=O into CHOH, giving CH₃CH(OH)CH₂CH₃.
  2. Complete deoxygenation changes the carbonyl carbon into CH₂, giving CH₃CH₂CH₂CH₃.
  3. Both preserve the four-carbon chain, but only the first retains oxygen.
  4. The deoxygenation method must be chosen for substrate compatibility with its acidic or basic conditions.

Answer Hydride: butan-2-ol. Deoxygenation: butane.

11 / UNDERSTAND & APPLY

11. Wittig olefination replaces oxygen by an ylide-derived carbon

A phosphorus ylide can convert an aldehyde or ketone into an alkene. The new C=C bond joins the original carbonyl carbon and the ylide carbon, while the carbonyl oxygen becomes part of a phosphine oxide product. Carbon accounting therefore requires keeping both reacting carbon centers visible.

For the simple methylene ylide Ph₃P=CH₂, the net transformation R₂C=O→R₂C=CH₂ installs a terminal methylene group. A substituted ylide can produce an alkene with E/Z possibilities. Selectivity depends on ylide stabilization, substrate and conditions; a generic Wittig arrow does not justify claiming one universal alkene geometry.

A useful mechanistic description involves formation and collapse of a four-membered oxaphosphetane-type intermediate, with detailed pathways depending on reagent class and conditions. The strong phosphorus–oxygen bond in the product contributes to the driving force. Do not draw the carbonyl oxygen vanishing or turning into water unless the actual reaction says so.

In a synthesis sequence, audit the carbon count after every carbon-adding step, the oxidation level after every redox step, and the survival of every other functional group. Aldol and Cannizzaro chemistry involve additional alpha-carbon or disproportionation logic and receive dedicated lessons rather than being conflated with all nucleophilic carbonyl additions.

  • Keep the C=O π-to-O arrow when adding a nucleophile.
  • Do not confuse a protonating workup with the strong-reagent reaction medium.
  • Count the new carbon in Grignard, cyanohydrin and Wittig products.
  • Distinguish hemiacetal OH/OR from acetal OR/OR connectivity.
  • An acetal must be removed if the target requires the original carbonyl.
  • Mild ketone reduction, ester reduction and carbonyl deoxygenation require different selectivity decisions.

ORIGINAL WORKED EXAMPLE

Map a Wittig product

Predict the alkene from cyclopentanone and the methylene ylide Ph₃P=CH₂ in the elementary product model.

  1. The carbonyl carbon remains part of the five-membered ring.
  2. Its oxygen is replaced by a double bond to the incoming CH₂ carbon.
  3. The product is methylenecyclopentane, with an exocyclic C=CH₂ bond.
  4. The ring is not enlarged; total carbon count increases from five to six, and phosphine oxide is the oxygen-containing coproduct.

Answer Methylenecyclopentane plus triphenylphosphine oxide.

12 / UNDERSTAND & APPLY

12. Eight solved product and mechanism checks

These are original conceptual reaction exercises. Reagent labels specify textbook transformations, not instructions for unsupervised chemical handling.

ORIGINAL PRACTICE · 1

What product results from ordinary NaBH₄ reduction of hexan-3-one followed by protonation?

Reveal answer & reasoning

Hexan-3-ol, racemic under achiral unbiased conditions.

The carbonyl carbon gains H and oxygen becomes OH. It bears ethyl and propyl groups, so the new alcohol carbon is stereogenic.

ORIGINAL PRACTICE · 2

Why is a C=O π-bond-to-oxygen arrow needed when Nu⁻ attacks carbonyl carbon?

Reveal answer & reasoning

It preserves the carbon octet while forming the new C–Nu bond.

The π electron pair becomes an oxygen lone pair, giving an alkoxide before protonation.

ORIGINAL PRACTICE · 3

Formaldehyde reacts with an ethyl organometallic carbon nucleophile and is then protonated. What alcohol skeleton forms?

Reveal answer & reasoning

Propan-1-ol, CH₃CH₂CH₂OH.

The two-carbon ethyl group bonds to the one-carbon formaldehyde center, making a three-carbon primary alcohol.

ORIGINAL PRACTICE · 4

Does a ketone converted to R₂C(OR′)₂ still contain a carbonyl at that carbon?

Reveal answer & reasoning

No; it is an acetal center.

The carbon has two C–O single bonds instead of a C=O bond. Aqueous acid can regenerate the carbonyl under suitable conditions.

ORIGINAL PRACTICE · 5

Why can excess aqueous acid reverse an acetal protecting group?

Reveal answer & reasoning

It catalyzes hydrolysis with water as reactant.

Protonation enables C–O bond cleavage and the reverse addition sequence, restoring carbonyl plus alcohol. Water-rich conditions favor deprotection.

ORIGINAL PRACTICE · 6

What is the net oxime product of propanone with hydroxylamine under suitable condensation conditions?

Reveal answer & reasoning

(CH₃)₂C=NOH, plus water.

The carbonyl C=O becomes C=N–OH through addition and dehydration; the hydroxylamine oxygen remains attached to nitrogen.

ORIGINAL PRACTICE · 7

A substrate contains an unprotected OH group and a ketone. Why might a proposed one-equivalent Grignard addition fail?

Reveal answer & reasoning

Acid–base reaction with OH can consume the reagent before carbonyl addition.

The carbon nucleophile is also a strong base. Protection, stoichiometry and compatibility must be addressed rather than assuming all reagent reaches C=O.

ORIGINAL PRACTICE · 8

What does methylene Wittig olefination do to propanone?

Reveal answer & reasoning

It gives 2-methylpropene, (CH₃)₂C=CH₂.

The carbonyl carbon retains its two methyl groups and forms a double bond to the ylide CH₂ carbon. The product has four carbons, one more than propanone.

KEEP THE BIG IDEAS

Quick revision

  • Carbonyl addition is driven by an electrophilic carbon and requires π electrons to move to oxygen.
  • Preparation routes must control oxidation level and preserve the intended carbon skeleton.
  • Hydride adds hydrogen; carbon nucleophiles add a new C–C bond.
  • Imine, oxime and acetal formation involve addition plus controlled elimination or substitution steps.
  • Protection has installation, compatible transformation and removal stages.
  • Every synthesis answer should check carbon count, charge, workup and competing functional groups.
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