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Aldehydes, Ketones and Carboxylic Acids · Class 12 Notes

Class 12 Chemistry notes on Aldehydes, Ketones and Carboxylic Acids: the carbonyl group, preparations and named reactions, nucleophilic addition and the reactivity order, the aldol and Cannizzaro fork, oxidation and reduction, the distinguishing tests, acidity, and a JEE Advanced tier on mechanism, tautomerism and the ortho effect.

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In short

One polar bond explains the whole chapter. Oxygen has already pulled the pi cloud away from the carbonyl carbon, so that carbon is electron poor and nucleophiles attack it — the opposite of an alkene. Reactivity falls as you add alkyl groups and bulk, which is why aldehydes react faster than ketones, and whether a carbonyl has an alpha hydrogen decides between aldol and Cannizzaro.

Contents
  1. ·How to Read This Set
  2. 1The Carbonyl Group one polar bond, and what follows
  3. 2Preparation routine routes and four named reactions
  4. 3Nucleophilic Addition the core reaction, and the reactivity order
  5. 4The Alpha Hydrogen Fork aldol against Cannizzaro
  6. 5Oxidation and Reduction how far each reagent goes
  7. 6Telling Them Apart the tests, in the order you run them
  8. 7Carboxylic Acids why they beat phenols and alcohols
  9. 8Mechanism and Stereochemistry JEE Advanced tier begins
  10. 9Keto and Enol Forms tautomerism and enol content
  11. 10The Ortho Effect the rule that ignores the group
  12. 11Acid Derivatives one order governs every interconversion
  13. ★Aldehydes, Ketones and Carboxylic Acids · Fact Sheet
0

How to Read This Set

This chapter is built in two tiers. Sections 1 to 7 are the full syllabus for NEET and JEE Main. Sections 8 to 11 are the JEE Advanced layer, where you are asked to explain the mechanism and predict the stereochemistry rather than just name a product.

The whole chapter in one line

The carbonyl carbon is electron poor, so nucleophiles attack it.

Alkenes react with electrophiles. Carbonyls do the exact opposite, because the oxygen has already pulled the pi cloud away from the carbon.

Think it through
How to use the two tiers
  • If you are sitting NEET or JEE Main, sections 1 to 7 plus the fact sheet are complete. Nothing in the Advanced tier will be asked of you.
  • If you are sitting JEE Advanced, read all eleven. The last four are where the marks separate candidates, because they cannot be answered from memory.
  • Read each section, then study the drawing beside it. The drawings carry information the prose deliberately does not repeat.
Trap alert
The one habit that fixes this chapter
  • Before writing any product, ask two questions in this order.
  • One: is this an aldehyde or a ketone? That fixes the reactivity.
  • Two: does it have an alpha hydrogen? That decides whether it can give an aldol at all, or must give Cannizzaro instead.
  • Almost every wrong answer in this chapter comes from skipping the second question.
1

The Carbonyl Group

The carbon and oxygen are joined by a double bond, exactly as two carbons are in an alkene. But oxygen is far more electronegative, so it drags the pi cloud towards itself. That single difference reverses the chemistry.

Same double bond as an alkene, opposite behaviour, purely because of polarity.THE CARBONYL GROUP: ONE POLAR BOND EXPLAINS THE WHOLE CHAPTERORHthe C=O bondδ−δ+O is far more electronegative,so it pulls the pi cloud acrossTHE CARBON IS ELECTRON POORso a NUCLEOPHILE attacks itNu− adds across the C=OTHE OXYGEN IS ELECTRON RICHso an ELECTROPHILE, often H+,attaches there firstWHY IT MATTERSThe carbon is sp2, so thegroup is TRIGONAL PLANARwith 120° angles.Alkenes react withELECTROPHILES.Carbonyls react withNUCLEOPHILES.Same bond, oppositebehaviour.
Same double bond as an alkene, opposite behaviour, purely because of polarity.

The carbon is sp2 hybridised, so the group is trigonal planar with bond angles near 120°. That flatness matters later, because it is why addition products come out racemic.

Think it through
Physical properties worth remembering
  • Carbonyl compounds have no O-H, so they cannot hydrogen bond to each other. Their boiling points sit between alkanes and alcohols of similar mass.
  • They can accept hydrogen bonds from water, so the lower members dissolve well.
  • Carboxylic acids form dimers through two hydrogen bonds, so they boil higher than alcohols of comparable mass.
2

Preparation

The routine routes

Starting materialReagentProduct
1° alcoholPCC, or Cu at 573 Kaldehyde
2° alcoholPCC or K2Cr2O7ketone
alkyneH2O / H2SO4, HgSO4ethyne gives ethanal, others give ketones
alkeneO3, then Zn / H2Oaldehyde and ketone (ozonolysis)
acyl chlorideH2 / Pd-BaSO4aldehyde (Rosenmund)
nitrileDIBAL-H, or SnCl2/HClaldehyde
esterDIBAL-Haldehyde

The four named preparations

Each of these is defined by what stops the reaction going too far.FOUR NAMED PREPARATIONS WORTH KNOWING COLDROSENMUNDreactionacyl chlorideH2 / Pd-BaSO4ALDEHYDEthe catalyst is poisoned on purpose, so it stops at the aldehydeSTEPHENreactionnitrileSnCl2 / HCl, then H2OALDEHYDEgoes through an imine, which is then hydrolysedETARDreactiontolueneCrO2Cl2, then H2OBENZALDEHYDEoxidises the methyl group but stops before the acidGATTERMANN-KOCHreactionbenzeneCO + HCl / AlCl3 + CuClBENZALDEHYDEputs a CHO straight onto the ring
Each of these is defined by what stops the reaction going too far.
Trap alert
Why the catalyst is poisoned on purpose

In Rosenmund reduction the palladium is deliberately poisoned with BaSO4 and sulfur. Without the poison the reaction would run straight past the aldehyde to the alcohol. The same idea appears in Lindlar's catalyst. A poisoned catalyst is not a weaker catalyst, it is a catalyst that has been told where to stop.

3

Nucleophilic Addition

The nucleophile attacks the carbon. The pi electrons move onto the oxygen, which becomes an alkoxide, and a proton then picks it up. The carbon changes from planar sp2 to tetrahedral sp3.

Aldehydes always beat ketones, and aliphatic always beats aromatic.REACTIVITY TOWARDS NUCLEOPHILIC ADDITIONHCHOtwo H, nothing in the wayCH3CHOone alkyl groupCH3COCH3two alkyl groupsC6H5CHOring donates by resonanceC6H5COCH3ring plus alkylC6H5COC6H5two rings, fully blockedreactivity falls this wayTwo reasons push the same way. STERIC: bigger groups block the nucleophile.ELECTRONIC: alkyl (+I) and aryl (resonance) both reduce the positive charge on the carbon.
Aldehydes always beat ketones, and aliphatic always beats aromatic.
ReagentProductNote
HCNcyanohydrinadds one carbon; hydrolysis then gives a hydroxy acid
NaHSO3bisulfite addition compoundcrystalline, so it is used to purify carbonyls
R-MgX, then H2OalcoholHCHO gives 1°, other aldehydes 2°, ketones 3°
alcohol / dry HClhemiacetal, then acetalacetals are used to protect a carbonyl group
NH2-OHoximeaddition, then elimination of water
NH2-NH2hydrazone2,4-DNP gives the orange test precipitate
Think it through
The bisulfite trick

The NaHSO3 adduct is a solid you can filter off, and dilute acid or alkali regenerates the original carbonyl. So it is a purification tool, not just a reaction. It works for aldehydes and for methyl ketones, but bulkier ketones are too hindered.

Reactions blurring together?
Sort the carbonyls by reactivity yourself.

Drag groups on and off the carbonyl and watch the reactivity bar move. The order stops being a list to memorise.

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4

The Alpha Hydrogen Fork

The hydrogen on the carbon next to the carbonyl is acidic, because the anion left behind is stabilised by the C=O. Whether a compound has one decides which reaction it can give at all.

Ask this one question before writing anything else.ONE QUESTION SPLITS THE WHOLE CHAPTER: IS THERE AN ALPHA HYDROGEN?count the H on the carbonNEXT to the C=OHAS an alpha hydrogenALDOL condensationdilute NaOHgives a beta-hydroxy carbonyl,which loses water on heatingNO alpha hydrogenCANNIZZARO reactionconcentrated NaOHone molecule is oxidised, anotherreduced: disproportionationHCHO and C6H5CHO have NO alpha hydrogen, so they give Cannizzaro and never aldol.That single fact answers a large share of the questions from this chapter.
Ask this one question before writing anything else.
Aldol condensation
2 CH3CHO  +  dil. NaOH  →  CH3-CH(OH)-CH2-CHO

The product is a beta-hydroxy aldehyde. On heating it loses water to give an alpha, beta-unsaturated carbonyl. A cross aldol between two different partners that both have alpha hydrogens gives four products, which is why it is rarely useful.

Cannizzaro reaction
2 HCHO  +  conc. NaOH  →  CH3OH  +  HCOONa

One molecule is oxidised to the acid salt, the other reduced to the alcohol. In a crossed Cannizzaro with HCHO present, formaldehyde is always the one oxidised, because it is the better hydride donor.

5

Oxidation and Reduction

ReagentDoes whatApplies to
Tollens' / Fehling'sgentle oxidation to the acidaldehydes only
KMnO4 or K2Cr2O7oxidation to the acidaldehydes; ketones need harsh conditions and cleave
NaBH4 or LiAlH4reduction to the alcoholboth aldehydes and ketones
Zn-Hg / conc. HClC=O all the way to CH2Clemmensen, acidic conditions
NH2NH2 / KOH, glycolC=O all the way to CH2Wolff-Kishner, basic conditions
Trap alert
Clemmensen or Wolff-Kishner: pick by what else is in the molecule

Both convert C=O into CH2, so the choice is never about the carbonyl. Use Wolff-Kishner if the molecule has an acid-sensitive group, and Clemmensen if it has a base-sensitive one. A question that mentions another functional group is telling you which to pick.

6

Telling Them Apart

Run the tests in this order. Each one narrows the field: carbonyl or not, then aldehyde or ketone, then aliphatic or aromatic.

Five tests, and what each one actually proves.WHICH TEST TELLS WHAT APARTTESTWHAT YOU SEEWHAT IT PROVES2,4-DNP (Brady's)orange or yellow pptany carbonyl: aldehyde OR ketoneTollens'silver mirroraldehyde only, aliphatic and aromaticFehling's / Benedict'sred Cu2OALIPHATIC aldehyde onlyIodoform, I2 + NaOHyellow CHI3 crystalsCH3CO- group, or CH3CH(OH)-NaHCO3brisk effervescencecarboxylic acid, NOT phenolThe pair examiners use most: Tollens' works for BOTH aliphatic and aromatic aldehydes,but Fehling's refuses benzaldehyde. That is how you separate the two aldehydes.
Five tests, and what each one actually proves.
Think it through
The iodoform test is not about aldehydes

It is positive for a CH3CO- group, or a CH3CH(OH)- group which oxidises to it. So ethanal gives it, propanone gives it and ethanol gives it, but propanal and methanol do not. Read the structure, not the family.

7

Carboxylic Acids

The whole order comes from how well the negative charge is spread once the proton leaves.WHY A CARBOXYLIC ACID BEATS A PHENOL, WHICH BEATS AN ALCOHOLtrichloroacetic acidpKa 0.65three -I groupsformic acidpKa 3.75no +I alkylbenzoic acidpKa 4.20ring is -Iacetic acidpKa 4.76referencephenolpKa 10.00charge on ONE oxygenethanolpKa 16.00barely acidic at allshorter bar = stronger acidCARBOXYLATEcharge shared by TWO equivalent oxygensboth C-O bonds become identicalPHENOXIDEcharge pushed onto ring CARBON atoms,which are far less electronegative
The whole order comes from how well the negative charge is spread once the proton leaves.
Think it through
The argument in three steps
  • An alkoxide has its charge on one oxygen with nothing to share it, so an alcohol is barely acidic.
  • A phenoxide spreads the charge into the ring, but onto carbon atoms, which hold it poorly.
  • A carboxylate spreads it over two equivalent oxygen atoms, and both C-O bonds become identical in length. That is the best stabilisation of the three.

What changes the strength

ChangeEffectBecause
add -I groups (Cl, NO2, F)strongerthe anion is stabilised
add more of themstronger stilltrichloroacetic beats dichloro beats chloro
move the -I group awayweakerinduction fades after about three carbons
add +I alkyl groupsweakeracetic acid is weaker than formic
-NO2 on a benzoic ringstronger-M and -I both pull
-OCH3 at the para positionweaker+M pushes density in

Reactions of the -COOH group

ReagentProductNote
NaHCO3salt + CO2 + H2Othe test that separates acids from phenols
alcohol / conc. H2SO4esteresterification, and it is reversible
PCl5, PCl3 or SOCl2acyl chlorideSOCl2 is preferred, since the by-products are gases
NH3, then heatamidegoes through the ammonium salt
LiAlH41° alcoholNaBH4 is too mild to touch a -COOH
Cl2 / red Palpha-chloro acidHell-Volhard-Zelinsky, needs an alpha hydrogen
soda lime, heatalkanedecarboxylation, loses CO2
Trap alert
Three things students lose marks on
  • NaBH4 does not reduce a carboxylic acid. Only LiAlH4 or diborane will.
  • HVZ needs an alpha hydrogen, so benzoic acid and formic acid cannot give it.
  • Formic acid has an aldehyde group hidden inside it, so it is the one carboxylic acid that reduces Tollens and Fehling reagents.
Tier 2
The JEE Advanced layer
Everything so far is complete for NEET and JEE Main. Advanced asks you to explain why, and to predict stereochemistry. That is what the next four sections add.
8

Mechanism and Stereochemistry

Advanced questions rarely ask for the product alone. They ask which face was attacked, what the intermediate was, and whether the product turns out optically active.

A planar carbonyl has two identical faces, so a new stereocentre forms both ways in equal amounts.THE MECHANISM: WHY THE PRODUCT OF CYANOHYDRIN FORMATION IS RACEMIC1. Nu attacks the planar carbonORHsp2, TRIGONAL PLANARCN−CN−the nucleophile can arrive fromEITHER face, equally often2. Two mirror-image products, in equal amountsOHRHCNROHRHCNSa 50:50 RACEMIC mixture, so it is optically inactiveWHY THIS MATTERSA planar sp2 carbon hastwo identical faces.Attack is equally likelyat each, so the newstereocentre is formedboth ways.No chiral influence meansno optical activity.
A planar carbonyl has two identical faces, so a new stereocentre forms both ways in equal amounts.
Think it through
Three mechanisms worth writing out once
  • Aldol: base removes the alpha hydrogen to give an enolate. That enolate is the nucleophile, and it attacks a second carbonyl.
  • Cannizzaro: hydroxide adds to the carbonyl, then a hydride shifts from that intermediate to a second molecule. It is an internal redox, which is exactly why it needs no alpha hydrogen.
  • Acetal formation: needs dry HCl. The acid protonates the oxygen first, which makes the carbon far more electrophilic. Water present would reverse the whole thing.
Trap alert
Why semicarbazide reacts at only one nitrogen

Semicarbazide has two NH2 groups, yet only one forms the semicarbazone. The other nitrogen has its lone pair delocalised into the neighbouring C=O, so it is no longer available to act as a nucleophile. Same molecule, two nitrogens, only one of them free.

9

Keto and Enol Forms

An alpha hydrogen can move to the carbonyl oxygen. The two forms are tautomers: real, separate structures in equilibrium, not resonance forms of one structure.

Enol content is tiny for a simple ketone, but rises sharply when the enol is stabilised.KETO AND ENOL: THE SAME MOLECULE, TWO FORMS IN EQUILIBRIUMOCH3CH2-HKETO formthe alpha H is acidictautomerismOHCH3CH2ENOL forman -ene plus an -olENOL CONTENTpropanone0.00025%cyclohexanone0.02%acetylacetone80%phenol100%Enol content rises when the enol is stabilised: conjugation, or an internal hydrogen bond.Phenol is 100 percent enol, because its keto form would destroy the aromatic ring.
Enol content is tiny for a simple ketone, but rises sharply when the enol is stabilised.
Think it through
Resonance and tautomerism are not the same thing
  • Resonance structures differ only in where the electrons are drawn. They are not real, separate species.
  • Tautomers differ in where an atom sits, here a hydrogen. Both are real and can in principle be isolated.
  • Acetylacetone is about 80 percent enol, because its enol is conjugated and held by an internal hydrogen bond. Phenol is 100 percent enol, since its keto form would break aromaticity.
10

The Ortho Effect

The proof is the contrast between the two methyl compounds: ortho strengthens, para weakens.THE ORTHO EFFECT: THE ONE ACIDITY RULE THAT IGNORES THE GROUPbenzoic acidH at orthopKa 4.20reference2-methylbenzoicCH3 at ortho, a +I grouppKa 3.91still STRONGER2-nitrobenzoicNO2 at ortho, a -I grouppKa 2.17stronger, as expected4-methylbenzoicCH3 at parapKa 4.37weaker, as expected2-methylbenzoic acid is STRONGER than benzoic acid, even though CH3 is electron donating.Any group at the ortho position twists the COOH out of the ring plane. It loses conjugationwith the ring, so the carboxylate is left better stabilised. Steric, not electronic.
The proof is the contrast between the two methyl compounds: ortho strengthens, para weakens.
Trap alert
The rule that ignores whether the group pushes or pulls

Every ortho-substituted benzoic acid is stronger than benzoic acid itself, whether the group is electron donating or withdrawing. A group at the ortho position twists the -COOH out of the ring plane, so it can no longer conjugate with the ring, and the carboxylate ends up better stabilised. The cause is steric, not electronic, which is precisely why the usual +I and -I reasoning fails here.

11

Acid Derivatives

One order governs every interconversion. You can move down it freely, never up.ACID DERIVATIVES: ONE ORDER EXPLAINS EVERY INTERCONVERSIONacyl chlorideR-COClCl is a weak +M donor and a great leaving groupanhydrideR-CO-O-CO-Rthe O is shared between two carbonylsesterR-COORO donates by +M, so the carbonyl is calmeramideR-CONH2N is the best +M donor of the fourreactivity towards nucleophilic acyl substitution falls this wayYou can always go DOWN this list, never up. An acyl chloride gives an ester or an amide easily.An amide will not give you an acyl chloride, because that is climbing against the order.
One order governs every interconversion. You can move down it freely, never up.
ReactionWhat it doesCondition
Baeyer-Villigerinserts an O next to the carbonyl, giving an estera peroxy acid, such as mCPBA
Perkinaromatic aldehyde to an unsaturated acidanhydride and its sodium salt
Benzoin condensationjoins two benzaldehydesaqueous ethanolic KCN
Hunsdieckersilver salt of an acid to an alkyl halideBr2; the chain shortens by one carbon
Kolbe electrolysiscarboxylate salt to an alkaneelectrolysis; the chain doubles
HVZalpha-halogenation of an acidCl2 / red P, needs an alpha hydrogen
Think it through
Baeyer-Villiger: which group migrates

The oxygen inserts on the side of the group that migrates best, and the order is tertiary alkyl > cyclohexyl > secondary > phenyl > primary > methyl. So an unsymmetrical ketone gives one ester in large excess rather than a mixture. Advanced questions turn on picking the right side.

Mechanisms still feel like memorising?
Push the arrows yourself and watch the intermediate form.

Run the aldol and Cannizzaro step by step, then flip a carbonyl over and see why the product comes out racemic.

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★ Aldehydes, Ketones and Carboxylic Acids · Fact Sheet

Every rule for revision day. Print this page alone.

THE CARBONYL

C is δ+, O is δ−

sp², trigonal planar, 120°
Nucleophiles attack the carbon.

REACTIVITY ORDER

HCHO > CH3CHO > CH3COCH3

> C6H5CHO > C6H5COCH3
Steric and electronic agree.

NAMED PREPS

Rosenmund: acyl chloride

Stephen: nitrile
Etard and Gattermann-Koch: ring.

THE ALPHA FORK

Has alpha-H: ALDOL, dil. NaOH

No alpha-H: CANNIZZARO, conc. NaOH
HCHO and PhCHO give Cannizzaro.

C=O TO CH2

Clemmensen: Zn-Hg / HCl, acidic

Wolff-Kishner: NH2NH2 / KOH, basic
Choose by the other group.

THE TESTS

2,4-DNP: any carbonyl

Tollens: any aldehyde
Fehling: ALIPHATIC aldehyde only.

IODOFORM

CH3CO- or CH3CH(OH)-

Ethanal, propanone, ethanol yes
Propanal and methanol no.

ACIDITY ORDER

Cl3CCOOH > HCOOH > C6H5COOH

> CH3COOH > phenol > ethanol
Carboxylate: charge on TWO O.

COOH REACTIONS

NaBH4 will NOT reduce it

HVZ needs an alpha-H
HCOOH reduces Tollens.

STEREOCHEMISTRY

Planar sp² has two faces

Cyanohydrin comes out RACEMIC
so it is optically inactive.

ENOL CONTENT

Propanone 0.00025%

Acetylacetone 80%
Phenol 100%, to keep aromaticity.

ORTHO EFFECT

ALL ortho-substituted benzoic

acids beat benzoic acid.
Steric, not electronic.

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