Aldehydes, Ketones and Carboxylic Acids | ISC Class 12 Chemistry Notes

This note covers the structures, names, preparation, physical properties and reactions of aldehydes, ketones and carboxylic acids, nucleophilic addition, carbonyl tests, condensation reactions, acidity, acid derivatives and uses.

How are aldehydes and ketones structured and named?

Definition: A functional group is an atom or group responsible for characteristic reactions. A carbonyl group is a carbon atom double-bonded to oxygen, C=O; a carboxyl group is COOH. An aldehyde has at least one hydrogen attached to its carbonyl carbon; a ketone has two carbon groups attached to that carbon.

In general formulae, R and R′ represent alkyl groups, which are carbon groups derived from alkanes, saturated hydrocarbons containing carbon-carbon single bonds; they may be identical or different. Ar represents an aryl group derived from an aromatic ring, such as a benzene ring. Aldehydes include RCHO and ArCHO; methanal is HCHO. Ketones contain R-C(=O)-R′, with alkyl or aryl groups.

What makes the carbonyl group polar?

The carbonyl carbon is sp²-hybridised: one s orbital and two p orbitals combine to form three equivalent bonding orbitals. An orbital describes where an electron is likely to be found. Carbon forms three sigma bonds by head-on orbital overlap. The remaining p orbital forms a pi bond with oxygen by sideways overlap.

The carbonyl carbon and its three attached atoms lie in one plane, with bond angles approximately 120°. Oxygen has two non-bonding electron pairs. Its greater electronegativity, or attraction for shared electrons, draws electron density towards oxygen.

Consequently, carbon is an electrophilic centre, able to accept an electron pair, while oxygen is a nucleophilic centre, able to donate one. The symbols δ⁺ and δ⁻ mean partial positive and partial negative charge. Resonance represents electron delocalisation using contributing structures with the same atomic arrangement.

The drawing shows sigma bonding in the molecular plane, sideways overlap forming a pi bond above and below that plane, and a planar carbonyl arrangement with angles labelled 120°.

See Fig. 8.1 in your NCERT textbook

How does systematic naming work?

IUPAC, the International Union of Pure and Applied Chemistry, provides systematic naming rules. Replace the final -e of the parent alkane with -al for an aldehyde or -one for a ketone. Number an aldehyde chain from CHO; number a ketone chain to give C=O the lowest possible position.

For CHO attached to a ring, use carbaldehyde. In common names, α, β and γ identify successive carbons away from the functional group. The α-carbon is directly beside the carbonyl carbon; an α-hydrogen is a hydrogen attached to that adjacent carbon.

How are carbonyl compounds prepared from alcohols and hydrocarbons?

What happens to alcohols?

Oxidation, here removal of hydrogen or addition of oxygen, generally converts primary alcohols, whose OH-bearing carbon is attached to one other carbon, into aldehydes. Secondary alcohols have two carbon groups at that carbon and give ketones. Controlled conditions are needed to stop a primary alcohol at the aldehyde stage.

Dehydrogenation means removal of hydrogen. Passing vapours of volatile alcohols over silver or copper catalysts produces aldehydes from primary alcohols and ketones from secondary alcohols. A catalyst changes reaction rate without being consumed overall.

Worked example 1. How can hexan-1-ol be converted into hexanal without continuing to the acid?

Answer: Use pyridinium chlorochromate, abbreviated PCC, for controlled oxidation. The terminal CH₂OH group becomes CHO while the six-carbon chain remains intact. Choosing a reagent that stops at the aldehyde is the essential step. Reduction reverses these types of change by adding hydrogen or removing oxygen.

How do alkenes and alkynes supply the carbonyl group?

Ozonolysis cleaves a carbon-carbon double bond using ozone, O₃. Subsequent treatment with zinc dust and water gives aldehydes, ketones or a mixture, depending on the groups attached to the original double-bonded carbons. Each of those carbons becomes a carbonyl carbon.

Worked example 2. Convert but-2-ene, CH₃CH=CHCH₃, into ethanal.

Answer: Treat with ozone, followed by zinc and water. Cleavage between the two double-bonded carbons produces two molecules of CH₃CHO. Each original double-bonded carbon retains its hydrogen and methyl group, explaining why the products are aldehydes.

Hydration is addition of water. Ethyne gives ethanal with sulphuric acid, H₂SO₄, and mercury(II) sulphate, HgSO₄. Other alkynes give ketones in this reaction. Propyne, CH₃C≡CH, gives propanone, CH₃COCH₃; the symbol ≡ denotes a triple bond.

Note: Oxidising an alcohol and reducing a carbonyl compound reverse the change between the alcohol and carbonyl functional groups. They require different reagents; the name of the starting material alone does not specify the reaction conditions.

Which selective methods prepare aldehydes and ketones?

How do acid chlorides, nitriles and esters react?

An acid chloride, RCOCl, is an acid derivative with chlorine replacing the OH of COOH. Here CO represents C=O in a condensed formula. In Rosenmund reduction, hydrogen over palladium supported on barium sulphate converts it into an aldehyde. For benzoyl chloride:

C₆H₅COCl + H₂ → C₆H₅CHO + HCl, using palladium on barium sulphate.

A nitrile contains C≡N. In the Stephen reaction, tin(II) chloride, SnCl₂, with hydrochloric acid reduces a nitrile to an imine, a compound containing C=N. Hydrolysis, bond cleavage by reaction with water, then gives the aldehyde.

Diisobutylaluminium hydride, abbreviated DIBAL-H, also selectively reduces nitriles and esters to aldehydes after hydrolysis. An ester contains COOR in place of the COOH of an acid. Thus ethanenitrile, CH₃CN, can be converted into ethanal using DIBAL-H followed by water.

A Grignard reagent is an organomagnesium halide, RMgX, where Mg is magnesium and X is a halogen. Reaction of a nitrile with this reagent, followed by hydrolysis, gives a ketone. Propanenitrile and phenylmagnesium bromide give propiophenone, C₆H₅COCH₂CH₃.

Acid chlorides react with dialkylcadmium compounds, R₂Cd, to give ketones. Here Cd denotes cadmium. Dialkylcadmium is prepared from a Grignard reagent and cadmium chloride. These routes distinguish selective aldehyde preparation from ketone preparation.

What other routes are useful?

Dry distillation, strong heating without added solvent, of calcium acetate gives acetone: (CH₃COO)₂Ca → CH₃COCH₃ + CaCO₃. Here Ca denotes calcium. Heating calcium formate with calcium acetate gives ethanal: (HCOO)₂Ca + (CH₃COO)₂Ca → 2CH₃CHO + 2CaCO₃.

Friedel-Crafts acylation introduces an acyl group, RCO, into an aromatic ring. Benzene reacts with an acid chloride in the presence of anhydrous aluminium chloride, AlCl₃, to give an aromatic ketone. Anhydrous means free from water.

Worked example 3. What is the product when benzene reacts with propanoyl chloride, CH₃CH₂COCl, using anhydrous AlCl₃?

Answer: The product is C₆H₅COCH₂CH₃, propiophenone or 1-phenylpropan-1-one. The propanoyl group replaces a ring hydrogen. The product has carbon groups on both sides of C=O and is therefore a ketone.

Why do carbonyl compounds show characteristic physical properties?

Methanal is a gas at room temperature, whereas ethanal is a volatile liquid. Other aldehydes and ketones are liquids or solids at room temperature. Volatility describes the tendency to enter the vapour state.

How do intermolecular forces affect boiling points?

Dipole-dipole interactions are attractions between the oppositely charged ends of polar molecules. Weak molecular association through these interactions makes aldehydes and ketones boil above hydrocarbons and ethers of comparable molecular masses.

Their boiling points are below those of alcohols of similar molecular masses because aldehydes and ketones lack intermolecular hydrogen bonding between their own molecules. A hydrogen bond is an attraction involving hydrogen bonded to a strongly electronegative atom and an electron-rich atom.

The values below show the comparison. K means kelvin, the unit of absolute temperature. Relative molecular mass compares molecular mass with one-twelfth of the mass of a carbon-12 atom and has no unit.

Why does solubility fall with chain length?

The lower members, such as methanal, ethanal and propanone, are miscible with water in all proportions: they form a single liquid phase with it. Their oxygen atoms accept hydrogen bonds from water molecules.

Solubility decreases rapidly as the alkyl chain becomes longer. Aldehydes and ketones are fairly soluble in organic solvents such as benzene, ether, methanol and chloroform. Lower aldehydes have sharp, pungent odours; increasing molecular size makes the odour less pungent and more fragrant.

How does nucleophilic addition occur at a carbonyl group?

A nucleophile donates an electron pair to form a bond. In a carbonyl compound, it attacks the partially positive carbon. Write Nu⁻ for a negatively charged nucleophile and H⁺ for a proton, the nucleus of a hydrogen atom. Addition places these species across C=O.

What are the mechanism steps?

1. The nucleophile approaches the carbonyl carbon approximately perpendicular to the plane of its three sigma bonds.
2. A new carbon-nucleophile bond forms while the carbonyl pi electrons move towards oxygen.
3. A tetrahedral alkoxide intermediate forms. An alkoxide contains negatively charged oxygen; tetrahedral describes four bonds directed towards the corners of a tetrahedron.
4. The oxygen accepts a proton from the medium, giving the neutral addition product. Protonation means addition of a proton.

The carbon changes from sp² to sp³ hybridisation, involving one s and three p orbitals. An intermediate is a species formed in one reaction step and consumed in a later step.

The figure labels a planar carbonyl starting structure, a slow first step leading to a tetrahedral intermediate, and a fast second step involving H⁺. The final structure has both Nu and OH attached to the former carbonyl carbon.

See Fig. 8.2 in your NCERT textbook

Why are aldehydes generally more reactive?

Steric hindrance is obstruction caused by surrounding groups. Ketones have two relatively large substituents that hinder approach more than the single such substituent in an aldehyde. Two alkyl groups also reduce the carbonyl carbon's electrophilicity more effectively through electron donation.

Benzaldehyde is less reactive than propanal towards nucleophilic addition because resonance with its benzene ring reduces carbonyl polarity. Therefore, the general aldehyde-ketone comparison must be combined with the actual groups present.

Hydrogen cyanide, HCN, gives a cyanohydrin, which has OH and CN on the same carbon. Pure HCN reacts very slowly. A base generates the stronger nucleophile cyanide, CN⁻; cyanide attack followed by protonation gives the product.

Sodium hydrogensulphite, NaHSO₃, also called sodium bisulphite, gives addition compounds by nucleophilic attack followed by proton transfer. The equilibrium lies largely towards products for most aldehydes, but towards reactants for most ketones because of steric effects.

The water-soluble hydrogensulphite compound releases the original carbonyl compound with dilute mineral acid or alkali. This reversible reaction is useful for separating and purifying aldehydes.

What products form with nitrogen reagents, Grignard reagents and reducing agents?

How do ammonia derivatives give carbonyl derivatives?

For ammonia derivatives, write H₂N-Z, where Z is the atom or group attached to nitrogen. Nitrogen attacks the carbonyl carbon; proton transfers produce an intermediate bearing OH and NH-Z. Elimination of water then forms C=N-Z. This reversible addition-elimination reaction is acid-catalysed.

2,4-Dinitrophenylhydrazine, abbreviated 2,4-DNP reagent here, forms yellow, orange or red solid derivatives useful for characterising aldehydes and ketones. A derivative is a compound made from another compound by a specified chemical change.

How does a Grignard reagent extend a carbon chain?

The carbon group of RMgX adds to the carbonyl carbon to form an alkoxide adduct, an addition product. Hydrolysis then gives an alcohol. Methanal yields a primary alcohol, other aldehydes yield secondary alcohols, and ketones yield tertiary alcohols, whose OH-bearing carbon has three carbon neighbours.

Worked example 4. Propanone reacts with methylmagnesium bromide and the adduct is hydrolysed. Identify the alcohol.

Answer: Addition of the methyl group gives a carbon bonded to three methyl groups. Hydrolysis gives (CH₃)₃COH, 2-methylpropan-2-ol, a tertiary alcohol. The carbonyl oxygen becomes the alcohol oxygen.

How do reductions differ?

Sodium borohydride, NaBH₄, and lithium aluminium hydride, LiAlH₄, reduce aldehydes to primary alcohols and ketones to secondary alcohols. Catalytic hydrogenation, addition of hydrogen using a catalyst, also gives these alcohols.

Clemmensen reduction uses zinc amalgam, zinc combined with mercury, and concentrated hydrochloric acid to replace carbonyl oxygen by two hydrogens. Wolff-Kishner reduction uses hydrazine followed by heating with sodium or potassium hydroxide in a high-boiling solvent such as ethylene glycol.

Heating with hydrogen iodide, HI, and red phosphorus also reduces the carbonyl group to a methylene group, CH₂. Thus these hydrocarbon-forming reductions differ from reductions that retain oxygen as OH.

How do oxidation and chemical tests distinguish carbonyl compounds?

Aldehydes are easily oxidised to carboxylic acids. Ketones are generally oxidised under vigorous conditions, using strong oxidising agents and elevated temperatures. Their carbon-carbon bonds break, producing a mixture of acids with fewer carbon atoms than the parent ketone.

What observations identify aldehydes?

Tollens’ reagent is freshly prepared ammoniacal silver nitrate solution. Warming it with an aldehyde produces a bright silver mirror. The aldehyde becomes a carboxylate ion, RCOO⁻, in this alkaline medium. A carboxylate is the ion formed when a carboxylic acid loses H⁺.

Fehling’s reagent is made by mixing equal amounts of aqueous copper sulphate solution and alkaline sodium potassium tartrate solution. Heating with an aliphatic aldehyde gives reddish-brown copper(I) oxide, Cu₂O. Aromatic aldehydes do not respond to this test.

What does the iodoform test establish?

A methyl ketone contains CH₃CO attached to another carbon group. Oxidation with a sodium hypohalite converts its methyl group into a haloform and leaves a carboxylate containing one fewer carbon atom. A hypohalite contains oxygen bonded to a halogen.

With iodine and sodium hydroxide the haloform is iodoform, CHI₃, a yellow precipitate, meaning an insoluble solid formed in solution. Ethanal also responds because it contains the required CH₃CO arrangement. The test also detects alcohols with CH₃CH(OH) that oxidise to that arrangement.

Worked example 5. A sample is either methanal or ethanal. It gives yellow iodoform when warmed with iodine and sodium hydroxide. Identify it.

Answer: It is ethanal. Ethanal has the required methyl group attached to the carbonyl carbon; methanal lacks it. A silver mirror alone would not distinguish these two aldehydes.

Derivation: Molecular formula from percentage composition

Worked example 7. A compound contains 69.77% carbon, 11.63% hydrogen and the rest oxygen, with molecular mass 86. It does not reduce Tollens’ reagent, forms a sodium hydrogensulphite addition compound and gives a positive iodoform test. Vigorous oxidation gives ethanoic and propanoic acids. Determine its molecular formula and structure.

1. Find the oxygen percentage by subtraction: $w_{\mathrm O}=100-69.77-11.63=18.60\%$.
2. Use a 100 g basis and relative atomic masses 12, 1 and 16 for carbon, hydrogen and oxygen. The amounts are $n_{\mathrm C}=\frac{69.77}{12}\,\mathrm{mol}\approx5.814\,\mathrm{mol}$, $n_{\mathrm H}=\frac{11.63}{1}\,\mathrm{mol}=11.63\,\mathrm{mol}$ and $n_{\mathrm O}=\frac{18.60}{16}\,\mathrm{mol}=1.1625\,\mathrm{mol}$.
3. Divide each amount by the smallest: $n_{\mathrm C}:n_{\mathrm H}:n_{\mathrm O}=5.814:11.63:1.1625\approx5:10:1$. The empirical formula is $\mathrm{C_5H_{10}O}$; the small departures from whole numbers result from rounding the percentages.
4. The empirical formula mass is $M_{\mathrm{emp}}=5(12)+10(1)+16=86$. The molecular formula multiplier is $k=\frac{86}{86}=1$, so the molecular formula is also $\mathrm{C_5H_{10}O}$.

Answer: The 100 g basis contains 69.77 g carbon, 11.63 g hydrogen and 18.60 g oxygen. The molecular formula is $\mathrm{C_5H_{10}O}$, and the structure is $\mathrm{CH_3COCH_2CH_2CH_3}$, pentan-2-one.

The hydrogensulphite addition compound indicates a carbonyl group. The negative Tollens’ test identifies a ketone; the positive iodoform test requires a methyl ketone. The oxidation products confirm the position: cleavage giving ethanoic and propanoic acids divides the five-carbon skeleton into fragments with two and three carbons.

Why do some compounds undergo aldol condensation and others Cannizzaro reaction?

The α-hydrogens of aldehydes and ketones are acidic because the carbonyl group withdraws electrons and the conjugate base is resonance-stabilised. A conjugate base is the species remaining after loss of H⁺. The resulting enolate ion has charge delocalised between oxygen and the α-carbon.

How does the aldol mechanism proceed?

1. Dilute hydroxide removes an α-hydrogen, giving an enolate ion.
2. The enolate carbon attacks the carbonyl carbon of another molecule and forms a carbon-carbon bond.
3. Protonation of the resulting alkoxide gives a β-hydroxy aldehyde or β-hydroxy ketone. Here β denotes the second carbon away from the carbonyl carbon.
4. Loss of water, called dehydration, produces an α,β-unsaturated carbonyl compound with a double bond between those two positions.

The initial addition is the aldol reaction; subsequent dehydration gives aldol condensation. Ethanal forms 3-hydroxybutanal, which loses water on heating to give but-2-enal:

2CH₃CHO → CH₃CH(OH)CH₂CHO → CH₃CH=CHCHO + H₂O.

The first step uses dilute sodium hydroxide. Ketones with α-hydrogen form β-hydroxy ketones, called ketols, and can also undergo condensation. Propanone gives 4-hydroxy-4-methylpentan-2-one before dehydration.

Cross aldol condensation involves two different aldehydes or ketones. When both have α-hydrogens, a mixture of four products results. Ethanal and propanal give the condensation products but-2-enal, 2-methylpent-2-enal, 2-methylbut-2-enal and pent-2-enal.

What changes in the Cannizzaro reaction?

Aldehydes without α-hydrogen undergo Cannizzaro reaction on heating with concentrated alkali. This is disproportionation: molecules of the same substance undergo oxidation and reduction. Benzaldehyde gives benzyl alcohol and sodium benzoate:

2C₆H₅CHO + NaOH → C₆H₅CH₂OH + C₆H₅COONa.

1. Hydroxide attacks one aldehyde molecule, forming a tetrahedral intermediate.
2. This intermediate transfers hydride, hydrogen with an electron pair, to a second aldehyde molecule.
3. The donor becomes a carboxylic acid while the acceptor becomes an alkoxide.
4. Proton transfer gives the carboxylate salt and alcohol.

Worked example 6. Compare ethanal and benzaldehyde in base.

Answer: Ethanal has α-hydrogens and undergoes aldol reaction with dilute alkali. Benzaldehyde lacks α-hydrogen and undergoes Cannizzaro reaction with concentrated alkali. Inspect the structure and the conditions together.

What additional reactions are important for benzaldehyde?

How is benzaldehyde prepared and how stable is it?

In the Etard reaction, toluene, C₆H₅CH₃, reacts with chromyl chloride, CrO₂Cl₂, to give a chromium complex. Hydrolysis of this intermediate produces benzaldehyde. Carbon disulphide, CS₂, is used as the solvent in this preparation.

Benzaldehyde is a colourless liquid that oxidises on exposure to air to benzoic acid. Reduction to the alcohol gives benzyl alcohol, C₆H₅CH₂OH; reduction of the carbonyl group to CH₂ gives toluene. These products represent different extents of reduction.

Its CHO group forms a cyanohydrin with HCN, a hydrogensulphite addition compound with NaHSO₃, and an oxime, hydrazone or phenylhydrazone with the corresponding nitrogen reagent. For example, hydroxylamine gives C₆H₅CH=N-OH after loss of water.

What are benzoin condensation and Perkin’s reaction?

Benzoin condensation joins two benzaldehyde molecules in the presence of a cyanide catalyst, commonly potassium cyanide in aqueous ethanol. It gives benzoin, an α-hydroxy ketone: 2C₆H₅CHO → C₆H₅CH(OH)COC₆H₅. This reaction does not require benzaldehyde to possess an α-hydrogen.

Perkin’s reaction heats benzaldehyde with ethanoic anhydride and anhydrous sodium ethanoate, the sodium salt of ethanoic acid. After hydrolysis the product is cinnamic acid, C₆H₅CH=CHCOOH. The net reaction is C₆H₅CHO + (CH₃CO)₂O → C₆H₅CH=CHCOOH + CH₃COOH.

Phosphorus pentachloride, PCl₅, replaces carbonyl oxygen with two chlorine atoms. Benzaldehyde gives benzal chloride: C₆H₅CHO + PCl₅ → C₆H₅CHCl₂ + POCl₃, where POCl₃ is phosphorus oxychloride. Aldehydes and ketones similarly form geminal dichlorides, with both chlorine atoms attached to the same carbon.

The CHO group is deactivating and meta-directing: it reduces the ring's reactivity towards electrophilic substitution and favours substitution at position 3 relative to itself. Halogenation, nitration and sulphonation respectively introduce a halogen, NO₂ or SO₃H group, giving mainly meta-substituted products. Typical reagents are bromine with iron(III) bromide, nitric acid with sulphuric acid, and fuming sulphuric acid respectively.

How are carboxylic acids classified, named and prepared?

A carboxyl group, COOH, consists of a carbonyl group joined to a hydroxyl group, OH. Carboxylic acids may be aliphatic, such as ethanoic acid, or aromatic, such as benzoic acid. Monocarboxylic acids contain one COOH group; dicarboxylic acids contain two.

Saturated, open-chain monocarboxylic acids have the molecular formula CₙH₂ₙO₂, where n is the total number of carbon atoms. The formula describes that class and does not apply unchanged to aromatic or dicarboxylic acids.

For simple open-chain acids, replace the alkane ending -e by -oic acid and number the carboxyl carbon as carbon 1. Bonds around the carboxyl carbon lie in one plane, separated by about 120°. Resonance makes this carbon less electrophilic than an aldehyde or ketone carbonyl carbon.

Which preparations preserve or extend the carbon chain?

Oxidation of primary alcohols and aldehydes gives carboxylic acids. Potassium permanganate, KMnO₄, or acidified potassium dichromate, K₂Cr₂O₇, can oxidise primary alcohols. Thus butan-1-ol and butanal both yield butanoic acid without changing the carbon count.

Nitrile hydrolysis proceeds through an amide, a compound containing CONH₂, to the acid. Mild conditions can stop at the amide stage. A Grignard reagent reacts with carbon dioxide, CO₂, to form a carboxylate salt; acidification gives the carboxylic acid.

Making a nitrile or Grignard reagent from an alkyl halide and then converting it into an acid adds one carbon to the original alkyl group. Alkyl halides have a halogen attached to an alkyl group. The added carbon comes from cyanide or carbon dioxide respectively.

Vigorous oxidation of alkylbenzenes can convert the side chain into COOH. Primary and secondary alkyl groups undergo this change; tertiary groups are not affected in this manner. Hydrolysis of acid chlorides, anhydrides or esters also provides acids.

Why are carboxylic acids high-boiling and acidic?

How do hydrogen bonds affect physical properties?

Aliphatic carboxylic acids containing up to nine carbon atoms are colourless liquids at room temperature with unpleasant odours. Higher acids are wax-like solids and practically odourless because of low volatility. Extensive intermolecular hydrogen bonding gives acids higher boiling points than aldehydes, ketones and even comparable alcohols.

Most carboxylic acids exist as dimers in the vapour phase or in aprotic solvents. A dimer is a pair of associated molecules; an aprotic solvent lacks a hydrogen that it readily donates in hydrogen bonding. Hydrogen bonds are not broken completely even in the vapour phase.

Draw two RCOOH molecules facing each other. Join each OH hydrogen to the other molecule's carbonyl oxygen using a dotted hydrogen bond. Label the cyclic pair “dimer” and mark both hydrogen bonds.

Simple aliphatic acids with up to four carbon atoms are miscible with water. Solubility decreases as the carbon chain grows. Higher acids are practically insoluble because their hydrocarbon portion is hydrophobic, meaning it interacts unfavourably with water. Benzoic acid is nearly insoluble in cold water.

How is acid strength explained?

Dissociation in water produces a resonance-stabilised carboxylate ion and hydronium ion, H₃O⁺: RCOOH + H₂O ⇌ RCOO⁻ + H₃O⁺. The symbol ⇌ indicates a reversible equilibrium, with forward and reverse reactions occurring at equal rates at equilibrium.

Derivation: Acid dissociation constant

The acid dissociation constant, $K_a$, measures the extent of acid dissociation in water. Square brackets indicate equilibrium concentrations.

1. For the equilibrium between the acid, water, carboxylate and hydronium ions, write $K_{\mathrm{eq}}=\frac{[\mathrm{H_3O^+}][\mathrm{RCOO^-}]}{[\mathrm{H_2O}][\mathrm{RCOOH}]}$.
2. In dilute aqueous solution, treat the water concentration as constant and incorporate it into the acid dissociation constant: $K_a=K_{\mathrm{eq}}[\mathrm{H_2O}]$.
3. Substitution cancels the water concentration, giving $K_a=\frac{[\mathrm{H_3O^+}][\mathrm{RCOO^-}]}{[\mathrm{RCOOH}]}$.

Result: Acid strength is also expressed as $\mathrm{p}K_a=-\log_{10}K_a$, where $\log_{10}$ is the base-ten logarithm. A smaller $\mathrm{p}K_a$ means a stronger acid.

Carboxylate has two equivalent resonance structures with negative charge distributed over two oxygen atoms. Phenoxide, the conjugate base of phenol, has non-equivalent contributors involving oxygen and less electronegative carbon. Carboxylate is better stabilised, explaining why carboxylic acids are stronger than alcohols and many simple phenols.

Electron-withdrawing groups stabilise carboxylate and increase acidity; electron-donating groups destabilise it and decrease acidity. The inductive effect is electron displacement through sigma bonds. Its influence decreases with distance, so a nearby halogen generally has a stronger effect than a more distant one.

Greater withdrawal explains the acidity sequence CF₃COOH > CCl₃COOH > CHCl₂COOH > CH₂ClCOOH. Here > means “more acidic than”. On an aromatic ring, a nitro group strengthens the acid, while a methoxy group can weaken it.

What reactions convert carboxylic acids into salts, derivatives and hydrocarbons?

How do acids form salts and derivatives?

Carboxylic acids react with active metals to release hydrogen, with alkalies to form salt and water, and with carbonates or hydrogencarbonates to release carbon dioxide. Sodium hydrogencarbonate is also called sodium bicarbonate. Representative equations are:

• 2RCOOH + 2Na → 2RCOONa + H₂; Na denotes sodium.
• RCOOH + NaOH → RCOONa + H₂O.
• 2RCOOH + Na₂CO₃ → 2RCOONa + H₂O + CO₂.
• RCOOH + NaHCO₃ → RCOONa + H₂O + CO₂.

Esterification is formation of an ester from an acid and an alcohol or phenol. Concentrated sulphuric acid or hydrogen chloride catalyses the reaction: RCOOH + R′OH ⇌ RCOOR′ + H₂O. Removing water or ester favours further ester formation.

In its mechanism, carbonyl oxygen is protonated, alcohol attacks, and proton transfer makes a water molecule available to leave. Water is eliminated and loss of a proton gives the ester. The overall process is nucleophilic acyl substitution, replacement of a group at an acyl carbon.

Heating acids with a dehydrating agent such as phosphorus pentoxide, P₂O₅, forms acid anhydrides, containing CO-O-CO. For ethanoic acid: 2CH₃COOH → (CH₃CO)₂O + H₂O.

Phosphorus pentachloride, phosphorus trichloride, PCl₃, or thionyl chloride, SOCl₂, converts COOH into COCl. Thionyl chloride is preferred because the other products escape as gases: RCOOH + SOCl₂ → RCOCl + SO₂ + HCl.

Ammonia first forms an ammonium carboxylate. Strong heating then gives an amide: CH₃COOH + NH₃ → CH₃COONH₄; CH₃COONH₄ → CH₃CONH₂ + H₂O.

How do reduction and decarboxylation differ?

LiAlH₄ reduces acids to primary alcohols; diborane, B₂H₆, is also effective. NaBH₄ does not reduce COOH. Decarboxylation removes carbon dioxide. Heating a sodium carboxylate with soda lime, NaOH and calcium oxide in a 3:1 ratio, gives a hydrocarbon with one fewer carbon atom.

CH₃COONa + NaOH → CH₄ + Na₂CO₃, on heating with soda lime.

Kolbe electrolysis passes electric current through an aqueous alkali-metal carboxylate solution. Decarboxylation and coupling give a hydrocarbon with twice the carbons of the acid's alkyl group. Sodium ethanoate gives ethane: 2CH₃COONa + 2H₂O → C₂H₆ + 2CO₂ + H₂ + 2NaOH.

Hell-Volhard-Zelinsky reaction, abbreviated HVZ, replaces an α-hydrogen by chlorine or bromine using a small amount of red phosphorus. Benzoic acid instead undergoes ring substitution: COOH is deactivating and meta-directing, so nitration with nitric acid and sulphuric acid, and sulphonation with fuming sulphuric acid, favour the meta position.

How are the acids distinguished and where are these compounds used?

Which observations separate formic, acetic and benzoic acids?

All three acids release carbon dioxide with sodium hydrogencarbonate. This identifies their carboxyl group but does not by itself distinguish them. The bubbling is effervescence caused by gas evolution.

Formic acid reduces Tollens’ reagent to give a silver mirror, while acetic and benzoic acids do not under this test. Formic acid is oxidised to carbon dioxide. This behaviour distinguishes it from the other two named acids.

For samples restricted to these three acids, cold-water solubility provides another distinction. Benzoic acid is nearly insoluble in cold water, whereas formic and acetic acids are miscible with water. Use solubility together with the chemical tests rather than treating any single observation as a universal identification.

What uses follow from this chemistry?

Formalin is the aqueous formaldehyde preparation used for preservation. Carbonyl compounds also supply starting materials for plastics, resins and other organic products. Keep the compound and its derivative distinct: the listed food-preservative use belongs specifically to sodium benzoate.

Glossary

• Carbonyl group — A carbon atom double-bonded to oxygen, forming the characteristic reactive group in aldehydes and ketones.
• Carboxyl group — The COOH functional group, containing a carbonyl group attached to a hydroxyl group.
• Nucleophile — An electron-pair donor that forms a bond by attacking an electron-deficient centre.
• Electrophile — An electron-pair acceptor that forms a bond with an electron-rich atom or species.
• Steric hindrance — Obstruction by neighbouring groups that makes approach to a reactive centre more difficult.
• Cyanohydrin — An addition product with hydroxyl and cyanide groups attached to the same carbon atom.
• Enolate ion — A resonance-stabilised conjugate base formed by removing an α-hydrogen from a carbonyl compound.
• Aldol condensation — Reaction forming a β-hydroxy carbonyl compound followed by dehydration to an α,β-unsaturated carbonyl compound.
• Cannizzaro reaction — Disproportionation of an aldehyde without α-hydrogen in concentrated alkali, producing alcohol and carboxylate.
• Conjugate base — The species remaining when an acid loses a proton during an acid-base reaction.
• Esterification — Formation of an ester by reacting a carboxylic acid with an alcohol or phenol.
• Decarboxylation — Removal of carbon dioxide from a carboxylic acid or its salt during a reaction.

Common errors and misconceptions

• Misconception: Every aldehyde gives Fehling’s test. Correct: Aromatic aldehydes such as benzaldehyde do not respond to this test.
• Misconception: Carbonyl compounds cannot hydrogen-bond with water. Correct: Their oxygen accepts hydrogen bonds from water, explaining the miscibility of lower members.
• Misconception: Every ketone gives iodoform. Correct: The test requires the appropriate CH₃CO arrangement; benzophenone does not contain it.
• Misconception: Aldol and Cannizzaro reactions need the same structures and conditions. Correct: Aldol requires α-hydrogen and dilute alkali; Cannizzaro uses aldehydes without α-hydrogen and concentrated alkali.
• Misconception: A larger pKₐ means a stronger acid. Correct: A smaller pKₐ corresponds to a stronger acid.
• Misconception: Sodium borohydride reduces carboxylic acids like aldehydes. Correct: It does not reduce COOH; lithium aluminium hydride can reduce acids to primary alcohols.
• Misconception: Soda-lime decarboxylation and Kolbe electrolysis give the same hydrocarbon. Correct: Sodium ethanoate gives methane with soda lime and ethane by Kolbe electrolysis.

Exam-style questions with model answers

Q1. Explain two reasons why aldehydes are generally more reactive than ketones towards nucleophilic addition. [2 marks]

1. Aldehydes offer less steric obstruction because they have fewer bulky carbon groups around the carbonyl carbon.
2. The two alkyl groups of a ketone donate electron density more effectively, reducing the carbonyl carbon's electrophilicity.

Q2. Separate samples are known to be methanal, ethanal and benzaldehyde. Describe a test sequence using iodine with sodium hydroxide and Fehling’s reagent to distinguish them, giving the observations. [3 marks]

1. Warm separate portions with iodine and sodium hydroxide. Ethanal gives a yellow iodoform precipitate because it contains the required methyl-carbonyl arrangement.
2. For the two samples that do not give iodoform, heat fresh portions with Fehling’s reagent. Methanal gives a reddish-brown precipitate.
3. Benzaldehyde does not respond to Fehling’s reagent. These observations distinguish all three within the stated set of possible samples.

Q3. The pKₐ values of 4-nitrobenzoic acid, benzoic acid and 4-methoxybenzoic acid are 3.41, 4.19 and 4.46 respectively. Arrange them in decreasing acid strength and explain both substituent effects. [3 marks]

1. Decreasing acid strength is 4-nitrobenzoic acid, benzoic acid, then 4-methoxybenzoic acid. Smaller pKₐ indicates a stronger acid, so the supplied values establish this order.
2. The nitro group withdraws electron density and stabilises the carboxylate conjugate base, increasing acidity relative to benzoic acid.
3. The methoxy substituent donates electron density in this comparison, destabilising the conjugate base and decreasing acidity relative to benzoic acid.

Q4. Explain the four main steps of base-catalysed aldol condensation of ethanal, CH₃CHO. Name the initial addition product and the product after heating. [4 marks]

1. Dilute hydroxide removes an α-hydrogen from ethanal. This produces a resonance-stabilised enolate ion, which acts as a carbon nucleophile.
2. The enolate attacks the carbonyl carbon of another ethanal molecule. A new carbon-carbon bond and an alkoxide intermediate form.
3. Protonation gives CH₃CH(OH)CH₂CHO, named 3-hydroxybutanal, the initial β-hydroxy aldehyde or aldol.
4. On heating, the aldol loses water to form CH₃CH=CHCHO, but-2-enal, the α,β-unsaturated condensation product.

Q5. Describe the Cannizzaro reaction of benzaldehyde, C₆H₅CHO: explain its structural suitability, state the conditions and equation, and outline the mechanism through the final products. [5 marks]

1. Benzaldehyde lacks an α-hydrogen. Its CHO group is attached directly to a benzene-ring carbon, so it is suitable for Cannizzaro reaction.
2. Heat it with concentrated sodium hydroxide. The balanced reaction is 2C₆H₅CHO + NaOH → C₆H₅CH₂OH + C₆H₅COONa.
3. Hydroxide attacks the carbonyl carbon of one molecule, producing a tetrahedral intermediate bearing negatively charged oxygen and a hydroxyl group.
4. The intermediate transfers hydride to another benzaldehyde molecule. The donor is oxidised towards the acid while the acceptor is reduced to an alkoxide.
5. Proton transfer gives benzyl alcohol and sodium benzoate. The same starting aldehyde undergoes both oxidation and reduction, so the overall reaction is disproportionation.

Q6. Describe five transformations of ethanoic acid, CH₃COOH: reaction with sodium hydrogencarbonate; esterification with ethanol; formation of ethanoyl chloride using thionyl chloride; conversion through ammonium ethanoate to ethanamide; and soda-lime decarboxylation of its sodium salt. Give equations and conditions where needed. [5 marks]

1. With sodium hydrogencarbonate, ethanoic acid gives carbon dioxide effervescence: CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂.
2. With ethanol and an acid catalyst, it forms ethyl ethanoate: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O. Concentrated sulphuric acid can catalyse the reaction.
3. Thionyl chloride gives ethanoyl chloride: CH₃COOH + SOCl₂ → CH₃COCl + SO₂ + HCl. The gaseous by-products leave the reaction mixture.
4. Ammonia forms ammonium ethanoate: CH₃COOH + NH₃ → CH₃COONH₄. Strong heating gives ethanamide: CH₃COONH₄ → CH₃CONH₂ + H₂O.
5. Heat sodium ethanoate with soda lime to obtain methane: CH₃COONa + NaOH → CH₄ + Na₂CO₃. The hydrocarbon has one fewer carbon atom than the acid.

Key takeaways

• The polar carbonyl group has an electrophilic carbon that undergoes nucleophilic addition through a tetrahedral intermediate.
• Aldehydes are generally more reactive than ketones because fewer bulky substituents and less alkyl electron donation favour attack at their carbonyl carbon.
• Reagent choice controls whether a carbonyl compound becomes an alcohol, a hydrocarbon or an oxidation product.
• Aldol condensation requires α-hydrogen; aldehydes without α-hydrogen undergo Cannizzaro reaction with concentrated alkali.
• Tollens’, Fehling’s and iodoform tests answer different structural questions and should be interpreted with their limitations.
• Carboxylate resonance stabilisation explains acid strength, while electron-withdrawing substituents generally strengthen the corresponding carboxylic acid.
• Carboxylic acids form salts and derivatives, undergo reduction and decarboxylation, and can react at their α-carbon or aromatic ring.
• Carbon counting distinguishes reactions that preserve the chain, add a carbon, remove a carbon or join two carbon groups.

Test yourself

Why does benzaldehyde react less readily than propanal in nucleophilic addition?

Resonance with the benzene ring reduces the carbonyl group's polarity and makes its carbon less electrophilic.

What distinguishes Rosenmund reduction from reduction with sodium borohydride?

Rosenmund reduction converts an acid chloride into an aldehyde; sodium borohydride converts aldehydes and ketones into alcohols.

What is obtained by hydrolysing the product of propanone and methylmagnesium bromide?

The product is 2-methylpropan-2-ol, a tertiary alcohol with three methyl groups on the OH-bearing carbon.

Why is a negative Fehling’s test insufficient to prove that a compound is a ketone?

Aromatic aldehydes, including benzaldehyde, also do not respond to Fehling’s test.

What does benzaldehyde form in benzoin condensation?

Two benzaldehyde molecules join under cyanide catalysis to form benzoin, C₆H₅CH(OH)COC₆H₅, an α-hydroxy ketone.

Why is thionyl chloride useful for preparing acid chlorides?

The other products, sulphur dioxide and hydrogen chloride, escape as gases and make product purification easier.

How does sodium ethanoate behave under two different decarboxylation conditions?

Heating with soda lime gives methane, whereas Kolbe electrolysis of its aqueous solution gives ethane.

Which structural condition is needed for the Hell-Volhard-Zelinsky reaction?

The carboxylic acid must have an α-hydrogen that chlorine or bromine can replace in the presence of red phosphorus.