Class 12Chemistry · Organic ChemistryFull chapter

Aldehydes, Ketones and Carboxylic Acids

The whole chapter in one place — read it, then test yourself. Clear notes, key equations, a practice quiz, and worked NCERT solutions & PYQs.

Structure, Nomenclature and Preparation of Aldehydes and Ketones

Quick answer The polar carbonyl group defines aldehydes and ketones; this section covers naming conventions and the standard laboratory routes used to make them.

The carbonyl group, C=O, is the defining feature of aldehydes and ketones. The carbonyl carbon is sp2 hybridised and lies at the centre of a trigonal planar arrangement, with bond angles close to 120°. In an aldehyde the carbonyl carbon is bonded to at least one hydrogen atom (the –CHO group), while in a ketone it is bonded to two carbon-containing groups (–CO–). Because oxygen is more electronegative than carbon, the C=O bond is strongly polarised, leaving the carbonyl carbon electrophilic (δ+) and the oxygen nucleophilic (δ−). This polarisation is the reason for almost all of the chemistry in this chapter.

In the IUPAC system, aldehydes are named by replacing the -e of the parent alkane with -al, and ketones by replacing it with -one. The longest chain containing the carbonyl carbon is the parent chain; for aldehydes the carbonyl carbon is always C1, while for ketones it is given the lowest possible locant. For example, CH3–CH(CH3)–CH2–CH2–CHO is numbered from the CHO carbon outward, giving 4-methylpentanal; CH3–CH2–CO–CH2–CH3 is pentan-3-one. Many common names (formaldehyde, acetaldehyde, acetone, benzaldehyde) remain in everyday use.

Several standard routes are used to prepare aldehydes and ketones:

  • Controlled oxidation of alcohols: primary alcohols give aldehydes and secondary alcohols give ketones. Strong oxidants (KMnO4, acidic K2Cr2O7) tend to over-oxidise primary alcohols to carboxylic acids, so a milder, selective reagent such as PCC (pyridinium chlorochromate) is used to stop cleanly at the aldehyde.
  • Rosenmund reduction: an acid chloride is hydrogenated over palladium on barium sulphate poisoned with sulphur/quinoline, which stops the reduction at the aldehyde stage instead of going on to the alcohol.
  • Stephen reaction: a nitrile is reduced with SnCl2/HCl to an imine salt, which is hydrolysed to the aldehyde.
  • Ozonolysis of alkenes: the C=C bond is cleaved by ozone followed by reductive work-up (Zn/H2O) to give two carbonyl fragments — useful both for synthesis and for structure determination.
  • Ketones from nitriles or acid chlorides: reaction of a nitrile with a Grignard reagent (followed by hydrolysis) or an acid chloride with a milder organometallic reagent gives a ketone; Friedel–Crafts acylation of an arene with an acid chloride/anhydride (AlCl3) is the standard route to aryl ketones.

Worked example: Predict the products of ozonolysis of 2-methylbut-2-ene, (CH3)2C=CH–CH3, followed by Zn/H2O work-up. The C=C bond is cleaved; each carbon of the former double bond becomes a carbonyl carbon. The more substituted carbon, (CH3)2C=, bears two methyl groups and becomes a ketone, (CH3)2C=O (acetone); the less substituted carbon, =CH–CH3, bears one hydrogen and becomes an aldehyde, CH3CHO (ethanal). So ozonolysis of 2-methylbut-2-ene gives a mixture of acetone and acetaldehyde.

General formula (aldehyde) CnH2nO (R–CHO) n = total carbons; the carbonyl carbon bears one H
General formula (ketone) CnH2nO (R–CO–R′) carbonyl carbon is bonded to two carbon groups
Rosenmund reduction R–COCl + H₂ →(Pd/BaSO₄, S-poisoned) R–CHO + HCl Selective reduction of an acid chloride to an aldehyde
Stephen reaction R–C≡N + SnCl₂/HCl → R–CH=NH →(H₂O) R–CHO Nitrile reduced to an imine salt, then hydrolysed to the aldehyde
Selective alcohol oxidation R–CH₂OH →(PCC) R–CHO PCC avoids further oxidation to the carboxylic acid
Remember
  • Carbonyl carbon is sp2, trigonal planar (~120 degrees); C=O is strongly polarised (delta+ at C, delta- at O)
  • Aldehyde: -CHO always at C1 of the chain; Ketone: -CO- within the chain; IUPAC suffixes -al and -one
  • PCC selectively oxidises 1 degree alcohols to aldehydes without over-oxidation to acids
  • Rosenmund reduction: acid chloride to aldehyde using H2/Pd-BaSO4 (poisoned catalyst)
  • Ozonolysis of alkenes cleaves C=C to give two carbonyl fragments
  • Friedel-Crafts acylation is the standard route to aryl ketones

Physical Properties and Nucleophilic Addition Reactions

Quick answer Carbonyl polarity governs boiling points and water solubility, and drives the characteristic nucleophilic addition chemistry of aldehydes and ketones.

Because the carbonyl group is polar but cannot hydrogen-bond with itself (there is no O–H or N–H), aldehydes and ketones have boiling points higher than hydrocarbons and ethers of comparable molar mass, but lower than alcohols of similar size. For example, propan-1-ol (molar mass 60, b.p. 97 °C) boils well above propanal (molar mass 58, b.p. 49 °C) or acetone (molar mass 58, b.p. 56 °C), even though the three have almost identical molar mass — only the alcohol can form intermolecular hydrogen bonds. The lower aldehydes and ketones (up to about 4 carbons) are miscible with water because the carbonyl oxygen can hydrogen-bond with water molecules; water solubility falls as the hydrocarbon part of the molecule grows.

The dominant reaction type of the carbonyl group is nucleophilic addition. A nucleophile attacks the electrophilic carbonyl carbon from a direction roughly perpendicular to the C=O plane; the π electrons move fully onto oxygen, generating a tetrahedral alkoxide intermediate, which is then protonated (by solvent or on aqueous work-up) to give the neutral addition product. Because this depends on how open and how electron-poor the carbonyl carbon is, reactivity decreases as alkyl substitution increases — both a steric effect (bulkier groups hinder the nucleophile's approach) and an electronic effect (alkyl groups are electron-donating and reduce the positive charge on carbon): HCHO > other aldehydes > ketones.

Typical nucleophilic addition reactions include:

  • Addition of HCN to give a cyanohydrin, R2C(OH)CN — adds one carbon and a nitrile that can later be hydrolysed to –COOH.
  • Addition of NaHSO3 to give a crystalline bisulphite addition compound, used to purify aldehydes and methyl ketones.
  • Addition of Grignard reagents (R′MgX) followed by acidic work-up, giving secondary or tertiary alcohols — one of the most important carbon–carbon bond-forming reactions in organic synthesis.
  • Addition of alcohols under acid catalysis to give a hemiacetal, then a second equivalent of alcohol gives the acetal (or ketal); this is reversible and acetals are used as protecting groups for carbonyls.
  • Addition of ammonia derivatives — hydroxylamine, hydrazine, phenylhydrazine, 2,4-dinitrophenylhydrazine — which add and then lose water to give an oxime, hydrazone or 2,4-DNP derivative; these are crystalline solids with sharp melting points, classically used to identify carbonyl compounds.

Worked example: Write the mechanism for base-catalysed addition of HCN to acetone. Step 1: CN attacks the electrophilic carbonyl carbon of (CH3)2C=O; the C=O π electrons shift onto oxygen. Step 2: this gives a tetrahedral alkoxide intermediate, (CH3)2C(O)(CN). Step 3: the alkoxide is protonated by HCN (regenerating CN) or by aqueous work-up, giving the cyanohydrin (CH3)2C(OH)CN, 2-hydroxy-2-methylpropanenitrile.

Cyanohydrin formation R₂C=O + HCN ⇌ R₂C(OH)CN Nucleophilic addition of CN; product is an alpha-hydroxynitrile
Bisulphite addition R₂C=O + NaHSO₃ → R₂C(OH)SO₃Na Crystalline addition compound with aldehydes/methyl ketones; used for purification
Acetal formation R–CHO + 2 R′OH →(H⁺, −H₂O) RCH(OR′)₂ Via a hemiacetal, RCH(OH)(OR′); acid-catalysed and reversible
Nucleophilic addition reactivity order HCHO > RCHO > R₂C=O Decreasing electrophilicity of carbonyl carbon with increasing alkyl substitution
Remember
  • No O-H/N-H on the carbonyl carbon, so aldehydes/ketones have boiling points between hydrocarbons/ethers and alcohols
  • Lower members (up to ~4 carbons) are water-soluble via H-bonding through the carbonyl oxygen
  • Mechanism: nucleophile attacks carbon, forms a tetrahedral alkoxide intermediate, then protonation
  • Reactivity order for nucleophilic addition: HCHO > aldehydes > ketones (steric + electronic reasons)
  • Cyanohydrin, bisulphite adduct, Grignard addition product, acetal/hemiacetal, and 2,4-DNP/oxime/hydrazone are key addition products

Reduction, Oxidation, Aldol Condensation and the Cannizzaro and Iodoform Reactions

Quick answer Carbonyl compounds can be reduced to alcohols or alkanes, oxidised (aldehydes only, easily), and combined with themselves via base-catalysed aldol or Cannizzaro reactions; the iodoform test identifies methyl ketones.

Aldehydes and ketones can be reduced in two distinct ways depending on the product needed. Reduction to the corresponding alcohol is achieved with hydride donors such as NaBH4 or LiAlH4 (or by catalytic hydrogenation, H2/Ni). Complete removal of the oxygen, reducing the carbonyl carbon all the way to a –CH2– group, is done by either the Clemmensen reduction (Zn–Hg amalgam with concentrated HCl; chosen for substrates that would be attacked by strong base) or the Wolff–Kishner reduction (hydrazine, NH2NH2, followed by strong base such as KOH in ethylene glycol, heat; chosen for substrates that would be damaged by strong acid).

Oxidation highlights a key chemical difference between aldehydes and ketones. Aldehydes have a hydrogen atom directly attached to the carbonyl carbon, so they are easily oxidised to carboxylic acids even by mild oxidants; this is the basis of two classic identification tests — Tollens' reagent (ammoniacal AgNO3), which deposits a bright silver mirror, and Fehling's solution (alkaline Cu2+ tartrate complex), which gives a brick-red precipitate of Cu2O. Ketones have no such hydrogen and resist mild oxidation, giving a negative result with both tests; only powerful oxidants under harsh conditions cleave a C–C bond of a ketone into a mixture of smaller carboxylic acids.

When an aldehyde or ketone has at least one alpha-hydrogen (a hydrogen on the carbon next to the carbonyl), dilute base can remove it to form a resonance-stabilised enolate, which attacks a second carbonyl molecule — the aldol reaction. The initial product is a beta-hydroxy aldehyde/ketone (an "aldol"); on warming, this readily loses water to give an alpha,beta-unsaturated carbonyl compound — the combined two-step sequence is aldol condensation. When one partner has no alpha-hydrogen (e.g. an aromatic aldehyde such as benzaldehyde) and is used with a ketone under base, the reaction is often called a Claisen–Schmidt condensation.

Carbonyl compounds with no alpha-hydrogen (e.g. HCHO, benzaldehyde, (CH3)3C–CHO) cannot undergo aldol condensation. Instead, with concentrated alkali they undergo the Cannizzaro reaction: two molecules of the aldehyde disproportionate, one being oxidised to a carboxylate salt and the other reduced to the alcohol.

Worked example (Cannizzaro stoichiometry): 60 g of formaldehyde, HCHO (molar mass 30 g/mol), is treated with excess concentrated NaOH. Moles of HCHO = 60/30 = 2 mol. The reaction is 2HCHO + NaOH → CH3OH + HCOONa, so 2 mol HCHO gives 1 mol CH3OH (molar mass 32) and 1 mol HCOONa (molar mass 68). Mass of methanol formed = 1 × 32 = 32 g; mass of sodium formate formed = 1 × 68 = 68 g. Check: 60 g HCHO + 40 g NaOH = 100 g reactants = 32 g + 68 g products, so mass is conserved.

Finally, methyl ketones (and any compound oxidisable to one, such as ethanol or propan-2-ol) give a positive iodoform test: I2/NaOH converts the CH3CO– group into a pale-yellow precipitate of triiodomethane (CHI3), leaving the rest of the molecule as the sodium salt of a carboxylic acid. This is both a qualitative test and a synthetic method for shortening a chain by one carbon.

Aldol condensation (example) 2 CH₃CHO →(dil. NaOH) CH₃CH(OH)CH₂CHO →(Δ, −H₂O) CH₃CH=CHCHO Enolate attacks a second carbonyl; the beta-hydroxy aldehyde dehydrates on heating
Cannizzaro reaction (general) 2 R–CHO + NaOH → R–CH₂OH + R–COONa Only for aldehydes with no alpha-H; an intermolecular disproportionation
Iodoform reaction CH₃–CO–R + 3I₂ + 4NaOH → CHI₃↓ + R–COONa + 3NaI + 3H₂O Positive for the CH3CO- or CH3CH(OH)- group; yellow precipitate of iodoform
Remember
  • NaBH4/LiAlH4 reduce C=O to alcohol; Clemmensen (Zn-Hg/HCl) or Wolff-Kishner (N2H4/KOH) give complete reduction to CH2
  • Aldehydes give positive Tollens' (silver mirror) and Fehling's (red Cu2O ppt) tests; ketones give negative results
  • Aldol condensation needs an alpha-H: base forms an enolate that adds to a second carbonyl, then dehydrates on heating
  • Cannizzaro reaction: carbonyl compounds with NO alpha-H disproportionate in conc. NaOH into an alcohol plus a carboxylate salt
  • Iodoform test (I2/NaOH giving yellow CHI3) is positive for the CH3CO- or CH3CH(OH)- structural unit

Carboxylic Acids: Structure, Nomenclature and Preparation

Quick answer The -COOH group combines carbonyl and hydroxyl character; this section covers its resonance-stabilised structure, IUPAC naming, and the main synthetic routes to carboxylic acids.

The functional group of a carboxylic acid is –COOH (the carboxyl group), formally a combination of a carbonyl (C=O) and a hydroxyl (–OH) on the same carbon. Saturated, straight-chain monocarboxylic acids have the general formula CnH2nO2. The carboxyl carbon is sp2 hybridised and planar. In the carboxylate ion (formed after loss of the acidic proton), both carbon–oxygen bonds are of equal, intermediate length — explained by resonance delocalisation of the negative charge equally over both oxygens. This delocalisation is the structural reason carboxylic acids are much more acidic than alcohols or phenols (covered in the next section).

In the IUPAC system, carboxylic acids are named by replacing the terminal -e of the corresponding alkane with -oic acid; the carboxyl carbon is always C1. For example, CH3–CH2–COOH is propanoic acid, and (CH3)2CH–CH2–COOH is 3-methylbutanoic acid. Common names — formic acid (HCOOH), acetic acid (CH3COOH), benzoic acid (C6H5COOH) — remain in wide use and are accepted by IUPAC.

Standard methods of preparation include:

  • Oxidation of primary alcohols or aldehydes with strong oxidants such as acidified KMnO4 or K2Cr2O7 — unlike the controlled oxidation used to stop at an aldehyde, here oxidation is taken to completion.
  • Oxidation of alkylbenzenes: hot alkaline KMnO4 oxidises a benzylic side chain, regardless of its length, all the way to –COOH, so toluene gives benzoic acid; this requires at least one benzylic hydrogen on the side chain.
  • Hydrolysis of nitriles, esters or amides under acidic or basic conditions regenerates the carboxylic acid (or its salt, which is then acidified).
  • Grignard reagent + CO2: a Grignard reagent adds to carbon dioxide to give a magnesium carboxylate, which on acidic work-up gives the carboxylic acid — this route adds exactly one carbon.

Worked example: 46 g of toluene (C6H5CH3, molar mass 92 g/mol) is refluxed with excess alkaline KMnO4 and then acidified, giving 80% of the theoretical yield of benzoic acid. Moles of toluene = 46/92 = 0.5 mol. The reaction C6H5CH3 + 3[O] → C6H5COOH + H2O is 1:1, so theoretical moles of benzoic acid (molar mass 122 g/mol) = 0.5 mol, i.e. theoretical mass = 0.5 × 122 = 61 g. At 80% yield, actual mass = 0.80 × 61 = 48.8 g.

General formula (saturated monocarboxylic acid) CnH2nO₂ (R–COOH)
Oxidation route R–CH₂OH →(KMnO₄ / K₂Cr₂O₇) R–CHO →([O]) R–COOH Strong oxidant taken to completion, unlike the PCC route which stops at the aldehyde
Grignard carboxylation R–MgX + CO₂ → R–COOMgX →(H₃O⁺) R–COOH Adds one carbon; acidic work-up liberates the free acid
Nitrile hydrolysis R–C≡N + 2H₂O →(H⁺/OH⁻, Δ) R–COOH + NH₃
Remember
  • -COOH combines a carbonyl and a hydroxyl on the same carbon; general formula CnH2nO2 for saturated monoacids
  • The carboxylate ion has two equal C-O bond lengths due to resonance delocalisation of the negative charge
  • IUPAC: suffix -oic acid; the carboxyl carbon is always numbered C1
  • Prepared by oxidation of 1 degree alcohols/aldehydes, oxidation of alkylbenzenes with hot KMnO4, hydrolysis of nitriles/esters/amides, and RMgX + CO2

Acidity of Carboxylic Acids and the Effect of Substituents

Quick answer Carboxylic acids are acidic because the carboxylate ion is resonance-stabilised; electron-withdrawing substituents strengthen this acidity further via the inductive effect.

Carboxylic acids are markedly acidic compared with alcohols and even phenols, because the carboxylate ion formed on loss of H+ is stabilised by resonance delocalisation of the negative charge over two equivalent oxygen atoms, whereas an alkoxide ion has no such delocalisation and a phenoxide ion delocalises charge less effectively (onto carbon, which is less electronegative than oxygen). Acid strength is expressed by the dissociation constant, Ka, or more conveniently by pKa = −log10Ka: a larger Ka (smaller pKa) means a stronger acid.

Substituents change acid strength mainly through the inductive effect. Electron-withdrawing groups (halogens, –NO2, –CN, –OR) pull electron density away from the carboxylate oxygens through the sigma-bond framework, further stabilising the negative charge and so increasing Ka (decreasing pKa). Electron-donating alkyl groups push electron density towards the carboxylate, destabilising the negative charge and so decreasing acid strength. The inductive effect falls off quickly with distance, so a halogen closer to –COOH increases acidity more than the same halogen farther away, and multiple electron-withdrawing groups have a cumulative (though diminishing) effect. For example, CH3COOH < ClCH2COOH < Cl2CHCOOH < Cl3CCOOH in acid strength (pKa approximately 4.76, 2.86, 1.29 and 0.66 respectively) shows both effects clearly.

Worked example: Calculate the pH of a 0.01 M solution of acetic acid (Ka = 1.8 × 10−5). Let [H3O+] = x at equilibrium. Then Ka = x²/(0.01 − x) ≈ x²/0.01 (assuming x is small compared with 0.01). So x² = 1.8 × 10−5 × 0.01 = 1.8 × 10−7, giving x = 4.24 × 10−4 mol/L. Checking: x/0.01 ≈ 4.2%, close to the usual 5% cut-off for the approximation; solving the exact quadratic gives x = 4.15 × 10−4 mol/L, essentially the same answer. So pH = −log(4.2 × 10−4) ≈ 3.38 — far more acidic than pure water (pH 7), but, as expected for a weak acid, well short of a strong acid at the same concentration (pH 2.00).

Acid dissociation constant Ka = [RCOO⁻][H₃O⁺] / [RCOOH] Larger Ka means a stronger acid
pKa pKa = −log₁₀(Ka) Smaller pKa means a stronger acid
Weak-acid [H3O+] approximation [H₃O⁺] ≈ √(Ka × C) mol L⁻¹ · Valid when degree of dissociation is small (below about 5%)
Remember
  • Carboxylic acids are more acidic than alcohols/phenols because the carboxylate ion is resonance-stabilised over two equal oxygens
  • Ka = acid dissociation constant; pKa = -log(Ka); a smaller pKa means a stronger acid
  • Electron-withdrawing groups (halogens, NO2) near -COOH increase acidity through the inductive effect; alkyl groups decrease it
  • The inductive effect weakens rapidly with distance from -COOH; multiple EWGs have a cumulative but diminishing effect
  • For a weak acid, [H3O+] is approximately the square root of (Ka times concentration) when dissociation is small

Chemical Reactions and Uses of Carboxylic Acids

Quick answer Carboxylic acids form esters, acid chlorides and anhydrides, undergo alpha-halogenation and decarboxylation, and direct electrophilic substitution to the meta position on aromatic rings.

The –COOH group undergoes several characteristic transformations. In esterification (Fischer esterification), a carboxylic acid reacts reversibly with an alcohol in the presence of a strong acid catalyst (conc. H2SO4 or dry HCl gas) and heat to give an ester and water: R–COOH + R′OH ⇌ R–COOR′ + H2O. Because this is an equilibrium, ester yield can be improved by using an excess of one reactant or by removing water as it forms. Carboxylic acids can also be converted to more reactive derivatives: acid chlorides with SOCl2 or PCl5, and anhydrides by dehydration or reaction with an acid chloride; both are far more electrophilic than the parent acid and are used as convenient acylating agents.

Worked example (equilibrium yield): 1 mol acetic acid is mixed with 1 mol ethanol and allowed to reach esterification equilibrium, for which Keq = 4. Let x mol of ester form at equilibrium. Then Keq = [ester][water] / [acid][alcohol] = x² / (1−x)² = 4. Taking the square root of both sides, x/(1−x) = 2, so x = 2(1−x) = 2 − 2x, giving 3x = 2 and x = 2/3 ≈ 0.667 mol of ethyl acetate at equilibrium — about 67% conversion, even though the reaction never goes to completion because it is a true equilibrium.

The alpha-carbon of a carboxylic acid can be halogenated by the Hell–Volhard–Zelinsky (HVZ) reaction: treatment with Cl2 or Br2 in the presence of a catalytic amount of red phosphorus converts the acid, via its acid halide (which is more enolisable than the acid itself), into the alpha-halo acid: R–CH2–COOH + X2 →(red P) R–CHX–COOH + HX (X = Cl, Br). Carboxylic acid salts undergo decarboxylation on heating with soda lime (NaOH/CaO), losing carbon as carbonate to give an alkane with one fewer carbon than the original acid — a classical method of shortening a carbon chain. Aromatic carboxylic acids undergo electrophilic aromatic substitution at the ring; because –COOH is electron-withdrawing, it deactivates the ring and directs incoming electrophiles to the meta position. Finally, –COOH is reduced all the way to –CH2OH by the strong hydride donor LiAlH4 (NaBH4 is not strong enough for this reduction).

Practically, low-molecular-weight carboxylic acids and their salts are used as food preservatives (e.g. acetic acid in vinegar, sodium benzoate), and esters of carboxylic acids are responsible for many natural fruit fragrances and are used as synthetic flavouring and fragrance agents.

Fischer esterification R–COOH + R′OH ⇌(H⁺, Δ) R–COOR′ + H₂O Acid-catalysed and reversible; equilibrium constant is typically only a few units for simple acid + alcohol pairs
Hell-Volhard-Zelinsky reaction R–CH₂–COOH + X₂ →(red P) R–CHX–COOH + HX Alpha-halogenation via the enol of the intermediate acid halide
Soda-lime decarboxylation R–COONa + NaOH →(CaO, Δ) R–H + Na₂CO₃ Converts the acid salt to an alkane with one fewer carbon
Acid to primary alcohol reduction R–COOH →(LiAlH₄) R–CH₂OH NaBH4 cannot reduce a free carboxylic acid
Remember
  • Fischer esterification: RCOOH + R'OH gives RCOOR' + H2O, acid-catalysed and reversible; excess reactant or water removal improves ester yield
  • SOCl2 or PCl5 convert -COOH to the far more reactive acid chloride
  • HVZ reaction (X2 with catalytic red P) installs a halogen at the alpha-carbon of the acid
  • Soda-lime decarboxylation removes -COOH as Na2CO3, giving an alkane with one fewer carbon
  • -COOH is a deactivating, meta-directing group on an aromatic ring; LiAlH4 (not NaBH4) reduces -COOH to -CH2OH

Key equations

Every formula in this chapter, in one place — screenshot it before your exam.

CnH2nO (R–CHO)
General formula (aldehyde)
CnH2nO (R–CO–R′)
General formula (ketone)
R–COCl + H₂ →(Pd/BaSO₄, S-poisoned) R–CHO + HCl
Rosenmund reduction
R–C≡N + SnCl₂/HCl → R–CH=NH →(H₂O) R–CHO
Stephen reaction
R–CH₂OH →(PCC) R–CHO
Selective alcohol oxidation
R₂C=O + HCN ⇌ R₂C(OH)CN
Cyanohydrin formation
R₂C=O + NaHSO₃ → R₂C(OH)SO₃Na
Bisulphite addition
R–CHO + 2 R′OH →(H⁺, −H₂O) RCH(OR′)₂
Acetal formation
HCHO > RCHO > R₂C=O
Nucleophilic addition reactivity order
2 CH₃CHO →(dil. NaOH) CH₃CH(OH)CH₂CHO →(Δ, −H₂O) CH₃CH=CHCHO
Aldol condensation (example)
2 R–CHO + NaOH → R–CH₂OH + R–COONa
Cannizzaro reaction (general)
CH₃–CO–R + 3I₂ + 4NaOH → CHI₃↓ + R–COONa + 3NaI + 3H₂O
Iodoform reaction
CnH2nO₂ (R–COOH)
General formula (saturated monocarboxylic acid)
R–CH₂OH →(KMnO₄ / K₂Cr₂O₇) R–CHO →([O]) R–COOH
Oxidation route
R–MgX + CO₂ → R–COOMgX →(H₃O⁺) R–COOH
Grignard carboxylation
R–C≡N + 2H₂O →(H⁺/OH⁻, Δ) R–COOH + NH₃
Nitrile hydrolysis
Ka = [RCOO⁻][H₃O⁺] / [RCOOH]
Acid dissociation constant
pKa = −log₁₀(Ka)
pKa
[H₃O⁺] ≈ √(Ka × C)
Weak-acid [H3O+] approximationmol L⁻¹
R–COOH + R′OH ⇌(H⁺, Δ) R–COOR′ + H₂O
Fischer esterification
R–CH₂–COOH + X₂ →(red P) R–CHX–COOH + HX
Hell-Volhard-Zelinsky reaction
R–COONa + NaOH →(CaO, Δ) R–H + Na₂CO₃
Soda-lime decarboxylation
R–COOH →(LiAlH₄) R–CH₂OH
Acid to primary alcohol reduction

Test yourself

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0 correct · 0/12 answered
Q1 Nomenclature easy

What is the IUPAC name of CH3-CHO?

Q2 Functional groups easy

Which functional group is present in all carboxylic acids?

Q3 Oxidation tests easy

Aldehydes give a positive Tollens' test (silver mirror) mainly because they:

Q4 Aldol condensation medium

Which of these carbonyl compounds cannot undergo aldol condensation?

Q5 Acid derivatives medium

Which reagent converts a carboxylic acid directly into the corresponding acid chloride?

Q6 Cannizzaro reaction medium

The Cannizzaro reaction is characteristic of aldehydes that:

Q7 Iodoform test medium

Which of the following gives a positive iodoform test?

Q8 Acidity and substituent effects medium

Arrange in increasing order of acid strength: acetic acid (I), chloroacetic acid (II), dichloroacetic acid (III).

Q9 Acidity (numerical) hard

The Ka of acetic acid is 1.8 x 10^-5. What is the approximate pH of a 0.01 M acetic acid solution?

Q10 HVZ reaction hard

In the Hell-Volhard-Zelinsky (HVZ) reaction, propanoic acid reacts with Br2 in the presence of catalytic red phosphorus to give:

Q11 Reduction of carbonyl compounds hard

Which statement correctly distinguishes the Clemmensen and Wolff-Kishner reductions of a carbonyl group to -CH2-?

Q12 Decarboxylation (numerical) hard

8.2 g of sodium acetate, CH3COONa (molar mass 82 g/mol), is heated with soda lime. What mass of methane gas is evolved, assuming complete decarboxylation?

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 Write the IUPAC names of: (i) CH3-CH(CH3)-CH2-CH2-CHO, (ii) C6H5-CO-CH2-CH3, (iii) CH3-CH2-COOH.Nomenclature

IUPAC names:

  • (i) CH3–CH(CH3)–CH2–CH2–CHO: the carbonyl carbon of an aldehyde is always C1. Numbering from CHO: C1(CHO)–C2H2–C3H2–C4H(CH3)–C5H3. This is a 5-carbon (pentanal) chain with a methyl branch at C4, so the name is 4-methylpentanal.
  • (ii) C6H5–CO–CH2–CH3 (propiophenone): the parent chain is propan-1-one with a phenyl group on C1, giving 1-phenylpropan-1-one.
  • (iii) CH3–CH2–COOH: a 3-carbon acid, propanoic acid.
2 How will you convert ethanal (acetaldehyde) into propanone (acetone)?Interconversion / synthesis

Ethanal (CH3CHO) has two carbons while propanone (CH3COCH3) has three, so a carbon must be added; this is done via a Grignard reagent.

  1. React ethanal with methylmagnesium iodide (CH3MgI): CH3CHO + CH3MgI → CH3CH(OMgI)CH3, then hydrolyse with dilute acid to give propan-2-ol, CH3CH(OH)CH3.
  2. Oxidise propan-2-ol with acidified K2Cr2O7 (or PCC): CH3CH(OH)CH3 →([O]) CH3COCH3 (propanone).

Since propan-2-ol is a secondary alcohol, its oxidation stops cleanly at the ketone stage, as it cannot be oxidised further without breaking a carbon-carbon bond.

3 Give a simple chemical test to distinguish between propanal (an aldehyde) and propanone (a ketone).Distinguishing tests

Add Tollens' reagent (ammoniacal AgNO3) to both compounds and warm gently.

  • Propanal (CH3CH2CHO) has a hydrogen atom on the carbonyl carbon, so it is readily oxidised to propanoic acid; Ag+ is reduced to metallic silver, which deposits as a bright silver mirror on the test tube.
  • Propanone (CH3COCH3) has no hydrogen on the carbonyl carbon, so it cannot be oxidised this way; no silver mirror forms.

Fehling's solution works equally well: propanal gives a brick-red precipitate of Cu2O, propanone gives no change.

4 Arrange the following in increasing order of boiling point, and explain: CH3CH2CH2CH3 (butane), CH3CH2CHO (propanal), CH3COCH3 (propanone), CH3CH2CH2OH (propan-1-ol).Physical properties

Increasing order of boiling point: butane (−0.5 °C) < propanal (49 °C) < propanone (56 °C) < propan-1-ol (97 °C).

All four compounds have very similar molar masses (58–60 g/mol), so the differences arise entirely from intermolecular forces. Butane is non-polar, held together only by weak London dispersion forces, so it boils lowest (it is a gas at room temperature). Propanal and propanone both have a polar C=O bond and experience dipole-dipole attraction, raising their boiling points well above butane's. Propan-1-ol boils highest because its –OH group lets its molecules hydrogen bond with each other, a much stronger intermolecular force than dipole-dipole attraction alone.

5 Arrange the following in increasing order of acid strength, and explain: acetic acid, chloroacetic acid, dichloroacetic acid, trichloroacetic acid.Acidity and substituent effects

Increasing acid strength: CH3COOH < ClCH2COOH < Cl2CHCOOH < Cl3CCOOH (approximate pKa values 4.76, 2.86, 1.29 and 0.66 respectively).

Chlorine is strongly electronegative and withdraws electron density through the sigma-bond framework (a −I inductive effect). This inductive pull stabilises the negative charge on the carboxylate ion once the acidic proton is lost, making the proton easier to lose, i.e. increasing Ka and decreasing pKa. Each additional chlorine atom on the alpha-carbon adds further electron withdrawal, so acidity rises as one, two, then three chlorines are introduced, though the increase becomes smaller with each successive chlorine.

6 Benzaldehyde is treated with concentrated NaOH. Explain why it does not undergo aldol condensation, name the reaction it undergoes instead, and give the products with a brief mechanism.Cannizzaro reaction

Benzaldehyde has no alpha-hydrogen (the carbon attached to CHO is part of the aromatic ring and has no replaceable H), so it cannot form an enolate and cannot undergo aldol condensation. Instead, with concentrated NaOH it undergoes the Cannizzaro reaction:

2 C6H5CHO + NaOH → C6H5CH2OH (benzyl alcohol) + C6H5COONa (sodium benzoate)

Mechanism (outline):

  1. Hydroxide ion adds to the carbonyl carbon of one molecule of benzaldehyde, giving a tetrahedral alkoxide intermediate, C6H5CH(O)(OH).
  2. This intermediate transfers a hydride ion to the carbonyl carbon of a second molecule of benzaldehyde.
  3. The molecule that donated the hydride is oxidised to benzoic acid, instantly deprotonated by excess NaOH to sodium benzoate; the molecule that accepted the hydride is reduced to the benzyl alkoxide ion, protonated on work-up to benzyl alcohol.

Overall, one molecule of aldehyde is oxidised and the other reduced — an intermolecular disproportionation.

Previous-year board questions 4

Q1 Write the mechanism of the nucleophilic addition of HCN to acetaldehyde (CH3CHO) to form a cyanohydrin. 2022 2 marks

HCN is only weakly ionised, so the reaction is catalysed by a trace of base (or by CN already present), which is essential for it to proceed at a useful rate.

  1. Step 1: A cyanide ion, CN, attacks the electrophilic carbonyl carbon of CH3CHO; the C=O π electrons shift completely onto the oxygen atom.
  2. Step 2: This produces a tetrahedral alkoxide intermediate, CH3CH(O)CN.
  3. Step 3: The alkoxide oxygen is protonated, either by another molecule of HCN (regenerating CN to continue the cycle) or during aqueous work-up, giving the neutral cyanohydrin, CH3CH(OH)CN (2-hydroxypropanenitrile).

The product is formed as a racemic mixture, because the trigonal planar carbonyl carbon can be attacked from either face with equal probability, generating a new stereocentre with no preference for either configuration.

Q2 An organic compound A (molecular formula C8H8O) gives an orange-red precipitate with 2,4-DNP reagent and a positive iodoform test, but does not reduce Tollens' reagent. Vigorous oxidation of A with hot KMnO4 gives an acid B (molecular formula C7H6O2). Identify A and B and write the relevant reactions. 2023 3 marks

A is acetophenone, C6H5COCH3 (molecular formula C8H8O: 6+1+1 = 8 carbons, 5+3 = 8 hydrogens, 1 oxygen).

  • It gives an orange-red precipitate with 2,4-DNP because it contains a carbonyl group — a positive 2,4-DNP test is characteristic of all aldehydes and ketones.
  • It gives a positive iodoform test because it contains a CH3CO– group directly attached to the ring: C6H5COCH3 + 3I2 + 4NaOH → C6H5COONa + CHI3↓ (yellow) + 3NaI + 3H2O.
  • It does not reduce Tollens' reagent because it is a ketone with no hydrogen on the carbonyl carbon, so it cannot be easily oxidised.

On vigorous oxidation with hot KMnO4, the methyl group attached to the ring is oxidatively cleaved away (lost as CO2/carbonate), leaving the ring-bound carboxyl group: C6H5COCH3 →(KMnO4, Δ) C6H5COOH. B is benzoic acid, C6H5COOH (molecular formula C7H6O2: 7 carbons, 6 hydrogens, 2 oxygens, matching the given formula).

Q3 Acetic acid reacts with ethanol in the presence of an acid catalyst: CH3COOH + C2H5OH is in equilibrium with CH3COOC2H5 + H2O, with equilibrium constant Keq = 4. Starting with 1 mol acetic acid and 1 mol ethanol, calculate the number of moles of ester present at equilibrium. 2023 4 marks

Let x mol of acetic acid react with x mol of ethanol at equilibrium. Then [acid] = (1−x), [alcohol] = (1−x), [ester] = x, [water] = x (using moles directly, since equal numbers of moles of reactants and products means the volume terms cancel in the equilibrium expression).

Keq = [ester][water] / [acid][alcohol] = x² / (1 − x)² = 4

Taking the square root of both sides (x and 1−x are both positive since x < 1): x / (1 − x) = 2

x = 2(1 − x) = 2 − 2x ⇒ 3x = 2 ⇒ x = 2/3 mol ≈ 0.667 mol

So about 2/3 mol (roughly 66.7%) of the acetic acid is converted to ethyl acetate at equilibrium; the reaction does not go to completion because it is a reversible equilibrium, not because of a kinetic limitation.

Q4 Propanoic acid is treated with bromine in the presence of a catalytic amount of red phosphorus. Name this reaction, write the product formed, and briefly explain the mechanism. 2024 3 marks

This is the Hell–Volhard–Zelinsky (HVZ) reaction, used to introduce a halogen selectively at the alpha-carbon of a carboxylic acid.

CH3CH2COOH + Br2 →(red P, catalytic) CH3CHBrCOOH + HBr

The product is 2-bromopropanoic acid.

Mechanism (outline): Red phosphorus reacts with Br2 in situ to generate phosphorus tribromide (PBr3), which converts a small amount of the carboxylic acid into the corresponding acid bromide, CH3CH2COBr. This acid bromide, being more enolisable than the parent acid, exists in a small amount of its enol form, which is far more nucleophilic at the alpha-carbon. This enol reacts rapidly with Br2 at the alpha-carbon to give the alpha-bromo acid bromide, CH3CHBrCOBr, which then exchanges with another unreacted molecule of the parent acid, regenerating the acid bromide catalyst and releasing the alpha-bromo acid, CH3CHBrCOOH, plus HBr. This exchange step keeps regenerating the acid bromide, so only a catalytic amount of red phosphorus is required even though every molecule of acid is eventually converted.

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