Carbonyl Group: C=O (carbon-oxygen double bond) — one of the most important functional groups in organic chemistry.
Aldehyde (RCHO): Carbonyl group bonded to one carbon (or H) and one hydrogen. The –CHO group is always at the end of the chain.
Ketone (RCOR'): Carbonyl group bonded to two carbon atoms. The C=O is within the chain.
Carboxylic Acid (RCOOH): Carbonyl group with an –OH group attached to the same carbon → –COOH (carboxyl group).
8.1.1 IUPAC Nomenclature of Aldehydes
IUPAC Rules — Aldehydes
Replace the ending '–e' of alkane with '–al'. (e.g., methane → methanal, ethane → ethanal)
The –CHO carbon is always C-1; numbering starts from the aldehyde end.
For aldehydes attached to a ring: add suffix 'carbaldehyde' after the full ring name. (e.g., cyclohexanecarbaldehyde)
Simplest aromatic aldehyde: benzenecarbaldehyde (IUPAC) or benzaldehyde (accepted IUPAC common name)
8.1.1 IUPAC Nomenclature of Ketones
IUPAC Rules — Ketones
Replace the ending '–e' of alkane with '–one'. (e.g., propane → propanone)
Number from the end nearer to the carbonyl group to give it the lowest locant.
For cyclic ketones: carbonyl carbon is C-1.
Common names: name both alkyl groups + "ketone". (e.g., dimethyl ketone = acetone)
Master Nomenclature Table
Structure
Common Name
IUPAC Name
HCHO
Formaldehyde
Methanal
CH3CHO
Acetaldehyde
Ethanal
CH3CH2CH2CH2CHO
Valeraldehyde
Pentanal
CH2=CHCHO
Acrolein
Prop-2-enal
C6H5CHO
Benzaldehyde
Benzenecarbaldehyde (Benzaldehyde)
CH3COCH3
Acetone (Dimethyl ketone)
Propanone
C6H5COCH3
Acetophenone / Methyl phenyl ketone
1-Phenylethan-1-one
C6H5COC6H5
Benzophenone
Diphenylmethanone
CH3COCH2CH2CH3
Methyl n-propyl ketone
Pentan-2-one
8.1.2 Structure of the Carbonyl Group
KEY CONCEPT — Carbonyl Group Structure
The carbonyl carbon is sp² hybridised — forms three σ bonds. The fourth electron remains in a p-orbital and forms a π bond with the oxygen p-orbital.
All three atoms bonded to carbonyl carbon lie in the same plane (trigonal planar, ~120° angles).
The π-electron cloud lies above and below this plane.
Polarity: C=O is highly polar — O is more electronegative → Cδ+, Oδ−. Carbonyl carbon acts as Lewis acid (electrophile); carbonyl oxygen acts as Lewis base (nucleophile).
Resonance: >C=O ↔ >C+–O–. The dipolar structure explains the high polarity.
Carbonyl compounds are more polar than ethers but form no H-bonds with each other (unlike alcohols).
⚛️
Orbital Diagram — Formation of Carbonyl Group (Fig. 8.1)
NCERT p.231 (Fig. 8.1)
Shows three stages: (1) sp² orbitals of carbon with one p-orbital perpendicular; (2) lateral overlap of p-orbital of C with p-orbital of O to form π bond; (3) final trigonal planar structure with 120° bond angles, showing π-cloud above and below the plane.
AI Image Prompt
Professional chemistry orbital diagram showing formation of carbonyl group. Fully white background, clean minimalistic. Three diagrams side by side: (1) sp² hybridised carbon with unhybridised p orbital shown as dumbbell perpendicular to the plane, bonded to oxygen showing lone pairs; (2) sideways overlap of p orbitals forming π bond shown as shaded lobes above and below; (3) final trigonal planar structure with 120° bond angles marked, two groups R attached to carbon, oxygen double-bonded. No background patterns, no text paragraphs.
8.2a Preparation of Aldehydes
General Methods (for both Aldehydes & Ketones)
1. Oxidation of Primary Alcohols → Aldehydes
RCH2OH —PCC / CrO₃ (anhyd.)→ RCHO (Aldehyde)
RCH2OH —KMnO₄ or K₂Cr₂O₇/H⁺→ RCOOH (over-oxidation to acid!) PCC = Pyridinium Chlorochromate — stops at aldehyde stage (no over-oxidation)
2. Dehydrogenation of Alcohols (Industrial Method)
RCH2OH —Ag or Cu catalyst, vapour, 573 K→ RCHO + H2
R–CHOH–R' —Cu, 573 K→ R–CO–R' (Ketone) + H2
3. From Hydrocarbons — Ozonolysis of Alkenes
Ozonolysis of alkenes with O3, followed by Zn/H2O (reductive workup), gives aldehydes, ketones or mixtures depending on substitution.
Acyl chloride (RCOCl) is hydrogenated with H2 over palladium-on-barium sulphate (Pd/BaSO₄) catalyst. BaSO₄ poisons the Pd catalyst to prevent over-reduction to primary alcohol.
Nitrile (RCN) is reduced with SnCl2 in anhydrous ether + dry HCl → stannous chloride imine complex → hydrolysis with water → aldehyde. Adds one carbon to the chain.
R–CHOH–R' —K₂Cr₂O₇/H⁺ or KMnO₄ or CrO₃→ R–CO–R' (Ketone)
★ 2. From Acyl Chlorides — Dialkylcadmium Reagent
Acyl chlorides react with dialkylcadmium (R2Cd) — prepared from Grignard reagent + CdCl2 — to give ketones. Cadmium reagent is less reactive than Grignard and doesn't over-react with the ketone product.
Benzene (or substituted benzene) + acyl chloride (RCOCl) or acid anhydride with anhyd. AlCl3 (Lewis acid) → aryl ketone. This is the best method for aromatic ketones.
Physical state: Methanal (HCHO) is a gas at room temperature. Ethanal is a volatile liquid. All others are liquids or solids.
Boiling points — ordered (same MW, ~58–60):
n-Butane 273 KAlkane (van der Waals only)
<
Methoxyethane 281 KEther (dipole-dipole)
<
Propanal 322 KAldehyde (stronger dipole)
<
Acetone 329 KKetone
<
Propan-1-ol 370 KAlcohol (H-bonding)
Why higher bp than hydrocarbons/ethers? Strong dipole-dipole interactions in the polar C=O group.
Why lower bp than alcohols? No intermolecular hydrogen bonding (no O–H bond in aldehydes/ketones).
Solubility: Lower members (methanal, ethanal, propanone) are miscible with water in all proportions — form H-bonds with water via the carbonyl oxygen. Solubility decreases with increasing chain length.
Odour: Lower aldehydes have pungent odours; higher members are more fragrant. Many natural fragrances are aldehydes and ketones (vanillin, cinnamaldehyde).
8.4a Chemical Reactions — Nucleophilic Addition
Why Nucleophilic Addition? (Mechanism)
Mechanism of Nucleophilic Addition
Carbonyl carbon is sp² (planar). A nucleophile attacks the electrophilic δ+ carbonyl carbon perpendicularly to the plane of the sp² orbitals (from above or below). Carbon rehybridises from sp² → sp³ → tetrahedral alkoxide intermediate forms → proton from medium → neutral product.
Aldehydes are more reactive than ketones in nucleophilic addition. Two reasons:
Steric: Ketones have two bulky alkyl groups hindering approach of nucleophile; aldehydes have only one alkyl group.
Electronic: Alkyl groups (electron-donating, +I effect) in ketones reduce the δ+ charge on carbonyl carbon → less electrophilic → less reactive. Aldehydes have only one +I group.
Reactivity order: HCHO > RCHO > RCOR'
Benzaldehyde is less reactive than propanal because the C6H5 group donates electrons to carbonyl carbon through resonance (π-π conjugation) → reduces electrophilicity of carbonyl C.
Mechanism: Pure HCN reacts slowly (weak nucleophile). Base catalyst (NaOH/KCN) generates CN⁻ ion (strong nucleophile). CN⁻ attacks carbonyl C → alkoxide → picks up H⁺ from HCN → cyanohydrin + CN⁻ (regenerated).
Importance: Cyanohydrins are useful synthetic intermediates — can be hydrolysed to α-hydroxy acids or reduced to amino alcohols.
Limitation (steric): Sterically hindered ketones (e.g., 2,2,6-trimethylcyclohexanone) give very low yield due to steric crowding around carbonyl carbon.
(b) Addition of NaHSO₃ — Bisulphite Addition
>C=O + NaHSO3 ⇌ >C(–OH)(–OSO2Na) (Bisulphite addition compound, crystalline)
To regenerate: treat with dil. HCl or NaOH
Key features:
Equilibrium favours product for aldehydes and methyl ketones; unfavourable for most ketones (steric reasons).
Addition compound is water-soluble and crystalline → useful for separation and purification of aldehydes from non-reacting impurities.
Reversible — original carbonyl compound easily regenerated with dilute acid or alkali.
(c) Addition of Alcohols — Acetal and Hemiacetal Formation
Acetals and ketals are hydrolysed back to aldehydes/ketones with aqueous mineral acid. They are used as protecting groups for carbonyl compounds in synthesis.
(e) Addition of Ammonia and its Derivatives — Nucleophilic Addition-Elimination
Nucleophiles of the type H2N–Z add to the carbonyl group. The addition is followed by loss of water (elimination) → C=N–Z double bond (imine-type products). Reaction is acid-catalysed and reversible.
★ 2,4-DNP derivatives are yellow, orange or red crystalline solids → used for characterisation and identification of aldehydes and ketones (sharp melting point).
8.4b Reduction Reactions of Aldehydes & Ketones
★ (i) Reduction to Alcohols
RCHO —NaBH₄ or LiAlH₄ or H₂/Pt→ RCH2OH (1° alcohol)
RCOR' —NaBH₄ or LiAlH₄→ R–CHOH–R' (2° alcohol)
★ (ii) Reduction to Hydrocarbons (C=O → CH₂)
Clemmensen Reduction: Zn-Hg amalgam + conc. HCl. For acid-stable substrates.
>C=O —Zn-Hg/conc. HCl→ >CH₂ + H₂O
Wolff-Kishner Reduction: NH2NH2 (hydrazine) → hydrazone → heat with KOH in ethylene glycol. For acid-sensitive substrates.
>C=O —NH₂NH₂→ >C=NNH₂ —KOH/Δ→ >CH₂ + N₂
🧠
Clemmensen vs Wolff-Kishner
Clemmensen: Acidic conditions (Zn-Hg/HCl) → for acid-stable molecules.
Wolff-Kishner: Alkaline conditions (KOH/hydrazine) → for acid-sensitive molecules (e.g., acetals, base-stable groups).
Both convert C=O → CH₂ completely (remove the oxygen entirely).
8.4c Oxidation Reactions — Distinguishing Aldehydes from Ketones
KEY CONCEPT — Oxidation Difference
Aldehydes are easily oxidised to carboxylic acids by mild or strong oxidising agents. They have a C–H bond on the carbonyl carbon which is easily broken.
R–CHO —[O]→ R–COOH
Ketones are generally resistant to mild oxidising agents. Under vigorous conditions (strong oxidising agents + high T), C–C bond cleavage occurs → mixture of carboxylic acids with fewer carbons.
→ This difference is exploited in two classic tests to distinguish aldehydes from ketones:
Positive test: Bright silver mirror deposited on the inner wall of the test tube (Ag metal formed). Aldehyde is oxidised to carboxylate anion (in alkaline medium).
Important: Aromatic aldehydes (benzaldehyde) do NOT respond to Fehling's test. Aliphatic aldehydes give positive test. Ketones: No reaction.
★ (iii) Iodoform Reaction (Haloform Reaction) — Oxidation by NaOX
Compounds having CH3CO– group (methyl ketones) or CH3CHOH– group (which oxidises to CH3CO–) react with iodine/NaOH (sodium hypoiodite) to give iodoform (CHI3) — a yellow precipitate with characteristic smell.
CH3CHOH– group: ethanol (CH3CH2OH), propan-2-ol (oxidised to CH3CO– then reacts)
Important: This is also used as a detection test for the CH3CO– or CH3CH(OH)– group.
8.4d Reactions Due to α-Hydrogen
Acidity of α-Hydrogen
The hydrogen atoms on the carbon adjacent to the C=O group (α-carbon) are acidic. Reasons:
Strong electron-withdrawing effect of the C=O group weakens the C–H bond.
The conjugate base (enolate ion) is resonance-stabilised: –C––C=O ↔ –C=C–O– (enolate)
★ Aldol Condensation — KEY NAMED REACTION
Aldehydes and ketones with at least one α-hydrogen undergo aldol reaction in the presence of dilute alkali (NaOH, Ba(OH)2) to form β-hydroxy carbonyl compounds (aldol or ketol).
On heating, aldol and ketol readily lose water (dehydration) → α,β-unsaturated carbonyl compound (aldol condensation product).
Mechanism of aldol: Base deprotonates α-H → enolate anion (nucleophile) → attacks carbonyl carbon of another molecule (electrophile) → β-hydroxy carbonyl compound.
Cross Aldol Condensation
Aldol condensation between two different aldehydes/ketones. If both have α-H, a mixture of 4 products forms (all combinations). Useful when one component has no α-H (acts only as electrophile).
Example: Benzaldehyde (no α-H) + Acetone (has α-H) → Benzalacetone + Dibenzalacetone (1,3-Diphenyl-2-propen-1-one / Benzalacetophenone) as major product.
Aldehydes which do NOT have α-hydrogen undergo self oxidation-reduction (disproportionation) on treatment with concentrated NaOH: one molecule is oxidised to the carboxylate (acid salt) and another is reduced to the alcohol.
Examples without α-H: HCHO, C6H5CHO, (CH3)3CCHO, 2,2-dimethylpropanal. Note: Formaldehyde in this reaction is always the reductant (gives methanol) because it has no alkyl group stabilisation.
Electrophilic Substitution of Aromatic Aldehydes and Ketones
The –CHO and –COR groups on benzene ring are deactivating and meta-directing (due to –M and –I effects of C=O which withdraws electrons from the ring). EAS occurs at meta position.
Compounds containing the carboxyl group (–COOH). The carboxyl group consists of a carbonyl (C=O) group attached to a hydroxyl (–OH) group.
Aliphatic: R–COOH (fatty acids C₁₂–C₁₈ occur naturally as glycerol esters)
Aromatic: Ar–COOH (e.g., benzoic acid)
IUPAC Rules for Carboxylic Acids
Replace the ending '–e' of alkane with '–oic acid'. (e.g., methane → methanoic acid)
The –COOH carbon is always C-1.
For dicarboxylic acids: retain 'e' of alkane + add 'dioic acid'. (e.g., ethanedioic acid)
For aromatic: benzenecarboxylic acid (IUPAC) or benzoic acid (accepted)
Common and IUPAC Names — Master Table
Structure
Common Name
IUPAC Name
Source/Origin
HCOOH
Formic acid
Methanoic acid
Red ants (Latin: formica)
CH3COOH
Acetic acid
Ethanoic acid
Vinegar (Latin: acetum)
CH3CH2COOH
Propionic acid
Propanoic acid
—
CH3CH2CH2COOH
Butyric acid
Butanoic acid
Rancid butter (Latin: butyrum)
(CH3)2CHCOOH
Isobutyric acid
2-Methylpropanoic acid
—
HOOC–COOH
Oxalic acid
Ethanedioic acid
—
HOOC–CH2–COOH
Malonic acid
Propanedioic acid
—
HOOC–(CH2)2–COOH
Succinic acid
Butanedioic acid
—
HOOC–(CH2)4–COOH
Adipic acid
Hexanedioic acid
Nylon-6,6 manufacture
C6H5COOH
Benzoic acid
Benzenecarboxylic acid
Benzoin resin
8.6.2 Structure of the Carboxyl Group
Structure of –COOH
The bonds to the carboxyl carbon lie in one plane, separated by ~120°. Three resonance structures stabilise the carboxyl group:
–C(=O)(O–H) ↔ –C+(–O–)(–OH) ↔ –C(–O)(=OH+)
Key consequence: Carboxyl carbon is less electrophilic than aldehyde/ketone carbonyl carbon because resonance involving the –OH lone pair reduces the δ+ charge on carbon. This is why carboxylic acids do NOT undergo typical nucleophilic addition reactions like aldehydes and ketones.
8.7 Methods of Preparation of Carboxylic Acids
1. Oxidation of Primary Alcohols and Aldehydes
RCH2OH —alk. KMnO₄ or Jones reagent (CrO₃–H₂SO₄)→ RCOOH
RCHO —Tollens' / Fehling's / KMnO₄→ RCOOH
2. Oxidation of Alkylbenzenes (Vigorous)
Entire alkyl side chain (regardless of length) is oxidised to –COOH. Primary and secondary alkyl groups react; tertiary groups are not affected.
Boiling points:Higher than alcohols of same molecular mass — due to more extensive intermolecular hydrogen bonding (two H-bond donors + two acceptors per molecule). Carboxylic acids exist as dimers in vapour phase or aprotic solvents via double hydrogen bonding (cyclic dimer structure).
Solubility: C₁–C₄ are fully miscible with water (form H-bonds). Solubility decreases with carbon chain length. Higher acids are nearly insoluble in water but soluble in organic solvents (benzene, ether, ethanol, CHCl3). Benzoic acid is nearly insoluble in cold water.
8.9 Chemical Reactions of Carboxylic Acids
8.9.1 Reactions Involving O–H Bond Cleavage (Acidity)
Acidity — Why Carboxylic Acids Are the Most Acidic
Why stronger than phenols? The carboxylate ion (RCOO⁻) is stabilised by two equivalent resonance structures with negative charge on both oxygens (both electronegative). The phenoxide ion has non-equivalent structures with negative charge on less electronegative ring carbons. Therefore, carboxylate is more stable → carboxylic acid is stronger.
2R–COOH + 2Na → 2R–COONa + H2↑
R–COOH + NaOH → R–COONa + H2O R–COOH + NaHCO3 → R–COONa + H2O + CO2↑ ★ (distinguishes RCOOH from phenol — phenol does NOT react with NaHCO₃)
Effect of Substituents on Acidity of Carboxylic Acids
Electron-withdrawing groups (EWG): Stabilise the carboxylate anion → increase acidity (decrease pKa). Effect increases with number of EWG and decreases with distance from –COOH.
Electron-donating groups (EDG): Destabilise the carboxylate anion → decrease acidity (increase pKa).
Acidity order of haloacetic acids: CF3COOH > CCl3COOH > CHCl2COOH > CH2ClCOOH > CH3COOH
Acidity order of chloroacetic acids by position: α-Cl > β-Cl > γ-Cl (inductive effect decreases rapidly with distance)
For aromatic acids: EWG at o/p → more acidic; EDG at o/p → less acidic. 4-Nitrobenzoic acid (pKa 3.41) > Benzoic acid (4.19) > 4-Methoxybenzoic acid (4.46)
EWG power: CF3 > NO2 > CN > F > Cl > Br > I > Ph
8.9.2 Reactions Involving C–OH Bond Cleavage
1. Formation of Acid Anhydride
2CH3COOH —H₂SO₄ or P₂O₅, Δ→ CH3CO–O–COCH3 (Ethanoic anhydride) + H2O
★ 2. Esterification (Fischer Esterification)
Carboxylic acid + alcohol with conc. H2SO4 catalyst. Reversible — remove water or ester to drive equilibrium right.
RCOOH + R'OH ⇌H⁺, Δ RCOOR' (Ester) + H2O
Mechanism (nucleophilic acyl substitution): H⁺ protonates carbonyl O → electrophilicity of carbonyl C increases → alcohol (nucleophile) attacks → tetrahedral intermediate forms → proton transfer → –OH₂ (good leaving group) departs → protonated ester → deprotonation → ester product.
3. Reactions with PCl₅, PCl₃, SOCl₂ (→ Acyl Chloride)
Carboxylic acids with α-hydrogen undergo halogenation at the α-position with Cl2 or Br2 in presence of small amount of red phosphorus (catalyst) → α-halocarboxylic acid.
R–CH2–COOH —(i) Cl₂ or Br₂ / red P (ii) H₂O→ R–CHX–COOH (α-Halocarboxylic acid, X = Cl or Br)
Example: CH3CH2COOH + Cl2/red P → CH3CHClCOOH (2-Chloropropanoic acid)
Importance: α-Halocarboxylic acids are useful synthetic intermediates (can be further modified at the α-position).
Ring Substitution of Aromatic Carboxylic Acids
The –COOH group is a deactivating and meta-directing group (–M effect withdraws electrons). Therefore, EAS in benzoic acid gives meta products.
Important: Benzoic acid does NOT undergo Friedel-Crafts reaction because the –COOH group is deactivating AND the Lewis acid AlCl₃ gets bonded to the –COOH group (forms a complex with it).
✏️ Practice Questions
Q1
Arrange the following in increasing order of boiling points and explain the reason: n-Butane, Ethoxyethane (diethyl ether), Butanal, Butan-1-ol
Reasons:
• n-Butane (alkane): Only weak van der Waals forces → lowest bp (273 K).
• Ethoxyethane (ether): Dipole-dipole interactions + van der Waals, but no H-bonding (no O–H) → 307.6 K.
• Butanal (aldehyde): Stronger dipole-dipole due to polar C=O; no H-bonding with itself → 348 K (higher than ether, lower than alcohol).
• Butan-1-ol (alcohol): Extensive intermolecular H-bonding via –OH → highest bp (390 K).
Q2
Explain why aldehydes are generally more reactive than ketones in nucleophilic addition reactions. Also explain why benzaldehyde is less reactive than propanal.
Aldehydes > Ketones:
1. Steric effect: Ketones have two alkyl/aryl groups flanking the carbonyl carbon → nucleophile approach is hindered. Aldehydes have only one such group → less steric crowding → easier nucleophilic attack.
2. Electronic effect: Alkyl groups have +I effect → donate electrons to carbonyl carbon in ketones → reduce δ+ charge → less electrophilic. Aldehydes have only one alkyl group → carbonyl carbon more electrophilic → more reactive.
Benzaldehyde < Propanal:
The phenyl ring (C₆H₅–) donates electrons to the carbonyl carbon through conjugation (π-π overlap, resonance). Lone pairs from benzene ring delocalise into the C=O π system → reduces δ+ on carbonyl C → less electrophilic → less reactive than aliphatic propanal.
Q3
An organic compound (A) with molecular formula C₈H₈O forms an orange-red precipitate with 2,4-DNP reagent but neither reduces Tollens' nor Fehling's reagent. It gives yellow precipitate with I₂/NaOH (iodoform test). On drastic oxidation with chromic acid, it gives compound (B) with formula C₇H₆O₂. Identify A and B.
Step-by-step analysis:
• Forms 2,4-DNP derivative → contains C=O group (aldehyde OR ketone).
• Does NOT reduce Tollens' or Fehling's → it's a ketone (not an aldehyde).
• Positive iodoform test → contains CH₃CO– group (methyl ketone).
• C₈H₈O = high degree of unsaturation → aromatic ring must be present (doesn't decolourise bromine water → not a C=C alkene).
• Oxidation product (B) = C₇H₆O₂ → this is benzoic acid (C₆H₅COOH).
Therefore: A = Acetophenone (C₆H₅–CO–CH₃) and B = Benzoic acid (C₆H₅COOH)
Reactions: A + I₂/NaOH → C₆H₅COONa + CHI₃↓ (iodoform). A + [O] (H₂CrO₄) → C₆H₅COOH (B).
Q4
Which compounds will undergo Aldol condensation, which will undergo Cannizzaro reaction, and which will undergo neither?
(i) Methanal (ii) 2-Methylpentanal (iii) Benzaldehyde (iv) Benzophenone (v) Cyclohexanone
Rule: Aldol → requires α-H. Cannizzaro → no α-H, aldehyde only. Neither → ketone with no α-H.
(i) Methanal (HCHO): No α-H (no carbon next to CHO except the aldehyde carbon itself) → Cannizzaro reaction → CH₃OH + HCOONa
(ii) 2-Methylpentanal: Has α-H (on C-3) → Aldol condensation
(iii) Benzaldehyde (C₆H₅CHO): No α-H (the CHO carbon is directly attached to the ring; ring H is not α-H) → Cannizzaro reaction → C₆H₅CH₂OH + C₆H₅COONa
(iv) Benzophenone (C₆H₅COC₆H₅): No α-H (both groups are aryl, no alkyl H adjacent to C=O) AND it's a ketone (not an aldehyde) → Neither
(v) Cyclohexanone: Has α-H (both α-carbons of the ring) → Aldol condensation
Q5
Arrange in increasing order of acid strength and explain:
(i) CF₃COOH, CCl₃COOH, CH₂ClCOOH, CH₃COOH, CH₃CH₂COOH
(ii) Benzoic acid, p-Nitrobenzoic acid, p-Methoxybenzoic acid
(i) Increasing acid strength: CH₃CH₂COOH < CH₃COOH < CH₂ClCOOH < CCl₃COOH < CF₃COOH Reasoning: More electron-withdrawing substituents on the α-carbon stabilise the carboxylate anion (RCOO⁻) by withdrawing electrons away from the negative charge. F has highest –I effect (most electronegative) → CF₃COOH is strongest. Three Cl (CCl₃) > one Cl (CH₂Cl). Methyl (+I, electron-donating) destabilises anion. Ethyl has larger +I than methyl → propanoic acid is weakest.
(ii) Increasing acid strength: p-Methoxybenzoic acid (pKa 4.46) < Benzoic acid (4.19) < p-Nitrobenzoic acid (3.41) Reasoning: –NO₂ at para position withdraws electrons from the ring by both –I and –M effects → stabilises carboxylate anion → strongest acid. –OCH₃ at para donates electrons by +M effect → destabilises anion → weakest acid among three. Benzoic acid is intermediate.
Q6
Although phenoxide ion has MORE resonance structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why? (Exercise 8.20)
The key is not the number of resonance structures but their quality and stability:
Carboxylate ion (RCOO⁻): Two resonance structures — negative charge is on both oxygen atoms (electronegative), and the two structures are equivalent in energy. Both oxygens share the charge equally. This is very effective stabilisation.
Phenoxide ion (C₆H₅O⁻): More resonance structures, but the negative charge in most structures is placed on the ring carbon atoms (which are less electronegative than oxygen). These are higher-energy, non-equivalent structures → less effective stabilisation. Also, the ring carbons are less electronegative than oxygen → less favourable for carrying negative charge.
Conclusion: The carboxylate ion is more effectively stabilised than the phenoxide ion because its charge is delocalised over two equivalent electronegative oxygen atoms. More stable conjugate base → stronger acid → carboxylic acids (pKa ~5) are stronger than phenols (pKa ~10).
Q7
Give simple chemical tests to distinguish between:
(i) Propanal vs Propanone (ii) Benzaldehyde vs Acetophenone (iii) Pentan-2-one vs Pentan-3-one
(i) Propanal vs Propanone:
Use Tollens' test (ammoniacal AgNO₃) or Fehling's test. Propanal (aldehyde) → silver mirror (Tollens') or red-brown ppt Cu₂O (Fehling's). Propanone (ketone) → no reaction.
Alternatively: Iodoform test with I₂/NaOH → both give yellow CHI₃ precipitate (propanone has CH₃CO–, propanal has CH₃CHO which also responds). So Tollens' is more definitive here.
(ii) Benzaldehyde vs Acetophenone:
Use Tollens' test: Benzaldehyde (aldehyde) → silver mirror. Acetophenone (ketone) → no reaction.
OR: Iodoform test: Acetophenone (CH₃CO–) → yellow iodoform precipitate. Benzaldehyde → no iodoform.
(iii) Pentan-2-one (CH₃COCH₂CH₂CH₃) vs Pentan-3-one (CH₃CH₂COCH₂CH₃):
Iodoform test with I₂/NaOH: Pentan-2-one has CH₃CO– group → gives yellow CHI₃ precipitate. Pentan-3-one has no CH₃CO– group → no iodoform.
Alternatively, treat with NaHSO₃: Pentan-3-one (less hindered symmetric ketone) gives better bisulphite addition product. (Less reliable test.)
Q8
How will you convert Ethanal into: (i) Butane-1,3-diol (ii) But-2-enal (iii) But-2-enoic acid
An organic compound contains 69.77% C, 11.63% H, and the rest is O. Molecular mass = 86. It does not reduce Tollens' reagent but forms an addition compound with NaHSO₃ and gives positive iodoform test. On vigorous oxidation, it gives ethanoic acid and propanoic acid. Identify the compound.
Step 1: Empirical formula
C: 69.77/12 = 5.81 mol; H: 11.63/1 = 11.63 mol; O: 18.60/16 = 1.16 mol
Ratio C:H:O = 5.81:11.63:1.16 = 5:10:1 → Empirical formula C₅H₁₀O (MW = 86) ✓ = Molecular formula
Step 2: Functional group analysis
• Does NOT reduce Tollens' → NOT an aldehyde → ketone
• Forms bisulphite addition compound → ketone (consistent)
• Positive iodoform → contains CH₃CO– group (methyl ketone)
Step 3: Oxidation products
Vigorous oxidation gives CH₃COOH (ethanoic, C₂) + CH₃CH₂COOH (propanoic, C₃). Total carbons: 2 + 3 = 5 ✓
Cleavage of C₂–C₃ bond in the ketone: C₂–CO–C₃ = pentan-2-one (CH₃CO–CH₂CH₂CH₃)
Answer: Compound A = Pentan-2-one (CH₃COCH₂CH₂CH₃)
Q10
How will you prepare Benzoic acid from: (i) Ethylbenzene (ii) Acetophenone (iii) Bromobenzene?
(i) Ethylbenzene → Benzoic acid:
C₆H₅CH₂CH₃ →(KMnO₄/KOH, heat)→ C₆H₅COOK →(H₃O⁺)→ C₆H₅COOH
(Entire alkyl side chain oxidised regardless of length)
(ii) Acetophenone → Benzoic acid:
C₆H₅COCH₃ →(KMnO₄/KOH, heat)→ C₆H₅COOK →(H₃O⁺)→ C₆H₅COOH
(Oxidation of the side chain –COCH₃ also proceeds to –COOH)
An organic compound (A) with molecular formula C₉H₁₀O forms 2,4-DNP derivative, reduces Tollens' reagent and undergoes Cannizzaro reaction. On vigorous oxidation it gives 1,2-benzenedicarboxylic acid (phthalic acid). Identify compound (A).
Analysis:
• Forms 2,4-DNP → aldehyde or ketone ✓
• Reduces Tollens' → aldehyde
• Undergoes Cannizzaro → no α-H (aldehyde without α-H)
• C₉H₁₀O with high unsaturation + doesn't decolourise bromine water → aromatic ring present
• Oxidation gives phthalic acid (1,2-C₆H₄(COOH)₂) → the compound must have substituents at 1,2-positions of benzene ring
• Molecular formula C₉H₁₀O = C₆H₄ (ring) + C₃H₆O (substituents). With an ortho-disubstituted ring and a –CHO group with no α-H...
Candidate: 2-Methylbenzaldehyde has α-H. Need ortho-CHO with no α-H.
A compound with –CHO directly on ring and another substituent at ortho with no α-H: C₆H₄(CH₂CH₃)(CHO) at 1,2 positions = 2-Ethylbenzaldehyde is not right as it has α-H on CH₂.
With C₉H₁₀O and Cannizzaro: 2-Methylbenzaldehyde does have α-H (–CH₃ is α to ring but NOT α to CHO). The CHO in an aryl aldehyde has no α-H because the adjacent carbon is the aromatic ring. ∴ A = 2-Methylbenzaldehyde (or o-Tolualdehyde), which oxidises to give phthalic acid when the –CH₃ is also oxidised under vigorous conditions.
🎯 Important Exam Points — Quick Reference
CONCEPTCarbonyl C is sp² hybridised, trigonal planar (~120°). C=O is polar (C is δ+, electrophile; O is δ−, nucleophile). Aldehydes more reactive than ketones in nucleophilic addition (steric + electronic).
CONCEPTReactivity in nucleophilic addition: HCHO > RCHO > RCOR'. Aldehydes have one alkyl group; ketones have two bulky groups hindering approach of nucleophile + more +I effect reducing δ+.
CONCEPTIodoform test: CH₃CO– or CH₃CH(OH)– group gives yellow CHI₃ (iodoform) precipitate with I₂/NaOH. Gives positive test with CH₃CHO, CH₃COCH₃, CH₃COR, C₂H₅OH, CH₃CHOHР.
CONCEPTCarboxylic acids: more acidic than phenols because carboxylate ion has charge on two equivalent electronegative O atoms (equivalent resonance). Phenoxide: charge on one O + ring C atoms (less electronegative, non-equivalent).
CONCEPTEWG increases acidity of RCOOH (stabilises RCOO⁻). EDG decreases acidity. EWG effect decreases with distance from COOH. CF₃COOH > CCl₃COOH > CH₂ClCOOH > CH₃COOH > CH₃CH₂COOH.
MCQNaHCO₃ test: RCOOH + NaHCO₃ → effervescence (CO₂). Phenol does NOT react with NaHCO₃. This is the key test to distinguish carboxylic acids from phenols. Both react with NaOH.
MCQNaBH₄ cannot reduce –COOH or –COOR. LiAlH₄ can reduce all carbonyl groups. Diborane reduces –COOH selectively without reducing –NO₂, –Cl, etc. PCC oxidises 1° alcohol → aldehyde only (not over-oxidised to acid).
MCQAromatic carboxylic acids (–COOH) are deactivating + meta-directing. Aromatic aldehydes and ketones (–CHO, –COR) are also deactivating + meta-directing. Neither undergoes Friedel-Crafts reaction (benzoic acid — AlCl₃ forms complex with COOH).