Class 12 Chemistry | Chapter 8
Aldehydes, Ketones & Carboxylic Acids
Carbonyl Compounds • Nucleophilic Addition • Name Reactions • Acidity
1. Introduction to Carbonyl Compounds
Carbonyl compounds strictly contain the carbon-oxygen double bond (>C=O). These include aldehydes (R-CHO), ketones (R-CO-R'), carboxylic acids (R-COOH), and acid derivatives like acid chlorides (R-COCl), amides (R-CONH₂), and esters (R-COOR').
1.1 Structure of the Carbonyl Group
The carbonyl carbon atom is sp² hybridized and forms three sigma (σ) bonds. The fourth valence electron forms a pi (π) bond with oxygen. The bond angles are approximately 120°, making the group planar.
Due to higher electronegativity of oxygen, the C=O bond is highly polarized: Carbon gets a partial positive charge (δ+) acting as an electrophile, and oxygen gets a partial negative charge (δ−) acting as a nucleophile.
2. Preparation of Aldehydes and Ketones
2.1 By Oxidation of Alcohols
- Aldehydes: Prepared by oxidation of 1° alcohols using mild oxidizing agents like PCC (Pyridinium chlorochromate) or CrO₃ in anhydrous medium.
- Ketones: Prepared by oxidation of 2° alcohols using CrO₃ or KMnO₄.
2.2 By Ozonolysis of Alkenes
Alkenes react with ozone (O₃) to form ozonides, which upon reductive cleavage with Zn/H₂O give aldehydes/ketones based on the substitution pattern of the alkene.
2.3 By Hydration of Alkynes
Addition of water to ethyne in presence of H₂SO₄ and HgSO₄ gives acetaldehyde. All other alkynes give ketones.
2.4 Special Methods for Aldehydes ONLY
- Rosenmund Reduction: Acyl chloride (Acid chloride) is hydrogenated over catalyst,
Palladium on barium sulphate (partially poisoned with sulphur or quinoline).R-COCl + H₂ → (Pd/BaSO₄) → R-CHO + HCl
- Stephen Reaction: Nitriles are reduced to corresponding imine with SnCl₂
+ HCl, which on hydrolysis give corresponding aldehyde.R-CN + SnCl₂ + HCl → R-CH=NH → (H₃O⁺) → R-CHO
- Etard Reaction: Chromyl chloride (CrO₂Cl₂) oxidizes the methyl group of toluene to a chromium complex, which on hydrolysis yields benzaldehyde.
2.5 Special Methods for Ketones ONLY
- From Dialkylcadmium: React Grignard reagent with CdCl₂ to get R₂Cd. Then react R₂Cd with acid chloride to get Ketone.
- Friedel-Crafts Acylation: Benzene + R-COCl + Anhydrous AlCl₃ → Aryl Ketone.
3. Chemical Reactions (Aldehydes & Ketones)
3.1 Nucleophilic Addition Reactions Mechanism
Because the carbonyl carbon is electron deficient (δ+), a nucleophile attacks it from a direction roughly perpendicular to the plane of sp² orbitals. The hybridization changes from sp² to sp³, forming a tetrahedral alkoxide intermediate, which then captures a proton to give the neutral product.
💡 Reactivity Order: Aldehydes are generally MORE reactive than ketones towards
nucleophilic addition due to:
1. Steric reasons: Two bulky alkyl groups in ketones hinder nucleophilic attack. Aldehydes have only one alkyl group.
2. Electronic reasons: Two alkyl groups (+I effect) in ketones reduce the electrophilicity of the carbonyl carbon more than one alkyl group in aldehydes.
1. Steric reasons: Two bulky alkyl groups in ketones hinder nucleophilic attack. Aldehydes have only one alkyl group.
2. Electronic reasons: Two alkyl groups (+I effect) in ketones reduce the electrophilicity of the carbonyl carbon more than one alkyl group in aldehydes.
3.2 Important Nucleophilic Addition Reactions
- Addition of HCN: Yields cyanohydrins. (Base catalyzed to generate CN⁻ ion).
- Addition of Alcohols: Aldehydes react with one equivalent of alcohol to give Hemiacetal (unstable), and with two equivalents to give Acetal (stable). Ketones form Ketals with ethylene glycol.
- Addition of Ammonia Derivatives (NH₂-Z): Reacts in weak acidic medium to form >C=N-Z compounds (Imine derivatives like hydrazones, oximes, semicarbazones) with loss of water.
3.3 Reduction Reactions
- To Alcohols: NaBH₄ or LiAlH₄ reduces aldehydes to 1° alcohols and ketones to 2° alcohols.
- To Hydrocarbons (Clemmensen Reduction): The carbonyl group >C=O is reduced to >CH₂ using Zinc amalgam (Zn-Hg) and Conc. HCl.
- To Hydrocarbons (Wolff-Kishner Reduction): Using hydrazine (NH₂NH₂) followed by heating with KOH in high boiling solvent like ethylene glycol.
3.4 Oxidation Reactions
Aldehydes are easily oxidized to carboxylic acids (even by mild oxidizing agents). Ketones are very difficult to oxidize (requires strong agents like KMnO₄/HNO₃ and high temps, which breaks C-C bonds).
- Tollens' Test (Silver Mirror Test): Ammoniacal silver nitrate. Aldehydes reduce Ag⁺ to metallic Ag (silver mirror). Ketones do not.
- Fehling's Test: Fehling A (aq. CuSO₄) + Fehling B (Rochelle salt). Aldehydes reduce Cu²⁺ to red precipitate of Cu₂O. (Benzaldehyde does NOT respond to this test).
- Haloform Reaction (Iodoform Test): Compounds with exactly a CH₃-CO- group (methyl ketones) react with NaOX (NaOH + I₂) to give a yellow precipitate of Iodoform (CHI₃). Acetaldehyde is the only aldehyde that gives this test.
3.5 Reactions due to α-Hydrogen
The α-hydrogen of carbonyl compounds is acidic due to the strong electron-withdrawing effect of the carbonyl group and resonance stabilization of the resulting enolate ion.
- Aldol Condensation: Aldehydes/Ketones having at least one α-hydrogen undergo a reaction in presence of dilute alkali (dil. NaOH or Ba(OH)₂) to form β-hydroxy aldehydes (aldol) or β-hydroxy ketones (ketol). On heating, they lose water to form α,β-unsaturated carbonyl compounds.
- Cross Aldol Condensation: When carried out between two different aldehydes/ketones.
- Cannizzaro Reaction: Aldehydes which DO NOT have an α-hydrogen (e.g., Formaldehyde HCHO, Benzaldehyde C₆H₅CHO) undergo self-oxidation and reduction (disproportionation) on treatment with Concentrated alkali (50% NaOH). One molecule is reduced to alcohol, the other oxidized to carboxylic acid salt.
4. Carboxylic Acids
4.1 Preparation
- From oxidation of 1° alcohols or aldehydes (using KMnO₄ / K₂Cr₂O₇).
- From Alkylbenzenes: Toluene/Ethylbenzene oxidized with KMnO₄-KOH (heat) yields Benzoic acid. (Note: The entire alkyl chain, regardless of length, is chopped off to become -COOH as long as it has at least one benzylic hydrogen).
- From Nitriles and Amides: Hydrolysis (H⁺ or OH⁻).
- From Grignard Reagents: RMgX + Solid CO₂ (Dry ice) followed by hydrolysis.
4.2 Physical Properties & Acidity
Carboxylic acids exist as dimers in vapor state or in aprotic solvents due to strong intermolecular hydrogen bonding. They have higher boiling points than aldehydes, ketones, and even alcohols of comparable masses.
Acidity: R-COOH -> R-COO⁻ + H⁺. The carboxylate ion is heavily stabilized by equivalent resonance structures. Therefore, they are stronger acids than phenols and alcohols.
- Electron Withdrawing Groups (EWG like -Cl, -F, -NO₂, -CN) increase acidity by stabilizing the (-) charge via -I effect. Order of -I effect: F > Cl > Br > I.
- Electron Releasing Groups (ERG like -CH₃) decrease acidity. Order: HCOOH > CH₃COOH > CH₃CH₂COOH.
4.3 Important Reactions of Carboxylic Acids
- Esterification: RCOOH + R'OH → (conc. H₂SO₄) → RCOOR' + H₂O.
- Reaction with PCl₅, PCl₃, SOCl₂: Yields Acid Chlorides (RCOCl). SOCl₂ is preferred because side products are gases (SO₂, HCl).
- HVZ Reaction (Hell-Volhard-Zelinsky): Carboxylic acids having an α-hydrogen are halogenated at the α-position by reacting with Cl₂/Br₂ in presence of small amount of Red Phosphorus.
- Decarboxylation: Sodium salts of carboxylic acids heated with sodalime (NaOH + CaO in 3:1 ratio) lose CO₂ to form alkanes having one carbon less.
🎓 NEET Previous Year Questions
Q1. [NEET 2022] Which of the following compounds gives a positive iodoform test? (a)
Pentan-3-one (b) Pentan-2-one (c) 3-Methylpentan-2-one (d) Both b & c
Answer Positive Iodoform test is given by compounds
with the methyl ketone group (-CO-CH₃). Pentan-3-one is ethyl ketone. Pentan-2-one and
3-Methylpentan-2-one both have a 2-one (methyl ketone) structure. Correct is (d) Both b &
c.
Q2. [NEET 2021] Reaction of a carbonyl compound with one of the following reagents
involves nucleophilic addition followed by elimination of water. The reagent is:
Answer Hydrazine (or any ammonia
derivative NH₂-Z) in presence of weak acid. Addition forms an intermediate which loses water to
form hydrazone (>C=N-NH₂).
Q3. [NEET 2020] An organic compound (A) with molecular formula C8H8O forms an
orange-red precipitate with 2,4-DNP reagent and gives yellow precipitate on heating with iodine in the
presence of sodium hydroxide. It neither reduces Tollens’ or Fehlings’ reagent. (A) is:
Answer 2,4-DNP positive means it's a carbonyl
(aldehyde/ketone). Tollens/Fehling negative means it's a Ketone. Iodoform positive
means it's a Methyl ketone. C8H8O fits exactly with Acetophenone
(C6H5-CO-CH3).
Q4. [NEET 2019] Carboxylic acids have higher boiling points than aldehydes, ketones
and even alcohols of comparable molecular mass. It is due to their:
Answer Formation of intermolecular hydrogen
bonding which is more extensive and stable than alcohols, allowing them to exist as cyclic
dimers even in vapor phase.
Q5. [NEET 2018] Carboxylic acids are more acidic than phenol and alcohol because of:
Answer Resonance stabilization of the
carboxylate ion. The negative charge is delocalized equally over two highly electronegative
oxygen atoms, making it much more stable than phenoxide (where negative charge is on less
electronegative carbons).
💡 Rapid Revision
- Aldol: Needs dilute OH⁻ and at least ONE α-hydrogen. Products: self-condensation Aldol/Ketol.
- Cannizzaro: Needs conc. OH⁻ (50%) and ZERO α-hydrogen (e.g., HCHO, PhCHO). Products: Alcohol + Carboxylate.
- Clemmensen vs Wolff-Kishner: Both convert >C=O to >CH₂. Clemmensen uses Acidic media (Zn-Hg/HCl). WK uses Basic media (NH₂NH₂/KOH). Choose depending on acid/base sensitive groups in molecule.
- Tollens': Detects ALL aldehydes (aliphatic and aromatic). Fehling's: Detects ONLY aliphatic aldehydes (benzaldehyde fails).
CLASS 12 CHEMISTRY | NCERT SOLUTIONS
Chapter 8 — Aldehydes, Ketones & Carboxylic Acids
22 Solved Questions — Distinguishing Tests, Cross Aldol, Acidity & Conversions
Note: Almost no mathematical numericals occur here. Focus relies heavily on Acidity (pKa)
order of substituted benzoic/acetic acids, Aldol/Cannizzaro product prediction, chemical tests
(Tollens/Fehling/Iodoform), and multi-step organic synthesis.
📝 Chemical Distinguishing Tests (Q1 – Q6)
2 MarksQ1. Give a chemical test to distinguish
between Propanal and Propanone.
✓ Solution
Tollens' Test: Add Tollens' reagent (ammoniacal silver nitrate) and warm.
Propanal (an aldehyde): Reduces Tollens' reagent forming a bright silver mirror on the walls of the test tube.
Propanone (a ketone): Does not give this test (no silver mirror formed).
Alternatively, Fehling's test can be used (Propanal gives red ppt, Propanone does not).
Tollens' Test: Add Tollens' reagent (ammoniacal silver nitrate) and warm.
Propanal (an aldehyde): Reduces Tollens' reagent forming a bright silver mirror on the walls of the test tube.
Propanone (a ketone): Does not give this test (no silver mirror formed).
Alternatively, Fehling's test can be used (Propanal gives red ppt, Propanone does not).
2 MarksQ2. Give a chemical test to distinguish
between Pentan-2-one and Pentan-3-one.
✓ Solution
Iodoform Test (Haloform Reaction): Add NaOH and I₂ and heat gently.
Pentan-2-one: Contains a methyl ketone group (-CO-CH₃). It responds to the test forming a yellow precipitate of Iodoform (CHI₃).
Pentan-3-one: Does not contain a methyl ketone group (-CO-CH₂CH₃ on both sides). It does not give a yellow precipitate.
Iodoform Test (Haloform Reaction): Add NaOH and I₂ and heat gently.
Pentan-2-one: Contains a methyl ketone group (-CO-CH₃). It responds to the test forming a yellow precipitate of Iodoform (CHI₃).
Pentan-3-one: Does not contain a methyl ketone group (-CO-CH₂CH₃ on both sides). It does not give a yellow precipitate.
2 MarksQ3. Distinguish between Benzaldehyde and
Acetophenone.
✓ Solution
Iodoform Test:
Acetophenone (C₆H₅COCH₃): Has a methyl ketone group, gives a yellow precipitate of Iodoform on heating with NaOI.
Benzaldehyde (C₆H₅CHO): Has no methyl ketone group, does not give the iodoform test.
Alternatively, Tollens' test (Benzaldehyde forms silver mirror, Acetophenone does not). Note: Benzaldehyde does NOT give Fehling's test.
Iodoform Test:
Acetophenone (C₆H₅COCH₃): Has a methyl ketone group, gives a yellow precipitate of Iodoform on heating with NaOI.
Benzaldehyde (C₆H₅CHO): Has no methyl ketone group, does not give the iodoform test.
Alternatively, Tollens' test (Benzaldehyde forms silver mirror, Acetophenone does not). Note: Benzaldehyde does NOT give Fehling's test.
2 MarksQ4. Distinguish between Phenol and Benzoic
Acid.
✓ Solution
Sodium Bicarbonate (NaHCO₃) Test:
Benzoic Acid: Reacts with aq. NaHCO₃ to produce brisk effervescence of CO₂ gas because it is a stronger acid than carbonic acid (H₂CO₃).
Phenol: Is a weaker acid than carbonic acid and does not react with NaHCO₃ (no effervescence).
Sodium Bicarbonate (NaHCO₃) Test:
Benzoic Acid: Reacts with aq. NaHCO₃ to produce brisk effervescence of CO₂ gas because it is a stronger acid than carbonic acid (H₂CO₃).
Phenol: Is a weaker acid than carbonic acid and does not react with NaHCO₃ (no effervescence).
3 MarksQ5. An organic compound (A) C8H8O responds to
the iodoform test and 2,4-DNP test. It doesn't reduce Tollens reagent. Identify A and write the chemical
equation for the iodoform test.
✓ Solution
1. Responds to 2,4-DNP: It is a carbonyl compound (aldehyde or ketone).
2. Does not reduce Tollens': It is a ketone.
3. Responds to iodoform test: Must contain a -CO-CH₃ group.
Given formula C8H8O. The benzene ring is C6H5. That leaves C2H3O, which precisely fits -CO-CH₃.
Therefore, A is Acetophenone (1-phenylethan-1-one, C₆H₅-CO-CH₃).
1. Responds to 2,4-DNP: It is a carbonyl compound (aldehyde or ketone).
2. Does not reduce Tollens': It is a ketone.
3. Responds to iodoform test: Must contain a -CO-CH₃ group.
Given formula C8H8O. The benzene ring is C6H5. That leaves C2H3O, which precisely fits -CO-CH₃.
Therefore, A is Acetophenone (1-phenylethan-1-one, C₆H₅-CO-CH₃).
C₆H₅-CO-CH₃ + 3NaOI → C₆H₅COONa (Sodium
benzoate) + CHI₃↓(yellow ppt) + 2NaOH
2 MarksQ6. Distinguish between Ethanal (Acetaldehyde)
and Propanal.
✓ Solution
Iodoform Test:
Ethanal (CH₃CHO): Is the ONLY aldehyde containing the CH₃-CO- structure. Gives a yellow precipitate of Iodoform on warming with NaOH and I₂.
Propanal (CH₃CH₂CHO): Does not have CH₃-CO- structure. Does not give the iodoform test.
Iodoform Test:
Ethanal (CH₃CHO): Is the ONLY aldehyde containing the CH₃-CO- structure. Gives a yellow precipitate of Iodoform on warming with NaOH and I₂.
Propanal (CH₃CH₂CHO): Does not have CH₃-CO- structure. Does not give the iodoform test.
💡 Mechanism, Reactivity & Acidity (Q7 – Q14)
3 MarksQ7. Arrange the following in increasing order
of their reactivity towards nucleophilic addition: Ethanal, Propanal, Propanone, Butanone.
✓ Solution
Rule 1 (Steric Factor): Aldehydes (one alkyl group) are more reactive than ketones (two bulky alkyl groups).
Rule 2 (+I Effect): More or larger alkyl groups increase the +I effect, which decreases the positive charge (electrophilicity) on the carbonyl carbon, making it less reactive.
Aldehydes: Ethanal (CH₃CHO) > Propanal (CH₃CH₂CHO).
Ketones: Propanone (CH₃COCH₃) > Butanone (CH₃CH₂COCH₃).
Overall increasing order: Butanone < Propanone < Propanal < Ethanal.
Rule 1 (Steric Factor): Aldehydes (one alkyl group) are more reactive than ketones (two bulky alkyl groups).
Rule 2 (+I Effect): More or larger alkyl groups increase the +I effect, which decreases the positive charge (electrophilicity) on the carbonyl carbon, making it less reactive.
Aldehydes: Ethanal (CH₃CHO) > Propanal (CH₃CH₂CHO).
Ketones: Propanone (CH₃COCH₃) > Butanone (CH₃CH₂COCH₃).
Overall increasing order: Butanone < Propanone < Propanal < Ethanal.
3 MarksQ8. Arrange the following acids in order of
their increasing acid strength: CH₃COOH, FCH₂COOH, ClCH₂COOH, CCl₃COOH.
✓ Solution
Acid strength increases with the presence of Electron Withdrawing Groups (EWG) due to the -I effect stabilizing the carboxylate anion.
1. CH₃COOH has no EWG (CH₃ is ERG, +I effect), so it is the weakest.
2. Fluorine is more electronegative than Chlorine, so its -I effect is stronger: FCH₂COOH > ClCH₂COOH.
3. Three chlorines exert a massive -I effect: CCl₃COOH is the strongest.
Increasing order: CH₃COOH < ClCH₂COOH < FCH₂COOH < CCl₃COOH.
Acid strength increases with the presence of Electron Withdrawing Groups (EWG) due to the -I effect stabilizing the carboxylate anion.
1. CH₃COOH has no EWG (CH₃ is ERG, +I effect), so it is the weakest.
2. Fluorine is more electronegative than Chlorine, so its -I effect is stronger: FCH₂COOH > ClCH₂COOH.
3. Three chlorines exert a massive -I effect: CCl₃COOH is the strongest.
Increasing order: CH₃COOH < ClCH₂COOH < FCH₂COOH < CCl₃COOH.
3 MarksQ9. Predict the products of Aldol condensation
of Propanal.
✓ Solution
Propanal: CH₃-CH₂-CHO. The α-carbon is the CH₂ group.
In presence of dilute NaOH, the α-carbon of one molecule attacks the carbonyl carbon of another.
Propanal: CH₃-CH₂-CHO. The α-carbon is the CH₂ group.
In presence of dilute NaOH, the α-carbon of one molecule attacks the carbonyl carbon of another.
CH₃-CH₂-CHO + CH₃-CH₂-CHO →
CH₃-CH₂-CH(OH)-CH(CH₃)-CHO (3-Hydroxy-2-methylpentanal)
On heating, it loses water (α,β-elimination):→ CH₃-CH₂-CH=C(CH₃)-CHO (2-Methylpent-2-enal)
3 MarksQ10. Explain Cannizzaro reaction with an
example. Which aldehyde undergoes this?
✓ Solution
Aldehydes which do not contain an α-hydrogen atom undergo Cannizzaro reaction on treatment with concentrated alkali (e.g., 50% NaOH or KOH).
It is a disproportionation reaction (redox) where one molecule is reduced to alcohol while another is oxidized to the salt of a carboxylic acid.
Example (Formaldehyde):
Aldehydes which do not contain an α-hydrogen atom undergo Cannizzaro reaction on treatment with concentrated alkali (e.g., 50% NaOH or KOH).
It is a disproportionation reaction (redox) where one molecule is reduced to alcohol while another is oxidized to the salt of a carboxylic acid.
Example (Formaldehyde):
2 HCHO + NaOH(conc.) → CH₃OH (Methanol) + HCOONa (Sodium formate)
Benzaldehyde (C₆H₅CHO) also undergoes this reaction to give Benzyl alcohol and Sodium
benzoate.2 MarksQ11. Why is the boiling point of carboxylic
acids higher than that of alcohols of comparable mass?
✓ Solution
Both form hydrogen bonds. However, carboxylic acids have more extensive intermolecular hydrogen bonding. Two molecules of a carboxylic acid bond to each other via two hydrogen bonds to form a stable cyclic dimer, which exists even in the vapor phase. Alcohols form only one hydrogen bond per molecule pairing (linear association).
Both form hydrogen bonds. However, carboxylic acids have more extensive intermolecular hydrogen bonding. Two molecules of a carboxylic acid bond to each other via two hydrogen bonds to form a stable cyclic dimer, which exists even in the vapor phase. Alcohols form only one hydrogen bond per molecule pairing (linear association).
2 MarksQ12. What is Stephen reaction? Write
equations.
✓ Solution
Stephen reaction converts Nitriles (Cyanides) to Aldehydes.
Nitriles are reduced to corresponding imines with stannous chloride (SnCl₂) in the presence of hydrochloric acid, which on hydrolysis give corresponding aldehyde.
Stephen reaction converts Nitriles (Cyanides) to Aldehydes.
Nitriles are reduced to corresponding imines with stannous chloride (SnCl₂) in the presence of hydrochloric acid, which on hydrolysis give corresponding aldehyde.
R-C≡N + SnCl₂ + HCl → R-CH=NH (Imine)
R-CH=NH + H₃O⁺ → R-CHO + NH₄⁺
R-CH=NH + H₃O⁺ → R-CHO + NH₄⁺
2 MarksQ13. Write the reaction showing
Hell-Volhard-Zelinsky (HVZ) reaction. What is its significance?
✓ Solution
Carboxylic acids having an α-hydrogen are halogenated at the α-position on treatment with chlorine or bromine in the presence of a small amount of red phosphorus.
Carboxylic acids having an α-hydrogen are halogenated at the α-position on treatment with chlorine or bromine in the presence of a small amount of red phosphorus.
R-CH₂-COOH + (i) X₂/Red P (ii) H₂O → R-CH(X)-COOH
Significance: It provides a method to introduce a halogen exclusively at the α-position, which can
further be replaced by nucleophiles (like -OH, -CN, -NH₂) to synthesize various α-substituted
acids (e.g., amino acids).3 MarksQ14. Arrange the following in decreasing order
of pKa values: Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid.
✓ Solution
Note carefully: HIGHER pKa = WEAKER acid. LOWER pKa = STRONGER acid.
To arrange in decreasing order of pKa means arranging from Weaker acid to Stronger acid.
Weakest acid: 4-Methoxybenzoic acid (Methoxy is ERG, +R effect destabilizes anion).
Next: Benzoic acid.
Next: 4-Nitrobenzoic acid (One EWG nitro group).
Strongest acid: 3,4-Dinitrobenzoic acid (Two strong EWG nitro groups).
Decreasing pKa order (Weak to Strong): 4-Methoxybenzoic acid > Benzoic acid > 4-Nitrobenzoic acid > 3,4-Dinitrobenzoic acid.
Note carefully: HIGHER pKa = WEAKER acid. LOWER pKa = STRONGER acid.
To arrange in decreasing order of pKa means arranging from Weaker acid to Stronger acid.
Weakest acid: 4-Methoxybenzoic acid (Methoxy is ERG, +R effect destabilizes anion).
Next: Benzoic acid.
Next: 4-Nitrobenzoic acid (One EWG nitro group).
Strongest acid: 3,4-Dinitrobenzoic acid (Two strong EWG nitro groups).
Decreasing pKa order (Weak to Strong): 4-Methoxybenzoic acid > Benzoic acid > 4-Nitrobenzoic acid > 3,4-Dinitrobenzoic acid.
📈 Organic Conversions (Q15 – Q22)
3 MarksQ15. How will you convert Propanone to
Propene?
✓ Solution
First, reduce ketone to alcohol. Then dehydrate it to an alkene.
1. CH₃-CO-CH₃ (Propanone) + NaBH₄ (or LiAlH₄) → CH₃-CH(OH)-CH₃ (Propan-2-ol).
2. CH₃-CH(OH)-CH₃ + Conc. H₂SO₄ (Heat, 443 K) → CH₃-CH=CH₂ (Propene) + H₂O.
First, reduce ketone to alcohol. Then dehydrate it to an alkene.
1. CH₃-CO-CH₃ (Propanone) + NaBH₄ (or LiAlH₄) → CH₃-CH(OH)-CH₃ (Propan-2-ol).
2. CH₃-CH(OH)-CH₃ + Conc. H₂SO₄ (Heat, 443 K) → CH₃-CH=CH₂ (Propene) + H₂O.
2 MarksQ16. Convert Benzoic acid to Benzaldehyde.
✓ Solution
Direct reduction from acid to aldehyde is tough. Convert to acid chloride first, then use Rosenmund reduction.
1. C₆H₅-COOH + SOCl₂ → C₆H₅-COCl (Benzoyl chloride).
2. C₆H₅-COCl + H₂ → (Pd/BaSO₄, Rosenmund Reduction) → C₆H₅-CHO (Benzaldehyde).
Direct reduction from acid to aldehyde is tough. Convert to acid chloride first, then use Rosenmund reduction.
1. C₆H₅-COOH + SOCl₂ → C₆H₅-COCl (Benzoyl chloride).
2. C₆H₅-COCl + H₂ → (Pd/BaSO₄, Rosenmund Reduction) → C₆H₅-CHO (Benzaldehyde).
2 MarksQ17. Benzene to Acetophenone.
✓ Solution
Use Friedel-Crafts Acylation.
Use Friedel-Crafts Acylation.
C₆H₆ + CH₃COCl (Acetyl chloride) → (Anhydrous
AlCl₃) → C₆H₅-CO-CH₃ (Acetophenone) + HCl
3 MarksQ18. Ethanal to But-2-enal.
✓ Solution
Carbon number doubles from 2 to 4, generating an unsaturated aldehyde → Aldol Condensation.
1. CH₃CHO + CH₃CHO → (Dil. NaOH) → CH₃-CH(OH)-CH₂-CHO (3-Hydroxybutanal or Aldol).
2. CH₃-CH(OH)-CH₂-CHO → (Heat, -H₂O) → CH₃-CH=CH-CHO (But-2-enal or Crotonaldehyde).
Carbon number doubles from 2 to 4, generating an unsaturated aldehyde → Aldol Condensation.
1. CH₃CHO + CH₃CHO → (Dil. NaOH) → CH₃-CH(OH)-CH₂-CHO (3-Hydroxybutanal or Aldol).
2. CH₃-CH(OH)-CH₂-CHO → (Heat, -H₂O) → CH₃-CH=CH-CHO (But-2-enal or Crotonaldehyde).
2 MarksQ19. Acetaldehyde to Lactic Acid
(2-Hydroxypropanoic acid).
✓ Solution
Need to add one carbon containing an acid group: Nucleophilic addition of HCN followed by hydrolysis.
1. CH₃-CHO + HCN(base catalyst) → CH₃-CH(OH)-CN (Acetaldehyde cyanohydrin).
2. CH₃-CH(OH)-CN + H₂O/H⁺ (Hydrolysis) → CH₃-CH(OH)-COOH (Lactic acid).
Need to add one carbon containing an acid group: Nucleophilic addition of HCN followed by hydrolysis.
1. CH₃-CHO + HCN(base catalyst) → CH₃-CH(OH)-CN (Acetaldehyde cyanohydrin).
2. CH₃-CH(OH)-CN + H₂O/H⁺ (Hydrolysis) → CH₃-CH(OH)-COOH (Lactic acid).
2 MarksQ20. Benzaldehyde to 3-Phenylpropan-1-ol.
✓ Solution
This requires cross aldol with acetaldehyde to elongate the chain, followed by complete reduction.
1. C₆H₅CHO + CH₃CHO → (Dil. NaOH/heat) → C₆H₅-CH=CH-CHO (Cinnamaldehyde).
2. C₆H₅-CH=CH-CHO + H₂/Ni (Catalytic Hydrogenation) → C₆H₅-CH₂-CH₂-CH₂OH (3-Phenylpropan-1-ol). Both C=C and C=O are reduced.
This requires cross aldol with acetaldehyde to elongate the chain, followed by complete reduction.
1. C₆H₅CHO + CH₃CHO → (Dil. NaOH/heat) → C₆H₅-CH=CH-CHO (Cinnamaldehyde).
2. C₆H₅-CH=CH-CHO + H₂/Ni (Catalytic Hydrogenation) → C₆H₅-CH₂-CH₂-CH₂OH (3-Phenylpropan-1-ol). Both C=C and C=O are reduced.
3 MarksQ21. Benzaldehyde to Benzophenone.
✓ Solution
1. Oxidation: C₆H₅CHO + KMnO₄/H⁺ → C₆H₅COOH (Benzoic Acid).
2. Chlorination: C₆H₅COOH + SOCl₂ → C₆H₅COCl (Benzoyl chloride).
3. Friedel-Crafts: C₆H₅COCl + Benzene + Anhyd. AlCl₃ → C₆H₅-CO-C₆H₅ (Benzophenone).
1. Oxidation: C₆H₅CHO + KMnO₄/H⁺ → C₆H₅COOH (Benzoic Acid).
2. Chlorination: C₆H₅COOH + SOCl₂ → C₆H₅COCl (Benzoyl chloride).
3. Friedel-Crafts: C₆H₅COCl + Benzene + Anhyd. AlCl₃ → C₆H₅-CO-C₆H₅ (Benzophenone).
2 MarksQ22. Toluene to Benzoic acid.
✓ Solution
Direct oxidation of the side chain.
Direct oxidation of the side chain.
C₆H₅-CH₃ + KMnO₄ + KOH(Heat) →
C₆H₅-COO⁻K⁺ (Potassium benzoate)
C₆H₅-COO⁻K⁺ + H₃O⁺ (Dil. HCl) → C₆H₅-COOH (Benzoic acid)
C₆H₅-COO⁻K⁺ + H₃O⁺ (Dil. HCl) → C₆H₅-COOH (Benzoic acid)
✍ Score Guide — 22 Questions
Questions carefully curated from core CBSE/NCERT patterns, heavily prioritizing Distinguishing Tests, pKa arrangements, and multi-step name-reaction conversions.
Questions carefully curated from core CBSE/NCERT patterns, heavily prioritizing Distinguishing Tests, pKa arrangements, and multi-step name-reaction conversions.
High-Yield Facts & Formulas: Aldehydes, Ketones & Acids
Carbonyl Hybridization
Carbonyl carbon is sp2 hybridized; C=O bond is strongly polar (Cδ+-Oδ-).
Carbonyl carbon is sp2 hybridized; C=O bond is strongly polar (Cδ+-Oδ-).
Aldehyde vs Ketone Reactivity
Aldehydes are more reactive than ketones towards nucleophilic addition due to steric and inductive effects.
Aldehydes are more reactive than ketones towards nucleophilic addition due to steric and inductive effects.
Rosenmund Reduction
RCOCl + H2 (Pd/BaSO4) → RCHO + HCl. Catalyst is poisoned by sulfur/quinoline to prevent further reduction.
RCOCl + H2 (Pd/BaSO4) → RCHO + HCl. Catalyst is poisoned by sulfur/quinoline to prevent further reduction.
Stephen Reaction
RCN + SnCl2 + HCl → RCH=NH → (H3O+) → RCHO.
RCN + SnCl2 + HCl → RCH=NH → (H3O+) → RCHO.
Etard Reaction
Toluene + CrO2Cl2 → Chromium complex → Benzaldehyde.
Toluene + CrO2Cl2 → Chromium complex → Benzaldehyde.
Gatterman-Koch Reaction
Benzene + CO + HCl (Anhyd. AlCl3/CuCl) → Benzaldehyde.
Benzene + CO + HCl (Anhyd. AlCl3/CuCl) → Benzaldehyde.
Cyanohydrin formation
Nucleophilic addition of HCN. Base catalyzed to increase nucleophilicity of CN-.
Nucleophilic addition of HCN. Base catalyzed to increase nucleophilicity of CN-.
Hemiacetal & Acetal
Aldehyde + 1 Alcohol → Hemiacetal. + 2 Alcohols → Acetal.
Aldehyde + 1 Alcohol → Hemiacetal. + 2 Alcohols → Acetal.
Ammonia Derivatives
Reaction with H2N-Z (Z = OH, NH2, PhNH, etc.) is acid catalyzed. Optimum pH 3.5.
Reaction with H2N-Z (Z = OH, NH2, PhNH, etc.) is acid catalyzed. Optimum pH 3.5.
Clemmensen Reduction
Carbonyl → CH2 using Zn-Hg and conc. HCl.
Carbonyl → CH2 using Zn-Hg and conc. HCl.
Wolff-Kishner Reduction
Carbonyl → CH2 using Hydrazine followed by KOH in ethylene glycol.
Carbonyl → CH2 using Hydrazine followed by KOH in ethylene glycol.
Tollen's Reagent
Ammoniacal silver nitrate [Ag(NH3)2]+. Oxidizes aldehydes to carboxylates.
Ammoniacal silver nitrate [Ag(NH3)2]+. Oxidizes aldehydes to carboxylates.
Fehling's Solution
A (aq. CuSO4) and B (Sodium potassium tartrate + NaOH).
A (aq. CuSO4) and B (Sodium potassium tartrate + NaOH).
Iodoform Reaction
Methyl ketones/alcohols give yellow ppt of CHI3 with I2/NaOH.
Methyl ketones/alcohols give yellow ppt of CHI3 with I2/NaOH.
Aldol Condensation
Requirement: Presence of at least one α-hydrogen atom.
Requirement: Presence of at least one α-hydrogen atom.
Cannizzaro Reaction
Requirement: Absence of α-hydrogen atom. (Disproportionation).
Requirement: Absence of α-hydrogen atom. (Disproportionation).
Boiling points
Acids > Alcohols > Aldehydes > Ethers > Hydrocarbons (comparable mass).
Acids > Alcohols > Aldehydes > Ethers > Hydrocarbons (comparable mass).
Acidic strength of Carboxylic acid
High due to resonance stabilization of carboxylate ion (identical structures).
High due to resonance stabilization of carboxylate ion (identical structures).
EWG vs ERG (Acidity)
EWG increases acidity by stabilizing the anion; ERG decreases acidity.
EWG increases acidity by stabilizing the anion; ERG decreases acidity.
HVZ Reaction
α-halogenation of carboxylic acids using X2/Red Phosphorus.
α-halogenation of carboxylic acids using X2/Red Phosphorus.
Urotropine formation
Formaldehyde + Ammonia → Hexamethylenetetramine (Urinary antiseptic).
Formaldehyde + Ammonia → Hexamethylenetetramine (Urinary antiseptic).
Schiff's Reagent
Rosaniline hydrochloride solution decolorized by SO2; restored by aldehydes.
Rosaniline hydrochloride solution decolorized by SO2; restored by aldehydes.
Formaldehyde oxidation with SeO2
Selenium dioxide used for specific oxidation of methyl or methylene groups to >C=O.
Selenium dioxide used for specific oxidation of methyl or methylene groups to >C=O.
Cooke's Reagent
For detection of carbon monoxide, but in context of organic, tollen's/fehling are key for aldehydes.
For detection of carbon monoxide, but in context of organic, tollen's/fehling are key for aldehydes.
Carboxylic acid + PCl5
Gives Acid chloride (RCOCl).
Gives Acid chloride (RCOCl).
Decarboxylation
Sodium salt of acid + Soda lime (NaOH+CaO) + Heat → Alkane (one carbon less).
Sodium salt of acid + Soda lime (NaOH+CaO) + Heat → Alkane (one carbon less).
Reaction with diazomethane
RCOOH + CH2N2 → RCOOCH3 (Methyl ester).
RCOOH + CH2N2 → RCOOCH3 (Methyl ester).
Esterification rate
Order of alcohol: MeOH > EtOH > 2-propanol > t-butanol (due to steric hindrance).
Order of alcohol: MeOH > EtOH > 2-propanol > t-butanol (due to steric hindrance).
Benzaldehyde + Cl2 (Anhyd.
AlCl3)
Gives m-chlorobenzaldehyde (-CHO is meta directing).
Gives m-chlorobenzaldehyde (-CHO is meta directing).
Grignard reagent + CO2
Forms Carboxylic acid after hydrolysis.
Forms Carboxylic acid after hydrolysis.
Beckmann Rearrangement
Oxime → Amide (using acid catalyst).
Oxime → Amide (using acid catalyst).
Fehling's test fail
Benzaldehyde does not give Fehling's test (due to stabilization by resonance).
Benzaldehyde does not give Fehling's test (due to stabilization by resonance).
Benedict's Solution
Similar to Fehling's but uses citrate instead of tartrate.
Similar to Fehling's but uses citrate instead of tartrate.
Paraformaldehyde
Linear polymer of formaldehyde.
Linear polymer of formaldehyde.
Paraldehyde
Cyclic trimer of acetaldehyde, used as a sedative.
Cyclic trimer of acetaldehyde, used as a sedative.
Carbonyl addition mechanism
Attack of nucleophile on sp2 C, forming tetrahedral intermediate (sp3).
Attack of nucleophile on sp2 C, forming tetrahedral intermediate (sp3).
Ketals
Ketone + Alcohol → Hemiketal → Ketal. Difficult to form compared to Acetals.
Ketone + Alcohol → Hemiketal → Ketal. Difficult to form compared to Acetals.
Cross-Aldol Condensation
Reaction between two different aldehydes/ketones. Gives mixture of products.
Reaction between two different aldehydes/ketones. Gives mixture of products.
DNP test
2,4-Dinitrophenylhydrazine (Brady's reagent) gives Orange/Yellow ppt with carbonyls.
2,4-Dinitrophenylhydrazine (Brady's reagent) gives Orange/Yellow ppt with carbonyls.
Acetone solubility
Miscible with water in all proportions. Use of dipole-dipole interactions.
Miscible with water in all proportions. Use of dipole-dipole interactions.
Carboxylic acid + SOCl2
Best for acid chlorides as products (SO2, HCl) are gases and escape.
Best for acid chlorides as products (SO2, HCl) are gases and escape.
Formaldehyde reduction
HCHO + H2/Ni → CH3OH.
HCHO + H2/Ni → CH3OH.
Acetic acid industrial prep
Methanol + CO (Rh catalyst/HI) → Acetic acid (Monsanto process).
Methanol + CO (Rh catalyst/HI) → Acetic acid (Monsanto process).
Formic acid reduction
Oxidized easily (unlike other acids) due to presence of -CHO type structure. Redues Tollen's.
Oxidized easily (unlike other acids) due to presence of -CHO type structure. Redues Tollen's.
Amide formation
Acid chloride + Ammonia/Amine.
Acid chloride + Ammonia/Amine.
Friedel-Crafts acylation catalyst
Need more than stoichiometric amount of AlCl3 due to complexation with carbonyl.
Need more than stoichiometric amount of AlCl3 due to complexation with carbonyl.
Oppenauer Oxidation
Specific oxidation of secondary alcohols to ketones using aluminum isopropoxide/acetone.
Specific oxidation of secondary alcohols to ketones using aluminum isopropoxide/acetone.
Acetaldehyde + SeO2
Gives Glyoxal (OHC-CHO).
Gives Glyoxal (OHC-CHO).
Acid anhydride formation
Two acid molecules - H2O (using P2O5).
Two acid molecules - H2O (using P2O5).
Nitration of Benzoic acid
Gives m-nitrobenzoic acid (-COOH is meta directing).
Gives m-nitrobenzoic acid (-COOH is meta directing).
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App