Educational Galaxy

Chapter 8: Aldehydes, Ketones and Carboxylic Acids — EG's own summary, in simple words

Book: Chemistry — Part II (CBSE, Class 12)  |  Chapter 8  |  Medium: English

The carbonyl group (C=O) is the most versatile functional group in organic chemistry: aldehydes and ketones add nucleophiles, oxidise and reduce in predictable ways, and condense with each other; carboxylic acids add acidity and a whole family of derivatives. This chapter teaches their names and preparations, the addition mechanism, the name reactions, and how to tell aldehydes, ketones and acids apart with chemical tests.

Diagram (labels on the image): Aldehyde (–CHO), Ketone (C=O), Carboxylic acid (–COOH) (labels as drawn). Labels added by Educational Galaxy on an originally generated illustration.

EG schematic (drawn and labelled by Educational Galaxy): Carbonyl map: nucleophilic addition to C=O (aldehydes more reactive than ketones), then aldol condensation, Cannizzaro disproportionation, Clemmensen/Wolff–Kishner reduction, and oxidation to acids.

8.1 Nomenclature and structure of the carbonyl group

Aldehydes (R–CHO) take the suffix ‘al’ (ethanal, benzaldehyde as a retained name, formaldehyde for methanal); ketones (R–CO–R′) take ‘one’ (propanone, butan-2-one). Carboxylic acids (R–COOH) take ‘oic acid’ (ethanoic acid, benzoic acid). The carbonyl carbon is sp², trigonal planar, and polar: oxygen pulls electrons, leaving carbon slightly positive and open to nucleophilic attack.

Aldehydes are more reactive than ketones toward nucleophiles for two reasons: ketones have two alkyl groups donating electrons (+I effect) that reduce the carbon’s positive charge, and two groups also block the nucleophile’s approach (steric hindrance). Formaldehyde (no alkyl group) is the most reactive; benzaldehyde is less reactive than aliphatic aldehydes because the ring donates by resonance.

8.2 Preparation of aldehydes and ketones

• From alcohols: oxidation of primary alcohols (acidified K₂Cr₂O₇ or PCC for stopping at the aldehyde) and secondary alcohols (to ketones); dehydrogenation of alcohols over hot copper (primary → aldehyde, secondary → ketone, tertiary → alkene).
• From hydrocarbons: ozonolysis of alkenes (breaks C=C into carbonyls); hydration of alkynes with HgSO₄/H₂SO₄ (ethyne → ethanal, other terminal alkynes → methyl ketones by Markovnikov addition).
• From acyl chlorides: Rosenmund reduction (H₂/Pd–BaSO₄, poisoned catalyst) stops at the aldehyde; acyl chloride + dialkylcadmium gives ketones.
• From nitriles: Stephen reduction (SnCl₂/HCl, then hydrolysis) gives aldehydes; Grignard + nitrile, then hydrolysis, gives ketones.
• Aromatic aldehydes/ketones: Friedel–Crafts acylation (RCOCl/AlCl₃) gives aryl ketones; the Gattermann–Koch reaction (CO + HCl/AlCl₃) gives benzaldehyde from benzene.

8.3 Nucleophilic addition and addition–elimination

The mechanism: a nucleophile attacks the carbonyl carbon, forming a tetrahedral alkoxide intermediate, which is protonated to the addition product. Because the carbonyl is planar, attack can come from either face (important for chirality in later chapters).

• HCN addition gives cyanohydrins — useful because the CN group can become an acid or amine later.
• NaHSO₃ addition gives a crystalline bisulphite adduct, used to purify aldehydes (and methyl ketones).
• Alcohols add to give hemiacetals and then acetals (with two alcohol molecules, acid-catalysed); acetals protect carbonyls during synthesis.
• Ammonia derivatives add and then lose water (addition–elimination) to give imines and related products: hydroxylamine → oxime, hydrazine → hydrazone, phenylhydrazine → phenylhydrazone, semicarbazide → semicarbazone, and 2,4-dinitrophenylhydrazine (Brady’s reagent) → orange 2,4-DNP precipitate — the test for a carbonyl group.
• Reactivity: HCN and NaHSO₃ add to most aldehydes and methyl ketones; bulky ketones (benzophenone) react slowly or not at all.

8.4 Reduction and oxidation of aldehydes and ketones

• Reduction to alcohols: LiAlH₄ or NaBH₄ reduces aldehydes to primary alcohols and ketones to secondary alcohols.
• Clemmensen reduction: Zn–Hg + concentrated HCl converts C=O fully to CH₂ (aldehyde → alkane, ketone → alkane). It survives acid-stable groups.
• Wolff–Kishner reduction: NH₂NH₂ + KOH in hot ethylene glycol also converts C=O to CH₂, but under basic conditions — choose it when the molecule is acid-sensitive.
• Oxidation: aldehydes oxidise easily to carboxylic acids (Tollens’ reagent, ammoniacal AgNO₃, gives a silver mirror; Fehling’s solution gives a red Cu₂O precipitate). Ketones resist mild oxidation and need vigorous conditions that break C–C bonds. This is the aldehyde-vs-ketone distinction: Tollens’ and Fehling’s tests are positive only for aldehydes (and α-hydroxy ketones).
• Haloform reaction: methyl ketones (and ethanal, and ethanol) with I₂/NaOH give yellow iodoform (CHI₃) precipitate — the iodoform test for the CH₃CO– group.

8.5 Aldol condensation and Cannizzaro reaction

Aldol condensation: aldehydes or ketones with α-hydrogens, in dilute alkali, condense: one molecule’s α-carbon attacks another’s carbonyl carbon, giving a β-hydroxy aldehyde/ketone (an ‘aldol’). On heating, water is lost to give an α,β-unsaturated carbonyl (crotonaldehyde from ethanal). Crossed aldol (two different carbonyls) gives a mixture unless one has no α-hydrogens (like formaldehyde or benzaldehyde).

Cannizzaro reaction: aldehydes with NO α-hydrogens (formaldehyde, benzaldehyde) cannot do aldol; in concentrated alkali they disproportionate — one molecule is oxidised to the acid (as its salt) and another reduced to the alcohol. Example: 2HCHO + NaOH → CH₃OH + HCOONa. The mechanism transfers a hydride from one molecule to another.

Electrophilic substitution of aromatic aldehydes/ketones: the carbonyl group is deactivating and meta-directing (it withdraws electrons), so nitration of benzaldehyde gives m-nitrobenzaldehyde.

8.6 Carboxylic acids — preparation, acidity and reactions

• Preparation: oxidation of primary alcohols and aldehydes; hydrolysis of nitriles (R–CN + H₂O/H⁺ → R–COOH) and of esters; Grignard reagent + CO₂, then hydrolysis (adds one carbon: R–MgX + CO₂ → R–COOH); aromatic acids from alkylbenzenes oxidised by KMnO₄.
• Acidity: carboxylic acids (pKa ≈ 4–5) are far more acidic than alcohols and phenols because the carboxylate ion is resonance-stabilised with the charge shared equally by two oxygens. Electron-withdrawing groups increase acidity (trichloroacetic acid is much stronger than acetic acid — the −I effect grows with more chlorines and shorter distance); electron-donating groups decrease it. Formic acid is stronger than acetic acid because the methyl group donates electrons.
• Reactions: with NaHCO₃ they release CO₂ (the acid test); with alcohols they form esters (Fischer esterification, H₂SO₄ catalyst); with PCl₅/SOCl₂ they form acyl chlorides; with NH₃ they form ammonium salts, which on heating give amides, which on dehydration give nitriles; with LiAlH₄ they reduce to primary alcohols.
• Hell–Volhard–Zelinsky (HVZ) reaction: a carboxylic acid with α-hydrogens, treated with Br₂ and red phosphorus, substitutes bromine at the α-carbon (once or more) — the route to α-halo acids used in amino-acid synthesis.
• Physical properties: acids hydrogen-bond strongly (even as dimers in the vapour), so they boil higher than alcohols of similar mass; the first four acids mix with water, solubility falling as the chain grows.

Distinguishing aldehydes, ketones and acids

Test	Aldehyde	Ketone	Carboxylic acid
Tollens’ (Ag mirror)	Positive	Negative	Negative*
Fehling’s (red ppt)	Positive	Negative	Negative
NaHCO₃ (CO₂ fizz)	No	No	Yes
Iodoform (yellow ppt)	Only ethanal	Only methyl ketones	No
2,4-DNP (orange ppt)	Yes	Yes	No

Key Terms / Important Terms

• Carbonyl group: C=O; polar, planar, attacked by nucleophiles at carbon.
• Cyanohydrin: Addition product of HCN to a carbonyl; CN can become acid or amine.
• Acetal: A carbonyl protected by two alkoxy groups (from aldehyde + 2 alcohols).
• Tollens’ reagent: Ammoniacal AgNO₃; silver mirror with aldehydes.
• Fehling’s solution: Alkaline Cu²⁺ complex; red Cu₂O with aldehydes.
• Aldol: β-Hydroxy aldehyde/ketone from condensation of two carbonyls with α-hydrogens.
• Cannizzaro reaction: Disproportionation of α-hydrogen-free aldehydes in concentrated alkali.
• HVZ reaction: α-Bromination of carboxylic acids with Br₂/red phosphorus.

Practice set — try it yourself

1. Why is ethanal more reactive than propanone toward HCN? Give two reasons.
2. What is Brady’s reagent? What colour precipitate does it give, and with what?
3. Distinguish ethanal from propanone using two chemical tests.
4. Why does benzaldehyde undergo Cannizzaro but not aldol condensation?
5. Arrange in increasing acidity: CH₃COOH, ClCH₂COOH, Cl₃CCOOH, HCOOH. Justify.

Questions with Answers — full answers

1. Write the mechanism of HCN addition to ethanal in words. Answer: CN⁻ attacks the carbonyl carbon, forming a tetrahedral alkoxide intermediate; the alkoxide is then protonated (by HCN or water) to give the cyanohydrin, 2-hydroxypropanenitrile.
2. How would you convert (a) benzoyl chloride to benzaldehyde, (b) propanone to propane? Answer: (a) Rosenmund reduction: H₂/Pd–BaSO₄ stops at the aldehyde. (b) Clemmensen (Zn–Hg/HCl) or Wolff–Kishner (NH₂NH₂/KOH) reduces C=O fully to CH₂.
3. Why does formaldehyde react fastest with nucleophiles among aldehydes? Answer: It has no alkyl group donating electrons or blocking attack, so its carbonyl carbon is the most positive and the most accessible.
4. Give the aldol product of ethanal and its dehydration product. Answer: Aldol addition gives 3-hydroxybutanal; heating loses water to give but-2-enal (crotonaldehyde), an α,β-unsaturated aldehyde.
5. A compound gives a silver mirror with Tollens’ reagent and a yellow precipitate with I₂/NaOH. Identify the functional group pattern. Answer: Silver mirror means an aldehyde; iodoform means the CH₃CO– pattern. The compound is ethanal (the only aldehyde with a methyl group next to the carbonyl).
6. Why is trichloroacetic acid stronger than acetic acid? Answer: Three chlorine atoms withdraw electrons by the −I effect, stabilising the carboxylate anion and making proton loss easier. More chlorines and closer chlorines both increase acidity.
7. Write the HVZ reaction of ethanoic acid with excess bromine and red phosphorus. Answer: CH₃COOH + Br₂/P → BrCH₂COOH (bromoethanoic acid), and with excess bromine, Br₂CHCOOH and then Br₃CCOOH.
8. How would you prepare benzoic acid from (a) toluene, (b) benzonitrile? Answer: (a) Oxidise toluene with alkaline KMnO₄, then acidify. (b) Hydrolyse benzonitrile with dilute acid or alkali, then acidify.

Extension Questions with Answers

1. Crossed aldol between ethanal and formaldehyde in dilute NaOH gives mainly one product. Which, and why? Answer: Formaldehyde has no α-hydrogens, so it can only be the electrophile (carbonyl) partner; ethanal provides the enolate. The main product is 3-hydroxypropanal (which dehydrates to propenal).
2. Why do ketones not give Fehling’s test, even though they have a carbonyl group? Answer: Fehling’s solution is a mild oxidant that oxidises aldehydes to acids but cannot oxidise ketones without breaking C–C bonds. Ketones would need vigorous oxidation.
3. Aromatic aldehydes undergo electrophilic substitution at the meta position. Why? Answer: The −CHO group withdraws electrons by resonance, destabilising the arenium ions for ortho/para attack more than for meta attack, so substitution goes meta.
4. Explain why carboxylic acids boil higher than alcohols of similar molar mass. Answer: Acid molecules pair into hydrogen-bonded dimers (two H-bonds per pair) and also associate in chains, needing much more energy to separate than alcohol molecules.
5. Convert: (a) benzoic acid to benzaldehyde, (b) ethanal to ethanoic acid, (c) propanone to propan-2-ol. Answer: (a) Benzoic acid → benzoyl chloride (SOCl₂) → Rosenmund reduction to benzaldehyde. (b) Oxidise ethanal with Tollens’ reagent or acidified K₂Cr₂O₇. (c) Reduce propanone with NaBH₄ or LiAlH₄.

Board-pattern Questions with Answers

1. Write the IUPAC names of CH₃CHO, CH₃COCH₃ and C₆H₅COOH. Answer: Ethanal, propanone, benzoic acid.
2. Give the test to distinguish (a) benzaldehyde from acetophenone, (b) ethanoic acid from ethanol. Answer: (a) Tollens’ test: benzaldehyde gives a silver mirror, acetophenone does not. (b) NaHCO₃: ethanoic acid fizzes (CO₂), ethanol does not.
3. Why is formic acid stronger than acetic acid? Answer: The methyl group in acetic acid donates electrons (+I), destabilising the carboxylate anion and making proton loss harder. Formic acid has only hydrogen, so its anion is more stable.
4. What is the product of acetophenone with I₂/NaOH? Why does benzophenone not react? Answer: Yellow iodoform (CHI₃) plus sodium benzoate — acetophenone is a methyl ketone. Benzophenone has no methyl group next to the carbonyl, so it cannot form iodoform.
5. Name the product of the Cannizzaro reaction of benzaldehyde. Answer: Benzyl alcohol + sodium benzoate (the salt of benzoic acid) — one molecule reduced, one oxidised.