The hybridisation of the carbonyl carbon and the approximate bond angle around it in propanone are:

Answer: B. The carbonyl carbon forms three σ bonds in one plane, which needs sp² hybridisation and gives about 120°. The unhybridised p orbital forms the π bond with oxygen. Option A describes a saturated carbon, option C a carbon in a triple bond or cumulene, and option D is not seen in organic carbon.

Which reagent converts an acyl chloride into an aldehyde without reducing it further to an alcohol?

Answer: C. This is the Rosenmund reduction, in which BaSO₄ poisons the palladium so the reduction stops at the aldehyde. LiAlH₄ and NaBH₄ would carry on to the alcohol, and Zn-Hg with HCl is the Clemmensen reagent that turns C=O into CH₂.

Match List I with List II. List I: (i) Etard reaction (ii) Gattermann–Koch reaction (iii) Stephen reaction (iv) Rosenmund reduction. List II: (p) SnCl₂ and HCl, then H₂O (q) CO and HCl with AlCl₃ (r) CrO₂Cl₂ followed by hydrolysis (s) H₂ over Pd on BaSO₄

Answer: A. Etard uses chromyl chloride on toluene, Gattermann–Koch uses CO and HCl with AlCl₃, Stephen uses SnCl₂ and HCl on a nitrile, and Rosenmund uses H₂ over poisoned palladium on an acyl chloride. Each of the other options swaps at least two of these pairings.

Aldehydes, Ketones and Carboxylic Acids — CUET Chemistry Notes & MCQs

📌 Snapshot

Why one C=O group runs so much of organic chemistry: named reactions, tests and acid strength made simple.
Naming, structure and every preparation method, from Rosenmund and Etard to Grignard and Friedel Crafts.
Why nucleophiles attack the carbonyl carbon, why aldehydes beat ketones, and how aldol and Cannizzaro differ.
Tests that tell compounds apart, pKa reasoning for carboxylic acids, worked examples and exam trap tables.

📖 Detailed Notes

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Chemistry Class XII, Chapter 8.

Almond essence, nail polish remover and vinegar: one group, three personalities

Benzaldehyde gives almonds their smell, acetone removes nail polish and ethanoic acid makes vinegar sour. Three very different substances, yet each has a carbon double bonded to oxygen, the carbonyl group (>C=O). What sits on the other two bonds of that carbon decides everything.

This chapter has one big idea: the carbonyl carbon is electron poor, so nucleophiles attack it. Once you see why that happens, you can predict why aldehydes react faster than ketones, why aldehydes are easy to oxidise, why some carbonyls undergo aldol and others Cannizzaro, and why a carboxylic acid is acidic while an aldehyde is not.

Family	Carbonyl carbon is bonded to	General formula	Example
Aldehyde	One H and one C (or H)	R–CHO	CH₃CHO
Ketone	Two carbons	R–CO–R′	CH₃COCH₃
Carboxylic acid	One OH and one C (or H)	R–COOH	CH₃COOH

Names and structure

Common names of aldehydes come from the parent acid: replace "ic acid" with "aldehyde". So formic acid gives formaldehyde, acetic acid gives acetaldehyde. Positions on the chain are marked with Greek letters α, β, γ, starting from the carbon next to the carbonyl.

IUPAC rules. Aldehydes end in –al and the chain is numbered from the CHO carbon, which is always C-1. Ketones end in –one and the numbering starts from the end nearer the carbonyl. Acids end in –oic acid, again with the COOH carbon as C-1. When CHO or COOH is attached to a ring, the suffix becomes carbaldehyde or carboxylic acid.

Formula	Common name	IUPAC name
HCHO	Formaldehyde	Methanal
CH₃CHO	Acetaldehyde	Ethanal
CH₃CH₂CHO	Propionaldehyde	Propanal
CH₃COCH₃	Acetone	Propanone
CH₃COCH₂CH₃	Ethyl methyl ketone	Butan-2-one
C₆H₅CHO	Benzaldehyde	Benzenecarbaldehyde
C₆H₅COCH₃	Acetophenone	1-Phenylethanone
C₆H₅COC₆H₅	Benzophenone	Diphenylmethanone
HCOOH	Formic acid (ants)	Methanoic acid
CH₃COOH	Acetic acid (vinegar)	Ethanoic acid
CH₃CH₂CH₂COOH	Butyric acid (rancid butter)	Butanoic acid
C₆H₅COOH	Benzoic acid	Benzenecarboxylic acid
HOOC–COOH	Oxalic acid	Ethanedioic acid
HOOC–CH₂–COOH	Malonic acid	Propanedioic acid

Worked example: naming. Name (CH₃)₂CHCH₂CHO, CH₃COCH₂CH(CH₃)₂, CH₃CH=CHCHO and ClCH₂CH₂COOH.

Step	What we do	Result
1	Find the longest chain containing the carbonyl or COOH carbon, and number from that carbon (aldehyde, acid) or from the nearer end (ketone)	CHO carbon is C-1 in the first, C=O is C-2 in the second
2	(CH₃)₂CHCH₂CHO: four carbon chain, methyl on C-3	3-Methylbutanal
3	CH₃COCH₂CH(CH₃)₂: five carbon chain, C=O at C-2, methyl on C-4	4-Methylpentan-2-one
4	CH₃CH=CHCHO: double bond starts at C-2	But-2-enal
5	ClCH₂CH₂COOH: chlorine on C-3	3-Chloropropanoic acid

The common wrong step: numbering a ketone from the wrong end. The carbonyl must get the lowest possible number, so 4-methylpentan-2-one is correct and 2-methylpentan-4-one is not.

Structure of the carbonyl group. The carbonyl carbon is sp² hybridised. It forms three σ bonds in one plane at about 120° (trigonal planar) and the fourth p orbital overlaps sideways with oxygen's p orbital to form the π bond, which lies above and below the plane. Oxygen is more electronegative than carbon, so the π electrons are pulled towards it. Two resonance forms describe this: the neutral C=O and a dipolar form with C⁺ and O⁻. The result is a polar bond with carbon δ⁺ (electrophilic) and oxygen δ⁻ (nucleophilic). Aldehydes and ketones therefore have large dipole moments and are more polar than ethers.

In a carboxylic acid the carbonyl carbon is less electrophilic than in an aldehyde. The lone pair on the neighbouring OH oxygen flows into the carbonyl by resonance and partly cancels its positive charge. That is why carboxylic acids do not show the typical addition reactions of aldehydes and ketones.

Making aldehydes and ketones

Four general routes work for both families.

Method	Reagents and conditions	Product	Remember
Oxidation of alcohols	1° alcohol with a mild oxidant such as PCC; 2° alcohol with CrO₃ or K₂Cr₂O₇	1° gives aldehyde; 2° gives ketone	Strong oxidants push an aldehyde on to the acid
Dehydrogenation	Alcohol vapour over Cu at 573 K	1° gives aldehyde; 2° gives ketone	3° alcohols give alkenes
Ozonolysis of alkenes	O₃, then Zn and H₂O	Aldehydes and/or ketones, depending on substitution	Breaks the C=C and puts C=O on each carbon
Hydration of alkynes	Dilute H₂SO₄ with HgSO₄	Ethyne gives ethanal; every other alkyne gives a ketone	Enol first, then it tautomerises to the carbonyl

Aldehydes only. Each of these stops at the aldehyde stage.

Named method	Reagents	Change	Example
Rosenmund reduction	H₂ over Pd on BaSO₄	Acyl chloride → aldehyde	CH₃COCl → CH₃CHO
Stephen reaction	SnCl₂ and HCl, then H₂O	Nitrile → aldehyde (via imine)	C₆H₅CN → C₆H₅CHO
DIBAL-H	DIBAL-H at low temperature, then H₂O	Nitrile or ester → aldehyde	RCN → RCHO
Etard reaction	CrO₂Cl₂ in CS₂, then H₃O⁺	Toluene → benzaldehyde	C₆H₅CH₃ → C₆H₅CHO
Side chain chlorination	Cl₂ with light, then hydrolysis	Toluene → benzal chloride → benzaldehyde	C₆H₅CHCl₂ → C₆H₅CHO
Gattermann–Koch	CO and HCl with anhydrous AlCl₃ and CuCl	Benzene → benzaldehyde	C₆H₆ → C₆H₅CHO

The Rosenmund catalyst is deliberately poisoned with BaSO₄. A full strength palladium catalyst would carry on and reduce the aldehyde to an alcohol. In Gattermann–Koch, CO and HCl together act like unstable formyl chloride (HCOCl), and the attacking electrophile is the formyl cation, HCO⁺. The reaction is an electrophilic substitution on the ring.

Ketones only.

Named method	Reagents	Change	Example
Acyl chloride with dialkylcadmium	R′₂Cd (made from RMgX and CdCl₂)	Acyl chloride → ketone	2CH₃COCl + (CH₃)₂Cd → 2CH₃COCH₃ + CdCl₂
Nitrile with Grignard	RMgX, then H₃O⁺	Nitrile → ketone	CH₃CN + C₆H₅MgBr → C₆H₅COCH₃
Friedel–Crafts acylation	RCOCl with anhydrous AlCl₃	Arene → aryl ketone	C₆H₆ + CH₃COCl → C₆H₅COCH₃ + HCl

Pause and predict: why must you use a dialkylcadmium and not a Grignard reagent to turn an acyl chloride into a ketone?

Check your answer: a Grignard reagent is so reactive that it attacks the ketone as soon as it forms and carries on to a tertiary alcohol. The cadmium reagent is gentler and stops at the ketone.

Making carboxylic acids

Method	Reagents	Key point
Oxidation of 1° alcohols or aldehydes	Alkaline KMnO₄ (then acidify), acidic K₂Cr₂O₇, or CrO₃ in H₂SO₄	Same number of carbons
Oxidation of alkylbenzenes	Hot KMnO₄ with KOH, then acid	Every alkyl side chain with a benzylic H becomes COOH. A 3° group such as tert-butyl is untouched
Hydrolysis of nitriles	Dilute acid or alkali, with heat	Goes nitrile → amide → acid. Mild conditions can stop at the amide
Grignard reagent with CO₂	RMgX with dry ice, then H₃O⁺	The acid has one more carbon than R
Hydrolysis of acid derivatives	Water on acyl chlorides and anhydrides, or acid or base on esters	Base hydrolysis of an ester gives the carboxylate salt, which is then acidified

Physical properties: the hydrogen bonding ladder

Methanal is a gas and ethanal is a volatile liquid. Other lower aldehydes and ketones are liquids, and the first few dissolve in water because the carbonyl oxygen accepts hydrogen bonds from water. Solubility falls as the alkyl part grows. Lower aldehydes smell sharp, while higher aldehydes and many ketones have pleasant smells and are used in perfumes and flavourings.

Boiling points follow the strength of intermolecular forces. Compare molecules of similar mass.

Compound (mass about 58–60)	Boiling point	Force that holds it together
n-Butane	273 K	Weak London forces only
Methoxyethane	281 K	Weak dipole–dipole
Propanal	322 K	Strong dipole–dipole (C=O is very polar)
Propanone	329 K	Strong dipole–dipole
Propan-1-ol	370 K	Hydrogen bonding between molecules
Ethanoic acid	391 K	Hydrogen bonding that links molecules in pairs

So aldehydes and ketones boil higher than hydrocarbons and ethers (dipoles) but lower than alcohols (they have no O–H, so they cannot hydrogen bond with each other). Carboxylic acids boil even higher than alcohols because each pair of molecules is held by two hydrogen bonds, forming a dimer.

Beyond the textbook. For the exam: carboxylic acids have higher boiling points than alcohols of similar mass because of stronger hydrogen bonding. In reality: the pairing is so strong that acids stay as dimers even in the vapour and in non-polar solvents like benzene. A solute that pairs up appears to have double its molar mass, which is one of the standard examples of abnormal molar mass in the Solutions chapter. Why it matters: a van't Hoff factor below 1 for ethanoic acid in benzene comes straight from this dimer.

Nucleophilic addition: the main reaction of C=O

Why it happens. The carbonyl carbon is δ⁺, so it attracts nucleophiles (Nu⁻). The nucleophile attacks from above or below the plane (where the π cloud is, and where there is no crowding), the π electrons move onto oxygen, and the carbon changes from sp² to sp³. The product alkoxide then picks up H⁺.

Step	What happens	Why
1	Nu⁻ attacks the carbonyl carbon, perpendicular to the plane	Carbon is δ⁺ and the π face is open
2	The C=O π electrons move onto oxygen, giving a tetrahedral alkoxide (sp³ carbon)	Oxygen can hold a negative charge because it is electronegative
3	The alkoxide takes H⁺ from the medium	Gives the neutral addition product, with Nu and H added across C=O

Why aldehydes beat ketones. Two reasons work together. Electronic: a ketone has two electron releasing alkyl groups pushing electrons onto the carbonyl carbon, reducing its δ⁺ charge, while an aldehyde has only one (or none in methanal). Steric: two bulky groups in a ketone crowd the carbon and block the nucleophile, and the tetrahedral product is also more crowded. Aromatic carbonyls are slower still because the ring donates electrons by resonance.

Reactivity order: HCHO > CH₃CHO > CH₃CH₂CHO > CH₃COCH₃ > CH₃COCH₂CH₃ > C₆H₅COCH₃.

Worked example: ranking reactivity. Arrange in increasing order of reactivity towards nucleophilic addition: (a) ethanal, (b) propanone, (c) propanal, (d) butanone.

Step	What we do	Result
1	Aldehydes are more reactive than ketones	(a) and (c) are above (b) and (d)
2	Within each family, a bigger alkyl group is more electron releasing and bulkier, so reactivity falls	Ethanal > propanal, and propanone > butanone
3	Write from least to most reactive	Butanone < propanone < propanal < ethanal, that is (d) < (b) < (c) < (a)

Worked example: substituents on benzaldehyde. Arrange p-nitrobenzaldehyde, benzaldehyde, p-tolualdehyde and acetophenone in increasing order of reactivity. The nitro group withdraws electrons and raises the δ⁺ on carbon, so p-nitrobenzaldehyde is the most reactive. The methyl group releases electrons, so p-tolualdehyde is slower than benzaldehyde. Acetophenone is a ketone, so it is slowest: acetophenone < p-tolualdehyde < benzaldehyde < p-nitrobenzaldehyde.

The common wrong step: ranking by size of molecule alone. What matters is how much positive charge the carbonyl carbon carries and how crowded it is.

Addition reactions at a glance.

Nucleophile	Reagent	Product	Notes
CN⁻	HCN with a trace of base	Cyanohydrin, R₂C(OH)CN	The base makes CN⁻, a far better nucleophile than HCN itself
HSO₃⁻	Saturated NaHSO₃	Crystalline bisulphite addition product	Works for aldehydes and small ketones. Acid or alkali regenerates the carbonyl, so it purifies carbonyls
ROH	2 ROH with dry HCl	Hemiacetal, then acetal RCH(OR′)₂	Acetals are stable to base and hydrolysed by dilute acid, so they protect C=O. A ketone with ethylene glycol gives a cyclic ketal
H₂N–Z	Weakly acidic medium	C=N–Z compound plus H₂O	Addition followed by loss of water

The H₂N–Z family is exam favourite material. The general change is >C=O + H₂N–Z → >C=N–Z + H₂O.

H₂N–Z reagent	Name of product
R–NH₂ (Z = alkyl or aryl)	Substituted imine (Schiff's base)
NH₂OH (Z = OH)	Oxime
NH₂NH₂ (Z = NH₂)	Hydrazone
C₆H₅NHNH₂ (Z = NHC₆H₅)	Phenylhydrazone
2,4-Dinitrophenylhydrazine	2,4-DNP derivative (yellow, orange or red solid)
NH₂NHCONH₂ (Z = NHCONH₂)	Semicarbazone

Because 2,4-DNP gives a coloured precipitate with any aldehyde or ketone, it is the standard test for the presence of a carbonyl group.

Reduction: stop at the alcohol or go all the way

Reagent	Change	Example
NaBH₄, LiAlH₄ or H₂ with Ni, Pd or Pt	C=O → CH–OH	Aldehyde gives 1° alcohol; ketone gives 2° alcohol
Clemmensen: Zn-Hg and conc. HCl	C=O → CH₂	C₆H₅COCH₃ → C₆H₅CH₂CH₃, acid stable compounds only
Wolff–Kishner: NH₂NH₂, then KOH in ethylene glycol, heat	C=O → CH₂	Same product, for compounds that cannot survive acid

Both Clemmensen and Wolff–Kishner remove the oxygen completely, giving the CH₂ group. The choice between them is simply whether your molecule tolerates acid or base.

Oxidation: the test that tells aldehydes from ketones

An aldehyde carries an H on the carbonyl carbon, so it is oxidised to the acid with the same number of carbons, even by very mild oxidants. A ketone has no such H, so only a harsh oxidant at high temperature breaks its C–C bonds and gives a mixture of acids with fewer carbons.

Test	Reagent	Aldehyde gives	Ketone gives	Trap
Tollens'	Ammoniacal AgNO₃, [Ag(NH₃)₂]⁺	Silver mirror	Nothing	Positive for aliphatic and aromatic aldehydes
Fehling's	Cu²⁺ in alkaline sodium potassium tartrate (solutions A and B mixed fresh)	Red brown Cu₂O precipitate	Nothing	Aromatic aldehydes do not respond

RCHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → RCOO⁻ + 2Ag + 4NH₃ + 2H₂O

RCHO + 2Cu²⁺ + 5OH⁻ → RCOO⁻ + Cu₂O + 3H₂O

Pause and predict: you have two unlabelled bottles, ethanal and benzaldehyde. Tollens' reagent gives a silver mirror with both. Which single test separates them?

Check your answer: Fehling's solution. Ethanal gives the red brown precipitate, benzaldehyde does not.

The haloform (iodoform) reaction. Compounds containing CH₃CO– (methyl ketones, ethanal) or CH₃CH(OH)– (ethanol, propan-2-ol, since NaOI first oxidises them to a carbonyl) react with NaOI (or I₂ with NaOH) to give a yellow precipitate of iodoform (CHI₃) and a sodium carboxylate with one carbon fewer.

CH₃COCH₃ + 3NaOI → CH₃COONa + CHI₃ + 2NaOH

Acetone passes, as do ethanal, ethanol, butan-2-one and acetophenone. Methanol, methanal, propanal, pentan-3-one, benzaldehyde and benzophenone all fail. The reaction touches only the methyl group next to the carbonyl, so any C=C in the molecule survives.

α-Hydrogen chemistry: aldol and Cannizzaro

An α-hydrogen is one on the carbon next to the carbonyl. It is unusually acidic because removing it leaves an enolate ion whose negative charge is shared with the electronegative oxygen. Whether a carbonyl has α-hydrogens decides which of two reactions it undergoes.

Feature	Aldol condensation	Cannizzaro reaction
Who reacts	Aldehydes and ketones with α-H	Aldehydes without α-H
Conditions	Dilute alkali (NaOH)	Concentrated alkali
What happens	Enolate of one molecule attacks C=O of another	Hydride transfer: one molecule is oxidised, the other reduced
Products	β-Hydroxy carbonyl; on heating, α,β-unsaturated carbonyl	Carboxylate salt plus alcohol
Examples	Ethanal, propanal, propanone	Methanal, benzaldehyde, 2,2-dimethylpropanal

Aldol mechanism.

Step	What happens	Why
1	OH⁻ removes an α-H, giving an enolate	α-H is acidic because the enolate is resonance stabilised
2	The enolate carbon attacks the C=O carbon of a second molecule	The carbon with the extra electrons attacks the δ⁺ carbon
3	The alkoxide takes H⁺ from water	Gives the β-hydroxy compound (aldol, from ald and ol)
4	On heating, water is lost between α and β carbons	Gives a conjugated α,β-unsaturated carbonyl, which is more stable

2CH₃CHO → (dil. NaOH) CH₃CH(OH)CH₂CHO (3-hydroxybutanal) → (heat, −H₂O) CH₃CH=CHCHO (but-2-enal)

If two different carbonyls that both have α-H are mixed, each can act as either partner and the crossed aldol gives four products. Ethanal with propanal gives 3-hydroxybutanal, 3-hydroxy-2-methylpentanal, 3-hydroxypentanal and 3-hydroxy-2-methylbutanal.

Cannizzaro. Hydroxide adds to one aldehyde molecule, and the resulting alkoxide passes a hydride ion (H⁻) to a second molecule. One molecule ends up as the carboxylate (oxidised), the other as the alcohol (reduced).

2HCHO + NaOH(conc.) → CH₃OH + HCOONa

2C₆H₅CHO + NaOH(conc.) → C₆H₅CH₂OH + C₆H₅COONa

Pause and predict: what do you get if benzaldehyde is warmed with concentrated NaOH, and what do you get from ethanal under dilute NaOH?

Check your answer: benzaldehyde has no α-H, so it disproportionates into benzyl alcohol and sodium benzoate. Ethanal has α-H, so it does aldol and gives 3-hydroxybutanal.

Electrophilic substitution in the ring. In benzaldehyde and acetophenone the carbonyl group withdraws electrons from the ring, so it is deactivating and meta directing. Nitration of benzaldehyde gives m-nitrobenzaldehyde.

Beyond the textbook. For the exam: only aldehydes without α-H undergo Cannizzaro. In reality: a mixture of methanal with another α-H-free aldehyde does a "crossed" Cannizzaro in which methanal, the better hydride donor, is almost always the one oxidised to formate, so the other aldehyde is reduced to its alcohol. Chemists use this to convert aromatic aldehydes into alcohols cleanly. Why it matters: it shows that Cannizzaro is a hydride transfer, which is why it needs two aldehyde molecules and strong base.

Carboxylic acids: why they are acidic

A carboxylic acid loses H⁺ to give a carboxylate ion, RCOO⁻. In the carboxylate the negative charge is shared equally by two oxygen atoms through two equivalent resonance structures, so both C–O bonds are the same length. In a phenoxide ion the charge is spread onto ring carbons, which are far less able to hold a negative charge. In an alkoxide it is stuck on one oxygen with no spreading at all. That is why the acidity order is carboxylic acid > phenol > alcohol (pKa about 4–5, 10 and 16).

Acid strength is read from the pKa:

$$\text{p}K_a = -\log K_a$$

A smaller pKa means a stronger acid, and each unit of pKa is a factor of 10 in $K_a$.

Acid	pKa	Why
CF₃COOH	0.23	Three F atoms pull electrons very strongly and stabilise the anion
HCOOH	3.75	No electron releasing alkyl group
C₆H₅COOH	4.19	The sp² carbon of the ring is slightly electron withdrawing
CH₃COOH	4.76	The CH₃ group releases electrons and destabilises the anion
p-Nitrobenzoic acid	3.41	NO₂ withdraws electrons
p-Methoxybenzoic acid	4.46	OCH₃ donates electrons

The rule: anything that pulls electron density away from the carboxylate stabilises it and strengthens the acid. Anything that pushes electron density in weakens it. Electron withdrawing groups strengthen with number and with closeness to the COOH. The order of groups is Ph < I < Br < Cl < F < CN < NO₂ < CF₃, so FCH₂COOH is stronger than ClCH₂COOH, and ClCH₂COOH is stronger than CH₃COOH.

Worked example: how much stronger? Compare trifluoroethanoic acid (pKa 0.23) with ethanoic acid (pKa 4.76).

Step	What we do	Result
1	Convert pKa to $K_a$ using $K_a = 10^{-\text{p}K_a}$	$K_a(\text{CH}_3\text{COOH}) = 10^{-4.76} \approx 1.7 \times 10^{-5}$ and $K_a(\text{CF}_3\text{COOH}) = 10^{-0.23} \approx 0.59$
2	Take the ratio, which is 10 raised to the pKa difference	$10^{4.53} \approx 3.4 \times 10^{4}$
3	Interpret	CF₃COOH is about 34,000 times stronger

The common wrong step: subtracting pKa values and calling 4.53 the ratio. pKa is a logarithm, so the ratio is 10 to the power of the difference.

The CO₂ test. Carboxylic acids are strong enough to react with NaHCO₃ and release CO₂ (brisk effervescence), but phenols are not. This is the standard way to tell a carboxylic acid from a phenol.

Beyond the textbook. For the exam: carboxylic acids do not reduce Tollens' reagent. In reality: methanoic acid is the exception. It still carries a hydrogen on the carbonyl carbon, so it behaves partly like an aldehyde and reduces Tollens' reagent and Fehling's solution. Why it matters: it fits the rule that the test reads the H on the carbonyl carbon, not the family name.

Reactions of carboxylic acids

Reaction	Reagents	Product	Key point
With metals and alkalis	Na, NaOH, Na₂CO₃, NaHCO₃	Carboxylate salt (H₂ with metal; CO₂ with carbonates)	CO₂ with NaHCO₃ is the –COOH test
Esterification (Fischer)	R′OH with conc. H₂SO₄ or dry HCl	Ester + H₂O	Reversible. Protonation, alcohol attack, loss of water
Acid chloride	SOCl₂, PCl₅ or PCl₃	RCOCl	SOCl₂ is preferred because SO₂ and HCl leave as gases
Anhydride	Heat with P₂O₅ or conc. H₂SO₄	(RCO)₂O	Two acid molecules lose one H₂O
Amide	NH₃ gives ammonium salt, then heat	RCONH₂	Heating drives out water
Reduction	LiAlH₄ or B₂H₆	RCH₂OH	NaBH₄ does not reduce –COOH
Decarboxylation	Sodalime (NaOH and CaO), heat	R–H (one carbon less)	CH₃COONa gives CH₄
Kolbe electrolysis	Electrolysis of aqueous RCOONa	R–R (double the carbons)	Ethane from ethanoate
Hell–Volhard–Zelinsky	Cl₂ or Br₂ with a little red phosphorus	α-Halo acid	Needs an α-H. Halogen goes to the α-carbon
Ring substitution (benzoic acid)	HNO₃ with H₂SO₄	m-Nitrobenzoic acid	COOH is deactivating and meta directing

Key equations:

CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O (conc. H₂SO₄)

CH₃COONa + NaOH → CH₄ + Na₂CO₃ (CaO, heat)

2CH₃COONa + 2H₂O → C₂H₆ + 2CO₂ + H₂ + 2NaOH (electrolysis)

CH₃CH₂COOH + Br₂ → CH₃CHBrCOOH + HBr (red P)

Benzoic acid does not undergo Friedel–Crafts reactions. Its ring is deactivated, and the AlCl₃ catalyst binds to the COOH group.

Beyond the textbook. For the exam: esterification is acid catalysed and reversible. In reality: aspirin is made by acetylating the OH group of salicylic acid (2-hydroxybenzoic acid) with ethanoic anhydride, which is an acid derivative made from exactly this chemistry. Why it matters: the same ester, anhydride and acid chloride reactions you memorise here build real medicines.

Choosing the right test or reagent

Task	Reagent	Positive result
Is a carbonyl present?	2,4-DNP	Yellow, orange or red precipitate
Aldehyde or ketone?	Tollens' or Fehling's	Silver mirror, or red precipitate, for aldehyde
Aliphatic or aromatic aldehyde?	Fehling's	Only the aliphatic aldehyde responds
Methyl ketone or other ketone (pentan-2-one vs pentan-3-one)?	I₂ with NaOH	Yellow CHI₃ from the methyl ketone
Acetone or benzophenone?	I₂ with NaOH	Only acetone gives CHI₃
Ethanol or methanol?	I₂ with NaOH	Only ethanol
Carboxylic acid or phenol?	NaHCO₃	CO₂ fizz from the acid
Aldehyde with or without α-H?	Concentrated NaOH	No α-H gives Cannizzaro; α-H gives aldol (resin)

Worked example: identify the compound. A compound C₈H₈O gives an orange precipitate with 2,4-DNP, a yellow precipitate with NaOI, no reaction with Tollens' or Fehling's, and on vigorous oxidation gives benzoic acid (C₇H₆O₂). Find it.

Step	What we do	Result
1	Degrees of unsaturation $= \dfrac{2 \times 8 + 2 - 8}{2}$	5, so a benzene ring (4) plus one C=O or C=C
2	2,4-DNP positive	A carbonyl compound
3	Tollens' and Fehling's negative	A ketone, not an aldehyde
4	NaOI positive	A methyl ketone (CH₃CO–)
5	Oxidation gives benzoic acid	The carbonyl is attached directly to the benzene ring
6	Assemble C₆H₅ + CO + CH₃	Acetophenone, C₆H₅COCH₃

The common wrong step: choosing phenylacetaldehyde (C₆H₅CH₂CHO). It has the right formula and gives a positive DNP test, but it is an aldehyde, so it would give a silver mirror and fail the NaOI test.

Reagent to product: quick recognition

Starting material	Reagent	Product
C₆H₅COCl	H₂, Pd/BaSO₄	C₆H₅CHO
CH₃CN	SnCl₂/HCl, then H₂O	CH₃CHO
C₆H₅CH₃	CrO₂Cl₂, then H₃O⁺	C₆H₅CHO
C₆H₆	CO, HCl, anhydrous AlCl₃	C₆H₅CHO
C₆H₅Br	Mg, then CO₂, then H₃O⁺	C₆H₅COOH
C₆H₅CH₂CH₃	Hot KMnO₄/KOH, then H⁺	C₆H₅COOH
CH₃CHO	Dilute NaOH	CH₃CH(OH)CH₂CHO
C₆H₅CHO	Conc. NaOH	C₆H₅CH₂OH and C₆H₅COONa
CH₃CH₂COOH	Br₂, red P	CH₃CHBrCOOH
C₆H₅COCH₃	Zn-Hg, conc. HCl	C₆H₅CH₂CH₃

Exam traps

Trap	Why it is tempting	The truth
Fehling's gives a positive test with benzaldehyde	Tollens' does	Fehling's fails for aromatic aldehydes
Ketones are more reactive than aldehydes in nucleophilic addition	Alkyl groups look "more reactive"	Two alkyl groups reduce δ⁺ and add crowding, so aldehydes win
Any aldehyde with α-H can do Cannizzaro	Cannizzaro is also done in alkali	Cannizzaro needs no α-H; with α-H you get aldol
Iodoform test works for all alcohols	Ethanol works	Only CH₃CH(OH)– alcohols; methanol and propan-1-ol fail
NaBH₄ reduces carboxylic acids	It reduces aldehydes and ketones	Use LiAlH₄ or B₂H₆ for –COOH
HVZ needs only halogen	The α-carbon is attacked	It needs red phosphorus and an α-H
Benzoic acid undergoes Friedel–Crafts	Benzene does	COOH deactivates the ring and complexes with AlCl₃
Rosenmund uses a Grignard or LiAlH₄	Both are reductions	Rosenmund is H₂ over poisoned Pd/BaSO₄; stronger reagents over-reduce

Reaction sheet

Reaction	Reagents	Change
Rosenmund	H₂, Pd/BaSO₄	RCOCl → RCHO
Stephen	SnCl₂/HCl, H₂O	RCN → RCHO
Etard	CrO₂Cl₂, H₃O⁺	Toluene → benzaldehyde
Gattermann–Koch	CO, HCl, AlCl₃/CuCl	Benzene → benzaldehyde
Friedel–Crafts acylation	RCOCl, AlCl₃	Arene → aryl ketone
Cyanohydrin formation	HCN, trace base	C=O → C(OH)CN
Clemmensen	Zn-Hg, conc. HCl	C=O → CH₂
Wolff–Kishner	NH₂NH₂, KOH, glycol, heat	C=O → CH₂
Haloform	NaOX	CH₃CO– → R–COONa + CHX₃
Aldol	Dilute NaOH	α-H carbonyl → β-hydroxy carbonyl
Cannizzaro	Conc. NaOH	No α-H aldehyde → alcohol + carboxylate
HVZ	X₂, red P	RCH₂COOH → RCHXCOOH
Kolbe electrolysis	Electrolysis of RCOONa	2 RCOO⁻ → R–R
Decarboxylation	Sodalime, heat	RCOONa → R–H

60-second revision

Topic	Remember
Carbonyl carbon	sp², 120°, δ⁺, attacked by nucleophiles
Reactivity	HCHO > RCHO > RCOR′ > aromatic carbonyls
Aldehyde-only preparations	Rosenmund, Stephen, DIBAL-H, Etard, Gattermann–Koch
Ketone-only preparations	R₂Cd with RCOCl, RMgX with nitrile, Friedel–Crafts acylation
Tests	Tollens' (all RCHO), Fehling's (aliphatic RCHO only), NaOI (CH₃CO– and CH₃CH(OH)–)
Aldol vs Cannizzaro	α-H present gives aldol (dilute base); α-H absent gives Cannizzaro (conc. base)
Boiling order	Alkane < ether < aldehyde or ketone < alcohol < carboxylic acid
Acid strength	RCOOH > phenol > alcohol; EWG raises acidity, EDG lowers it
Reduce –COOH	LiAlH₄ or B₂H₆, not NaBH₄
Benzoic acid	Meta directing, no Friedel–Crafts

Where this chapter connects

The oxidation of alcohols that makes carbonyls is covered in Alcohols, Phenols and Ethers, which also shows why phenol is weaker than a carboxylic acid. Grignard reagents and nitriles start from haloalkanes, so revisit Haloalkanes and Haloarenes. Acid amides from this chapter become primary amines in Amines, and the carbonyl chemistry here explains why glucose in Biomolecules reacts with Tollens' reagent and hydroxylamine. Ozonolysis and alkyne hydration come from Hydrocarbons, and the inductive and resonance arguments used for acid strength are built in Organic Chemistry: Some Basic Principles and Techniques.

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