Aldehydes & Ketones — Nucleophilic Addition & Named Reactions

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Almost everything an aldehyde or ketone does starts the same way: a nucleophile attacks the δ+ carbonyl carbon. Two extra ideas cover the rest — the α-hydrogen is acidic (aldol, haloform), and aldehydes are easily oxidised while ketones are not (Tollens’, Fehling’s). Hold those three ideas and the named reactions become one story.

What you will be able to do

• Prepare aldehydes and ketones by oxidation, ozonolysis, hydration of alkynes and the named reductions (Rosenmund, Stephen, Etard, Gattermann–Koch)
• Explain nucleophilic addition and rank carbonyl compounds by reactivity
• Predict products with HCN, NaHSO₃, Grignard reagents, alcohols and ammonia derivatives
• Use Tollens’, Fehling’s, iodoform and 2,4-DNP tests to distinguish carbonyl compounds
• Predict aldol, cross-aldol, Claisen–Schmidt and Cannizzaro products, and solve conversions

The big picture

Three ideas run the whole chapter.

In one line: The polar C=O makes the carbonyl carbon electrophilic (nucleophilic addition); the α-hydrogen is acidic (aldol, haloform); and aldehydes are easily oxidised while ketones are not (Tollens’, Fehling’s).

Idea	Reactions it explains
Nucleophile attacks C of C=O	HCN, NaHSO₃, Grignard, alcohols (acetals), ammonia derivatives, reductions
α-H is acidic	Aldol, cross-aldol, Claisen–Schmidt, haloform
No α-H + strong base	Cannizzaro
Aldehyde H is easily removed in oxidation	Tollens’, Fehling’s, Benedict’s; aldehydes → acids
–CHO / –COR deactivates a benzene ring	meta substitution in benzaldehyde and acetophenone

Think of it like this: The carbonyl carbon is like a shop door with a bright “Open” sign (δ+): nucleophiles walk straight in. In a ketone, two alkyl “bouncers” stand at the door and make it harder to enter (steric hindrance), and they also dim the sign by pushing electrons in. That is why aldehydes react faster than ketones.

Structure and physical properties

A polar, planar group.

• The carbonyl carbon is sp²; the C=O group and its two attached atoms are trigonal planar (bond angles about 120°). The C=O π bond is polarised: Cᵟ⁺=Oᵟ⁻.
• Resonance: C=O ↔ C⁺–O⁻ explains the large dipole moment (about 2.3–2.8 D).
• Boiling points lie between hydrocarbons/ethers and alcohols of similar mass: dipole–dipole attractions but no H bonding between molecules.
• Lower members are water-soluble (H bonding with water); solubility falls as the carbon chain grows. Methanal is a gas; its 40% solution is formalin.

Compound	Molar mass	b.p. / K
n-Butane	58	273
Methoxyethane	60	281
Propanal	58	322
Acetone	58	329
Propan-1-ol	60	370

Preparing aldehydes and ketones

The routes you must know.

Route	Reagent	Gives	Note
Oxidation of alcohols	PCC; K₂Cr₂O₇/H⁺	1° → aldehyde; 2° → ketone	Strong oxidants take aldehydes on to acids; PCC stops at the aldehyde
Dehydrogenation	Cu or Ag, 573 K	1° → aldehyde; 2° → ketone	Industrial route; vapours over hot metal
Ozonolysis	O₃, then Zn/H₂O	Aldehydes and/or ketones	Cuts C=C into two C=O
Hydration of alkynes	dil. H₂SO₄ + HgSO₄	Ethyne → ethanal; other alkynes → ketones	Markovnikov; via the enol
Rosenmund reduction	RCOCl + H₂, Pd/BaSO₄	Aldehyde	BaSO₄ poisons Pd so it stops at the aldehyde
Stephen reaction	RCN + SnCl₂/HCl, then H₃O⁺	Aldehyde	Via the imine
DIBAL-H	RCN or RCOOR′, then H₂O	Aldehyde	Selective partial reduction; C=C untouched
Grignard + nitrile	RCN + R′MgX, then H₃O⁺	Ketone R–CO–R′	Via the imine
Acyl chloride + dialkylcadmium	2RCOCl + R′₂Cd	Ketone	R′₂Cd from R′MgX + CdCl₂

Aromatic aldehydes and ketones

Reaction	Reagent	Product
Etard reaction	Toluene + CrO₂Cl₂ (in CS₂), then H₃O⁺	Benzaldehyde
Chromic oxide method	Toluene + CrO₃ in acetic anhydride (gives benzylidene diacetate), then hydrolysis	Benzaldehyde
Side-chain chlorination	Toluene + Cl₂ (light) → C₆H₅CHCl₂; hydrolysis at 373 K	Benzaldehyde
Gattermann–Koch	Benzene + CO + HCl, anhydrous AlCl₃/CuCl	Benzaldehyde
Friedel–Crafts acylation	Benzene + RCOCl, anhydrous AlCl₃	Aryl ketone (e.g. acetophenone)

Solved example 01 — Choosing a route

Give a method to prepare (a) propanal from propanoyl chloride, (b) butan-2-one from propanenitrile.

1. (a) Rosenmund: CH₃CH₂COCl + H₂ over Pd/BaSO₄ → CH₃CH₂CHO.
2. (b) CH₃CH₂CN + CH₃MgBr → imine salt; H₃O⁺ → CH₃CH₂COCH₃.

Answer: (a) Rosenmund reduction. (b) Grignard addition to the nitrile, then hydrolysis.

Solved example 02 — Ozonolysis backwards

An alkene C₆H₁₂ on ozonolysis gives only propanone. Identify it.

1. Propanone is (CH₃)₂C=O; joining two at their carbonyl carbons gives (CH₃)₂C=C(CH₃)₂.

Answer: 2,3-Dimethylbut-2-ene.

Check yourself

• C1. Why is BaSO₄ added to the catalyst in the Rosenmund reduction?
• C2. Which alkyne gives an aldehyde on hydration?

Answers at the end.

Nucleophilic addition

The core mechanism.

1. The nucleophile attacks the electrophilic carbonyl carbon from above or below the planar group.
2. The carbon changes from sp² (trigonal) to sp³ (tetrahedral); the π electrons move onto oxygen, giving an alkoxide.
3. The alkoxide takes H⁺ from the medium to give the neutral addition product.

Key idea: Reactivity: HCHO > CH₃CHO > CH₃COCH₃; aldehydes > ketones; aliphatic > aromatic. Alkyl groups donate electrons (lowering δ+) and crowd the carbon. In benzaldehyde the ring’s resonance with C=O lowers δ+ on the carbon, so it is less reactive than ethanal.

Reagent	Product	Note
HCN (trace base)	Cyanohydrin R₂C(OH)CN	Base generates CN⁻, the real nucleophile; cyanohydrins → α-hydroxy acids
NaHSO₃ (saturated)	Bisulphite adduct (crystalline)	With aldehydes and methyl ketones; regenerated with dilute acid or alkali — used to separate them
RMgX, then H₃O⁺	Alcohol	HCHO → 1°; RCHO → 2°; R₂CO → 3°
ROH, dry HCl	Hemiacetal, then acetal	Ketones with ethylene glycol give cyclic ketals — protects C=O during other steps
H₂O	Gem-diol (hydrate)	Mostly reversible; chloral hydrate is stable

Solved example 03 — Ranking reactivity

Arrange HCHO, CH₃CHO, CH₃COCH₃ and C₆H₅CHO in decreasing reactivity toward HCN.

1. HCHO: no alkyl groups — most electrophilic, least hindered.
2. CH₃CHO: one methyl donor.
3. C₆H₅CHO: ring resonance lowers δ+.
4. CH₃COCH₃: two methyls — electronic and steric effects.

Answer: HCHO > CH₃CHO > C₆H₅CHO > CH₃COCH₃ (benzaldehyde and acetone are close; acetone is usually placed last).

Reactions with ammonia derivatives

Addition followed by loss of water.

R₂C=O + H₂N–Z ⇌ R₂C=N–Z + H₂O

Best at a weakly acidic pH (about 3.5): enough H⁺ to activate C=O, not so much that the amine is fully protonated

Reagent H₂N–Z	Name	Product
H₂N–OH	Hydroxylamine	Oxime
H₂N–NH₂	Hydrazine	Hydrazone
H₂N–NHC₆H₅	Phenylhydrazine	Phenylhydrazone
H₂N–NH–C₆H₃(NO₂)₂	2,4-Dinitrophenylhydrazine (2,4-DNP, Brady’s reagent)	2,4-DNP derivative — yellow/orange/red precipitate
H₂N–NHCONH₂	Semicarbazide	Semicarbazone
H₂N–R	Primary amine	Imine (Schiff base)

Key idea: 2,4-DNP gives a coloured precipitate with every aldehyde and ketone — the general test for a carbonyl group. Only one NH₂ of semicarbazide reacts: the other is part of an amide and its lone pair is delocalised onto C=O.

Check yourself

• C3. Why is the reaction with ammonia derivatives carried out at a weakly acidic pH?
• C4. Name the product of acetone with hydroxylamine.

Answers at the end.

Reduction

Three outcomes: alcohol, CH₂ or pinacol.

Reagent	Product	Note
NaBH₄ or LiAlH₄; H₂/Ni, Pt or Pd	Alcohol (aldehyde → 1°, ketone → 2°)	NaBH₄ does not reduce C=C
Clemmensen: Zn–Hg / conc. HCl	C=O → CH₂ (alkane)	For substrates stable to acid
Wolff–Kishner: NH₂NH₂, then KOH in ethylene glycol, heat	C=O → CH₂ (alkane)	For substrates stable to base; via the hydrazone

Solved example 04 — Clemmensen or Wolff–Kishner?

Convert acetophenone to ethylbenzene.

1. The C=O must become CH₂ without touching the ring.
2. Clemmensen: Zn–Hg / conc. HCl, heat (or Wolff–Kishner).

Answer: C₆H₅COCH₃ → C₆H₅CH₂CH₃.

Oxidation and the classic tests

Aldehydes are easily oxidised; ketones are not.

Test	Reagent	Positive result	Given by
Tollens’	[Ag(NH₃)₂]⁺ (ammoniacal AgNO₃)	Silver mirror	All aldehydes (aliphatic and aromatic)
Fehling’s	Cu²⁺ tartrate complex (Fehling A + B), alkaline	Red-brown Cu₂O precipitate	Aliphatic aldehydes (not benzaldehyde)
Benedict’s	Cu²⁺ citrate complex	Red-brown Cu₂O	Aliphatic aldehydes, reducing sugars
Iodoform	I₂ + NaOH	Yellow CHI₃ (antiseptic smell)	CH₃CO– compounds and CH₃CH(OH)– alcohols (incl. ethanal, ethanol)
2,4-DNP	Brady’s reagent	Yellow–red precipitate	All aldehydes and ketones
Schiff’s reagent	Decolourised rosaniline	Pink colour restored	Aldehydes

• Aldehydes → carboxylic acids with HNO₃, KMnO₄, K₂Cr₂O₇ and even mild oxidants.
• Ketones are oxidised only by strong oxidants at high temperature, with C–C bond cleavage, giving a mixture of acids with fewer carbons.

Haloform reaction

R–CO–CH₃ + 3NaOI → R–COONa + CHI₃↓ + 2NaOH

Methyl ketones (and ethanal) are converted to the acid with one carbon fewer; the CH₃ becomes CHX₃

• The α-H of the CH₃ group is replaced by halogen three times, then OH⁻ cleaves the C–CX₃ bond.
• Alcohols with CH₃CH(OH)– give a positive test because NaOI first oxidises them to methyl ketones (or ethanal).
• The C=C of α,β-unsaturated methyl ketones is not attacked.

Solved example 05 — Distinguishing pairs

Give one test to distinguish (a) propanal and propanone, (b) acetophenone and benzophenone, (c) pentan-2-one and pentan-3-one.

1. (a) Tollens’ (or Fehling’s): propanal gives a silver mirror; propanone does not.
2. (b) Iodoform: acetophenone (C₆H₅COCH₃) gives yellow CHI₃; benzophenone does not.
3. (c) Iodoform: pentan-2-one (a methyl ketone) is positive; pentan-3-one is not.

Answer: (a) Tollens’ test. (b) and (c) Iodoform test.

Reactions at the α-carbon

The acidic α-hydrogen.

The α-hydrogen is acidic (pKa about 19–20 for acetone) because the anion left behind (the enolate) is stabilised by resonance with C=O.

Aldol condensation

2CH₃CHO → CH₃CH(OH)CH₂CHO → CH₃CH=CHCHO + H₂O

dil. NaOH: aldol (3-hydroxybutanal); heat: dehydration to but-2-enal (crotonaldehyde)

1. OH⁻ removes an α-H, giving the enolate ion.
2. The enolate attacks the carbonyl carbon of a second molecule.
3. The alkoxide takes H⁺ from water: β-hydroxy aldehyde (aldol).
4. On heating, loss of water gives the α,β-unsaturated carbonyl compound (conjugated, so very favourable).

• Ketones give aldols too: acetone → 4-hydroxy-4-methylpentan-2-one → 4-methylpent-3-en-2-one (mesityl oxide).
• Cross aldol of two different compounds that both have α-H gives four products — not useful.
• Claisen–Schmidt: an aldehyde without α-H (benzaldehyde) + a ketone or aldehyde with α-H gives one main product: C₆H₅CHO + CH₃COCH₃ → C₆H₅CH=CHCOCH₃ (benzalacetone); with acetophenone → C₆H₅CH=CHCOC₆H₅ (chalcone).

Solved example 06 — Counting cross-aldol products

How many aldol products (β-hydroxy aldehydes, ignoring stereoisomers) form from a mixture of ethanal and propanal with dilute NaOH?

1. Both have α-H, so each can act as the enolate (donor) and as the acceptor.
2. Donor/acceptor pairs: ethanal/ethanal, propanal/propanal, ethanal enolate + propanal, propanal enolate + ethanal.

Answer: Four: 3-hydroxybutanal, 3-hydroxy-2-methylpentanal, 3-hydroxypentanal and 3-hydroxy-2-methylbutanal.

Cannizzaro reaction

No α-H plus strong base: one molecule oxidised, one reduced.

2HCHO + conc. NaOH → CH₃OH + HCOONa
2C₆H₅CHO + conc. NaOH → C₆H₅CH₂OH + C₆H₅COONa

Disproportionation by hydride transfer

• Only aldehydes without α-H: HCHO, C₆H₅CHO, (CH₃)₃CCHO, furfural.
• Mechanism: OH⁻ adds to C=O; the tetrahedral intermediate transfers a hydride ion to a second aldehyde molecule.
• Crossed Cannizzaro: with HCHO and another aldehyde without α-H, HCHO is always oxidised (to formate) and the other is reduced to the alcohol: C₆H₅CHO + HCHO → C₆H₅CH₂OH + HCOONa.

Solved example 07 — Crossed Cannizzaro

Predict the products when benzaldehyde and formaldehyde are treated with conc. NaOH.

1. Neither has α-H, so Cannizzaro occurs.
2. HCHO is the more reactive and is preferentially attacked by OH⁻; it donates the hydride and is oxidised.
3. Benzaldehyde accepts the hydride and is reduced.

Answer: Benzyl alcohol + sodium formate.

Check yourself

• C5. Which of these gives Cannizzaro: ethanal, benzaldehyde, acetone?
• C6. What is the product of acetone with dilute NaOH on heating?

Answers at the end.

Ring substitution

The carbonyl group is a meta director.

• –CHO and –COR withdraw electrons (–R, –I), deactivating the ring and directing electrophiles to the meta position.
• Benzaldehyde + conc. HNO₃/H₂SO₄ (273–283 K) → m-nitrobenzaldehyde.
• Friedel–Crafts reactions fail on these rings (deactivated, and AlCl₃ complexes with C=O).

Uses

Where these compounds turn up.

• Formalin (40% HCHO) preserves biological specimens; HCHO is used to make Bakelite and urea–formaldehyde resins.
• Acetaldehyde makes acetic acid, ethyl acetate and drugs; acetone is a common solvent.
• Benzaldehyde (bitter almond), vanillin (vanilla), cinnamaldehyde (cinnamon) and camphor are flavours and fragrances.

Conversions

Multi-step problems.

Solved example 08 — Ethanal to but-2-enal

Convert ethanal into but-2-enal.

1. Aldol: 2CH₃CHO + dil. NaOH → CH₃CH(OH)CH₂CHO.
2. Heat: loss of water → CH₃CH=CHCHO.

Answer: Aldol condensation followed by dehydration.

Solved example 09 — Benzene to m-nitroacetophenone

Convert benzene into m-nitroacetophenone.

1. Friedel–Crafts acylation: C₆H₆ + CH₃COCl / AlCl₃ → C₆H₅COCH₃.
2. Nitration: the acetyl group directs meta → m-nitroacetophenone.

Answer: Acylate first, then nitrate (the order matters: nitrobenzene will not undergo Friedel–Crafts).

Solved example 10 — Propanone to propene

Convert propanone into propene.

1. Reduce: CH₃COCH₃ + NaBH₄ → CH₃CH(OH)CH₃.
2. Dehydrate: conc. H₂SO₄, heat → CH₃CH=CH₂.

Answer: Reduction to propan-2-ol, then dehydration.

Solved example 11 — Ethanol to 3-hydroxybutanal

Convert ethanol into 3-hydroxybutanal.

1. Oxidise ethanol with PCC → ethanal.
2. Aldol with dilute NaOH (no heating) → 3-hydroxybutanal.

Answer: PCC oxidation, then aldol addition.

Advanced problems

Identify the compound from its reactions.

Solved example 12 — Identify A

Compound A (C₃H₆O) gives a 2,4-DNP precipitate and a positive iodoform test, but does not reduce Tollens’ reagent. Identify A.

1. 2,4-DNP positive: A is an aldehyde or ketone.
2. Tollens’ negative: a ketone.
3. Iodoform positive: a methyl ketone with three carbons.

Answer: A is propanone, CH₃COCH₃.

Solved example 13 — Identify B and its products

B (C₇H₆O) gives a silver mirror but no Fehling’s test and no iodoform. With conc. NaOH it gives C₇H₈O and the sodium salt of C₇H₆O₂. Identify all three.

1. Tollens’ positive, Fehling’s negative: an aromatic aldehyde.
2. C₇H₆O = C₆H₅CHO (benzaldehyde).
3. Conc. NaOH: Cannizzaro → C₆H₅CH₂OH (C₇H₈O) + C₆H₅COONa.

Answer: B = benzaldehyde; products = benzyl alcohol and sodium benzoate.

Solved example 14 — Identify C from ozonolysis

An alkene C₅H₁₀ on ozonolysis gives ethanal and propanone. Identify it.

1. Join the carbonyl carbons: CH₃CH= + =C(CH₃)₂.

Answer: 2-Methylbut-2-ene, CH₃CH=C(CH₃)₂.

Exam trap gallery

Where marks are lost in this chapter.

#	What students write	What is actually correct	Why
1	Ketones are more reactive than aldehydes toward nucleophiles.	Aldehydes are more reactive.	Ketones have two electron-donating, bulky alkyl groups.
2	Benzaldehyde gives Fehling’s test.	It gives Tollens’ but not Fehling’s.	Fehling’s works for aliphatic aldehydes.
3	Acetone gives the Cannizzaro reaction.	It has α-H, so it gives aldol condensation.	Cannizzaro needs an aldehyde without α-H.
4	In crossed Cannizzaro with HCHO, benzaldehyde is oxidised.	HCHO is oxidised (to formate); benzaldehyde is reduced to benzyl alcohol.	HCHO is the better hydride donor once OH⁻ adds.
5	Pentan-3-one gives the iodoform test.	Only methyl ketones (CH₃CO–) do; pentan-3-one has two ethyl groups.	The CH₃ next to C=O is needed.
6	Ethanol gives a negative iodoform test because it is an alcohol.	Ethanol is positive (oxidised to ethanal first).	CH₃CH(OH)– alcohols count.
7	Clemmensen and Wolff–Kishner give alcohols.	Both reduce C=O to CH₂.	NaBH₄ or LiAlH₄ give alcohols.
8	Hydration of propyne gives propanal.	It gives propanone (Markovnikov).	Only ethyne gives an aldehyde.
9	Both NH₂ groups of semicarbazide react with C=O.	Only the NH₂ attached to the other NH reacts.	The amide NH₂ lone pair is delocalised.
10	Cross aldol of two enolisable aldehydes gives one product.	Four products.	Each can be donor and acceptor.
11	The carbonyl group directs ortho/para in electrophilic substitution.	It is a meta director and deactivates the ring.	Electron-withdrawing by resonance.
12	Rosenmund reduction goes all the way to the alcohol.	It stops at the aldehyde.	BaSO₄ poisons the Pd catalyst.

Memory hook: “Nucleophile knocks, carbon opens” — nucleophilic addition. · “Tollens’ for all aldehydes, Fehling’s for aliphatic only.” · “No α-H, Cannizzaro; with α-H, aldol.” · “Methyl next to C=O → yellow iodoform.”

Practice MCQs

Levels: Easy, Moderate, Tough.

Q01. Hybridisation of the carbonyl carbon: (Easy)

• (a) sp
• (b) sp²
• (c) sp³
• (d) sp³d

Q02. A general test for both aldehydes and ketones: (Easy)

• (a) Tollens’
• (b) Fehling’s
• (c) 2,4-DNP
• (d) Schiff’s

Q03. Tollens’ reagent is: (Easy)

• (a) alkaline Cu²⁺
• (b) ammoniacal AgNO₃
• (c) I₂ + NaOH
• (d) acidified KMnO₄

Q04. Hydration of ethyne with HgSO₄/H₂SO₄ gives: (Easy)

• (a) ethanol
• (b) ethanal
• (c) acetone
• (d) acetic acid

Q05. Most reactive toward nucleophilic addition: (Easy)

• (a) HCHO
• (b) CH₃CHO
• (c) CH₃COCH₃
• (d) C₆H₅COCH₃

Q06. Iodoform test is given by: (Easy)

• (a) methanol
• (b) propan-1-ol
• (c) ethanal
• (d) benzaldehyde

Q07. Cannizzaro reaction is given by: (Easy)

• (a) CH₃CHO
• (b) HCHO
• (c) CH₃COCH₃
• (d) CH₃CH₂CHO

Q08. Rosenmund reduction converts: (Moderate)

• (a) RCN → RCHO
• (b) RCOCl → RCHO
• (c) RCOOH → RCH₂OH
• (d) RCHO → RCH₃

Q09. Toluene + CrO₂Cl₂, then hydrolysis, gives: (Moderate)

• (a) benzoic acid
• (b) benzaldehyde
• (c) benzyl alcohol
• (d) phenol

Q10. Acetophenone → ethylbenzene uses: (Moderate)

• (a) NaBH₄
• (b) Clemmensen reduction
• (c) Tollens’ reagent
• (d) Cannizzaro

Q11. Benzaldehyde gives a negative test with: (Moderate)

• (a) Tollens’
• (b) Fehling’s
• (c) 2,4-DNP
• (d) Schiff’s

Q12. Product of HCHO + CH₃MgBr, then H₃O⁺: (Moderate)

• (a) methanol
• (b) ethanol
• (c) propan-2-ol
• (d) ethanal

Q13. Acetone with hydroxylamine forms: (Moderate)

• (a) a hydrazone
• (b) an oxime
• (c) a semicarbazone
• (d) an imine of ammonia

Q14. Ethanal with dil. NaOH gives: (Moderate)

• (a) ethanol + acetate
• (b) 3-hydroxybutanal
• (c) but-1-ene
• (d) acetone

Q15. Benzaldehyde + acetone with dil. NaOH gives: (Moderate)

• (a) benzalacetone
• (b) benzoin
• (c) benzyl alcohol
• (d) benzoic acid

Q16. Nitration of benzaldehyde gives mainly: (Moderate)

• (a) o-nitrobenzaldehyde
• (b) m-nitrobenzaldehyde
• (c) p-nitrobenzaldehyde
• (d) nitrobenzene

Q17. Correct order of boiling points: (Moderate)

• (a) propanal > propan-1-ol > butane
• (b) propan-1-ol > propanal > butane
• (c) butane > propanal > propan-1-ol
• (d) propanal > butane > propan-1-ol

Q18. Methyl ketones on haloform reaction give an acid with: (Moderate)

• (a) the same number of C
• (b) one C fewer
• (c) one C more
• (d) two C fewer

Q19. Benzaldehyde + HCHO + conc. NaOH gives: (Tough)

• (a) benzoic acid + methanol
• (b) benzyl alcohol + formate
• (c) benzyl alcohol + methanol
• (d) benzoic acid + formate

Q20. Number of aldol products from ethanal + propanal (no stereoisomers): (Tough)

• (a) 2
• (b) 3
• (c) 4
• (d) 6

Q21. The best pH for oxime formation is about: (Tough)

• (a) 1
• (b) 3.5
• (c) 7
• (d) 12

Q22. Which gives both iodoform and Tollens’ tests? (Tough)

• (a) acetone
• (b) ethanal
• (c) propanal
• (d) benzaldehyde

Q23. Ozonolysis of 2-methylbut-2-ene gives: (Tough)

• (a) acetone + ethanal
• (b) two acetone
• (c) propanal + methanal
• (d) butanone + methanal

Q24. DIBAL-H converts an ester into: (Tough)

• (a) an alcohol only
• (b) an aldehyde
• (c) an acid
• (d) an alkane

Q25. Pinacolone-type ketone CH₃COC(CH₃)₃ with NaOI gives: (Tough)

• (a) CHI₃ + (CH₃)₃CCOONa
• (b) no reaction
• (c) CHI₃ + CH₃COONa
• (d) an aldol

Answer key with explanations

1. (b) sp² — Trigonal planar.
2. (c) 2,4-DNP — Coloured precipitate.
3. (b) ammoniacal AgNO₃ — [Ag(NH₃)₂]⁺.
4. (b) ethanal — Via vinyl alcohol.
5. (a) HCHO — No alkyl groups.
6. (c) ethanal — CH₃CO–.
7. (b) HCHO — No α-H.
8. (b) RCOCl → RCHO — H₂, Pd/BaSO₄.
9. (b) benzaldehyde — Etard reaction.
10. (b) Clemmensen reduction — C=O → CH₂.
11. (b) Fehling’s — Aromatic aldehyde.
12. (b) ethanol — 1° alcohol.
13. (b) an oxime — NH₂OH.
14. (b) 3-hydroxybutanal — Aldol.
15. (a) benzalacetone — Claisen–Schmidt.
16. (b) m-nitrobenzaldehyde — meta director.
17. (b) propan-1-ol > propanal > butane — H bonding > dipole > dispersion.
18. (b) one C fewer — CH₃ leaves as CHX₃.
19. (b) benzyl alcohol + formate — HCHO is oxidised.
20. (c) 4 — Each acts as donor and acceptor.
21. (b) 3.5 — Balance of activation and free amine.
22. (b) ethanal — CH₃CHO.
23. (a) acetone + ethanal — (CH₃)₂C= + =CHCH₃.
24. (b) an aldehyde — Partial reduction at low temperature.
25. (a) CHI₃ + (CH₃)₃CCOONa — Methyl ketone → acid with one C fewer.

Check-yourself answers

• C1. BaSO₄ poisons (partly deactivates) the palladium so reduction stops at the aldehyde instead of continuing to the alcohol.
• C2. Ethyne (gives ethanal).
• C3. In strong acid the nitrogen nucleophile is protonated and cannot attack; in neutral or basic solution the C=O is not activated. Around pH 3.5 both conditions are balanced.
• C4. Acetone oxime, (CH₃)₂C=NOH.
• C5. Benzaldehyde (no α-H).
• C6. Mesityl oxide, 4-methylpent-3-en-2-one (via the aldol).

One-page revision

Read this page the night before the exam.

Preparation

• 1° alcohol + PCC → aldehyde; 2° → ketone
• Ozonolysis; HgSO₄ hydration (only ethyne → aldehyde)
• Rosenmund RCOCl; Stephen / DIBAL-H RCN
• Etard, Gattermann–Koch, Friedel–Crafts acylation

Nucleophilic addition

• HCHO > RCHO > R₂CO; aliphatic > aromatic
• HCN → cyanohydrin; NaHSO₃ adduct
• RMgX: HCHO 1°, RCHO 2°, R₂CO 3°
• ROH/HCl → acetal; glycol protects C=O

Ammonia derivatives

• pH ≈ 3.5; addition then −H₂O
• NH₂OH oxime; NH₂NH₂ hydrazone
• 2,4-DNP: coloured ppt with all carbonyls
• Semicarbazide: one NH₂ reacts

Reduction and oxidation

• NaBH₄, LiAlH₄ → alcohols
• Clemmensen (acid), Wolff–Kishner (base) → CH₂
• Tollens’ all aldehydes; Fehling’s aliphatic
• Iodoform: CH₃CO– and CH₃CH(OH)–

α-H chemistry

• Aldol → β-hydroxy carbonyl → α,β-unsaturated
• Cross aldol: 4 products
• Claisen–Schmidt: PhCHO + ketone
• Acetone → mesityl oxide

No α-H and ring

• Cannizzaro: HCHO, PhCHO + conc. NaOH
• Crossed: HCHO oxidised
• –CHO, –COR: meta directors
• Order: acylate before nitrating