S5 Chemistry – Unit 6: Aldehydes and Ketones

6.1

Nomenclature of Aldehydes and Ketones

Aldehydes Compounds containing the aldehyde functional group (–CHO) at the end of a carbon chain. General formula: RCHO. Named using the suffix –al. The carbonyl carbon is always C1.

Ketones Compounds containing the carbonyl group (C=O) within a carbon chain (flanked by two carbon atoms). General formula: RCOR'. Named using the suffix –one. The smallest ketone has 3 carbons (propanone).

The Carbonyl Group

Both aldehydes and ketones contain the carbonyl group (C=O): a carbon doubly bonded to oxygen. The carbon is sp² hybridised (trigonal planar, ~120° bond angles). The C=O bond is polar (C^δ+=O^δ−) because oxygen is more electronegative than carbon, making the carbonyl carbon susceptible to nucleophilic attack.

IUPAC Rules for Aldehydes (–al)

Find the longest chain containing the –CHO group. The –CHO carbon is always C1.
Replace –e of the parent alkane with –al.
No locant needed for –CHO (always C1).
Number other substituents from C1 (the carbonyl end).

IUPAC Rules for Ketones (–one)

Find the longest chain containing the C=O group. Replace –e with –one.
Number the chain to give the C=O the lowest possible locant.
State the locant before –one: e.g. pentan-2-one, pentan-3-one.

Name	Formula	Structure	Common Name
Methanal	HCHO	H–CHO	Formaldehyde
Ethanal	CH₃CHO	CH₃–CHO	Acetaldehyde
Propanal	C₂H₅CHO	CH₃CH₂–CHO	Propionaldehyde
Butanal	C₃H₇CHO	CH₃CH₂CH₂–CHO	Butyraldehyde
Propanone	CH₃COCH₃	CH₃–CO–CH₃	Acetone
Butan-2-one	CH₃COC₂H₅	CH₃–CO–CH₂CH₃	Methyl ethyl ketone (MEK)
Pentan-2-one	CH₃CO(CH₂)₂CH₃	CH₃–CO–(CH₂)₂CH₃	—
Pentan-3-one	(C₂H₅)₂CO	C₂H₅–CO–C₂H₅	Diethyl ketone

Example 1

Naming Carbonyl Compounds

Name: (a) CH₃CH₂CH₂CHO (b) CH₃COCH₂CH₂CH₃ (c) CH₃CH(CH₃)CHO

Chain length = 4 C including CHO → butanal. CHO is always C1; no locant needed.

Chain = 5 C. C=O on C2 (from left: CH₃–C=O–...); from right = C4. Lower locant = C2 → pentan-2-one.

Longest chain containing CHO = 3 C (propanal). Branch: methyl on C2 → 2-methylpropanal.

6.2

Isomerism

Functional Group Isomerism: Aldehydes vs Ketones

Aldehydes and ketones with the same molecular formula are functional group isomers. They have the same molecular formula C_nH_2nO but different functional groups (–CHO vs C=O in chain).

Example: C₃H₆O → propanal (CH₃CH₂CHO) and propanone (CH₃COCH₃).

Position Isomerism in Ketones

For chains of 5+ carbons, the C=O can be at different positions. Example: C₅H₁₀O ketones include pentan-2-one and pentan-3-one.

Chain Isomerism

Aldehydes and ketones may also have chain isomers where the carbon skeleton differs. Example: C₄H₈O includes butanal, 2-methylpropanal, butan-2-one.

Formula	Aldehyde(s)	Ketone(s)
C₂H₄O	Ethanal	— (no 2-C ketone possible)
C₃H₆O	Propanal	Propanone
C₄H₈O	Butanal; 2-methylpropanal	Butan-2-one
C₅H₁₀O	Pentanal; 2-methylbutanal; 3-methylbutanal; 2,2-dimethylpropanal	Pentan-2-one; pentan-3-one; 3-methylbutan-2-one

Example 2

All Carbonyl Isomers of C₄H₈O

Draw and name all carbonyl compound isomers of C₄H₈O.

Aldehydes (CHO at end):
• Butanal: CH₃CH₂CH₂CHO (4-C straight chain)
• 2-methylpropanal: (CH₃)₂CHCHO (branched 3-C chain + methyl)

Ketones (C=O in chain, minimum 3-C chain):
• Butan-2-one: CH₃COCH₂CH₃ (only one possible ketone position for 4-C)

✓

Total: 3 carbonyl isomers of C₄H₈O.

6.3

Physical Properties

Boiling Points

Aldehydes and ketones have higher boiling points than alkanes of similar M_r (due to polar C=O dipole–dipole interactions) but lower boiling points than alcohols of similar M_r (no O–H group → cannot H-bond with each other).

Example: propanal (M_r=58, B.P.=49°C) vs propane (M_r=44, B.P.=−42°C) vs propan-1-ol (M_r=60, B.P.=97°C).

Solubility in Water

Short-chain aldehydes and ketones are miscible with water. The C=O oxygen can act as an H-bond acceptor with water’s O–H groups: C=O···H–O. As chain length increases, solubility decreases. Long-chain carbonyl compounds are insoluble in water.

Methanal and ethanal are completely miscible; propanone (acetone) is completely miscible and is an excellent polar aprotic solvent.

Compound	Formula	M_r	B.P. (°C)	Solubility
Methanal	HCHO	30	−19	Miscible
Ethanal	CH₃CHO	44	+20	Miscible
Propanal	C₂H₅CHO	58	+49	Miscible
Butanal	C₃H₇CHO	72	+75	Slightly soluble
Propanone	CH₃COCH₃	58	+56	Miscible
Butan-2-one	CH₃COC₂H₅	72	+80	Miscible
Pentan-2-one	CH₃CO(CH₂)₂CH₃	86	+102	Slightly soluble

📌

Smell Lower aldehydes (methanal, ethanal) have pungent, unpleasant smells. Higher aldehydes develop pleasant, fruity aromas. Ketones often have pleasant smells: propanone (nail polish remover), butan-2-one (solvent). Many natural fragrances are aldehydes or ketones.

6.4

Preparation

Preparation of Aldehydes

Method 1: Oxidation of Primary Alcohols (controlled)

Use acidified K₂Cr₂O₇ (or KMnO₄), limited amount, and distil off the aldehyde as it forms to prevent further oxidation to the acid.

RCH2OH --K2Cr2O7/H2SO4(lim), distil--> RCHO + H2O e.g. CH3CH2OH --[O], distil--> CH3CHO (ethanal) (ethanol) orange --> green

Method 2: Reduction of Acyl Chlorides (Rosenmund Reduction)

RCOCl + H2 --Pd/BaSO4 (poisoned catalyst)--> RCHO + HCl

The poisoned (deactivated) catalyst prevents further reduction to the primary alcohol.

Preparation of Ketones

Method 1: Oxidation of Secondary Alcohols

R2CHOH --K2Cr2O7/H2SO4, heat--> RCOR' + H2O e.g. CH3CH(OH)CH3 --[O]--> CH3COCH3 (propanone) (propan-2-ol) orange --> green

Ketones cannot be easily oxidised further, so excess oxidant does not cause over-oxidation.

Method 2: Dry Distillation of Calcium Salts of Carboxylic Acids

(RCOO)2Ca --strong heat--> RCOR + CaCO3 e.g. (CH3COO)2Ca --> CH3COCH3 + CaCO3 (calcium ethanoate) (propanone)

Example 3

Preparing Propanal from Propan-1-ol

Describe how you would prepare a pure sample of propanal from propan-1-ol.

Reagent: acidified K₂Cr₂O₇ (orange), use a limited quantity.

Set up apparatus for distillation: add propan-1-ol dropwise to the warm oxidising mixture and collect the distillate immediately.

The propanal (B.P. 49°C) distils off before it can be further oxidised to propanoic acid.

Equation: CH₃CH₂CH₂OH →^[O] CH₃CH₂CHO + H₂O

✓

Observe: orange → green (Cr⁶⁺ → Cr³⁺).

6.5

Reactions of Aldehydes and Ketones

📌

Reactivity Summary The polar C^δ+=O^δ− group makes carbonyl compounds susceptible to nucleophilic addition at the carbon. Aldehydes are generally more reactive than ketones because: (1) the aldehyde carbonyl carbon is less sterically hindered (only one R group); (2) alkyl groups in ketones donate electrons to C=O, reducing the δ+ charge.

Reaction 1: Nucleophilic Addition of HCN → Hydroxynitrile

Aldehydes and ketones react with HCN (or KCN + dilute H₂SO₄) to give hydroxynitriles (cyanohydrins). This reaction extends the carbon chain and introduces a new chiral centre.

RCHO + HCN --> RCH(OH)CN (aldehyde) (2-hydroxynitrile) e.g. CH3CHO + HCN --> CH3CH(OH)CN (ethanal) (2-hydroxypropanenitrile)

Mechanism: CN⁻ (nucleophile) attacks the δ+ carbonyl carbon → alkoxide ion → protonated by HCN (or H⁺) to give the hydroxynitrile.

Reaction 2: Nucleophilic Addition of NaBH₄ / LiAlH₄ → Alcohol

Aldehyde + 2[H] --> Primary alcohol RCHO + 2[H] --NaBH4 or LiAlH4--> RCH2OH Ketone + 2[H] --> Secondary alcohol RCOR' + 2[H] --NaBH4 or LiAlH4--> RCH(OH)R' e.g. CH3CHO + 2[H] --> CH3CH2OH (ethanol, primary) CH3COCH3 + 2[H] --> CH3CH(OH)CH3 (propan-2-ol, secondary)

Reaction 3: Condensation with 2,4-Dinitrophenylhydrazine (2,4-DNPH)

2,4-DNPH (Brady’s reagent) reacts with both aldehydes and ketones to give a yellow or orange precipitate called a 2,4-dinitrophenylhydrazone. This is a general test for the carbonyl group.

RCOR' + H2N-NHC6H3(NO2)2 --> RC(=N-NHC6H3(NO2)2)R' + H2O (carbonyl) (2,4-DNPH) (2,4-dinitrophenylhydrazone, yellow/orange ppt)

The melting point of the purified derivative is used to identify the specific aldehyde or ketone.

Reaction 4: Oxidation of Aldehydes (not ketones)

Aldehydes are easily oxidised to carboxylic acids by mild oxidising agents. Ketones are not oxidised under normal conditions. This difference is used to distinguish them.

RCHO + [O] --K2Cr2O7/H+, reflux--> RCOOH (aldehyde) (carboxylic acid) Ketones: RCOR' + [O] --> no reaction under mild conditions

Reaction 5: Aldol Condensation (Self-Condensation of Aldehydes)

Aldehydes (and some ketones) with an α-hydrogen undergo aldol condensation in the presence of dilute base (NaOH) or acid:

2 CH3CHO --dil. NaOH, 20 degC--> CH3CH(OH)CH2CHO (ethanal) (3-hydroxybutanal = aldol product) With heating, dehydration occurs: CH3CH(OH)CH2CHO --heat--> CH3CH=CHCHO + H2O (but-2-enal = crotonaldehyde)

Reaction 6: Iodoform Reaction

Ethanal and methyl ketones (CH₃COR) react with I₂/NaOH to give a yellow precipitate of CHI₃ (iodoform / triiodomethane).

CH3COR + 3I2 + 4NaOH --> CHI3(s) + RCOONa + 3NaI + 3H2O (methyl ketone) (yellow ppt)

Also positive for: ethanol and secondary alcohols of type CH₃CH(OH)R (which are first oxidised to methyl ketones in situ).

Reaction 7: Combustion

CH3CHO + 5/2 O2 --> 2CO2 + 2H2O CH3COCH3 + 4O2 --> 3CO2 + 3H2O

Example 4

Nucleophilic Addition of HCN to Propanone

Write the equation for the reaction of propanone with HCN. Name the product and state its significance.

Propanone: CH₃COCH₃. HCN adds across the C=O bond.

CH₃COCH₃ + HCN → (CH₃)₂C(OH)CN

Product: 2-hydroxy-2-methylpropanenitrile (acetone cyanohydrin).

✓

Significance: The –CN group can be hydrolysed to –COOH (carboxylic acid) or reduced to –CH₂NH₂ (amine), providing routes to α-hydroxy acids and amino acids. Chain length increased from 3 C to 4 C.

6.6

Distinguishing Tests

💡

Key Distinction: Aldehydes vs Ketones Both react with 2,4-DNPH (Brady’s reagent). To distinguish them, use a mild oxidising agent: aldehydes are oxidised; ketones are not.

Test	Reagent	Aldehyde Result	Ketone Result
Tollens’ reagent (Silver mirror test)	Ammoniacal AgNO₃ (Ag(NH₃)₂⁺), warm	Silver mirror on test tube wall (Ag deposited)	No change (no silver mirror)
Fehling’s solution	Fehling’s A (CuSO₄) + B (NaOH/KNa tartrate), heat	Brick-red precipitate of Cu₂O	No change (solution remains blue)
Acidified K₂Cr₂O₇	K₂Cr₂O₇/H₂SO₄, warm	Orange → green (aldehyde oxidised to acid)	No colour change (stays orange)
2,4-DNPH (Brady’s reagent)	2,4-dinitrophenylhydrazine solution	Yellow/orange precipitate	Yellow/orange precipitate
Iodoform test	I₂/NaOH, warm	Yellow CHI₃ ppt (ethanal only)	Yellow CHI₃ ppt (methyl ketones only)

Tollens’ Test (Silver Mirror)

RCHO + 2[Ag(NH3)2]+ + 2OH- --> RCOO- + 2Ag(s) + 4NH3 + H2O (aldehyde) (Tollens' reagent) (carboxylate) (silver mirror)

The aldehyde is oxidised to a carboxylate; Ag⁺ is reduced to Ag metal which coats the tube as a mirror.

Fehling’s Test

RCHO + 2Cu2+(aq) + 5OH- --> RCOO- + Cu2O(s) + 3H2O (aldehyde) (blue, Fehling's) (carboxylate) (brick-red ppt)

Note: Fehling’s reagent does not oxidise aromatic aldehydes (e.g. benzaldehyde) — only aliphatic aldehydes. Tollens’ reagent oxidises both.

Example 5

Identifying an Unknown Carbonyl Compound

An unknown compound gives a yellow precipitate with 2,4-DNPH, a silver mirror with Tollens’ reagent, but no yellow precipitate with I₂/NaOH. Identify the compound type and suggest a specific possibility if M_r = 58.

Yellow ppt with 2,4-DNPH → contains a carbonyl group (aldehyde or ketone).

Silver mirror with Tollens’ → aldehyde (not a ketone).

No iodoform ppt → NOT ethanal (CH₃CHO would give iodoform).

M_r = 58 for an aldehyde: C_nH_2nO = 58 → n=4 → C₄H₈O. Aldehyde = butanal (or 2-methylpropanal — also C₄H₈O). Neither gives iodoform. Most likely: butanal.

6.7

Uses of Aldehydes and Ketones

Compound	Use	Notes
Methanal (formaldehyde)	Preservation of biological specimens (formalin = 40% aq. solution)	Denatures proteins; kills microbes
Methanal	Manufacture of Bakelite and other phenol-formaldehyde resins	Thermosetting plastic
Methanal	Disinfectant and germicide	Used in hospitals (toxic; carcinogenic)
Ethanal (acetaldehyde)	Manufacture of ethanoic acid (acetic acid); ethanol; pyridine	Key industrial intermediate
Propanone (acetone)	Solvent for paints, lacquers, varnishes, nail polish remover	Miscible with water and organic solvents
Propanone	Manufacture of Perspex (polymethylmethacrylate, PMMA)	Via acetone cyanohydrin → methacrylate monomer
Butan-2-one (MEK)	Industrial solvent (glues, coatings, cleaning)	Good solvent for plastics and resins
Higher aldehydes	Perfumery and flavouring agents	Cinnamaldehyde (cinnamon), vanillin (vanilla), benzaldehyde (almond)
Glucose (polyhydroxyaldehyde)	Energy source in biology; food	Contains –CHO group; gives positive Fehling’s test
Cyclohexanone	Manufacture of nylon-6 (via caprolactam)	Key precursor in polymer industry

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✏️

Exercises

Name the following compounds:
(a) CH₃CH₂CHO (b) CH₃COCH₂CH₃ (c) (CH₃)₂CHCHO (d) CH₃CH₂COCH₂CH₃

(a) Propanal (3-C aldehyde; CHO at C1)
(b) Butan-2-one (4-C; C=O at C2; also called methyl ethyl ketone)
(c) 2-methylpropanal (3-C chain + methyl branch at C2; CHO at C1)
(d) Pentan-3-one (5-C; C=O at C3; two ethyl groups flanking C=O)
Draw and name all carbonyl isomers of C₃H₆O.

Two isomers:
1. Propanal: CH₃CH₂CHO (aldehyde)
2. Propanone: CH₃COCH₃ (ketone)
Both have molecular formula C₃H₆O — they are functional group isomers.
Explain why aldehydes are more reactive than ketones in nucleophilic addition reactions.

1. Steric effect: Aldehydes have only one alkyl group on the carbonyl carbon (less steric hindrance), making it more accessible to nucleophilic attack. Ketones have two alkyl groups, creating more steric hindrance.
2. Electronic effect: Alkyl groups are electron-donating (inductive effect), reducing the δ+ charge on the carbonyl carbon in ketones. Aldehydes have a less electron-rich carbonyl carbon and are therefore more electrophilic and more reactive toward nucleophiles.
Describe the Tollens’ test. Write the equation for the reaction of ethanal with Tollens’ reagent. What would you observe with propanone?

Tollens’ test: Add ammoniacal silver nitrate (Ag(NH₃)₂⁺) solution to the compound in a clean test tube. Warm gently in a water bath at ~60°C.

With ethanal:
CH₃CHO + 2Ag(NH₃)₂⁺ + 2OH⁻ → CH₃COO⁻ + 2Ag(s) + 4NH₃ + H₂O
Observation: silver mirror forms on test tube wall.

With propanone: No reaction — no silver mirror forms. Solution remains colourless.
Write the equation for the reaction of butanone (butan-2-one) with HCN. Name the product and explain what type of reaction this is.

CH₃COC₂H₅ + HCN → CH₃C(OH)(CN)C₂H₅
Product: 2-hydroxy-2-methylbutanenitrile
Type: Nucleophilic addition. CN⁻ acts as the nucleophile, attacking the electrophilic δ+ carbon of C=O. The product is a hydroxynitrile (cyanohydrin). Chain length extended from 4 to 5 carbons.
Which of the following compounds would give a positive iodoform test? Justify your answer.
(a) Propanone (b) Butanal (c) Butan-2-one (d) Pentan-3-one

Iodoform test is positive for methyl ketones (CH₃CO–R) and ethanal (CH₃CHO).

(a) Propanone (CH₃COCH₃): YES — has CH₃CO– group ✔
(b) Butanal (CH₃CH₂CH₂CHO): NO — not a methyl ketone or ethanal ✗
(c) Butan-2-one (CH₃COC₂H₅): YES — has CH₃CO– group ✔
(d) Pentan-3-one (C₂H₅COC₂H₅): NO — no CH₃CO– group ✗

📝

Unit Test

ℹ️

Instructions Total: 50 marks | Time: 50 minutes | Attempt all questions | Show all working.

Section A — Short Answer

30 marks

Q1 [4 marks]

Name the following compounds:
(a) CH₃CHO (b) CH₃CH₂CH₂CHO (c) CH₃COCH₂CH₂CH₃ (d) (CH₃)₂CHCOCH₃

(a) Ethanal
(b) Butanal
(c) Pentan-2-one (C=O at C2; 5-carbon chain)
(d) 3-methylbutan-2-one (5-C chain; C=O at C2; methyl on C3)

Q2 [4 marks]

Draw and name all carbonyl isomers of C₄H₈O. Classify each as an aldehyde or a ketone.

Three isomers:
1. Butanal: CH₃CH₂CH₂CHO — aldehyde
2. 2-methylpropanal: (CH₃)₂CHCHO — aldehyde
3. Butan-2-one: CH₃COCH₂CH₃ — ketone

Q3 [6 marks]

Describe, with observations and equations, how you would use chemical tests to distinguish between three unlabelled bottles containing: propanone, propanal, and propan-1-ol.

Step 1 — 2,4-DNPH (Brady’s reagent):
Propanone: yellow/orange ppt ✔
Propanal: yellow/orange ppt ✔
Propan-1-ol: no ppt ✗ → propan-1-ol identified

Step 2 — Tollens’ reagent (warm):
Propanal: silver mirror ✔ (aldehyde) → propanal identified
Propanone: no silver mirror ✗ → propanone identified

Equations:
Propanal + 2Ag(NH₃)₂⁺ + 2OH⁻ → C₂H₅COO⁻ + 2Ag(s) + 4NH₃ + H₂O

Q4 [5 marks]

Describe the preparation of butanal from butan-1-ol. Include reagents, conditions, and the equation. Explain how you would prevent over-oxidation to butanoic acid.

Reagent: Acidified potassium dichromate (K₂Cr₂O₇/H₂SO₄), limited quantity.
Conditions: Warm gently; distil off the product immediately as it forms.

Equation:
CH₃CH₂CH₂CH₂OH →^[O] CH₃CH₂CH₂CHO + H₂O

Preventing over-oxidation: Butanal (B.P. 75°C) is more volatile than butanoic acid (B.P. 164°C). By distilling the product as it forms, it is removed from the oxidising mixture before it can be further oxidised. Using a limited amount of oxidising agent also helps.

Q5 [5 marks]

Write equations for the following reactions of propanone:
(a) Reaction with HCN (name the product) (b) Reduction with NaBH₄ (name the product) (c) Reaction with 2,4-DNPH (d) Iodoform test with I₂/NaOH (e) Complete combustion

(a) CH₃COCH₃ + HCN → (CH₃)₂C(OH)CN — 2-hydroxy-2-methylpropanenitrile
(b) CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃ — propan-2-ol
(c) CH₃COCH₃ + 2,4-DNPH → 2,4-dinitrophenylhydrazone (yellow/orange ppt) + H₂O
(d) CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃(s) + CH₃COONa + 3NaI + 3H₂O — yellow ppt
(e) CH₃COCH₃ + 4O₂ → 3CO₂ + 3H₂O

Q6 [6 marks]

Explain, using intermolecular force theory, why:
(a) Ethanal (B.P. 20°C) has a higher boiling point than ethane (M_r ≈ 30, B.P. −89°C) despite similar molar masses. [2]
(b) Ethanal (B.P. 20°C) has a lower boiling point than ethanol (B.P. 78°C) despite similar molar masses. [2]
(c) Propanone (B.P. 56°C) is miscible with water. [2]

(a) Ethanal has a polar C=O group giving permanent dipole–dipole interactions between molecules, in addition to van der Waals forces. These are stronger than the pure van der Waals forces in ethane → higher B.P.

(b) Ethanol has an O–H group and forms strong intermolecular hydrogen bonds (O–H···O). Ethanal has no O–H and cannot H-bond with itself (only dipole-dipole + vdW) → much lower B.P. than ethanol despite similar M_r.

(c) Propanone’s C=O oxygen is an H-bond acceptor: it can form H-bonds with water’s O–H groups (C=O···H–O–H). This allows propanone to dissolve freely in water → miscible.

Section B — Extended Response

20 marks

Q7 [10 marks]

(a) Compare and contrast the reactions of ethanal and propanone with the following reagents, writing equations where possible and stating observations: (i) 2,4-DNPH; (ii) Tollens’ reagent; (iii) Fehling’s solution; (iv) NaBH₄; (v) HCN. [8 marks]

(b) How could you determine the identity of the specific aldehyde or ketone from its 2,4-DNPH derivative? [2 marks]

(a)(i) 2,4-DNPH:
Both: yellow/orange precipitate (2,4-dinitrophenylhydrazone). No distinction possible.

(ii) Tollens’:
Ethanal: CH₃CHO + 2Ag(NH₃)₂⁺ + 2OH⁻ → CH₃COO⁻ + 2Ag(s) + 4NH₃ + H₂O (silver mirror)
Propanone: no reaction (no silver mirror)

(iii) Fehling’s:
Ethanal: CH₃CHO + 2Cu²⁺ + 5OH⁻ → CH₃COO⁻ + Cu₂O(s) + 3H₂O (brick-red ppt)
Propanone: no reaction (solution stays blue)

(iv) NaBH₄:
Ethanal: CH₃CHO + 2[H] → CH₃CH₂OH (ethanol, primary alcohol)
Propanone: CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃ (propan-2-ol, secondary alcohol)

(v) HCN:
Ethanal: CH₃CHO + HCN → CH₃CH(OH)CN (2-hydroxypropanenitrile)
Propanone: CH₃COCH₃ + HCN → (CH₃)₂C(OH)CN (2-hydroxy-2-methylpropanenitrile)
Both react; ethanal is more reactive (less steric hindrance, more δ+ on C).

(b) The 2,4-DNPH derivative (hydrazone) is purified by recrystallisation and its melting point is measured. This is compared against a data book of known melting points for hydrazones of different carbonyl compounds. Each specific aldehyde/ketone gives a hydrazone with a characteristic, sharp melting point.

Q8 [10 marks]

(a) Describe the mechanism of nucleophilic addition of HCN to ethanal. Identify the nucleophile, electrophile, and name the product. [4 marks]

(b) The hydroxynitrile product from ethanal can be converted into two further products. Name these and give equations. [3 marks]

(c) Starting from butan-2-ol, describe with equations and conditions how you would prepare: (i) butan-2-one; (ii) but-2-ene; (iii) 3-hydroxybutanenitrile (a hydroxynitrile). [3 marks]

(a) Mechanism:
Electrophile: carbonyl carbon of ethanal (C^δ+)
Nucleophile: CN⁻ ion

Step 1: CN⁻ attacks the δ+ carbon of C=O: CH₃CHO + CN⁻ → CH₃CH(O⁻)CN (alkoxide intermediate)
Step 2: Protonation by HCN (or H₂O/H⁺): CH₃CH(O⁻)CN + H⁺ → CH₃CH(OH)CN
Product: 2-hydroxypropanenitrile

(b) Further conversions:
(i) Hydrolysis of CN to COOH: CH₃CH(OH)CN + H₂O →^H⁺/heat CH₃CH(OH)COOH (lactic acid / 2-hydroxypropanoic acid)
(ii) Reduction of CN to CH₂NH₂: CH₃CH(OH)CN + 2H₂ →^LiAlH₄ CH₃CH(OH)CH₂NH₂ (2-amino-1-propanol; an amino alcohol)

(c)
(i) Butan-2-one: CH₃CH(OH)CH₂CH₃ →^{K₂Cr₂O₇/H⁺} CH₃COCH₂CH₃ + H₂O (oxidation of 2° alcohol)
(ii) But-2-ene: CH₃CH(OH)CH₂CH₃ →^{conc. H₂SO₄, 170°C} CH₃CH=CHCH₃ + H₂O (dehydration)
(iii) 3-hydroxybutanenitrile: First oxidise butan-2-ol → butan-2-one (step i), then: CH₃COCH₂CH₃ + HCN → CH₃C(OH)(CN)CH₂CH₃ (3-hydroxy-3-methylbutanenitrile... or directly: butan-2-one is not 3-hydroxybutanenitrile precursor — need to use butanal instead)
Corrected: Use butanal + HCN → CH₃CH₂CH₂CH(OH)CN = 3-hydroxybutanenitrile (prepare butanal first from butan-1-ol by oxidation).

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