Edexcel International A Level Chemistry: Carbonyl compounds (YCH11) – Revision Notes

These notes condense Topic 15B: Carbonyl compounds (outcomes 15.6 to 15.8) as set out in Issue 1 (September 2017) of Pearson Edexcel’s International Advanced Subsidiary/Advanced Level in Chemistry specification, YCH11. Unit 4 (Rates, Equilibria and Further Organic Chemistry), an International A2 (IA2) unit, is where it is examined. The carbonyl compounds study guide explains each idea in full, with worked examples.

Also useful: the practice questions, the topic checklist, a free diagnostic and the course hub. For the neighbouring parts of Topic 15, see the chirality study guide (15A) and the carboxylic acids study guide (15C).

Key definitions

• Carbonyl group: the C=O group. It is polar, with δ+ on carbon and δ− on oxygen.
• Nucleophile: a species with a lone pair that it donates to an electron-poor atom to form a new covalent bond. CN⁻ uses the lone pair on its carbon.
• Nucleophilic addition: a nucleophile attacks the δ+ carbonyl carbon and the C=O π bond breaks, so two groups add across the double bond and nothing is lost.
• Hydroxynitrile: a compound with –OH and –C≡N on the same carbon, made from a carbonyl compound and HCN. The nitrile carbon counts as C1 in the name.
• Enantiomers: non-superimposable mirror-image molecules with a chiral carbon.
• Racemic mixture: equal amounts of two enantiomers. Their rotations cancel, so it is optically inactive.
• Oxidation in organic terms: gain of oxygen or loss of hydrogen, shown with [O]. Reduction: gain of hydrogen, shown with [H].

15.6 Naming in four lines

• Aldehyde: C=O at the end of the chain, written –CHO, suffix -al. The CHO carbon is C1, so there is no locant: hexanal, 2-methylbutanal.
• Ketone: C=O between two carbons, suffix -one, with a locant from C5 upwards: hexan-2-one, hexan-3-one.
• Choose the longest chain through the C=O carbon. Give C=O the lowest number. Side chains go in alphabetical order.
• Skeletal formulae: zigzag of carbon, no C or H shown, =O drawn on its carbon. An aldehyde’s =O sits at a chain end.

Formula	Name	Family
CH₃CH₂CH₂CH₂CHO	pentanal	aldehyde
CH₃CH₂CH(CH₃)CH₂CHO	3-methylpentanal	aldehyde
CH₃COCH₂CH(CH₃)₂	4-methylpentan-2-one	ketone
CH₃CH₂COCH₂CH₂CH₃	hexan-3-one	ketone

15.7 Physical properties

Must-know distinction: with each other versus with water.

• With each other: no hydrogen bonds, because no H is bonded to O. Forces are London forces plus permanent dipole–dipole (polar C=O).
• With water: the δ− carbonyl O uses a lone pair to accept a hydrogen bond from a water H.

Consequences:

• Boiling temperature, at similar Mr: alkane < aldehyde or ketone < alcohol.
• Small aldehydes and ketones dissolve well in water; propanone is fully miscible. Solubility falls as the carbon chain gets longer, because the non-polar part cannot hydrogen bond with water.

15.8 Reactions summary

Reagent and conditions	Aldehyde	Ketone	Equation or note
Fehling’s or Benedict’s solution, warm	Blue → red precipitate (Cu₂O)	Stays blue	Cu(II) reduced to Cu(I)
Tollens’ reagent, warm water bath	Silver mirror	No change	Ag⁺ + e⁻ → Ag
K₂Cr₂O₇ / dilute H₂SO₄, warm	Orange → green	Stays orange	RCHO + [O] → RCOOH
LiAlH₄ in dry ether, then dilute acid	Primary alcohol	Secondary alcohol	RCHO + 2[H] → RCH₂OH; RCOR’ + 2[H] → RCH(OH)R’
HCN with KCN	Hydroxynitrile	Hydroxynitrile	Nucleophilic addition
2,4-DNPH	Yellow-orange precipitate	Yellow-orange precipitate	Equation not required
I₂ with NaOH(aq), warm	Positive for ethanal only	Positive for CH₃CO– ketones	Pale yellow CHI₃

Mole ratios worth knowing

• Tollens’: 1 mol aldehyde gives 2 mol Ag (the aldehyde loses two electrons).
• Fehling’s or Benedict’s: 1 mol aldehyde gives 1 mol Cu₂O.
• Iodoform: CH₃COR + 3I₂ + 4NaOH → CHI₃ + RCOONa + 3NaI + 3H₂O. 1 mol of the methyl carbonyl gives 1 mol CHI₃.

Method boxes

Drawing the HCN mechanism (15.8 iii)

1. Mark the C=O dipole: δ+ on C, δ− on O.
2. Draw :CN⁻ with the lone pair on carbon. Arrow from that lone pair to the carbonyl C.
3. Arrow from the C=O double bond to O.
4. Draw the intermediate with O⁻ and the new C–CN bond.
5. Arrow from an O⁻ lone pair to the H of H–CN; arrow from the H–C bond to the CN carbon.
6. Product: the hydroxynitrile, plus CN⁻ regenerated.

Why KCN? HCN is a weak acid, so it gives too few CN⁻ ions on its own.

Optical activity argument (15.8 iii)

1. The carbonyl carbon and its three attached atoms are coplanar.
2. CN⁻ is equally likely to attack either face.
3. When the new carbon carries four different groups, both enantiomers form in equal amounts.
4. A racemic mixture forms: no overall rotation of plane-polarised light.
5. If two groups on the carbonyl carbon are the same (methanal, propanone, pentan-3-one), the product is not chiral at all.

Identifying a carbonyl with 2,4-DNPH (15.8 iv)

1. Add 2,4-DNPH: a yellow-orange precipitate shows C=O.
2. Filter, recrystallise from a minimum volume of hot solvent, filter again and dry.
3. Measure the melting temperature; a sharp value shows purity.
4. Match it to a table of derivative melting temperatures.
5. If two candidates are close, use Tollens’, Fehling’s or the iodoform test to decide.

Testing an unknown (15.8 i and v)

1. 2,4-DNPH: carbonyl present or not.
2. Tollens’ or Fehling’s: aldehyde (positive) or ketone (negative).
3. Iodine with alkali: CH₃CO– group present or not.

Writing a full test answer

A test answer needs three parts: the reagent, the condition and the observation for each compound. For example: “Warm each with Benedict’s solution. The aldehyde gives a red precipitate; the ketone stays blue.” “No reaction” for the ketone is weaker than naming the colour it stays.

Boiling temperature template (15.7 i)

1. Name the forces in each compound.
2. Compare London forces (same or different numbers of electrons).
3. Say which extra force is present: dipole–dipole for C=O, hydrogen bonds for O–H.
4. Link: stronger or more intermolecular forces need more energy to overcome, so the boiling temperature is higher.

Small worked reminders

• Butanal + LiAlH₄ → butan-1-ol (primary). Butanone + LiAlH₄ → butan-2-ol (secondary).
• Hexan-2-one gives iodoform; hexan-3-one does not (its C=O carries ethyl and propyl, no methyl).
• Propanal with acidified dichromate under reflux → propanoic acid, and the mixture turns green.

Must-know distinctions

• Aldehyde vs ketone: the aldehyde has H on the C=O carbon, so mild oxidising agents oxidise it.
• 2,4-DNPH vs Tollens’: 2,4-DNPH shows any C=O; Tollens’ picks out aldehydes.
• Racemic vs achiral: a racemic product contains chiral molecules in a 50 : 50 mix; an achiral product has no chiral centre.
• [O] vs [H]: [O] for oxidation to the acid; 2[H] for reduction to the alcohol.

Quick self-test

1. Name (CH₃)₂CHCH₂COCH₂CH₃.
2. Name the straight-chain aldehyde with six carbon atoms.
3. Explain why butanone (Mr 72) boils higher than pentane (Mr 72).
4. Why is butanone more soluble in water than pentane?
5. State the colour change when propanal is warmed with acidified potassium dichromate(VI).
6. Name the product when heptan-4-one is reduced with LiAlH₄, and classify it.
7. Which of these give a positive iodoform test: ethanal, propanal, pentan-3-one, hexan-2-one?
8. Name the product of HCN with propanal. Will the sample rotate plane-polarised light?
9. Why is KCN added in the reaction with HCN?
10. 2.00 g of 3-methylpentanal reacts completely with Fehling’s solution. What mass of Cu₂O forms? (C = 12.0, H = 1.0, O = 16.0, Cu = 63.5)
11. 2.50 g of hexan-2-one is fully reduced by LiAlH₄. What is the maximum mass of alcohol formed? (C = 12.0, H = 1.0, O = 16.0)
12. Why is a 2,4-DNPH derivative recrystallised before its melting temperature is measured?

Answers

1. 5-methylhexan-3-one (C=O at C3 counting from the ethyl end).
2. Hexanal.
3. Similar London forces (similar number of electrons), but butanone also has permanent dipole–dipole forces from C=O, so more energy is needed to separate its molecules.
4. The δ− carbonyl O accepts hydrogen bonds from water; pentane cannot form hydrogen bonds with water.
5. Orange to green.
6. Heptan-4-ol, a secondary alcohol.
7. Ethanal and hexan-2-one (both have CH₃CO–).
8. 2-Hydroxybutanenitrile. No: the chiral product forms as a racemic mixture.
9. HCN is a weak acid; KCN supplies the CN⁻ nucleophile.
10. Mr(C₆H₁₂O) = 100.0; n = 2.00 ÷ 100.0 = 0.0200 mol; 1 : 1 ratio, Mr(Cu₂O) = 143.0; mass = 0.0200 × 143.0 = 2.86 g.
11. Hexan-2-ol, C₆H₁₄O, Mr 102.0; n = 2.50 ÷ 100.0 = 0.0250 mol; mass = 0.0250 × 102.0 = 2.55 g.
12. To remove impurities, which lower the melting temperature and widen its range, so it would not match the table.

Where marks are usually lost

• Writing ketone names without a locant from C5 upwards (hexanone is ambiguous).
• Claiming carbonyl molecules hydrogen bond to each other.
• Explaining solubility by “carbonyls are polar” alone, without the hydrogen bond to water.
• Describing Tollens’ or Fehling’s results without warming the tubes, or giving the wrong colours.
• Forgetting “dry” for LiAlH₄ in ether, or writing [H] instead of 2[H].
• Curly arrows that start at the N of cyanide, or miss the C=O to O arrow.
• Saying the hydroxynitrile “is optically active” when it forms as a racemic mixture.
• Using 2,4-DNPH to tell an aldehyde from a ketone.
• Stating that all aldehydes, or all ketones, give the iodoform test.

Official syllabus

Source: Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry, specification XCH11/YCH11, Issue 1, September 2017 (Pearson Education Limited). These notes cover Topic 15B: Carbonyl compounds, outcomes 15.6 to 15.8, in Unit 4.