Edexcel International A Level Chemistry: Carbonyl compounds (YCH11)

This guide teaches 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. It is International A2 (IA2) content: Unit 4 (Rates, Equilibria and Further Organic Chemistry) is where you meet it. Within that unit it forms part of Topic 15: Organic Chemistry: Carbonyls, Carboxylic Acids and Chirality.

Linked pages: shorter revision notes and a practice set for this topic; the hub page for this course with its matching checklist; and, to find weak spots first, the free diagnostic quizzes.

Coverage of Topic 15B

Spec ref	You should be able to
15.6	Name aldehydes and ketones; draw structural, displayed and skeletal formulae
15.7 i	Explain why aldehydes and ketones have no intermolecular hydrogen bonds, and what that does to boiling temperature
15.7 ii	Explain their solubility in water through hydrogen bonds to water molecules
15.8 i	Oxidation: Fehling’s or Benedict’s solution, Tollens’ reagent, acidified dichromate(VI)
15.8 ii	Reduction with LiAlH₄ in dry ether
15.8 iii	Nucleophilic addition of HCN (with KCN): mechanism and optical activity
15.8 iv	2,4-DNPH test and identification from derivative melting temperatures
15.8 v	The iodoform test (iodine with alkali)

The specification lists the reactions in 15.8 i, iv and v as a further suggested practical.

15.6 Naming and drawing aldehydes and ketones

Aldehydes and ketones share one functional group: C=O, the carbonyl group.

• In an aldehyde the carbonyl carbon is at the end of the chain and carries a hydrogen atom. The group is written –CHO. The suffix is -al, and the CHO carbon is always C1, so no locant is needed.
• In a ketone the carbonyl carbon is bonded to two carbon atoms. The suffix is -one, with a locant for the carbonyl carbon (propanone and butanone need none, because there is only one possible position).

Name the longest chain containing the carbonyl carbon, give the carbonyl the lowest number, then add side chains alphabetically.

Worked example 1: naming

(a) (CH₃)₃CCHO. The chain containing CHO has three carbons: propanal. C2 carries two methyl groups. Answer: 2,2-dimethylpropanal.

(b) CH₃COCH(CH₃)CH₂CH₃. The longest chain through C=O has five carbons. From the left, C=O is C2; from the right, C4. Take C2. The methyl group is on C3. Answer: 3-methylpentan-2-one.

(c) A skeletal formula shows a zigzag of four lines. The =O is drawn on the carbon at the first bend, next to one end. That is a five-carbon chain with C=O at C2: pentan-2-one.

Three ways to draw them

Structural formula: CH₃CH₂COCH₃ (butanone). A displayed formula draws every bond:

    H   H       H
    |   |       |
H — C — C — C — C — H
    |   |   ‖   |
    H   H   O   H

Skeletal formula: draw the carbon zigzag, leave out C and H atoms, and draw the =O on the correct corner. For an aldehyde, the =O sits on the end of a line, and you may write the end group as CHO.

15.7 Physical properties

i No hydrogen bonds between carbonyl molecules

Hydrogen bonding needs a hydrogen atom bonded directly to N, O or F. In aldehydes and ketones every hydrogen is bonded to carbon, so the molecules cannot hydrogen bond to each other. They have London forces plus permanent dipole–permanent dipole forces from the polar C=O bond.

Worked example 2: comparing boiling temperatures

Use these values. All four compounds have Mr close to 58–60.

Compound	Mr	Boiling temperature / °C
Butane	58	−1
Propanal	58	48
Propanone	58	56
Propan-1-ol	60	97

The London forces are similar in size, because the molecules have similar numbers of electrons. Butane has only London forces, so it boils lowest. Propanal and propanone add permanent dipole–dipole attractions from C=O, which need more energy to overcome. Propan-1-ol boils highest because its O–H group lets its molecules form hydrogen bonds with each other; these are the strongest of the intermolecular forces here. No covalent bonds break on boiling.

ii Hydrogen bonds with water

A carbonyl oxygen has two lone pairs and a δ− charge. It can accept a hydrogen bond from the δ+ hydrogen of a water molecule (O–H····O=C). That is why the small members dissolve well: propanone mixes with water in all proportions. As the hydrocarbon chain grows, more of the molecule cannot hydrogen bond, and breaking water’s own hydrogen bonds to fit it in is not repaid, so solubility falls.

15.8 i Oxidation: telling aldehydes from ketones

An aldehyde has a hydrogen on the carbonyl carbon, so it is oxidised easily to a carboxylic acid. Ketones lack this hydrogen, so these mild reagents leave them unchanged. The specification lets you write the oxidising agent as [O]:

RCHO + [O] → RCOOH

Reagent and conditions	Aldehyde	Ketone
Fehling’s or Benedict’s solution, warm	Blue solution gives a red precipitate (copper(I) oxide)	Stays blue
Tollens’ reagent (ammoniacal silver nitrate), warm in a water bath	Silver mirror forms on the tube (or a grey-black precipitate)	No change
Potassium dichromate(VI) with dilute sulfuric acid, warm	Orange to green	Stays orange

What is reduced? In Fehling’s and Benedict’s, copper(II) is reduced to copper(I). In Tollens’ reagent, Ag⁺ + e⁻ → Ag. With acidified dichromate, orange Cr₂O₇²⁻ ions become green Cr³⁺ ions. Fehling’s, Benedict’s and Tollens’ are alkaline, so the acid formed there is present as its carboxylate ion.

To make Tollens’ reagent, add a drop of sodium hydroxide solution to silver nitrate solution, then dilute ammonia dropwise until the brown precipitate just dissolves.

Worked example 3: how much silver?

1.71 g of heptanal, CH₃(CH₂)₅CHO, reacts completely with Tollens’ reagent. Each aldehyde molecule loses two electrons when it is oxidised, so 2 mol Ag form per mole of aldehyde. Use Ar: C 12.0, H 1.0, O 16.0, Ag 107.9.

Mr(C₇H₁₄O)  = 114.0
n(heptanal) = 1.71 ÷ 114.0 = 0.0150 mol
n(Ag)       = 2 × 0.0150 = 0.0300 mol
m(Ag)       = 0.0300 × 107.9 = 3.24 g

15.8 ii Reduction with LiAlH₄

Lithium tetrahydridoaluminate(III), LiAlH₄ (lithium aluminium hydride), in dry ether (ethoxyethane) reduces carbonyl compounds to alcohols. The ether must be dry because LiAlH₄ reacts violently with water. Water or dilute acid is added only after the reduction is over. The reducing agent can be written as [H]:

• aldehyde → primary alcohol: RCHO + 2[H] → RCH₂OH
• ketone → secondary alcohol: RCOR’ + 2[H] → RCH(OH)R’

Worked example 4: reduction and yield

2.15 g of 2,2-dimethylpropanal, (CH₃)₃CCHO, is reduced. Give the product, then the maximum mass, and the yield if 1.76 g is collected.

The product is a primary alcohol, 2,2-dimethylpropan-1-ol, (CH₃)₃CCH₂OH.

(CH₃)₃CCHO + 2[H] → (CH₃)₃CCH₂OH
Mr: C₅H₁₀O = 86.0   C₅H₁₂O = 88.0
n = 2.15 ÷ 86.0 = 0.0250 mol
maximum mass = 0.0250 × 88.0 = 2.20 g
yield = 1.76 ÷ 2.20 × 100 = 80.0%

15.8 iii Nucleophilic addition of HCN

Hydrogen cyanide adds across C=O to give a hydroxynitrile. HCN is a weak acid and barely ionises, so KCN is added to supply cyanide ions, the nucleophile. HCN is extremely toxic, so the reaction needs a fume cupboard.

The C=O bond is polar: C is δ+, O is δ−. The mechanism, drawn with curly arrows, using butanone:

1. Draw a curly arrow that starts at the lone pair on the C atom of :CN⁻ and ends on the δ+ carbonyl carbon. Another arrow takes the C=O π bond pair onto the oxygen.
2. This gives an intermediate with O⁻ on the former carbonyl carbon. A curly arrow goes from a lone pair on O⁻ to the H of an HCN molecule, and an arrow from the H–C bond to the C of that CN.
3. The product is 2-hydroxy-2-methylbutanenitrile, and CN⁻ is regenerated.

      δ+ δ−                   O⁻                       OH
CH₃CH₂–C=O  + :CN⁻  →  CH₃CH₂–C–CH₃  + H–CN  →  CH₃CH₂–C–CH₃  + CN⁻
       |                      |                        |
      CH₃                     CN                       CN

Show the dipoles, the lone pair on carbon and the negative charge in your drawing.

Optical activity as evidence

The carbonyl carbon is trigonal planar: it and its three neighbours sit flat. Cyanide can attack from above or below that plane with equal probability. When the new chiral carbon carries four different groups, top and bottom attack each make one enantiomer, so they form 50 : 50: a racemic mixture, which does not rotate plane-polarised light. So an optically inactive product supports this mechanism.

Worked example 5: is the product chiral?

(a) Butanone + HCN → CH₃CH₂C(OH)(CN)CH₃. The central carbon has OH, CN, CH₃ and C₂H₅: four different groups, so it is chiral. The product forms as a racemic mixture, so the sample is not optically active.

(b) Pentan-3-one + HCN → (CH₃CH₂)₂C(OH)CN, 2-ethyl-2-hydroxybutanenitrile. Two of the groups are ethyl, so there is no chiral centre and no optical isomers at all.

Both samples are inactive, for different reasons. Say which applies.

15.8 iv The 2,4-DNPH test

2,4-Dinitrophenylhydrazine (2,4-DNPH, Brady’s reagent) gives a yellow-orange precipitate with any aldehyde or ketone. It shows a carbonyl group is present but not which kind. You do not need the equation.

To identify the compound, filter off the precipitate (the derivative), recrystallise it from a minimum of hot solvent, dry it, and measure its melting temperature. Compare with a table of known values. A pure solid melts sharply at its true value; impurities lower the melting temperature and spread it over a range.

Worked example 6: identifying a carbonyl compound

Use these values for the 2,4-DNPH derivatives.

Carbonyl compound	Derivative melting temperature / °C
Ethanal	168
Propanal	156
Propanone	126
Butanal	123
Butanone	115

A purified derivative of compound Y melts at 125 °C. That sits between butanal and propanone, so the melting temperature alone does not decide. Y gives no silver mirror with Tollens’ reagent, so it is not an aldehyde. Y is propanone. Always back a close melting-temperature match with a second test.

15.8 v The iodoform test

Warm the compound with iodine solution and sodium hydroxide solution. A pale yellow precipitate of triiodomethane (iodoform, CHI₃), with an antiseptic smell, shows a CH₃CO– group. Positive carbonyl compounds are methyl ketones (CH₃COR) and ethanal, the only aldehyde with this group.

The rest of the molecule leaves as a carboxylate ion with one fewer carbon atom.

Worked example 7: equation and mass

3-Methylbutan-2-one, (CH₃)₂CHCOCH₃, has a CH₃CO– group, so it gives a positive test.

(CH₃)₂CHCOCH₃ + 3I₂ + 4NaOH → CHI₃ + (CH₃)₂CHCOONa + 3NaI + 3H₂O

The organic by-product is sodium 2-methylpropanoate. From 2.58 g of the ketone (Mr 86.0): n = 2.58 ÷ 86.0 = 0.0300 mol, so the maximum mass of CHI₃ (Mr 393.7) is 0.0300 × 393.7 = 11.8 g.

Common errors

• Numbering a ketone from the wrong end, giving pentan-4-one instead of pentan-2-one.
• Saying aldehydes hydrogen bond to each other. They hydrogen bond only to water.
• Writing “red precipitate” for a ketone with Fehling’s solution, or “silver mirror” with a ketone.
• Using LiAlH₄ in water, or forgetting that a ketone gives a secondary, not primary, alcohol.
• Starting the curly arrow from the negative charge on N rather than the lone pair on the carbon of CN⁻.
• Saying the HCN product is optically active. A racemic mixture is not.
• Claiming 2,4-DNPH tells aldehydes from ketones. It only shows C=O; use Tollens’ or Fehling’s for that.
• Saying every aldehyde gives iodoform. Only ethanal does.

Next steps

Next, read the condensed revision notes and then try the practice questions. Use the checklist to tick off each outcome, and take a free diagnostic.

Official syllabus

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