OCR A Level Chemistry A: Carbonyl compounds (H432)

Section 6.1.2, Carbonyl compounds, is taught here as set out in the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. Within Module 6 (Organic chemistry and analysis), it has lettered outcomes 6.1.2(a) through 6.1.2(e)(ii); each one is explained below with its additional guidance and PAG7 link. H432 is the full A level, and components 02 and 03 both examine Module 6.

Every H432 topic is reachable from the OCR chemistry hub. Carbonyl outcomes are listed on the printable checklist; the free diagnostics show gaps elsewhere. Recall drills follow in the revision notes, exam-style items in the practice questions.

Outcomes in 6.1.2

Outcome	You must be able to	Additional guidance
6.1.2(a)	Oxidise aldehydes to carboxylic acids with Cr₂O₇²⁻/H⁺ (K₂Cr₂O₇/H₂SO₄)	Use [O] and [H] in organic redox equations; PAG7 (see also 6.3.1 c)
6.1.2(b)(i)	Reduce carbonyl compounds to alcohols with NaBH₄ (nucleophilic addition)	None
6.1.2(b)(ii)	Add HCN, as NaCN(aq)/H⁺(aq), to form hydroxynitriles	See also 6.2.4 b
6.1.2(c)	Draw the nucleophilic addition mechanisms with NaBH₄ and with HCN	H⁻ is the nucleophile, then protonation from H₂O; CN⁻ attacks first, then protonation by H₂O or H⁺
6.1.2(d)(i)	Use 2,4-DNP to detect a carbonyl group	Equation and derivative structure not required; PAG7
6.1.2(d)(ii)	Identify a carbonyl compound from the melting point of its 2,4-DNP derivative	PAG7
6.1.2(e)(i)	Use Tollens’ reagent to detect an aldehyde group	[O] is acceptable in equations; PAG7
6.1.2(e)(ii)	Distinguish aldehydes from ketones, explained by oxidation of the aldehyde and reduction of Ag⁺ to Ag	PAG7

The carbonyl group

Aldehydes and ketones both contain C=O. In an aldehyde the carbonyl carbon carries a hydrogen, so the group ends a chain (–CHO). In a ketone it is joined to two carbons, inside the chain (–CO–).

Examples on this page: octanal, CH₃(CH₂)₆CHO, and benzaldehyde, C₆H₅CHO (aldehydes); 4-methylpentan-2-one, (CH₃)₂CHCH₂COCH₃, and heptan-4-one, CH₃CH₂CH₂COCH₂CH₂CH₃ (ketones).

Two features explain this section:

• The C=O bond is polar. Oxygen is more electronegative than carbon, so the carbon is δ+ and the oxygen δ−. Electron-pair donors (nucleophiles) are drawn to the δ+ carbon.
• The C=O double bond contains a π bond, which can break while the σ bond stays, so a nucleophile can add across C=O. Compare alkenes: non-polar C=C is attacked by electrophiles.

6.1.2(a): Oxidising aldehydes

Acidified dichromate, Cr₂O₇²⁻/H⁺ (K₂Cr₂O₇ with dilute H₂SO₄), oxidises an aldehyde to a carboxylic acid. Reflux with excess oxidising agent so the volatile aldehyde cannot escape unreacted. As the aldehyde is oxidised, the mixture turns from orange (Cr₂O₇²⁻) to green (Cr³⁺).

Write the oxidising agent as [O]. One oxygen atom is inserted into the C–H bond of the –CHO group:

RCHO + [O] → RCOOH

Ketones have no H on the carbonyl carbon, so they are not oxidised and the reagent stays orange. The same reagent oxidised alcohols in 4.2.1; this colour change is also a PAG7 test (6.3.1(c)(vi)).

Worked example 1: making octanoic acid

3.84 g of octanal is heated under reflux with excess acidified potassium dichromate. 3.37 g of pure octanoic acid is obtained. Write the equation and calculate the percentage yield. (Mᵣ: C₈H₁₆O 128.0, C₈H₁₆O₂ 144.0)

1. CH₃(CH₂)₆CHO + [O] → CH₃(CH₂)₆COOH
2. n(octanal) = 3.84 ÷ 128.0 = 0.0300 mol
3. One mole of octanal gives one mole of acid: at most 0.0300 mol
4. Theoretical mass = 0.0300 × 144.0 = 4.32 g
5. Percentage yield = 3.37 ÷ 4.32 × 100 = 78.0%

6.1.2(b): Nucleophilic addition

(b)(i) Reduction with NaBH₄

Sodium tetrahydridoborate(III), NaBH₄, reduces carbonyl compounds to alcohols. It is used in aqueous solution (or in methanol) and supplies hydride ions, H⁻. In equations write the reducing agent as [H]; two [H] are added across C=O.

• An aldehyde: RCHO + 2[H] → RCH₂OH, a primary alcohol
• A ketone: RCOR′ + 2[H] → RCH(OH)R′, a secondary alcohol

Worked example 2: two reductions

Benzaldehyde gives phenylmethanol:

C₆H₅CHO + 2[H] → C₆H₅CH₂OH

4-Methylpentan-2-one gives 4-methylpentan-2-ol:

(CH₃)₂CHCH₂COCH₃ + 2[H] → (CH₃)₂CHCH₂CH(OH)CH₃

(b)(ii) Addition of HCN

Adding HCN across C=O produces a hydroxynitrile: one carbon ends up carrying both an –OH group and a –C≡N group. HCN is very toxic and volatile, so the specification gives the reagent as NaCN(aq)/H⁺(aq): sodium cyanide supplies CN⁻ ions and the acid supplies H⁺ ions.

RCHO + HCN → RCH(OH)CN

The new bond is between carbon atoms, so the chain grows by one carbon. This is one of the C–C bond-forming reactions used in synthesis (6.2.4(b)(ii)), where the nitrile group can then be converted further.

Naming the product. The –C≡N carbon is carbon 1 of the nitrile chain, so the former carbonyl carbon becomes carbon 2 and the name starts “2-hydroxy…”.

Worked example 3: a hydroxynitrile from benzaldehyde

5.30 g of benzaldehyde reacts with NaCN(aq)/H⁺(aq). 4.79 g of product is isolated. Name the product and find the percentage yield. (Mᵣ: C₇H₆O 106.0, C₈H₇NO 133.0)

1. C₆H₅CHO + HCN → C₆H₅CH(OH)CN, 2-hydroxy-2-phenylethanenitrile
2. n(benzaldehyde) = 5.30 ÷ 106.0 = 0.0500 mol
3. Maximum n(product) = 0.0500 mol (1 : 1)
4. Theoretical mass = 0.0500 × 133.0 = 6.65 g
5. Percentage yield = 4.79 ÷ 6.65 × 100 = 72.0%

6.1.2(c): The two mechanisms

Both mechanisms have two stages: a nucleophile attacks the δ+ carbonyl carbon, then the O⁻ is protonated. Curly arrows start at a lone pair or a bond.

NaBH₄ with 4-methylpentan-2-one

1. Show the C=O dipole: δ+ on C, δ− on O.
2. The hydride ion, H⁻, uses its lone pair to attack the carbonyl carbon (curly arrow from that lone pair to C).
3. The π-bond electrons of C=O move onto O, shown by a curly arrow that starts on the double bond and ends on the O atom.
4. The intermediate is (CH₃)₂CHCH₂CH(CH₃)–O⁻, with the negative charge on oxygen.
5. The O⁻ uses a lone pair to take H from a water molecule (arrow to that H), while the water’s O–H bonding pair moves back onto its oxygen.
6. The products are 4-methylpentan-2-ol and OH⁻.

HCN with benzaldehyde

1. Show the C=O dipole.
2. The nucleophile is the cyanide ion, :C≡N⁻. Its carbon atom forms the new bond, so the curly arrow begins at the carbon’s lone pair.
3. As before, the C=O π-bond pair moves onto O.
4. The intermediate is C₆H₅CH(CN)–O⁻.
5. Protonate the oxygen with H⁺ (arrow from the O⁻ lone pair to H⁺) or with H₂O (as in step 5 above); the specification allows either.
6. The product is 2-hydroxy-2-phenylethanenitrile.

Nothing leaves in either mechanism: two groups add across C=O, which is why both are nucleophilic additions.

6.1.2(d): 2,4-Dinitrophenylhydrazine (2,4-DNP)

(d)(i) Detecting a carbonyl group

Add a few drops of the unknown to a solution of 2,4-DNP. An orange (or yellow-orange) precipitate shows a C=O group in an aldehyde or ketone. No precipitate means no aldehyde or ketone is present. Carboxylic acids and esters also contain C=O, but they do not give this precipitate. You do not need the equation or the structure of the precipitate (the derivative).

(d)(ii) Identifying the compound by melting point

The orange solid is a derivative of the particular carbonyl compound, and each derivative has its own melting point. The method, which is part of PAG7:

1. Filter off the orange precipitate.
2. Recrystallise it: dissolve in the minimum volume of hot solvent, cool so crystals form, filter and wash with a little cold solvent.
3. Dry the crystals.
4. Measure the melting point.
5. Compare with a table of known derivative melting points.

An impure solid starts melting too low and melts gradually, so a crude sample can match the wrong entry in the table. A pure sample melts sharply.

Worked example 4: identifying a C₇H₁₄O ketone

With 2,4-DNP, compound K (C₇H₁₄O) forms an orange precipitate; with Tollens’ reagent it forms no silver mirror. Its recrystallised derivative melts at 74–75 °C. Use these values for the derivatives: heptanal 108 °C, heptan-2-one 89 °C, heptan-3-one 81 °C, heptan-4-one 75 °C.

1. Orange precipitate: K is an aldehyde or a ketone.
2. No silver mirror: K is not an aldehyde, so heptanal is ruled out before you look at the melting point.
3. The sharp melting point of 74–75 °C matches 75 °C.
4. K is heptan-4-one, CH₃CH₂CH₂COCH₂CH₂CH₃.

6.1.2(e): Tollens’ reagent

(e)(i) Detecting an aldehyde

Tollens’ reagent is ammoniacal silver nitrate: aqueous ammonia is added to silver nitrate solution until the brown precipitate that forms first just redissolves. Add the unknown and warm in a water bath (not a flame, because many organic compounds are flammable). An aldehyde gives a silver mirror on the tube, or a grey-black precipitate of silver.

(e)(ii) Telling aldehydes from ketones

• Aldehyde: silver mirror forms. The aldehyde is oxidised to a carboxylic acid, and the silver ions are reduced to silver metal.
• Ketone: no change. A ketone cannot be oxidised this way, so the Ag⁺ ions are not reduced.

The equations, with [O] accepted for the organic change:

RCHO + [O] → RCOOH (oxidation)

Ag⁺ + e⁻ → Ag (reduction)

The ratio follows from the electrons: 1 mol of –CHO groups gives up 2 mol of electrons, enough to reduce 2 mol of Ag⁺.

Worked example 5: mass of silver

1.34 g of 3-phenylpropanal, C₆H₅CH₂CH₂CHO, reacts completely with excess Tollens’ reagent. Find the mass of silver deposited. (Mᵣ C₉H₁₀O 134.0; Aᵣ Ag 107.9)

1. n(aldehyde) = 1.34 ÷ 134.0 = 0.0100 mol
2. n(Ag) = 2 × 0.0100 = 0.0200 mol
3. Mass of Ag = 0.0200 × 107.9 = 2.16 g

The three tests side by side

Reagent	Aldehyde	Ketone	What it shows
2,4-DNP	Orange precipitate	Orange precipitate	C=O of an aldehyde or ketone present
Tollens’ reagent, warm	Silver mirror	No change	Aldehyde, not ketone
Cr₂O₇²⁻/H⁺, warm	Orange to green	Stays orange	Aldehyde, not ketone (primary and secondary alcohols also turn it green)

Use 2,4-DNP first, then Tollens’ reagent. The full scheme for unknowns is in 6.3.1(c).

Common errors

• Starting the cyanide arrow at the nitrogen. The bond forms from carbon, so the arrow starts at the carbon lone pair.
• Missing the arrow from the C=O bond to oxygen, or putting the negative charge on carbon in the intermediate.
• Writing H₂ instead of 2[H] in NaBH₄ equations.
• Describing the 2,4-DNP result as a colour change. The positive result is a precipitate.
• Naming hydroxynitriles from the wrong end. The nitrile carbon is carbon 1.
• Using the melting point of an unrecrystallised derivative.

Where next

Carry on with the revision notes and practice set. Yield methods are in amount of substance, half-equations in redox, and PAG work in practical skills for the endorsement.

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 6.1.2 Carbonyl compounds, learning outcomes (a) to (e). Published by OCR.