OCR A Level Chemistry A: Carbonyl compounds (H432) – Revision Notes

The study guide explains this content step by step; start there if any line below is unclear. Here, tables and short lists boil down outcomes 6.1.2(a) to (e)(ii), Carbonyl compounds, with their additional guidance and PAG7 links, from the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. The course is the full A level, and Module 6 is examined in components 02 and 03.

Then try the practice questions. For other H432 sections go to the OCR chemistry hub; log carbonyl outcomes on the printable checklist, and run the free diagnostics to find weaker areas.

Key terms

Term	Meaning
Carbonyl group	C=O; polar, with C δ+ and O δ−
Aldehyde	Carbonyl carbon bonded to at least one H: –CHO, at the end of a chain
Ketone	Carbonyl carbon bonded to two carbons: –CO– inside a chain
Nucleophile	Electron-pair donor, e.g. H⁻ or CN⁻
Nucleophilic addition	A nucleophile attacks the δ+ carbon and a group adds across C=O; nothing leaves
Hydroxynitrile	Product of HCN addition: –OH and –C≡N on the same carbon
[O], [H]	Oxidising and reducing agents in organic redox equations
Derivative	The solid formed from 2,4-DNP and a particular carbonyl compound

Reaction summary: 6.1.2(a) and (b)

Outcome	Reagent and conditions	Aldehyde gives	Ketone gives	Equation pattern
(a)	Cr₂O₇²⁻/H⁺ (K₂Cr₂O₇/H₂SO₄), heat under reflux	Carboxylic acid	No reaction	RCHO + [O] → RCOOH
(b)(i)	NaBH₄ (aqueous)	Primary alcohol	Secondary alcohol	add 2[H] across C=O
(b)(ii)	NaCN(aq)/H⁺(aq), i.e. HCN	Hydroxynitrile	Hydroxynitrile	add H and CN across C=O

Colour for (a): orange Cr₂O₇²⁻ to green Cr³⁺ when an aldehyde is present. A ketone leaves it orange.

The HCN reaction forms a new C–C bond and adds one carbon to the chain (see also 6.2.4(b)(ii)).

Method: write a product formula

1. Find the C=O carbon.
2. NaBH₄: put H on that carbon and H on the oxygen.
3. HCN: put CN on that carbon and H on the oxygen.
4. [O] with an aldehyde: change –CHO to –COOH.
5. Check: carbon skeleton unchanged (NaBH₄, [O]) or one carbon longer (HCN).

Naming hydroxynitriles

The nitrile carbon is C1, so the former carbonyl carbon is C2.

• Ethanal → 2-hydroxypropanenitrile
• Butanal → 2-hydroxypentanenitrile, C₅H₉NO
• Propanone → 2-hydroxy-2-methylpropanenitrile

6.1.2(c): Mechanisms in steps

Stage 1, nucleophilic attack (both reactions)

• Draw the δ+ and δ− on C=O.
• Arrow from the nucleophile’s lone pair to the carbonyl carbon.
  • NaBH₄: the nucleophile is H⁻.
  • HCN: the nucleophile is CN⁻; the arrow starts at the lone pair on carbon.
• The C=O π pair moves onto O (arrow starts on the double bond).
• Intermediate: O⁻ with a lone pair, on a carbon that now has four bonds.

Stage 2, protonation

• NaBH₄: arrow from the O⁻ lone pair to H of H₂O, plus an arrow from that O–H bond onto oxygen. By-product OH⁻.
• HCN: protonate with H₂O (as above) or with H⁺ (one arrow, O⁻ lone pair to H⁺). The specification accepts either.

Must-know distinctions

Compare	Point
C=C vs C=O	C=C is non-polar and undergoes electrophilic addition; C=O is polar and undergoes nucleophilic addition
NaBH₄ product from aldehyde vs ketone	Primary alcohol vs secondary alcohol
[H] vs H₂	Use 2[H] for NaBH₄ equations, not H₂
HCN vs NaCN/H⁺	Same reaction; HCN is very toxic and volatile, so the specification gives the reagent as NaCN(aq)/H⁺(aq)
CN⁻ arrow	Starts at carbon, never at nitrogen

6.1.2(d): 2,4-DNP

Step	Detail
Test (d)(i)	Add the unknown to 2,4-DNP solution
Positive result	Orange (yellow-orange) precipitate: aldehyde or ketone present
Not required	The equation and the structure of the derivative
Identify (d)(ii)	Filter, recrystallise, dry, measure melting point, compare with a data table
Why recrystallise	An impure solid melts low and over a range

Carboxylic acids and esters contain C=O but give no precipitate with 2,4-DNP.

6.1.2(e): Tollens’ reagent

• Ammoniacal silver nitrate; Ag⁺ ions are the oxidising agent.
• Warm in a water bath, not a flame (flammable organics).
• Aldehyde: silver mirror (or grey-black precipitate).
• Ketone: no change.

Explanation for (e)(ii):

• The aldehyde is oxidised to a carboxylic acid: RCHO + [O] → RCOOH.
• Each Ag⁺ ion gains an electron and becomes silver metal: Ag⁺ + e⁻ → Ag.
• One mole of aldehyde gives two moles of Ag (two electrons lost per –CHO).

Small worked reminder

0.290 g of propanal, CH₃CH₂CHO (Mᵣ 58.0), reacts fully with Tollens’ reagent.

• n(propanal) = 0.290 ÷ 58.0 = 0.00500 mol
• n(Ag) = 2 × 0.00500 = 0.0100 mol
• mass Ag = 0.0100 × 107.9 = 1.08 g

Test map for unknowns (PAG7)

Reagent	Aldehyde	Ketone	Primary or secondary alcohol
2,4-DNP	orange precipitate	orange precipitate	no precipitate
Tollens’, warm	silver mirror	no change	no change
Cr₂O₇²⁻/H⁺, warm	orange to green	stays orange	orange to green

Order: 2,4-DNP first (is C=O present?), then Tollens’ reagent (aldehyde or ketone?).

Links to other sections

• 4.2.1 Alcohols: primary alcohols are oxidised to aldehydes (distil) and secondary alcohols to ketones. Section 6.1.2(b)(i) reverses this: NaBH₄ turns the carbonyl compound back into the alcohol.
• 6.2.4 Carbon–carbon bond formation: the hydroxynitrile from 6.1.2(b)(ii) is one way to lengthen a chain; the nitrile can then be taken further.
• 6.3.1(c) Qualitative analysis: the 2,4-DNP, Tollens’ and dichromate tests here are part of the full scheme for identifying unknown organic compounds.

A two-step route to remember: propan-2-ol → (Cr₂O₇²⁻/H⁺, reflux) → propanone → (NaCN(aq)/H⁺(aq)) → 2-hydroxy-2-methylpropanenitrile.

Quick self-test

1. State the reagents written as Cr₂O₇²⁻/H⁺.
2. Write the equation, using [O], for the oxidation of butanal.
3. Name the product when pentan-3-one is reduced by NaBH₄, and state its class.
4. Name the nucleophile in the NaBH₄ mechanism and the species that protonates the intermediate.
5. Give the formula of the product when butanal reacts with HCN, and its Mᵣ (H 1.0, C 12.0, O 16.0, N 14.0).
6. Why does the cyanide arrow start at carbon?
7. What do you see when 2,4-DNP is added to a ketone?
8. Give two reasons the derivative must be recrystallised before its melting point is measured.
9. Explain why a ketone gives no silver mirror with Tollens’ reagent.
10. 2.10 g of cyclopentanone (Mᵣ 84.0) is reduced with NaBH₄, giving 1.72 g of cyclopentanol (Mᵣ 86.0). Calculate the percentage yield.
11. Compound J gives an orange precipitate with 2,4-DNP and turns acidified dichromate green. Is J an aldehyde or a ketone?
12. Which carbonyl compound reacts with HCN to give CH₃CH(OH)CN?

Answers

1. K₂Cr₂O₇ (potassium dichromate) acidified with dilute H₂SO₄.
2. CH₃CH₂CH₂CHO + [O] → CH₃CH₂CH₂COOH (butanoic acid).
3. Pentan-3-ol; a secondary alcohol.
4. The hydride ion, H⁻; water, H₂O.
5. CH₃CH₂CH₂CH(OH)CN, C₅H₉NO; Mᵣ = 5(12.0) + 9(1.0) + 14.0 + 16.0 = 99.0.
6. The new bond forms between the cyanide carbon and the carbonyl carbon, so the electron pair comes from the carbon’s lone pair.
7. An orange (yellow-orange) precipitate.
8. An impure sample melts below the true value, and over a spread of temperatures rather than sharply; either effect can give a false match.
9. Tollens’ reagent works by oxidising the compound; a ketone is not oxidised by it, so Ag⁺ is not reduced to Ag.
10. n = 2.10 ÷ 84.0 = 0.0250 mol; theoretical mass = 0.0250 × 86.0 = 2.15 g; yield = 1.72 ÷ 2.15 × 100 = 80.0%.
11. An aldehyde: the precipitate shows C=O, and only the aldehyde is oxidised by dichromate (a ketone would leave it orange).
12. Ethanal, CH₃CHO: remove the CN carbon and turn the C(OH) back into C=O.

Where marks are usually lost

• Writing H₂ or 2H in place of 2[H] for NaBH₄ reductions.
• Saying a ketone is oxidised by acidified dichromate, or giving no observation for the ketone (it stays orange).
• Starting the CN⁻ arrow at nitrogen rather than at the lone pair on carbon.
• Missing the arrow from the C=O bond to oxygen, so the oxygen never gains its negative charge.
• Leaving out the protonation step, or not showing OH⁻ formed when water protonates the intermediate.
• Describing 2,4-DNP as a colour change of the solution rather than a precipitate.
• Comparing an unpurified derivative’s melting point with the data.
• Explaining the Tollens’ test without saying which species is oxidised and which is reduced.
• Heating Tollens’ reagent over a flame instead of a water bath.
• Numbering a hydroxynitrile from the wrong end.

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.