Whiteboard Chemistry with Joe White

Aldehydes & Ketones

The carbonyl group and why it is attacked by nucleophiles, oxidation and reduction, the two nucleophilic-addition mechanisms (NaBH4 and HCN) in full, and the tests that identify a carbonyl compound.

AQA 7404/7405 Paper 2 A-level only
C O δ+ δ−
Where this sits

You met aldehydes and ketones as the oxidation products of alcohols at AS, and their tests in organic analysis. Here they get their own reactions, all driven by one feature: the polar C=O double bond. Because the carbon is electron-poor (δ+), carbonyl compounds are attacked by nucleophiles — and two nucleophilic-addition mechanisms are the heart of the topic.

The carbonyl group: oxidation & reduction

Aldehydes (R–CHO) and ketones (R–CO–R′) both contain the carbonyl group, C=O. Oxygen is far more electronegative than carbon, so the double bond is polar: the carbon is δ+ and the oxygen δ−. That δ+ carbon is the target for nucleophiles.

Oxidation. An aldehyde is readily oxidised to a carboxylic acid (by warming with acidified potassium dichromate(VI), orange solution → green solution). A ketone is not oxidised this way — the basis of the tests that tell them apart.

Reduction. Both are reduced by NaBH4 (sodium tetrahydridoborate) in aqueous solution to alcohols: an aldehyde gives a primary alcohol, a ketone a secondary alcohol. Using [H] for the reductant:

C δ+ O δ− Nu nucleophiles attack the δ+ carbon acidified K₂Cr₂O₇ · warm NaBH₄ KCN, then dilute acid carboxylic acid aldehydes only — ketones resist alcohol 1° from aldehyde · 2° from ketone hydroxynitrile chain one carbon longer
Everything follows from the δ+ carbon — that is where nucleophiles attack.
Exam questions
Q1a [2 marks]

This question is about the structural isomers shown.

PQR STU OH OH OH HO O O

Identify the isomer(s) that would react when warmed with acidified potassium dichromate(VI). State the expected observation when acidified potassium dichromate(VI) reacts.

Show answer

M1 — Q, R, S, T. 1 mark

M2 — (orange solution) turns green. 1 mark

M2 is independent of M1. Q is a primary alcohol, R and S are secondary alcohols and T is an aldehyde — all are oxidised. P (a tertiary alcohol) and U (a ketone) are not.

Q1b [2 marks]

Identify the isomer(s) that would react with Tollens’ reagent. State the expected observation when Tollens’ reagent reacts.

Show answer

M1 — T. 1 mark

M2 — silver mirror. 1 mark

A grey/black precipitate is also allowed. Only the aldehyde reacts — Tollens’ does not oxidise the alcohols or the ketone.

Parts (c)–(f) of this question move on to esters and spectroscopy — that chemistry lives in organic analysis and carboxylic acids (3.3.9).

Source: AQA A-Level Chemistry past papers.

Nucleophilic addition

Both key reactions add a nucleophile across the C=O by the same three moves: the nucleophile attacks the δ+ carbon, the C=O double bond breaks onto the oxygen (forming a negative alkoxide), and the oxygen is then protonated. Learn one mechanism and you have both.

Reduction by NaBH4

NaBH4 delivers a hydride ion, H, which acts as the nucleophile:

The NaBH4 mechanism, arrow by arrow
  • Arrow 1 — from the H (a B–H bond of BH4) to the δ+ carbon of C=O.
  • Arrow 2 — from the C=O double bond to the oxygen, forming a negatively charged alkoxide intermediate.
  • Arrow 3 — a lone pair on the alkoxide O takes an H+ (from water / dilute acid), giving the alcohol.

NaBH4 reduces the C=O but not a C=C double bond: H is a nucleophile, and only the C=O is polar with a δ+ carbon to attack. A C=C is non-polar and electron-rich, so it does not attract the nucleophile.

STEP 1 — attack, and the C=O breaks H₃C C O H H δ+ δ− H₃C C O H H STEP 2 — the alkoxide is protonated H₃C C O H H H + H₃C C OH H H
Nucleophilic addition of hydride: attack, break, protonate.

Addition of HCN (using KCN)

The cyanide ion, CN, is the nucleophile. Addition across the C=O gives a hydroxynitrile (a 2-hydroxynitrile), which is one carbon longer than the carbonyl — a useful way to extend a carbon chain. The overall equation is written with HCN:

Aldehydes and unsymmetrical ketones form a mixture of enantiomers this way — the new carbon carries four different groups, and the planar C=O is attacked from both sides equally. A symmetrical ketone such as propanone gives an achiral product (two identical CH3 groups), so there are no enantiomers to mix.

The HCN / KCN mechanism, arrow by arrow
  • Arrow 1 — from a lone pair on the C of CN to the δ+ carbon of C=O.
  • Arrow 2 — from the C=O double bond to the oxygen, forming the alkoxide.
  • Arrow 3 — the alkoxide O takes an H+ (from the dilute acid / HCN), giving the hydroxynitrile.
STEP 1 — CN⁻ attacks, the C=O breaks H₃C C O H C N δ+ δ− H₃C C O H CN STEP 2 — the alkoxide is protonated H₃C C O H CN H + H₃C C OH H CN
A chiral hydroxynitrile forms as a racemate — the planar C=O is attacked from both sides equally.
Precision points
  • KCN is very toxic (it releases toxic HCN). It is used instead of HCN because HCN is a volatile, extremely toxic gas and a weak acid, so KCN provides a higher, controlled concentration of the CN nucleophile.
  • Every curly arrow must start from a lone pair or a bond and finish on an atom or bond — the nucleophile’s lone pair to the δ+ carbon, the C=O to the oxygen.
Worked example — naming the hydroxynitrile

Name the product of 3-methylbutan-2-one, (CH3)2CHCOCH3, with KCN followed by dilute acid.

Step 1 — add CN and OH to the carbonyl carbon:

(CH3)2CHCOCH3 → (CH3)2CHC(OH)(CN)CH3

Step 2 — number from the nitrile carbon. The CN carbon is C1 (it counts as part of the chain — the reaction has extended the chain by one carbon). The longest chain through it is four carbons: butanenitrile.

Step 3 — place the substituents. The old carbonyl carbon is C2, so it carries the OH (2-hydroxy) and one methyl; the CH of the old (CH3)2CH group is C3 with the other methyl.

2-hydroxy-2,3-dimethylbutanenitrile

The OH always lands on C2 — nucleophilic addition puts it on the old carbonyl carbon, which sits next to the nitrile carbon.

Exam questions
Q2a [5 marks]

Aqueous NaBH4 reduces aldehydes but does not reduce alkenes. Show the first step of the mechanism of the reaction between NaBH4 and 2-methylbutanal. You should include two curly arrows. Explain why NaBH4 reduces 2-methylbutanal but has no reaction with 2-methylbut-1-ene.

Show answer

M1 — correct structure of 2-methylbutanal. 1 mark

M2 — two curly arrows and the lone pair on the hydride ion: 1 mark

CH₃CH₂ C CH₃ H C O H H

C2H5 is allowed for CH3CH2. M2 is penalised if wrong partial charges are put on the C=O; the product is ignored.

M3 — the H ion / nucleophile is attracted to the δ+ carbon. 1 mark

M4 — the C=C is electron-rich. 1 mark

M5 — the H ion / nucleophile is repelled by the C=C, OR a C=C is only attacked by electrophiles. 1 mark

Q2b [2 marks]

A student attempted to reduce a sample of 2-methylbutanal but added insufficient NaBH4. The student confirmed that the reduction was incomplete by using a chemical test. Give the reagent and observation for the chemical test.

Show answer

Reagent: Tollens’ (reagent) OR ammoniacal silver nitrate OR a description of making Tollens’. 1 mark

Observation: silver mirror / silver precipitate OR a black solid / precipitate / deposit. 1 mark

Fehling’s / Benedict’s solution with a red precipitate (allow orange or brown) also scores. NOT acidified dichromate — the unreacted aldehyde and the alcohol product cannot be told apart with it, because both are oxidised.

Q3 [4 marks]

Figure 3 shows the reactant species involved in the first step of a mechanism.

O H

Complete Figure 3 to show the structure of the intermediate formed with curly arrows involved in its formation. Give the name of the reaction mechanism.

Show answer
O H O

M1 — arrow from the lone pair on the hydride to the C. 1 mark

M2 — arrow from the C=O to the O. 1 mark

M3 — the intermediate structure (displayed or abbreviated structures allowed). 1 mark

M4 — nucleophilic addition. 1 mark

Any attempt to show further correct steps is ignored; further incorrect steps are penalised. The list principle applies to M4. Parts (a) and (b) of this question ask the same for an addition–elimination intermediate (carboxylic acid derivatives, 3.3.9) and a nitration intermediate (aromatic chemistry, 3.3.10).

Source: AQA A-Level Chemistry past papers.

Tests for carbonyl compounds

AQA names two tests for telling an aldehyde from a ketone — Tollens’ reagent and Fehling’s solution — and warming with acidified potassium dichromate(VI) does the same job through oxidation. In all three, only the easily-oxidised aldehyde reacts.

ReagentAldehydeKetone
Tollens’ reagentsilver mirrorno change
Fehling’s solutionbrick-red pptno change
Acidified K2Cr2O7orange soln → green solnno change
Tollens’ Fehling’s acidified K₂Cr₂O₇ silver mirror brick-red precipitate orange soln → green soln aldehyde only aldehyde only aldehyde only
All three are positive for the aldehyde only — a ketone gives no change with any of them.
Exam questions
Q4a [2 marks]

Propanone (CH3COCH3) reacts with the weak acid HCN to form a hydroxynitrile. This hydroxynitrile is usually made by reaction of propanone with KCN followed by dilute acid, instead of with HCN. State the hazard associated with the use of KCN. Suggest a reason, other than safety, why KCN is used instead of HCN.

Show answer

M1 — toxic / poisonous. 1 mark

M2 — HCN is weak / [CN] is too low, or the reverse argument: KCN dissociates (better than HCN) to provide the CN nucleophile. 1 mark

For M1, “can produce toxic fumes/gas” or “corrosive” is allowed.

Q4b [4 marks]

Outline the mechanism for the reaction of propanone with KCN followed by dilute acid.

Show answer
H₃C C O CH₃ C N H₃C C O CN H CH₃ + H₃C C OH CN CH₃

M1 — cyanide ion with the lone pair on its C and the negative charge, and a curly arrow from the lone pair to the C of the C=O. 1 mark

M2 — curly arrow from the double bond to the O. 1 mark

M3 — the intermediate anion — the new bond must be to the C of the CN. 1 mark

M4 — curly arrow from a lone pair on the O to H+. 1 mark

M1 is not given if a K–CN bond is shown breaking; M2 is not given if the dipole is drawn incorrectly. The arrow to H+ may instead go to the H of HCN.

Source: AQA A-Level Chemistry past papers.

Identify & react

Bring it together. First find out what you have, then use its reactions.

The method
  • Aldehyde or ketone? Tollens’ (silver mirror) or Fehling’s (brick-red precipitate) — positive for aldehydes only.
  • Its reactions: reduce with NaBH4 (→ alcohol), or add HCN/KCN (→ hydroxynitrile, extending the chain).
Precision points
  • Give the reagent and the observation for every test — “Tollens’ → silver mirror”, not just “Tollens’”.
  • In the mechanisms, the nucleophile’s arrow starts from its lone pair and the C=O arrow finishes on the oxygen; show the negative alkoxide intermediate.
  • A chiral hydroxynitrile forms as a racemate — the planar C=O is attacked from both sides equally.

Capstone quiz — four past-paper questions

Four real AQA multiple-choice questions on this topic, in the style that opens Paper 3. Pick one answer each — the reasoning appears once you commit.

Q1 [1 mark]

In which conversion does a nucleophile attack the organic reactant?

  1. CH3CH2CH3 → CH3CHClCH3
  2. CH3CH=CH2 → CH3CHBrCH3
  3. CH3CH2CH2OH → CH3CH=CH2
  4. CH3CH2CHO → CH3CH2CH(OH)CN
Q2 [1 mark]

What is the product when 3-methylbutan-2-one reacts with acidified KCN?

  1. 2-hydroxy-2,3-dimethylbutanenitrile
  2. 3-hydroxy-2,3-dimethylbutanenitrile
  3. 2-hydroxy-3-methylpentanenitrile
  4. 3-hydroxy-2-methylpentanenitrile
Q3 [1 mark]

The skeletal formulas of two compounds are shown.

O O

Which method would distinguish between samples of these compounds?

  1. comparing fingerprint regions of their infrared spectra
  2. obtaining molecular masses from their high resolution mass spectra
  3. warming with acidified potassium dichromate(VI) solution
  4. warming with Tollens’ reagent
Q4 [1 mark]

Which reaction results in an overall change in shape around a carbon atom?

  1. oxidation of propanal with acidified potassium dichromate(VI)
  2. polymerisation of tetrafluoroethene
  3. reaction of bromoethane with an excess of concentrated ammonia
  4. reaction of methane with an excess of chlorine in ultraviolet radiation

Source: AQA A-Level Chemistry past papers.

3.3.8 Aldehydes & ketones — Quick-reference summary
  • Carbonyl C=O is polar (Cδ+=Oδ−), so it undergoes nucleophilic addition.
  • Oxidation: aldehyde → carboxylic acid (acidified K2Cr2O7, orange solution → green solution); ketone not oxidised.
  • Reduction (NaBH4): nucleophilic addition of hydride (H) — aldehyde → 1° alcohol; ketone → 2° alcohol.
  • HCN / KCN: nucleophilic addition of CN gives a hydroxynitrile (extends the chain by one carbon); a chiral product forms as a racemate (planar C=O attacked from both sides).
  • Tests: Tollens’ (silver mirror) and Fehling’s (brick-red precipitate) are positive for aldehydes only.

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