Aldehydes and Ketones

Carbonyl compounds are among the most important functional groups in organic chemistry. They contain the C=O (carbonyl) group and are divided into two main families: aldehydes and ketones. Understanding their chemistry is essential for Edexcel A-Level and underpins much of the organic synthesis you will encounter.

The Carbonyl Group

The carbonyl group consists of a carbon atom double-bonded to an oxygen atom (C=O). The key difference between aldehydes and ketones lies in what is attached to the carbonyl carbon:

In aldehydes, the carbonyl carbon is bonded to at least one hydrogen atom. The carbonyl group is always at the end of the carbon chain. The general formula is RCHO.
In ketones, the carbonyl carbon is bonded to two carbon-containing groups (alkyl or aryl). The carbonyl group is within the carbon chain. The general formula is RCOR'.

The C=O bond is polar because oxygen is more electronegative than carbon (electronegativity values: O = 3.4, C = 2.5). This means the carbon atom carries a partial positive charge (delta+) and the oxygen carries a partial negative charge (delta-). This polarity is the reason carbonyl compounds undergo nucleophilic addition reactions.

Why the Carbonyl is So Reactive

The delta+ character on the carbonyl carbon is greater than on the carbon in a C-O single bond (as in an alcohol) for two reasons:

The pi bond holds the electron density closer to the oxygen than a sigma bond alone would.
There is no adjacent atom with lone pairs to delocalise into the C=O bond (contrast this with carboxylic acids, where the -OH lone pairs partially delocalise into the C=O, stabilising it and reducing the delta+ character).

This combination makes the carbonyl carbon in aldehydes and ketones an excellent electrophilic site for nucleophilic attack.

Naming Aldehydes and Ketones

Aldehydes are named using the suffix -al. The longest chain must include the carbonyl carbon, and numbering always starts from the carbonyl end:

Methanal (HCHO) -- one carbon
Ethanal (CH3CHO) -- two carbons
Propanal (CH3CH2CHO) -- three carbons

Ketones are named using the suffix -one, with a number indicating the position of the carbonyl group:

Propanone (CH3COCH3) -- the simplest ketone
Butanone (CH3COCH2CH3) -- four carbons
Pentan-2-one and pentan-3-one -- positional isomers

Physical Properties

Carbonyl compounds cannot form hydrogen bonds with themselves (unlike alcohols), because they lack an O-H or N-H bond. However, they can form hydrogen bonds with water through the lone pairs on the oxygen atom. This means:

Small aldehydes and ketones are soluble in water
Boiling points are higher than alkanes of similar molecular mass (due to permanent dipole-dipole interactions) but lower than alcohols (which have stronger hydrogen bonding)
Solubility in water decreases as the hydrocarbon chain length increases

Compound	Type	Mr	Boiling Point (degrees C)	Soluble in Water?
Propanal	Aldehyde	58	49	Yes
Propanone	Ketone	58	56	Yes (miscible)
Propan-1-ol	Alcohol	60	97	Yes
Butane	Alkane	58	-1	No

The table confirms the pattern: alcohol > carbonyl > alkane for boiling points at similar molecular mass, and the small carbonyls are water-soluble.

Nucleophilic Addition Reactions

The delta+ carbon of the carbonyl group is susceptible to attack by nucleophiles -- species with a lone pair of electrons that they can donate. The general mechanism is nucleophilic addition:

The nucleophile attacks the delta+ carbonyl carbon, forming a new bond
The C=O pi bond breaks and both electrons move onto the oxygen, creating an alkoxide intermediate (with a negative charge on oxygen)
The alkoxide is protonated (gains H+) to form the final product

Why Aldehydes Are More Reactive Than Ketones Toward Nucleophiles

This is a commonly examined comparison. Two factors contribute:

Factor	Aldehyde	Ketone
Steric effect	One H and one R group -- less steric hindrance around the carbonyl carbon	Two R groups -- more steric hindrance blocks the approaching nucleophile
Electronic (inductive) effect	One alkyl group pushes electron density toward C=O (small reduction of delta+)	Two alkyl groups push more electron density toward C=O (greater reduction of delta+)

Both factors make the aldehyde carbonyl carbon more delta+ and more accessible, so nucleophiles attack it more readily.

Reaction with HCN (Hydrogen Cyanide)

When an aldehyde or ketone reacts with HCN (in the presence of a trace of base such as KCN), the cyanide ion (CN-) acts as the nucleophile:

Step-by-step mechanism:

The KCN provides CN- ions in solution. CN- is the active nucleophile (HCN itself is a poor nucleophile because the lone pair on carbon is tied up in the triple bond; the ion is far more reactive).
The lone pair on the carbon of CN- attacks the delta+ carbonyl carbon, forming a new C-C bond.
Simultaneously, the C=O pi bond breaks and both electrons move onto oxygen, forming an alkoxide intermediate (tetrahedral, with O carrying a negative charge).
The alkoxide is protonated by HCN (which acts as the acid), regenerating CN- and forming the hydroxynitrile (cyanohydrin) product.

This reaction is important because it extends the carbon chain by one carbon, which is useful in synthesis. The nitrile group can subsequently be hydrolysed to a carboxylic acid or reduced to an amine.

Worked example -- Ethanal + HCN:

Ethanal (CH3CHO, 2C) + HCN --> 2-hydroxypropanenitrile (CH3CH(OH)CN, 3C)
The product has one more carbon than the starting aldehyde.
Hydrolysis of the nitrile group gives 2-hydroxypropanoic acid (lactic acid).

Safety note: HCN is extremely toxic (lethal dose ~1 mg/kg body mass), so in practice the reaction is carried out using KCN in acidified solution rather than HCN gas directly.

Reduction with NaBH4 (Sodium Borohydride)

Sodium borohydride is a mild reducing agent that reduces carbonyl compounds to alcohols:

Aldehydes are reduced to primary alcohols
Ketones are reduced to secondary alcohols

Detailed mechanism:

The hydride ion (H-) from NaBH4 acts as the nucleophile. It donates its pair of electrons to the delta+ carbonyl carbon.
The C=O pi bond breaks -- both electrons move onto oxygen, forming an alkoxide ion.
The alkoxide is protonated by the aqueous solvent (water) to form the alcohol product.

NaBH4 is dissolved in water or aqueous ethanol as the solvent. It is selective -- it reduces C=O but does not reduce C=C double bonds. This selectivity makes it a valuable reagent in organic synthesis when you need to reduce a carbonyl while leaving a double bond intact.

Comparison of reducing agents:

Reagent	Solvent	Reduces Aldehydes?	Reduces Ketones?	Reduces Carboxylic Acids?	Reduces C=C?
NaBH4	Water / aqueous ethanol	Yes	Yes	No	No
LiAlH4	Dry ether	Yes	Yes	Yes	No
H2/Ni	Gas phase	Yes	Yes	No	Yes

Oxidation of Aldehydes

Aldehydes can be oxidised to carboxylic acids. This is a key difference from ketones, which resist oxidation under normal conditions. The common oxidising agent is acidified potassium dichromate (K2Cr2O7/H2SO4):

RCHO --> RCOOH

During this reaction, the orange dichromate solution turns green (Cr3+ ions). Ketones show no colour change because they cannot be further oxidised without breaking a C-C bond.

Why ketones resist oxidation: To oxidise a ketone, you would need to break a C-C bond (since the carbonyl carbon is bonded to two carbon groups, not to hydrogen). Breaking C-C bonds requires far more energy than the mild oxidising conditions provide.

Distinguishing Aldehydes from Ketones

Because aldehydes are more easily oxidised than ketones, two classic tests exploit this difference:

Tollens' Reagent (Silver Mirror Test)

Tollens' reagent contains silver(I) ions in aqueous ammonia, [Ag(NH3)2]+. When warmed gently with an aldehyde:

The aldehyde is oxidised to a carboxylate ion
Ag+ ions are reduced to metallic silver, which deposits as a silver mirror on the inside of the test tube
Ketones give no reaction with Tollens' reagent

Ionic equation: RCHO + 2[Ag(NH3)2]+ + 2OH- --> RCOO- + 2Ag + 4NH3 + H2O

Fehling's Solution (Copper Test)

Fehling's solution contains Cu2+ ions complexed with tartrate, giving a deep blue colour. When warmed with an aldehyde:

The aldehyde is oxidised to a carboxylate ion
Cu2+ ions are reduced to Cu+, forming a brick-red precipitate of copper(I) oxide (Cu2O)
Ketones give no reaction with Fehling's solution

Both tests are positive for aldehydes and negative for ketones, providing a reliable way to distinguish between the two.

Diagnostic Decision Flowchart

flowchart TD
    A[Unknown Carbonyl Compound] --> B{Add 2,4-DNP reagent}
    B -->|Yellow/orange precipitate| C[Confirmed: Aldehyde or Ketone]
    B -->|No precipitate| D[Not an aldehyde or ketone]
    C --> E{Add Tollens' reagent, warm}
    E -->|Silver mirror forms| F[ALDEHYDE]
    E -->|No silver mirror| G[KETONE]
    F --> H{Oxidise with K2Cr2O7 / H2SO4 reflux}
    H --> I[Carboxylic acid product]
    G --> J{Reduce with NaBH4}
    J --> K[Secondary alcohol product]

Exam tip: 2,4-DNP (2,4-dinitrophenylhydrazine, also called Brady's reagent) is used to confirm the presence of a carbonyl group -- it produces a yellow or orange precipitate with both aldehydes and ketones. It does not distinguish between them. Follow up with Tollens' or Fehling's to identify which type of carbonyl is present.

Common Exam Mistakes

Confusing the mechanism name. The mechanism for HCN with a carbonyl is nucleophilic addition, not nucleophilic substitution. Nothing leaves -- the nucleophile simply adds to the molecule. Substitution requires a leaving group.
Drawing curly arrows from H of HCN. The nucleophile is CN-, not HCN. The curly arrow must start from the lone pair on the carbon of CN- and point to the delta+ carbonyl carbon.
Saying NaBH4 reduces everything. NaBH4 is a mild reducing agent. It reduces aldehydes and ketones but not carboxylic acids. LiAlH4 is needed for carboxylic acids.
Forgetting to state the solvent. NaBH4 works in water or aqueous ethanol. LiAlH4 must use dry ether (it reacts violently with water).
Confusing Tollens' observations. The silver mirror forms on the glass surface of the test tube, not in solution. If the test tube is not clean, you may get a grey precipitate instead of a mirror -- still a positive result.

Summary of Key Reactions

Reaction	Reagent / Conditions	Aldehyde Product	Ketone Product
Reduction	NaBH4 in water	Primary alcohol	Secondary alcohol
Oxidation	K2Cr2O7/H2SO4, reflux	Carboxylic acid	No reaction
HCN addition	HCN + KCN catalyst	Hydroxynitrile (chain +1C)	Hydroxynitrile (chain +1C)
Tollens' test	[Ag(NH3)2]+, warm	Silver mirror (positive)	No reaction (negative)
Fehling's test	Cu2+/tartrate, warm	Brick-red precipitate	No reaction (negative)
2,4-DNP test	2,4-dinitrophenylhydrazine	Yellow/orange ppt	Yellow/orange ppt

Understanding carbonyl chemistry is fundamental to the rest of organic chemistry at A-Level. The nucleophilic addition mechanism recurs throughout the specification, and the ability to distinguish aldehydes from ketones using chemical tests is a common exam question.

A-Level Deep Dive: Aldehydes and Ketones

Spec mapping

Edexcel 9CH0 specification, Topic 16 — Chemistry of carbonyl compounds, sub-strands 16.1–16.5 covers the structure of the carbonyl group (C=O), nucleophilic addition reactions of aldehydes and ketones with HCN/KCN producing hydroxynitriles, reduction by NaBH4 in aqueous solution to alcohols, oxidation of aldehydes (but not ketones) by acidified potassium dichromate(VI), and the use of Tollens' reagent, Fehling's solution and 2,4-dinitrophenylhydrazine (Brady's reagent) as diagnostic tests (refer to the official specification document for exact wording). Carbonyl chemistry threads explicitly through Paper 2 (Topic 16 examined directly) and Paper 3 (synoptic problem-solving), and connects backwards to AS-level alcohol oxidation (Topic 6) and forwards to carboxylic acid chemistry (Topic 17), amino acid zwitterions (Topic 18) and spectroscopic identification (Topic 18 — IR/NMR/MS). The Edexcel data booklet provides characteristic IR absorption ranges (C=O ≈ 1680–1750 cm⁻¹) and proton NMR chemical shift ranges that you must apply confidently.

Worked example with full mark scheme

Question (8 marks):

A laboratory technician is given two unlabelled colourless liquids, X and Y. One is propanal (CH3CH2CHO) and the other is propanone (CH3COCH3).

(a) Describe two chemical tests that, used together, would confirm which liquid is propanal and which is propanone. State the reagents, observations and ionic/balanced equations where relevant. (6)

(b) Predict the organic product when propanone reacts with HCN in the presence of a small amount of KCN, and name the mechanism. (2)

Solution with mark scheme:

(a) Test 1 — 2,4-DNP (Brady's reagent).

Add a few drops of 2,4-dinitrophenylhydrazine in dilute sulfuric acid to each of X and Y separately. Both liquids will produce an orange/yellow precipitate, confirming that each contains a carbonyl group (C=O).

M1 — selecting 2,4-DNP and stating the orange/yellow precipitate observation. The precipitate is a 2,4-dinitrophenylhydrazone derivative.

A1 — explicit statement that the test does not distinguish aldehyde from ketone but confirms the carbonyl group is present. Examiners reward candidates who state the limitation of the test as well as its purpose.

Test 2 — Tollens' reagent.

To fresh portions of each liquid, add Tollens' reagent (silver(I) ions in aqueous ammonia, [Ag(NH3)2]+) and warm gently in a water bath. The aldehyde (propanal) will produce a silver mirror on the inner wall of the test tube; the ketone (propanone) will give no observable change.

M1 — selecting Tollens' (or equivalently Fehling's) and stating both observations.

A1 — equation: CH3CH2CHO + 2[Ag(NH3)2]+ + 3OH⁻ → CH3CH2COO⁻ + 2Ag + 4NH3 + 2H2O.

M1 — correct identification: X = propanal if mirror forms in X; Y = propanone (no reaction).

A1 — link to mechanism: the aldehyde is oxidised to the carboxylate; Ag+ is reduced to Ag, completing the redox half-equations. Ketones cannot be oxidised under these conditions because no C–H bond is attached to the carbonyl carbon.

(b) Step 1 — identify the nucleophile and product.

CN⁻ (from KCN) attacks the δ+ carbonyl carbon of propanone. The resulting alkoxide is protonated by HCN to give the hydroxynitrile 2-hydroxy-2-methylpropanenitrile, (CH3)2C(OH)CN.

M1 — correct product structure and name.

A1 — mechanism named as nucleophilic addition.

Total: 8 marks (M4 A4).

Specimen question modelled on the Edexcel 9CH0 Paper 2 format

Question (8 marks): Compound Z has molecular formula C4H8O. Its IR spectrum shows a strong absorption at 1715 cm⁻¹. Compound Z does not produce a silver mirror with Tollens' reagent but does form an orange precipitate with 2,4-DNP.

(a) Deduce the functional group present in Z and identify Z. Justify your reasoning. (3)

(b) Z is reduced using NaBH4 in aqueous ethanol. Draw the mechanism for this reduction, including curly arrows. (4)

Mark scheme decomposition by AO:

Part	AO	Marks	Earned by
(a)	AO2	1	Identifying C=O at 1715 cm⁻¹ — within ketone/aldehyde range.
(a)	AO2	1	Tollens' negative + DNP positive → ketone (not aldehyde).
(a)	AO1	1	Naming Z as butanone (CH3COCH2CH3).
(b)	AO1	1	Curly arrow from H⁻ (of [BH4]⁻) to δ+ carbonyl C.
(b)	AO1	1	Curly arrow from C=O π bond to oxygen.
(b)	AO2	1	Tetrahedral alkoxide intermediate drawn with negative charge on O.
(b)	AO2	1	Protonation of alkoxide by H2O/H+ to give butan-2-ol.
(c)	AO3	1	NaBH4 is mild/selective/works in water; LiAlH4 reacts violently with H2O so requires dry ether (any one valid reason).

Total: 8 marks split AO1 = 3, AO2 = 4, AO3 = 1.

Synoptic links

Connects to:

Topic 6 — Alcohol oxidation: primary alcohols oxidise to aldehydes (then carboxylic acids); secondary alcohols oxidise to ketones. Reduction of carbonyls by NaBH4 reverses this — the redox cycle is symmetrical and Edexcel exploits this synoptically.
Topic 17 — Carboxylic acids: the further oxidation of an aldehyde gives a carboxylic acid, linking nucleophilic addition chemistry (Topic 16) to acid–base chemistry (Topic 17). Hydrolysis of the hydroxynitrile produced from RCHO + HCN gives a 2-hydroxycarboxylic acid (e.g. lactic acid from ethanal).
Topic 12 — Acid–base equilibria: Tollens' relies on the basic ammoniacal solution maintaining [Ag(NH3)2]+; the equilibrium between Ag+ + 2NH3 ⇌ [Ag(NH3)2]+ uses ligand-exchange reasoning from Topic 15 (transition metals) and equilibrium constants (Topic 11).
Topic 18 — Spectroscopy and chromatography: identification of unknown carbonyls relies on IR (C=O ≈ 1680–1750 cm⁻¹), proton NMR (no peak at δ 9–10 ppm rules out aldehyde-CHO) and mass spectrometry (loss of CHO = 29; loss of COCH3 = 43). The 2,4-DNP derivative has a sharp melting point that historically identified the parent carbonyl.
Topic 5 — Kinetics: addition of HCN is base-catalysed (KCN provides CN⁻); the rate law is first order in carbonyl and first order in CN⁻, with the slow step being the nucleophilic attack — this connects mechanism to rate equation (Topic 11).

Mark-scheme literacy

Carbonyl questions on 9CH0 split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge)	35–45%	Recall of reagents, conditions, observations, naming
AO2 (application)	35–45%	Selecting the correct test; deducing structure from spectroscopic data; predicting products
AO3 (analysis/evaluation)	15–25%	Justifying choice of reducing agent; explaining differences in reactivity; designing a synthesis

Examiner-rewarded phrasing: "the C=O carbon is δ+ because oxygen is more electronegative than carbon"; "CN⁻ is the active nucleophile because it has a lone pair on carbon and a negative charge"; "the alkoxide intermediate is then protonated by HCN/water to give the alcohol/hydroxynitrile"; "warm gently in a water bath" (not "heat strongly", which decomposes Tollens' to explosive silver fulminate). Phrases that lose marks: "HCN attacks the carbonyl" (incorrect — CN⁻ attacks); "the silver mirror forms in solution" (no — on the glass); "ketones give a positive Tollens' test if heated long enough" (false); "NaBH4 reduces alkenes" (false — selectivity is the whole point).

A specific Edexcel pattern: questions that ask "give the mechanism" require curly arrows from lone pair/bond to electrophile and an alkoxide intermediate explicitly drawn with the negative charge. Missing the negative charge on the intermediate loses the A1 even if the arrows are correct.

Grade-band model answers

3-mark question

Question: State the reagents and observations for a chemical test that would distinguish propanal from propanone.

Grade C response (~110 words):

Use Tollens' reagent. Add a few drops to each compound and warm in a water bath. With propanal (an aldehyde), a silver mirror will form on the inside of the test tube. With propanone (a ketone), there will be no reaction. This is because aldehydes are oxidised to carboxylate ions by Tollens' reagent, while ketones cannot be oxidised under these conditions.

Examiner commentary: Full marks (3/3). Reagent named correctly; observations stated for both liquids (this is essential — only stating the positive observation usually loses a mark); brief justification given. The phrase "warm in a water bath" rather than "heat" is a discriminator — strong heating decomposes Tollens' to explosive silver fulminate, a safety point examiners reward.

Grade A response (~150 words):*

Add Tollens' reagent (silver(I) in aqueous ammonia, [Ag(NH3)2]+) to fresh portions of each liquid in clean test tubes and warm gently in a water bath at approximately 60 °C. With propanal, a silver mirror will deposit on the inner wall of the test tube as the aldehyde is oxidised to the propanoate ion (CH3CH2COO⁻) and Ag+ is reduced to metallic Ag. With propanone, no observable change occurs because ketones cannot be oxidised without breaking a C–C bond, which Tollens' cannot achieve. Equation: CH3CH2CHO + 2[Ag(NH3)2]+ + 3OH⁻ → CH3CH2COO⁻ + 2Ag + 4NH3 + 2H2O.

Examiner commentary: Full marks (3/3). The equation is unnecessary for 3 marks but signals confidence and adds redundancy if any other line is mismarked. Naming the temperature shows lab awareness; the explanation "without breaking a C–C bond" is the precise mechanistic justification examiners reward.

6-mark question

Question: Compound P (C4H8O) does not react with Tollens' reagent. P reacts with NaBH4 in aqueous ethanol to give compound Q. Q reacts with hot acidified potassium dichromate(VI) to give P unchanged on warming. Identify P and Q, and explain the redox relationship between them.

Grade B response (~210 words):

P is butanone, CH3COCH2CH3, because it has formula C4H8O, contains a carbonyl group (C=O) but is not oxidised by Tollens', so it must be a ketone. Q is butan-2-ol, CH3CH(OH)CH2CH3, because NaBH4 reduces ketones to secondary alcohols. When Q is oxidised by hot acidified K2Cr2O7, it gives back the ketone P. The orange dichromate turns green as Cr2O7²⁻ is reduced to Cr3+. The reaction is reversible in the sense that reduction (NaBH4) and oxidation (K2Cr2O7) interconvert ketone and secondary alcohol.

Examiner commentary: 5/6. Correctly identifies P, Q, names both, and notes the colour change. Loses one mark for not stating that secondary alcohols cannot be oxidised further (whereas primary alcohols would oxidise on past the aldehyde to the carboxylic acid). The logic is sound but underdeveloped — the question asks to explain the redox relationship, which means more than naming reagents.

Grade A response (~280 words):*

P has molecular formula C4H8O and contains a carbonyl group (C=O — confirmed by its potential to react with 2,4-DNP, though that test is not stated here). The negative Tollens' result rules out an aldehyde, so P is a ketone. With four carbons there are two possible ketone isomers: butanone (CH3COCH2CH3) is the only one with C4H8O — methylpropanone is not possible because the carbonyl would be at the end (which would be an aldehyde). Therefore P = butanone.

Reduction of P by NaBH4 (mild, aqueous) proceeds via nucleophilic addition: H⁻ from [BH4]⁻ attacks the δ+ carbonyl carbon, the C=O π bond breaks heterolytically, and the resulting alkoxide is protonated by water. The product is Q = butan-2-ol, a secondary alcohol.

Oxidation of Q by hot acidified K2Cr2O7 regenerates P: the secondary alcohol is oxidised back to the ketone. The orange dichromate (Cr2O7²⁻) is reduced to green Cr3+. Crucially, secondary alcohols stop at the ketone stage — they cannot be oxidised further without C–C cleavage (in contrast to primary alcohols, which oxidise through aldehyde to carboxylic acid).

The redox relationship is therefore symmetric: ketone ⇌ secondary alcohol, interconverted by NaBH4 (forward, reduction) and K2Cr2O7/H+ (reverse, oxidation). The overall change in oxidation number of the central carbon is +2 (alcohol C is at OS −1; ketone C is at OS +1).

Examiner commentary: Full marks (6/6). Explicit ruling-out of alternative isomers; mechanistic detail tied to the redox argument; oxidation-state arithmetic supplied as evidence. This is examination craft.

9-mark question

Question: Compound R (C5H10O) has the following properties: it forms an orange precipitate with 2,4-DNP, gives a silver mirror with Tollens' reagent, and on reduction with NaBH4 forms compound S which can be dehydrated to give compound T (C5H10). T shows two peaks in its proton NMR spectrum.

(a) Identify R, S and T. (6) (b) Draw the mechanism for the reduction of R to S. (3)

Grade A response (~360 words):*

(a) The orange 2,4-DNP precipitate confirms a C=O group; the silver mirror with Tollens' confirms R is an aldehyde. So R has the form CnH(2n+1)CHO. With 5 carbons total, R is a C5 aldehyde.

S forms by NaBH4 reduction, so S is the corresponding primary alcohol C4H9CH2OH. T is the alkene formed by dehydration of S (loss of H2O), giving C5H10. Crucially, T shows only two distinct proton environments — this requires high symmetry.

The candidates: pentanal → pentan-1-ol → pent-1-ene shows 5 NMR environments — too many. The symmetric alternative is 3,3-dimethylpropanal? No — try 2,2-dimethylpropanal (pivaldehyde, (CH3)3CCHO). Reduction gives 2,2-dimethylpropan-1-ol, (CH3)3CCH2OH. Dehydration would normally fail (no β-hydrogen on the quaternary carbon), so this fails the dehydration test.

Try 3-methylbutanal (CH3)2CHCH2CHO. Reduction gives 3-methylbutan-1-ol (CH3)2CHCH2CH2OH. Dehydration gives 3-methylbut-1-ene with four environments — too many.

The only C5 aldehyde whose alcohol dehydrates to a C5 alkene with 2 environments is one giving 2,3-dimethylbut-2-ene, (CH3)2C=C(CH3)2, which has only methyl protons (12 equivalent H, one environment) — so we need one environment, not two. Two NMR environments suggests cyclopent-1-ene or 2-methylbut-2-ene (CH3)2C=CHCH3 (2 environments: vinyl H and methyl Hs). The latter comes from dehydration of 3-methylbutan-2-ol — but that's a secondary alcohol, requiring a ketone precursor, contradicting the Tollens' result.

A clean candidate is 2,2-dimethylpropanal giving neopentanol; but neopentanol cannot dehydrate without rearrangement. Therefore the question is consistent only if rearrangement is allowed, giving 2-methyl-2-butene as T.

So: R = 2,2-dimethylpropanal, S = 2,2-dimethylpropan-1-ol, T = 2-methylbut-2-ene (after carbocation rearrangement during E1 dehydration).

Examiner commentary: 7/9 — the candidate works through plausible structures and arrives at a defensible answer with explicit reasoning, but loses marks because the rearrangement argument is hand-waved (full credit would require an explicit hydride/methyl shift mechanism). For (b), expect the standard NaBH4 mechanism with H⁻ attack, alkoxide intermediate and protonation step (3/3).

A-Level-depth misconceptions

The errors that distinguish A from A* on carbonyl questions:

"HCN is the nucleophile." Wrong — CN⁻ is. HCN is a covalent, weakly polar molecule; the cyanide ion (with a lone pair on C and a negative charge) is what attacks. Curly arrows must originate from CN⁻.
Drawing the dipole on C=O the wrong way round. Oxygen is more electronegative (3.4) than carbon (2.5), so δ+ is on C and δ− is on O. Reversing this is a fatal mechanistic error.
Confusing nucleophilic addition with nucleophilic substitution. No leaving group leaves — the nucleophile adds. Substitution requires a leaving group (e.g. halide).
Stating that ketones are reduced to primary alcohols. Aldehydes → primary; ketones → secondary. The pattern must be memorised.
Claiming NaBH4 reduces carboxylic acids. It does not — only LiAlH4 (in dry ether) does. NaBH4 is selective for aldehydes and ketones; this selectivity is the point.
Forgetting that 2,4-DNP cannot distinguish aldehyde from ketone. Both give the orange precipitate. 2,4-DNP confirms the carbonyl group is present; Tollens'/Fehling's distinguishes which type.
Treating Tollens' silver mirror as "Ag2O precipitate". It is metallic Ag(0), deposited as a thin film on glass. Writing Ag2O loses the redox-half-equation mark.

Common errors and mark-loss patterns

Three patterns repeatedly cost candidates marks on Paper 2 carbonyl questions.

Missing the negative charge on the alkoxide intermediate. When drawing the mechanism for nucleophilic addition (HCN or NaBH4), the tetrahedral intermediate after C=O bond breakage must show O⁻. Examiners deduct marks even if all the curly arrows are correct.
Incorrect curly-arrow origin. Arrows must start from a lone pair, π-bond or σ-bond — not from atoms. Drawing an arrow starting on H of HCN (instead of from the lone pair on the C of CN⁻) is a categorical error.
Vague observations. "It changes colour" loses marks; "the orange dichromate turns green as Cr3+ ions form" earns full marks. Specificity counts.

Going further — university and academic signposting

Carbonyl chemistry connects directly to several undergraduate trajectories:

Pericyclic and stereoselective synthesis (Year 2/3 organic chemistry): the Mukaiyama aldol, Evans-asymmetric aldol, and modern catalytic enantioselective reductions (Corey–Bakshi–Shibata) all hinge on selective addition to C=O. The CIP descriptor of the new stereocentre formed during NaBH4 reduction is determined statistically (50:50 enantiomers) unless a chiral reagent is used.
Spectroscopy (Year 1 inorganic/organic): the C=O stretch in IR is a quantum-mechanically calculable observable (Hooke's law approximation). Conjugation lowers the wavenumber (esters > ketones > amides) by reducing the force constant.
Bioorganic chemistry (Year 2/3): glucose, fructose, ribose are all polyhydroxy aldehydes (aldoses) or ketones (ketoses). The Tollens' test classically identifies "reducing sugars" — sugars that can act as aldehydes in their open-chain form.
Carbonyl photochemistry (Year 3): UV irradiation of ketones produces n→π* excited states that undergo Norrish Type I and II reactions — a foundational topic in photochemistry research.

Oxbridge interview prompt: "The C=O bond in formaldehyde absorbs at 1745 cm⁻¹; in acetone it is at 1715 cm⁻¹; in ethyl acetate at 1735 cm⁻¹. Rationalise these shifts using bonding arguments."

Required practical reference (~150 words)

This lesson connects directly to CP7: The preparation of a pure organic solid and a test of its purity, in which a 2,4-dinitrophenylhydrazone derivative of a carbonyl compound is prepared, recrystallised and its melting point measured to identify the parent carbonyl. The key skills are: (i) measuring carbonyl and 2,4-DNP solution accurately; (ii) collecting the precipitate by filtration under reduced pressure (Buchner funnel); (iii) recrystallising from ethanol/water to remove impurities; and (iv) determining a sharp melting point and comparing to a literature table of derivative melting points. The melting point of a pure 2,4-DNP derivative is a fingerprint for the parent carbonyl. This practical also indirectly underpins CP6 (recrystallisation skill) and connects to CP14 (aspirin synthesis) via the broader skill of solid organic preparation. Examiners frequently ask about the rationale for recrystallisation, the meaning of a "sharp" melting point and the identification of impurities from melting-point depression.

Edexcel A-Level alignment footer

This content is aligned with the Pearson Edexcel GCE A Level Chemistry (9CH0) specification, Paper 2 — Core Organic and Physical Chemistry, and Paper 3 — General and Practical Principles in Chemistry. Topic 16: Chemistry of carbonyl compounds. For the most accurate and up-to-date information, please refer to the official Pearson Edexcel specification document.

Visual summary

graph TD
    A["Unknown C=O compound<br/>RCHO or RCOR'"] --> B{"2,4-DNP test"}
    B -->|"Orange ppt"| C["Carbonyl confirmed<br/>(aldehyde or ketone)"]
    B -->|"No ppt"| D["Not a carbonyl"]
    C --> E{"Tollens' reagent<br/>warm gently"}
    E -->|"Silver mirror"| F["ALDEHYDE<br/>RCHO"]
    E -->|"No change"| G["KETONE<br/>RCOR'"]
    F --> H["Oxidise:<br/>K2Cr2O7/H+ reflux<br/>→ RCOOH"]
    F --> I["Reduce: NaBH4<br/>→ primary alcohol<br/>RCH2OH"]
    G --> J["Reduce: NaBH4<br/>→ secondary alcohol<br/>RCH(OH)R'"]
    F --> K["+ HCN/KCN<br/>→ hydroxynitrile<br/>(chain +1C)"]
    G --> K
    K --> L["Hydrolysis:<br/>→ 2-OH carboxylic acid"]

    style F fill:#27ae60,color:#fff
    style G fill:#3498db,color:#fff
    style L fill:#e74c3c,color:#fff