Aldehydes and ketones

1. Overview

Aldehydes and ketones both contain the carbonyl group, C=O. This section covers making them by oxidising alcohols, reducing them back to alcohols, and adding hydrogen cyanide across the C=O bond by nucleophilic addition, with its mechanism. It also covers the test-tube reactions that detect a carbonyl group, tell an aldehyde from a ketone and reveal a CH₃CO- group: the tools for identifying an unknown compound.

Key Definitions

• Carbonyl group: the C=O group, a carbon atom joined to an oxygen atom by a double bond made of one σ bond and one π bond.
• Aldehyde: a carbonyl compound in which the carbonyl carbon is bonded to at least one hydrogen atom, giving a -CHO group at the end of a chain, for example ethanal, CH₃CHO.
• Ketone: a carbonyl compound in which the carbonyl carbon is bonded to two carbon atoms, so the C=O is inside the chain, for example propanone, CH₃COCH₃.
• Nucleophile: an electron-pair donor; a species with a lone pair that it uses to form a new covalent bond with an electron-deficient (δ+) atom, for example the cyanide ion, CN⁻.
• Nucleophilic addition: a reaction in which a nucleophile attacks the δ+ carbon atom of a C=O bond, the π bond breaks, and two groups add across the double bond to give a single product.
• Hydroxynitrile: a compound that contains both a hydroxy group, -OH, and a nitrile group, -C≡N; those made from aldehydes and ketones have both groups on the same carbon atom.
• 2,4-DNPH reagent: an acidified solution of 2,4-dinitrophenylhydrazine that gives a yellow, orange or red precipitate with aldehydes and ketones, so it detects their carbonyl group.
• Tollens' reagent: an alkaline solution of the diamminesilver(I) ion, [Ag(NH₃)₂]⁺; on warming, an aldehyde reduces it to silver (a silver mirror or black precipitate), but a ketone does not.
• Fehling's reagent: a blue alkaline solution of complexed copper(II) ions; on warming, an aldehyde reduces it to an orange-red precipitate of copper(I) oxide, Cu₂O, but a ketone does not.
• Tri-iodomethane (iodoform) test: warming a compound with aqueous iodine and aqueous sodium hydroxide; a pale yellow precipitate of CHI₃ shows a CH₃CO- group in an aldehyde or ketone (or CH₃CH(OH)- in an alcohol).

Content

Aldehydes, ketones and the C=O bond

What is joined to the carbonyl carbon decides the family:

	Aldehyde	Ketone
Carbonyl C bonded to	at least one H atom	two C atoms
Written as	-CHO, at a chain end	-CO-, inside the chain
Name ends in	-al: propanal, CH₃CH₂CHO	-one: propanone, CH₃COCH₃

• The aldehyde carbon is always carbon 1, so it needs no number: butanal, not butan-1-al.
• From five carbons up, a ketone needs a number. Pentan-2-one, CH₃COCH₂CH₂CH₃, and pentan-3-one, CH₃CH₂COCH₂CH₃, are different compounds.
• Write an aldehyde group as -CHO, never -COH, which looks like an alcohol.

Polarity. Oxygen is more electronegative than carbon (3.5 against 2.5 in the Data Booklet), so the bonding electrons in C=O are pulled towards oxygen: the carbon carries a partial positive charge, δ+, and the oxygen a partial negative charge, δ−.

Shape. The carbonyl carbon and the three atoms bonded to it lie in one plane (trigonal planar), with bond angles of about 120°.

So the δ+ carbon attracts nucleophiles, which add across the C=O (the electron-rich C=C of an alkene attracts electrophiles instead, topic 14.2). Because the group is flat, a nucleophile can approach from either side of the plane.

Making aldehydes and ketones by oxidising alcohols

Reagent: acidified potassium dichromate(VI), K₂Cr₂O₇ with dilute sulfuric acid, or acidified potassium manganate(VII), KMnO₄. Conditions: heat, and distil the product out of the mixture.

Alcohol	Product (distil)	K₂Cr₂O₇ / H⁺ colour
primary, RCH₂OH	aldehyde, RCHO	orange to green
secondary, RCH(OH)R′	ketone, RCOR′	orange to green
tertiary	no reaction	stays orange

Here R and R′ are alkyl groups. The green colour is Cr³⁺(aq), formed as dichromate(VI) is reduced. With acidified KMnO₄ the change is purple to colourless.

In organic redox equations, [O] stands for one oxygen atom from the oxidising agent:

Propan-1-ol to propanal:

CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O

Propan-2-ol to propanone:

CH₃CH(OH)CH₃ + [O] → CH₃COCH₃ + H₂O

Why distil to get an aldehyde. An aldehyde can be oxidised further, to a carboxylic acid. It has no O-H group, so its molecules cannot hydrogen bond to one another, and it boils lower than the alcohol it came from. Distilling, with the condenser sloping down to a receiver, removes the aldehyde as it forms, before it meets more oxidising agent. Under reflux, with the condenser vertical above the flask, the aldehyde would run back and be oxidised to the carboxylic acid (topic 18.1).

Ketones resist further oxidation by these reagents, so distillation simply collects them. Tertiary alcohols have no hydrogen atom on the carbon carrying the -OH, so they are not oxidised.

Reducing aldehydes and ketones to alcohols

Reduction reverses the oxidation: aldehydes give primary alcohols and ketones give secondary alcohols.

Use NaBH₄ (sodium borohydride) in water or ethanol, or LiAlH₄ (lithium aluminium hydride) in dry ether, since LiAlH₄ reacts violently with water. [H] stands for one hydrogen atom from the reducing agent; each C=O needs two:

Propanal to propan-1-ol:

CH₃CH₂CHO + 2[H] → CH₃CH₂CH₂OH

Propanone to propan-2-ol:

CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃

• They are reducing agents, used up in the reaction, not catalysts.
• Neither reduces the C=C bond of an alkene, so in a molecule with both groups only the C=O is reduced.
• LiAlH₄ is stronger: it also reduces carboxylic acids (topic 18.1); NaBH₄ does not.
• Both act as a source of hydride ions, H⁻, which add to the δ+ carbon from either side of the flat C=O: another nucleophilic addition.

Adding hydrogen cyanide: making hydroxynitriles

Reagents and conditions: HCN, with KCN as a catalyst, and heat. The product is a hydroxynitrile, with -OH and -C≡N on the carbon atom that was the carbonyl carbon.

Ethanal gives 2-hydroxypropanenitrile:

CH₃CHO + HCN → CH₃CH(OH)CN

Propanone gives 2-hydroxy-2-methylpropanenitrile:

CH₃COCH₃ + HCN → (CH₃)₂C(OH)CN

• The chain grows by one carbon, because the new bond joins the carbonyl carbon to the carbon of CN. The nitrile carbon is carbon 1 in the name, so ethanal (2 carbons) gives a propanenitrile.
• Why KCN? HCN is a weak acid, so it barely ionises and gives very few CN⁻ ions. KCN is ionic and supplies CN⁻, the nucleophile. CN⁻ is re-formed in the last step, so KCN is a catalyst.
• Safety: HCN is extremely toxic, so use a fume cupboard.

Optical isomers. In CH₃CH(OH)CN, the carbon joined to -OH carries four different groups (H, CH₃, OH and CN), so it is a chiral centre. CN⁻ is equally likely to attack the flat C=O from either side, so the two optical isomers form in equal amounts. The product from propanone has two CH₃ groups on that carbon, so it has no chiral centre.

The mechanism of nucleophilic addition

The reaction has two steps. First draw the C=O with δ+ on carbon and δ− on oxygen, and the cyanide ion as ⁻:C≡N, with the lone pair and the negative charge on the carbon atom. Each curly arrow moves an electron pair; it starts at a lone pair or a bond, never at a minus sign or an atom symbol:

Arrow	Starts at	Ends at
1 (step 1)	lone pair on C of CN⁻	δ+ carbonyl carbon
2 (step 1)	π bond of C=O	the O atom
3 (step 2)	lone pair on O⁻	H atom of H-CN
4 (step 2)	the H-C bond in HCN	the C atom of CN

Step 1: nucleophilic attack. Arrows 1 and 2 happen together: as the new C-C bond forms, the π bond breaks and its pair moves onto oxygen. The intermediate is an anion: the old carbonyl carbon has four single bonds, and the oxygen has a negative charge and three lone pairs (draw at least one).

Step 2: protonation. The O⁻ takes H⁺ from HCN (drawn as Hδ+-CNδ−), and the H-C bonding pair stays with the CN, re-forming CN⁻. You may also see this step drawn with H⁺ or a water molecule instead of HCN. The arrow still runs from a lone pair on O⁻ to the H, never from H⁺ to O⁻.

Worked example 1: a two-step synthesis

Question: Suggest how 2-hydroxyhexanenitrile, CH₃CH₂CH₂CH₂CH(OH)CN, can be made in two steps from an alcohol. Give the reagents and conditions and an equation for each step, and explain why the conditions in the first step matter.

Step 1: work backwards. The carbon carrying both -OH and -CN was the carbonyl carbon, and the CN was added by HCN. Taking HCN away leaves CH₃CH₂CH₂CH₂CHO, pentanal. Its C=O is at the end of the chain, so it is an aldehyde, made from the primary alcohol with the same chain: pentan-1-ol, CH₃CH₂CH₂CH₂CH₂OH. (Check the carbons: 5 in the alcohol, 5 in the aldehyde, 6 in the hydroxynitrile.)

Step 2: first reaction, oxidation. Heat pentan-1-ol with acidified potassium dichromate(VI) (K₂Cr₂O₇ and dilute H₂SO₄) and distil off the pentanal as it forms. The mixture turns from orange to green.

CH₃CH₂CH₂CH₂CH₂OH + [O] → CH₃CH₂CH₂CH₂CHO + H₂O

Step 3: second reaction, nucleophilic addition. Heat the pentanal with HCN and a little KCN as catalyst.

CH₃CH₂CH₂CH₂CHO + HCN → CH₃CH₂CH₂CH₂CH(OH)CN

Step 4: why distil. Pentanal boils lower than pentan-1-ol, so it leaves the flask before it can be oxidised further. Under reflux it would become pentanoic acid, which does not add HCN, so the second step would fail:

CH₃CH₂CH₂CH₂CHO + [O] → CH₃CH₂CH₂CH₂COOH

Answer: (1) Heat pentan-1-ol with acidified K₂Cr₂O₇, distilling off the pentanal as it forms so it is not oxidised to pentanoic acid. (2) Heat the pentanal with HCN and KCN as catalyst. The product's carbon 2 is a chiral centre, so it forms as an equal mixture of two optical isomers.

Worked example 2: the mechanism with a ketone

Question: Describe the mechanism of the reaction between pentan-3-one, CH₃CH₂COCH₂CH₃, and HCN in the presence of KCN. Name the product and state whether it has a chiral centre.

Step 1: attack by CN⁻. HCN adds across C=O and the attacking species is CN⁻, so this is nucleophilic addition. Draw Cδ+=Oδ− and ⁻:C≡N, with the lone pair and charge on carbon. Arrow 1 runs from the lone pair on the carbon of CN⁻ to the carbonyl carbon; arrow 2 runs from the π bond of C=O to the oxygen. The intermediate is the anion (CH₃CH₂)₂C(O⁻)CN, with three lone pairs on the O⁻.

Step 2: protonation. Arrow 3 runs from a lone pair on O⁻ to the H of Hδ+-CNδ−; arrow 4 runs from the H-C bond to the carbon of CN, re-forming CN⁻.

Step 3: the product. (CH₃CH₂)₂C(OH)CN has a four-carbon chain, counting the nitrile carbon as carbon 1, with -OH and an ethyl group on carbon 2: 2-ethyl-2-hydroxybutanenitrile. Carbon 2 carries two identical ethyl groups, so it is not a chiral centre.

Answer: Nucleophilic addition, with the four arrows in Steps 1 and 2. The product, 2-ethyl-2-hydroxybutanenitrile, has no chiral centre.

Testing for a carbonyl group: 2,4-DNPH

Method: add a few drops of the unknown (or a solution of it, if it is a solid) to 2,4-DNPH reagent and shake.

• Positive: a yellow, orange or red precipitate, usually orange crystals. An aldehyde or a ketone is present.
• Negative: no precipitate. Carboxylic acids and esters contain C=O but give none; nor do alcohols.
• What it cannot show: whether the compound is an aldehyde or a ketone, because both give the precipitate.

The reaction is a condensation: the carbonyl compound and 2,4-DNPH join and a water molecule is lost. The precipitate is an organic solid (a 2,4-dinitrophenylhydrazone), not an inorganic ion like the green Cr³⁺(aq) from acidified dichromate(VI).

Aldehyde or ketone? Ease of oxidation

An aldehyde has a hydrogen atom on its carbonyl carbon, so it is easily oxidised to a carboxylic acid:

RCHO + [O] → RCOOH

A ketone has no such hydrogen atom; oxidising it would mean breaking a C-C bond, which mild oxidising agents cannot do. Warm each test in a hot water bath, not over a flame, because the compounds are flammable.

Reagent (warm)	Aldehyde	Ketone
Tollens'	silver mirror, or black precipitate	no change: stays colourless
Fehling's	blue solution gives an orange-red precipitate	no change: stays blue
K₂Cr₂O₇ / H⁺	orange to green	no change: stays orange
KMnO₄ / H⁺	purple to colourless	no change: stays purple

In each positive result the reagent is the oxidising agent, so it is itself reduced:

Ag⁺ + e⁻ → Ag (Tollens': silver(I) to silver)

Cu²⁺ + e⁻ → Cu⁺ (Fehling's: copper(II) to copper(I), which precipitates as Cu₂O)

Both reagents are alkaline, so the acid formed ends up as its anion, RCOO⁻.

Making Tollens' reagent. Add aqueous ammonia to aqueous silver nitrate until the brown precipitate that forms first just dissolves. A brown solid in the test itself usually means too little ammonia was added, not a positive result.

Choose the reagent with care. Acidified dichromate(VI) and manganate(VII) also oxidise primary and secondary alcohols (and manganate(VII) reacts with alkenes), so their colour change means an aldehyde only once 2,4-DNPH has shown a carbonyl compound. Tollens' and Fehling's reagents leave alcohols alone.

The tri-iodomethane test for a CH₃CO- group

Method: add aqueous iodine (brown) to a few drops of the compound. Then add aqueous sodium hydroxide slowly, drop by drop, until the brown colour just fades. Warm gently and leave for a few minutes.

• Positive: a pale yellow precipitate of tri-iodomethane, CHI₃, which has an antiseptic smell.
• What it shows: an aldehyde or ketone with the structure CH₃CO-R, where R is H or a carbon group.

What happens. The CH₃ next to the C=O ends up as CHI₃, and the rest of the molecule becomes the carboxylate ion RCO₂⁻, with one carbon atom fewer than the starting compound. Iodine is reduced to iodide ions.

CH₃COR + 3I₂ + 4OH⁻ → CHI₃ + RCO₂⁻ + 3I⁻ + 3H₂O

For example, pentan-2-one, CH₃COCH₂CH₂CH₃, gives the butanoate ion, CH₃CH₂CH₂CO₂⁻.

Compound	CH₃CO-R?	Precipitate?
ethanal, CH₃CHO	yes (R = H)	yes
propanone, CH₃COCH₃	yes	yes
propanal, CH₃CH₂CHO	no	no
pentan-3-one	no	no
ethanoic acid, CH₃COOH	no (R is OH)	no

• Ethanal is the only aldehyde with a single C=O group that gives a yellow precipitate.
• Alcohols with a CH₃CH(OH)- group, such as ethanol and propan-2-ol, also give it, because the alkaline iodine first oxidises them to CH₃CO- compounds (topic 16.1). So test with 2,4-DNPH first to know you have an aldehyde or ketone.

Worked example 3: identifying an unknown carbonyl compound

Question: Compound W has an unbranched chain with no C=C bonds and no ring. Each molecule contains carbon, hydrogen and a single oxygen atom, and its $M_{\text{r}}$ is 114.0. The results of three tests are shown.

Test on W	Observation
2,4-DNPH reagent	orange precipitate
Tollens' reagent, warmed	no change: stays colourless
aqueous I₂, then NaOH(aq), warmed	pale yellow precipitate

(a) Identify W. (b) Write an equation for the last test and name the organic ion formed. (c) W has two other unbranched ketone isomers. Name them and explain why these three tests cannot tell them apart.

Step 1: 2,4-DNPH and the formula. The orange precipitate shows that W is an aldehyde or ketone. With one C=O, no C=C and no ring, its general formula is CₙH₂ₙO. With $A_{\text{r}}$ values C = 12.0, H = 1.0 and O = 16.0:

$$M_{\text{r}} = 12.0n + 2n(1.0) + 16.0 = 14.0n + 16.0$$

$$14.0n + 16.0 = 114.0 \quad\Rightarrow\quad n = 7$$

So W is C₇H₁₄O.

Step 2: the Tollens' result. No change, so W is not easily oxidised: it is a ketone, not heptanal. The unbranched ketones of C₇H₁₄O are heptan-2-one, heptan-3-one and heptan-4-one.

Step 3: the iodine result. A pale yellow precipitate shows a CH₃CO- group. Only heptan-2-one, CH₃COCH₂CH₂CH₂CH₂CH₃, has one.

Step 4: the equation. Here R is CH₃(CH₂)₄-, so the ion is CH₃(CH₂)₄CO₂⁻, with six carbons, one fewer than W.

CH₃CO(CH₂)₄CH₃ + 3I₂ + 4OH⁻ → CHI₃ + CH₃(CH₂)₄CO₂⁻ + 3I⁻ + 3H₂O

Step 5: the other two ketones. Heptan-3-one and heptan-4-one each give an orange precipitate, no change with Tollens' reagent and no yellow precipitate (neither has a CH₃CO- group), so their results are identical.

Answer: (a) W is heptan-2-one, CH₃COCH₂CH₂CH₂CH₂CH₃. (b) Equation as in Step 4; the precipitate is CHI₃ and the organic ion is hexanoate, CH₃(CH₂)₄CO₂⁻. (c) Heptan-3-one and heptan-4-one, ketones with no CH₃CO- group.

Key Equations

State symbols are left out of these organic equations. R and R′ are alkyl groups; for an aldehyde, R′ is H.

Oxidising alcohols (acidified K₂Cr₂O₇ or KMnO₄, heat, distil; [O] is one oxygen atom from the oxidising agent):

RCH₂OH + [O] → RCHO + H₂O

RCH(OH)R′ + [O] → RCOR′ + H₂O

Reducing aldehydes and ketones (NaBH₄ or LiAlH₄; [H] is one hydrogen atom from the reducing agent):

RCHO + 2[H] → RCH₂OH

RCOR′ + 2[H] → RCH(OH)R′

Adding HCN (KCN as catalyst, heat; nucleophilic addition):

RCOR′ + HCN → RC(OH)(CN)R′

Oxidising an aldehyde (Tollens', Fehling's or acidified dichromate(VI); in the alkaline reagents the product is RCOO⁻):

RCHO + [O] → RCOOH

Tri-iodomethane reaction (R is H or an alkyl group; the ion RCO₂⁻ has one carbon fewer than CH₃COR):

CH₃COR + 3I₂ + 4OH⁻ → CHI₃ + RCO₂⁻ + 3I⁻ + 3H₂O

Common Mistakes to Avoid

• Mistake: starting the first curly arrow at the nitrogen of CN⁻ or at its minus sign, or drawing the last arrow from H⁺ to O⁻. Fix: start every arrow at electrons: the first at the lone pair on the carbon of CN⁻ (so the new bond is C-C, as in CH₃CH(OH)CN), the last at a lone pair on O⁻, pointing to the H of HCN or H⁺.
• Mistake: giving the intermediate a positive charge, or leaving out the δ+ and δ− on C=O. Fix: the π electrons move onto oxygen, so the intermediate is an anion. For propanone it is (CH₃)₂C(O⁻)CN.
• Mistake: mixing up the two reactions of cyanide. Fix: a carbonyl compound needs HCN with KCN as catalyst, and heat (addition, giving a hydroxynitrile). A halogenoalkane needs KCN in ethanol, and heat (substitution, topic 15.1): 1-chloropentane gives hexanenitrile, CH₃CH₂CH₂CH₂CH₂CN, with no -OH.
• Mistake: writing "turns orange" for 2,4-DNPH or "goes yellow" for the iodine test, or describing a negative result as a change ("the blue colour disappears"). Fix: both positive results are solids: write "orange precipitate" and "pale yellow precipitate" (the 2,4-DNPH reagent is already coloured). For a negative result write "no change" and what stays the same: propanone with Fehling's reagent "stays a blue solution, no precipitate".
• Mistake: swapping the Tollens' and Fehling's observations, or explaining them with "Fehling's only reacts with aldehydes". Fix: link each result to the metal ion the aldehyde reduces: Tollens' (silver ions) gives a silver mirror; Fehling's (Cu²⁺) gives orange-red Cu₂O. A ketone is not easily oxidised, so propanone gives neither.
• Mistake: thinking every ketone, or every aldehyde, gives a yellow precipitate with alkaline iodine. Fix: only compounds with CH₃CO- (or alcohols with CH₃CH(OH)-) do. Hexan-2-one and ethanal give it; heptan-4-one and hexanal do not.

Exam Tips

• State the reagents and conditions in full: "acidified potassium dichromate(VI), heat, distil the aldehyde off as it forms"; "HCN with KCN as catalyst, heat".
• Describe the mechanism by drawing it with every piece of evidence (dipoles, lone pairs, charges, curly arrows, the O⁻ intermediate) and naming it: nucleophilic addition.
• Construct equations with [O] or [H] by balancing them like any other: reduction needs 2[H]. Tri-iodomethane is CHI₃, not CH₃I (iodomethane).
• State observations for every compound, including the negative ones ("no change, stays orange"), and in a practical name both reagents for the tri-iodomethane test: aqueous iodine and aqueous sodium hydroxide.
• Deduce or identify by using the results in order (2,4-DNPH, then Tollens' or Fehling's, then alkaline iodine), and give a name or a structure, not just "a ketone". If the results fit more than one isomer, name each one.