Last updated: 24 August 2026
Aldehydes sit at an important crossroads in Organic Chemistry. They can be made by oxidising primary alcohols, converted back into primary alcohols by reduction, oxidised onwards to carboxylic acids, and attacked by nucleophiles at their carbonyl group. Once you understand why the C=O bond behaves as it does, many apparently separate reactions begin to fit into one connected picture.
We will use ethanal, CH₃CHO, as the main example. Rather than memorising each reaction independently, keep following the same carbonyl carbon and ask two questions: where is the electron-poor centre, and what change is happening to the carbon atom?
Try these five questions before reading the chapter. Use the structures and reaction schemes carefully — Organic Chemistry often becomes much easier once you learn to read what is happening to the molecule.
- 5 diagnostic questions
- Structures and reaction schemes where useful
- Instant explanation for every answer option
1. What is an aldehyde?
An aldehyde is an organic compound containing a carbonyl group at the end of a carbon chain. The carbonyl carbon is bonded to at least one hydrogen atom, so the functional group is written as –CHO.
The simplest aldehyde is methanal, HCHO. The main example in this chapter is ethanal, CH₃CHO, which contains two carbon atoms.
Recognition rule: an aldehyde has a terminal carbonyl group, R–CHO. A ketone instead has its carbonyl carbon bonded to two carbon-containing groups, R–CO–R′.
Structure and bonding of the carbonyl group
The carbonyl carbon is sp² hybridised. The three regions of σ bonding around this carbon are arranged approximately trigonal planar, with bond angles close to 120°.
The C=O double bond consists of one σ bond and one π bond. The π bond is formed by sideways overlap of unhybridised p orbitals on carbon and oxygen.
Why is the carbonyl group polar?
Oxygen is more electronegative than carbon, so it attracts the bonding electrons in the C=O bond more strongly. Electron density is therefore drawn towards oxygen.
This polarity is central to aldehyde chemistry. The carbonyl carbon is electron-deficient and therefore acts as an electrophilic centre. Nucleophiles, which contain an available electron pair, can attack this carbon.
Carbonyl carbon: carries δ⁺ and is attacked by nucleophiles.
Carbonyl oxygen: carries δ⁻ because electron density is drawn towards oxygen.
2. Preparing ethanal from ethanol
Aldehydes can be prepared by the controlled oxidation of primary alcohols. To prepare ethanal, ethanol is oxidised using acidified potassium dichromate(VI).
Why must ethanal be distilled off?
An aldehyde is itself readily oxidised. If ethanal remains in contact with excess acidified potassium dichromate(VI), it can be oxidised further to ethanoic acid.
The reaction mixture is therefore warmed and the ethanal is distilled off as it forms. Removing the aldehyde from the oxidising mixture limits further oxidation.
A useful way to think about this oxidation sequence is: primary alcohol → aldehyde → carboxylic acid. Distillation allows us to stop at the aldehyde stage; prolonged heating under reflux favours continued oxidation to the carboxylic acid.
Observation with acidified potassium dichromate(VI)
During oxidation, the orange dichromate(VI) ions, Cr₂O₇²⁻, are reduced to green Cr³⁺ ions. The solution therefore changes from orange to green.
Exam distinction: primary alcohol → aldehyde requires controlled oxidation and distillation. Primary alcohol → carboxylic acid normally uses excess oxidising agent and heating under reflux.
3. Nucleophilic addition of hydrogen cyanide
One of the most important reactions of aldehydes is nucleophilic addition. Hydrogen cyanide adds across the C=O bond to form a hydroxynitrile.
Although HCN appears in the overall equation, the species that actually attacks the carbonyl group is CN⁻. Understanding the different roles of HCN and CN⁻ makes the mechanism much easier to follow.
Why is a small amount of KCN or NaCN added?
HCN is a weak acid, so it ionises only slightly:
HCN alone therefore provides only a low concentration of CN⁻. Since CN⁻ is the nucleophile that attacks the δ⁺ carbonyl carbon, a small amount of KCN or NaCN is added to provide the initial CN⁻ ions needed to start the reaction.
KCN and NaCN are ionic salts and dissociate readily in solution. For example:
Only a small amount of cyanide salt is required because CN⁻ is regenerated later in the mechanism. The salt therefore supplies the initial nucleophile without being required in a stoichiometric amount.
Key idea: the cyanide salt provides the initial CN⁻ nucleophile, while regeneration of CN⁻ during the second step explains why only a small amount is needed.
Step 1: attack by CN⁻
The C=O bond is polar, so the carbonyl carbon is δ⁺. The lone pair on CN⁻ is attracted to this electron-deficient carbon.
CN⁻ donates its electron pair to the carbonyl carbon. At the same time, the π electrons of the C=O bond move onto oxygen.
The carbonyl carbon changes from trigonal planar to tetrahedral, and a negatively charged alkoxide intermediate is formed.
Step 2: protonation of the intermediate
The O⁻ ion in the intermediate is strongly basic. It accepts a proton directly from an HCN molecule.
It is better to describe this step as proton transfer from HCN rather than saying that HCN provides a large concentration of free H⁺ ions. HCN is a weak acid, so it is only slightly ionised.
When HCN donates the proton, CN⁻ is formed again. The nucleophile required for the first step is therefore regenerated.
What is consumed overall?
The two mechanism steps reveal an important distinction between HCN and CN⁻:
CN⁻ appears on both sides when the two steps are combined, so it cancels. The overall reaction is therefore:
HCN is consumed overall, whereas CN⁻ is regenerated. One molecule of HCN is used for each molecule of hydroxynitrile formed.
Follow one reaction cycle carefully. A CN⁻ ion attacks ethanal and becomes bonded to the carbonyl carbon. HCN then donates a proton to the O⁻ intermediate, leaving behind a new CN⁻ ion. The identity of the CN⁻ ion has changed, but the CN⁻ species has been regenerated and can continue the reaction cycle.
Why does HCN addition increase the carbon chain length?
The carbon atom of CN⁻ becomes bonded to the original carbonyl carbon. Nucleophilic addition of HCN therefore increases the carbon chain length by one carbon atom.
Ethanal contains two carbon atoms, whereas 2-hydroxypropanenitrile contains three. The additional carbon comes from the cyanide group.
Formation of a chiral product
The carbonyl carbon in ethanal is trigonal planar. CN⁻ can therefore approach from either side of the C=O group.
After addition, the former carbonyl carbon is bonded to CH₃, H, OH and CN — four different groups. It is therefore a chiral centre.
Attack from opposite sides gives a pair of enantiomers. In an achiral reaction environment, both are formed equally, producing a racemic mixture.
4. Reduction of aldehydes
Reduction reverses the oxidation relationship between a primary alcohol and an aldehyde. The C=O group of an aldehyde is reduced to give a primary alcohol.
Common reducing agents
| Reducing agent | Typical condition | Product from an aldehyde |
|---|---|---|
| NaBH₄ | Aqueous or alcoholic medium | Primary alcohol |
| LiAlH₄ | Dry ether, followed by hydrolysis | Primary alcohol |
| H₂ / Ni | Catalytic hydrogenation | Primary alcohol |
For ethanal, reduction gives ethanol. Notice that the carbon skeleton does not change: only the oxidation level and bonding around the carbonyl carbon change.
5. Oxidation of aldehydes
Aldehydes are readily oxidised to carboxylic acids. Ethanal, for example, is oxidised to ethanoic acid.
Suitable oxidising agents
Acidified potassium dichromate(VI), K₂Cr₂O₇ / H⁺, can be used and changes from orange to green as dichromate(VI) ions are reduced to Cr³⁺ ions.
Acidified potassium manganate(VII), KMnO₄ / H⁺, can also oxidise aldehydes. The purple colour of MnO₄⁻ is discharged as it is reduced.
Controlled oxidation of ethanol: warm and distil → ethanal.
Further oxidation: oxidise the aldehyde → ethanoic acid.
6. Testing for aldehydes
Several qualitative tests are useful when identifying carbonyl compounds. The important point is that the tests do not all answer the same question. Some establish that an aldehyde or ketone carbonyl group is present, whereas others can distinguish an aldehyde from a ketone.
2,4-DNPH: testing for an aldehyde or ketone carbonyl group
Add 2,4-dinitrophenylhydrazine, 2,4-DNPH, to the unknown compound. Aldehydes and ketones react to form a coloured precipitate of a 2,4-dinitrophenylhydrazone.
Positive result: an orange or yellow-orange precipitate indicates the presence of an aldehyde or ketone carbonyl group.
A positive 2,4-DNPH test therefore tells you that an aldehyde or ketone is present, but does not by itself distinguish an aldehyde from a ketone.
Tollens’ reagent: distinguishing aldehydes from ketones
On gentle warming with Tollens’ reagent, an aldehyde reduces Ag⁺ to metallic silver while the aldehyde itself is oxidised.
Positive result: a silver mirror, or sometimes a grey deposit of silver, forms.
Ketones do not normally give this reaction. Tollens’ reagent can therefore be used to distinguish an aldehyde from a ketone.
Why is an aldehyde described as a reducing agent?
During the Tollens’ test, Ag⁺ is reduced to Ag. The aldehyde causes this reduction, so the aldehyde acts as a reducing agent. At the same time, the aldehyde itself is oxidised.
Fehling’s solution
Fehling’s solution contains Cu²⁺ ions in alkaline solution. On warming with ethanal, the aldehyde reduces Cu²⁺ and forms copper(I) oxide, Cu₂O.
Positive result for ethanal: the blue solution produces a brick-red precipitate of Cu₂O.
Typical aliphatic aldehydes such as ethanal give a positive Fehling’s test, whereas ketones do not.
The iodoform test and ethanal
Ethanal has one unusual qualitative property: it is the only aldehyde that gives a positive iodoform test.
Warm ethanal with iodine in alkaline solution, commonly written as I₂(aq) / NaOH(aq). A positive result produces a yellow precipitate of triiodomethane, CHI₃.
This behaviour is special to ethanal among aldehydes. Do not generalise the positive iodoform test to all compounds containing –CHO.
7. Choosing the correct qualitative test
Students often memorise the observations but overlook what each test actually proves. A better approach is to match the reagent to the chemical question being asked.
| Test | Positive observation | What the result tells you |
|---|---|---|
| 2,4-DNPH | Orange or yellow-orange precipitate | An aldehyde or ketone carbonyl group is present |
| Tollens’ reagent | Silver mirror or silver deposit | An aldehyde is present |
| Fehling’s solution | Blue solution gives brick-red Cu₂O precipitate | Positive for ethanal and typical aliphatic aldehydes |
| I₂ / NaOH, warm | Yellow CHI₃ precipitate | Ethanal gives a positive result among aldehydes |
Suppose two compounds both produce an orange precipitate with 2,4-DNPH. You cannot yet decide which one is an aldehyde. Warm separate samples with Tollens’ reagent: the sample producing a silver mirror is the aldehyde.
8. Seeing ethanal as part of a reaction pathway
Ethanal is easier to remember when its reactions are connected into a reaction map rather than treated as separate facts.
The unifying idea: oxidation and reduction change the oxidation level of the carbonyl carbon, while nucleophilic addition uses the δ⁺ carbonyl carbon as an electrophilic centre and can create a new C–C bond.
Common Mistakes
| Common mistake | Why it is wrong | Better approach |
|---|---|---|
| Using reflux when preparing an aldehyde from a primary alcohol | Reflux keeps the aldehyde in contact with the oxidising agent and favours further oxidation. | Warm and distil the aldehyde from the reaction mixture as it forms. |
| Saying HCN alone provides a high concentration of CN⁻ | HCN is a weak acid and ionises only slightly. | A small amount of KCN or NaCN supplies the initial CN⁻ nucleophile. |
| Saying HCN simply provides lots of free H⁺ | HCN is weakly ionised, so this gives the wrong picture of the protonation step. | State that the O⁻ intermediate accepts a proton directly from HCN. |
| Thinking CN⁻ is consumed overall | CN⁻ is used in the first step but regenerated in the second. | Add the two mechanism equations and show that CN⁻ cancels, leaving aldehyde + HCN → hydroxynitrile. |
| Saying 2,4-DNPH specifically identifies an aldehyde | Both aldehydes and ketones give a positive 2,4-DNPH test. | Use 2,4-DNPH to establish an aldehyde or ketone, then use Tollens’ reagent to distinguish an aldehyde. |
| Assuming every aldehyde gives a positive iodoform test | Ethanal is the special aldehyde that gives the iodoform reaction. | Remember the yellow CHI₃ precipitate as a distinctive reaction of ethanal among aldehydes. |
What you should be able to do
Recognise an aldehyde
Identify the terminal –CHO functional group and distinguish an aldehyde from a ketone.
Explain carbonyl reactivity
Use the polarity Cδ⁺=Oδ⁻ to explain why nucleophiles attack the carbonyl carbon.
Prepare ethanal
State the reagent, observation and need for distillation when ethanol is oxidised to ethanal.
Explain HCN addition
Explain why CN⁻ is needed, why a small amount of KCN or NaCN is added, and how HCN protonates the intermediate.
Follow the CN⁻ cycle
Show that CN⁻ is regenerated while HCN is consumed overall in forming the hydroxynitrile.
Interpret qualitative tests
Use 2,4-DNPH, Tollens’, Fehling’s and the iodoform test appropriately and state the expected observations.
These ten questions test whether you can apply the chapter to structures, reactions and unfamiliar examples.
- 10 mastery questions
- Structures and reaction schemes where useful
- Instant explanation for every answer option
Frequently asked questions
Questions students often ask about aldehydes
Most difficulties with aldehydes come from mixing up oxidation conditions, qualitative tests or the roles of HCN and CN⁻ in nucleophilic addition. These questions bring the main ideas together.
Why is the carbonyl carbon δ⁺ rather than the oxygen?
Oxygen is more electronegative than carbon, so the C=O bonding electrons are drawn towards oxygen. Oxygen therefore becomes δ⁻ and the carbonyl carbon becomes δ⁺. This electron deficiency is why nucleophiles attack the carbonyl carbon.
Why do we distil ethanal instead of heating the reaction under reflux?
Ethanal can itself be oxidised to ethanoic acid. Distillation removes ethanal from the oxidising mixture as it forms, reducing the opportunity for further oxidation. Reflux would keep ethanal in contact with the oxidising agent.
Why is only a small amount of KCN or NaCN needed?
HCN is a weak acid and therefore provides only a low concentration of CN⁻ by ionisation. A small amount of KCN or NaCN supplies the initial CN⁻ required for nucleophilic attack. CN⁻ is then regenerated when HCN protonates the alkoxide intermediate, so a stoichiometric amount of cyanide salt is not required.
Does HCN supply the proton during nucleophilic addition?
Yes. The negatively charged O⁻ intermediate accepts a proton directly from HCN. It is better to describe this as proton transfer from HCN rather than saying that the solution contains a large concentration of free H⁺. When HCN donates the proton, CN⁻ is regenerated.
Is HCN or CN⁻ consumed overall?
HCN is consumed overall. CN⁻ is used during the nucleophilic-attack step but regenerated when HCN protonates the alkoxide intermediate. Adding the two mechanism steps gives the overall equation RCHO + HCN → RCH(OH)CN, with CN⁻ cancelling from the equation.
Aldehyde chemistry becomes much easier once the reactions are traced back to the polar C=O bond. The δ⁺ carbonyl carbon explains nucleophilic addition, while oxidation and reduction connect aldehydes naturally to carboxylic acids and primary alcohols. For HCN addition, keep the roles especially clear: a small amount of cyanide salt supplies the initial CN⁻ nucleophile, HCN protonates the intermediate and is consumed overall, and CN⁻ is regenerated as the reaction proceeds.