Chemistry · Lesson

Aldehydes, ketones and carbonyl reactivity

Read the C=O motif through its neighbours and electronic context. Compare addition, hydration and tautomerism while preserving atom attachments, charge and equilibrium assumptions.

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By the end, you can

Distinguish an aldehyde from a ketone by carbonyl-carbon attachments.
Keep partial-charge reasoning separate from formal charge.
Balance a reversible water-addition net and normalize a stipulated hydration equilibrium.
Distinguish keto–enol tautomerism from resonance.

Look at the carbonyl carbon’s remaining attachments

A carbonyl has a C=O motif. In the simple neutral examples here, an aldehyde carbonyl carbon carries an H, whereas a ketone carbonyl carbon joins two carbon groups. Propanal CH₃CH₂CHO and propanone CH₃COCH₃ share C₃H₆O but differ in connectivity. Formaldehyde, with two attached H at its carbonyl carbon, is the smallest aldehyde example and shows why a carbon-group attachment is not compulsory for every aldehyde.

An ester or amide also contains a carbonyl but has another characteristic atom directly attached to its carbonyl carbon. That neighbour changes the reaction possibilities. Identifying C=O without inspecting the rest of the graph therefore gives an incomplete functional-group classification.

Carbonyl polarity guides an electron-flow hypothesis

In a familiar neutral carbonyl Lewis contributor, C and O can both have formal charge zero while the bond is polarized toward oxygen. Partial positive at carbon means an electron-distribution tendency rather than formal C⁺. A contributing polarized representation can help describe electron density, but the real molecule is one electronic entity, not a population switching between drawings.

A conceptual two-electron nucleophilic addition to carbonyl carbon must account for the C=O π pair moving toward O as the new carbon–nucleophile bond forms; otherwise the ordinary carbon octet would be exceeded. An anionic nucleophile can lead to an alkoxide form, with proton transfer producing a neutral alcohol-containing product. Acidic and basic conditions can change the sequence and catalysis. These are electron-accounting ideas, not a practical setup or universal relative aldehyde/ketone rate ranking.

Hydration changes the formula and bonding pattern

A symbolic hydration of ethanal is CH₃CHO+H₂O⇌CH₃CH(OH)₂. The product is a geminal diol, with both OH groups attached to the former carbonyl carbon. Its formula is C₂H₆O₂. That carbon no longer has a C=O bond in the product representation; the oxygen addition and proton transfers must be retained.

Hydration is reversible and its equilibrium depends on the molecular structure and conditions. Acid/base catalysis can alter forward and reverse rates without changing the same thermodynamic equilibrium constant at unchanged temperature, solvent and reference convention. No measured hydrate percentage is assigned to ethanal or propanone in this lesson.

A two-species model needs a ratio and a conserved total

For an unnamed invented carbonyl/hydrate pair, define K=a_hydrate/(a_carbonyl a_water). At stipulated fixed conditions, set a_water=1 and ideal dilute organic activities a_i≈c_i/c°, with common c°=1 mol L⁻¹. Then K=c_hydrate/c_carbonyl. If those are the only two organic species, c_total=c_hydrate+c_carbonyl and f_hydrate=K/(1+K).

The chosen numerical case has K=0.200 and c_total=0.0100 mol L⁻¹. The graph varies the stipulated parameter K across a family of such mathematical models; it is not a measured scan with changing solvent or temperature, nor a time trace. Finite thermodynamic states use positive K; the zero endpoint is only the formal K→0 limit of the fraction relation. Other reactions, aggregation and nonideal activities would require a different model.

A stipulated equilibrium ratio determines the hydrate fraction

Equilibrium hydrate fraction

Stipulated dimensionless equilibrium ratio K

Selected ratios; zero is a formal limit
— Two-species fraction relation

Underlying data

Stipulated dimensionless equilibrium ratio K	Equilibrium hydrate fraction
0	0 · Selected ratios; zero is a formal limit
0.05	0.047619047619047616 · Selected ratios; zero is a formal limit
0.2	0.16666666666666669 · Selected ratios; zero is a formal limit
1	0.5 · Selected ratios; zero is a formal limit
2	0.6666666666666666 · Selected ratios; zero is a formal limit
5	0.8333333333333334 · Selected ratios; zero is a formal limit

Original family of invented two-organic-species equilibrium models with a_water=1, compatible ideal dilute activities, common reference concentration and no other organic species. Positive K is the stipulated dimensionless ratio at each finite model state; the zero endpoint represents only the formal K→0 limit. This is not a measured temperature/solvent scan. Markers and dashed curve show f_hydrate=K/(1+K), not a formation time or real named-carbonyl hydrate percentage.

Moving a hydrogen nucleus is not resonance

Ethanal CH₃CHO and ethenol CH₂=CHOH share C₂H₄O, but a hydrogen changes its nuclear attachment from carbon to oxygen and the C–C/C–O bond orders change. They are keto–enol tautomers, linked by a chemical interconversion rather than two resonance contributors. Conditions and mechanism determine interconversion rates and equilibrium.

By contrast, the acetate resonance contributors retain the same atom-specific hydrogens and nuclear framework while changing electron bookkeeping. Formula equality is a necessary comparison step but is not enough to establish resonance. Label the atoms and attached hydrogens when a visually similar redraw might conceal a nuclear change.

Worked examples

Retain every atom during water addition

Balance the symbolic hydration CH₃CHO+H₂O⇌CH₃CH(OH)₂.

Ethanal supplies C₂H₄O and water supplies H₂O.
The product therefore needs C₂H₆O₂.
Its central carbon joins CH₃, H and two OH groups through four single bonds.

One water gives one gem-diol in the balanced net. No equilibrium percentage follows from this balance.

Normalize the unnamed hydration toy

For K=0.200 and c_total=0.0100 mol L⁻¹ under the stated assumptions, find both species concentrations.

f_hydrate=0.200/(1+0.200)=1/6.
c_hydrate=(1/6)×0.0100≈0.00166667 mol L⁻¹.
c_carbonyl=(5/6)×0.0100≈0.00833333 mol L⁻¹.
Their sum is 0.0100 mol L⁻¹ and their ratio 0.200.

About 0.00167 M hydrate and 0.00833 M carbonyl; the hydrate fraction is 16.7% in this invented model.

Track which nucleus changed attachment

Compare CH₃CHO with CH₂=CHOH using corresponding first C, second C and O labels.

Ethanal attached-H counts on those atoms are 3,1,0.
Ethenol attached-H counts are 2,1,1.
A hydrogen has changed attachment from the first carbon to oxygen.

These are tautomers, not resonance contributors, despite the same C₂H₄O formula.

Try it yourself

Does partial positive carbonyl carbon necessarily have formal charge +1 in the ordinary neutral contributor?
Show the answer and reasoning
No.
A polarized electron distribution and formal electron-allocation charge are different descriptions.
When a new two-electron bond forms at ordinary carbonyl carbon, why must the π pair be accounted for?
Show the answer and reasoning
Retaining the original double bond as well would exceed the ordinary carbon octet.
The elementary addition bookkeeping moves the π pair toward oxygen.
Under the stipulated two-species assumptions, which K gives hydrate fraction 1/2?
Show the answer and reasoning
K=1.
The ratio is one, so the two concentrations are equal.
At unchanged conditions, does a catalyst automatically increase the thermodynamic hydration K?
Show the answer and reasoning
No.
Catalysis changes compatible rates and the approach to equilibrium rather than that fixed equilibrium constant.

What to carry forward

Inspect all carbonyl-carbon attachments.
Separate partial charge from formal charge.
Water addition changes formula and bonding.
Normalize an equilibrium fraction under its stated species model.
Tautomerism moves nuclei; resonance does not.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.