Virginia Research Institute
Virginia Research Institute
Virtual Laboratory  ·  Built by E2 Innovations
← Back to Virtual Labs | Aldehydes & Ketones
Organic Chemistry · Carbonyl Chemistry I

Aldehydes & Ketones

The carbonyl group (C=O) is the most important functional group in organic chemistry — it is the gateway to almost every transformation that follows. This lab covers naming and structure, the major reactions of aldehydes and ketones (reduction, addition, acetal formation, imine formation, oxidation, Wittig), reactivity comparisons, and the diagnostic tests used to identify them in the laboratory.

Theory — Aldehydes & Ketones

Both aldehydes and ketones contain the carbonyl group — a carbon doubly bonded to oxygen (C=O). The carbonyl carbon is sp² hybridised, planar, and partially positive (δ+); the oxygen carries a partial negative charge (δ−). This dipole, combined with the empty π* orbital, makes the carbonyl carbon a very effective electrophile. Almost every reaction of aldehydes and ketones involves attack on this carbon by a nucleophile.

1. Naming

Aldehydes have the carbonyl on a terminal carbon (C1) with at least one H attached: R–CHO. The IUPAC suffix is -al: methanal (HCHO), ethanal (CH₃CHO), propanal, butanal, and so on. C1 of the parent chain is always the CHO carbon, so no locant is needed for the carbonyl. Aromatic aldehyde benzaldehyde (C₆H₅CHO) keeps its retained name. Other retained names: cinnamaldehyde, vanillin, citral.

Ketones have the carbonyl in the middle of the chain (between two C atoms): R–CO–R'. The IUPAC suffix is -one with a locant indicating the carbonyl carbon: propan-2-one (acetone), butan-2-one, pentan-3-one, cyclohexanone (no locant — symmetric), acetophenone (1-phenylethan-1-one). The locant is placed before the -one suffix in current IUPAC convention.

2. Structure and reactivity

The carbonyl C is sp² with three substituents in a plane (~120° bond angles); the C=O bond is shorter (1.22 Å) and stronger (~750 kJ/mol) than a typical C–C single bond. The high polarity (μ ≈ 2.7 D for acetone) makes the C electrophilic enough that even weak nucleophiles attack it readily, especially under acid catalysis (which protonates O, making C even more positive).

Two factors make aldehydes more reactive than ketones toward nucleophilic addition:

Order of reactivity (toward nucleophilic addition): HCHO > RCHO > R₂CO. Aromatic carbonyls (benzaldehyde, acetophenone) are less reactive than aliphatic because the aryl ring conjugates with C=O, delocalising the δ+.

3. Major reactions

(a) Reduction. NaBH₄ (mild) and LiAlH₄ (strong) both reduce aldehydes and ketones to alcohols. NaBH₄ is selective: it reduces aldehydes/ketones but leaves esters, carboxylic acids, and amides alone. LiAlH₄ reduces almost all carbonyl compounds and must be handled in dry, aprotic solvent under inert atmosphere (it reacts violently with water).

Reduction R-CHO + NaBH₄ / MeOH → R-CH₂OH (1° alcohol)
R-CO-R' + NaBH₄ / MeOH → R-CHOH-R' (2° alcohol)
Aldehyde → 1° alcohol; ketone → 2° alcohol

(b) Grignard addition. Grignard reagents (RMgX) add to carbonyls to give alcohols after aqueous workup. The carbon nucleophile (R⁻ equivalent) attacks the carbonyl C; the resulting alkoxide is then protonated. Reaction with HCHO gives 1° alcohol; with RCHO gives 2° alcohol; with R₂C=O gives 3° alcohol.

Grignard addition R-CHO + R'MgBr → R'-CHR-OMgBr (alkoxide intermediate)
Then H₂O / H⁺ workup → R'-CHR-OH
RCHO + R'MgX → 2° alcohol after workup

(c) Cyanohydrin formation. HCN (or NaCN/H⁺) adds across the C=O to give a cyanohydrin — useful as a one-carbon-extension and as a precursor to α-hydroxy acids. Equilibrium-controlled; aldehydes give better yields than ketones.

(d) Hydration and acetal formation. Water adds reversibly to a carbonyl to give a gem-diol (the hydrate). For most aldehydes/ketones, the equilibrium favours the carbonyl. Exceptions: HCHO is >99% hydrate in water; chloral (CCl₃CHO) is fully hydrated; cyclopropanone is fully hydrated.

Two equivalents of alcohol with acid catalysis give an acetal: a key reaction because acetals are stable to base and to nucleophiles but are easily hydrolysed by aqueous acid. Acetals are widely used as protecting groups: convert C=O → acetal, do reactions elsewhere, then hydrolyse the acetal back to C=O.

Acetal formation (and reverse) R-CHO + 2 R'-OH ⇌ R-CH(OR')₂ + H₂O  [H⁺ cat.; -H₂O drives forward]
Hydrolysis (reverse): R-CH(OR')₂ + H₂O / H⁺ → R-CHO + 2 R'-OH
Acetals are STABLE to base, LABILE to acid → ideal protecting group

(e) Imine formation. A primary amine (RNH₂) condenses with an aldehyde or ketone to give an imine (Schiff base, RCH=NR') with loss of water. Mechanism: nucleophilic addition of N to C=O → carbinolamine → loss of water (acid-catalysed) → imine. With secondary amines (R₂NH), the product is an enamine (R₂N-C=C-) instead of an imine, because there's no H on N to lose.

Imine vs enamine formation R-CHO + R'-NH₂ ⇌ R-CH=N-R' + H₂O  (IMINE — primary amine)
R-CO-CH₂R'' + R'₂NH ⇌ R'₂N-C(R)=CH-R'' + H₂O  (ENAMINE — secondary amine)
1° amine → imine; 2° amine → enamine

(f) Wittig reaction. A phosphonium ylide (Ph₃P=CHR, prepared from Ph₃P + RCH₂X then strong base) reacts with an aldehyde or ketone to give an alkene with concurrent loss of triphenylphosphine oxide (Ph₃P=O). The position of the new C=C is precisely controlled — it forms between the original carbonyl C and the ylide C. Stabilised ylides (those with an EWG on C) tend to give E-alkenes; unstabilised ylides tend to give Z-alkenes (Schlosser variant gives E).

(g) Oxidation. Aldehydes are easily oxidised to carboxylic acids by mild oxidants (Tollens' reagent — Ag(NH₃)₂⁺/OH⁻, Fehling's — Cu²⁺/citrate/OH⁻) and stronger ones (Jones reagent CrO₃/H₂SO₄/acetone, KMnO₄). Ketones do NOT oxidise under any of these conditions — the carbonyl C has no H to lose. This is the basis of the Tollens' and Fehling's tests, which distinguish aldehydes from ketones.

4. Diagnostic tests (see Section IV)

TestPositive resultDetectsNotes
2,4-DNP (Brady's)Yellow-orange-red precipitateAldehydes & ketonesGeneral carbonyl test; m.p. of derivative IDs the compound
Tollens'Silver mirror on flask wallAldehydes ONLYKetones do not react
Fehling'sBrick-red Cu₂O precipitateAliphatic aldehydesAromatic aldehydes give weak/negative result
IodoformYellow CHI₃ precipitateMethyl ketones; CH₃CHOR-CO-CH₃ pattern; also positive for RCH(OH)CH₃
Schiff'sMagenta colour returnsAldehydes onlyDecolourised fuchsin; aldehyde restores colour

5. α-Hydrogen acidity

The hydrogens on a carbon adjacent to a C=O (α-carbon) are unusually acidic. Why? Because deprotonation gives an enolate — a resonance-stabilised carbanion where the negative charge is delocalised onto oxygen. Typical pKa values:

This acidity means that bases as mild as NaOH can partially enolise α,β-functionalised carbonyls; with stronger bases (LDA, NaH), full enolisation is possible. Enolate chemistry — alkylation, aldol condensation, Michael addition — is the subject of "Carbonyl Chemistry II" and is not covered in this lab.

Instructions

This lab's Simulation section has four parts. Complete them in order.

1
Section I — Naming & Structure. Eight carbonyl compounds are shown as structural drawings. For each, identify the IUPAC name, classify as aldehyde or ketone, and identify the hybridisation of the carbonyl carbon.
2
Section II — Reaction Bench. Six reactions of aldehydes and ketones, presented as an interactive flask with reagent dispensers. Click a reagent bottle to add it to the flask, watch the reaction, then identify the product from four options.
3
Section III — Reactivity & Mechanism. Eight problems testing relative reactivity (aldehyde vs ketone, aliphatic vs aromatic, electronic effects), pKa of α-H, and arrow-pushing in nucleophilic addition.
4
Section IV — SDS & Microscale Tests. Read SDS extracts for four reagents (16 questions on hazards, PPE, disposal, first aid). Then run six microscale diagnostic tests (2,4-DNP, Tollens', Fehling's, iodoform, Schiff's, chromic acid) to identify unknown carbonyl compounds.
5
Prepare your lab notebook. Use the Example Report as your template. Document each interaction, your interpretation, and the conclusions you draw from the diagnostic tests.

Prerequisite: Ensure you have completed (or are familiar with) the Lab Skills & Safety lab, the Functional Group Tests lab, the Mechanisms lab, and the Alcohols lab. The carbonyl chemistry in this lab assumes you can recognise glassware, read an SDS, and predict mechanisms.

Simulation

Four interactive parts. Use the ↺ Reset Simulation button at any time to clear all answers and start over.

Carbonyl Workbench Section I — Naming & Structure

Eight carbonyl compounds. For each: (a) IUPAC name, (b) aldehyde or ketone, (c) hybridisation of the carbonyl carbon.

Score: 0 / 24 (3 questions × 8 compounds)

Six reactions of aldehydes & ketones. For each: read the prompt, click the reagent in the dispenser shelf to add it to the flask, then click the predicted product from the four options.

Score: 0 / 6

Eight conceptual problems on reactivity ordering, pKa values, and arrow-pushing. Choose the best answer for each.

Score: 0 / 8

Round 1 — SDS interpretation

Four key reagents used in carbonyl chemistry. Each has 4 questions.

SDS score: 0 / 16

Round 2 — Microscale diagnostic tests

Six unknown samples are presented. For each, run the indicated test and identify the functional group present based on the result.

Microscale score: 0 / 6

Team Questions

Discuss with your team before answering. Type a brief response into each box.

Question 1 — Naming. What is the IUPAC name of CH₃CH₂COCH₂CH₃?
Question 2 — Reactivity. Which is more reactive toward NaBH₄: propanal or propan-2-one? Why?
Question 3 — Diagnostic test. A student has two unlabelled bottles, both giving an orange precipitate with 2,4-DNP. Bottle A gives a silver mirror with Tollens'; Bottle B does not. Which bottle contains an aldehyde, and which a ketone?
Question 4 — Acetal protection. You want to reduce an ester group on a molecule that also contains a ketone, using LiAlH₄. The ketone would also be reduced. How can you prevent this?
Question 5 — Wittig. What alkene results from the Wittig reaction between cyclohexanone and Ph₃P=CH₂?
Question 6 — α-acidity. Why is the α-H of pentane-2,4-dione (pKa ≈ 9) so much more acidic than the α-H of acetone (pKa ≈ 20)?

Example Lab Notebook Entry

Use the format below as a template. Document each interaction in the simulation, then synthesise observations into a discussion.

Aldehydes & Ketones — Lab Notebook Entry

Submitted by: [Student Name]

Course: Organic Chemistry I · Section: 201-A · Date: April 25, 2026

Objective

To recognise aldehydes and ketones by structure and IUPAC name; to predict the products of common carbonyl reactions (reduction, Grignard addition, acetal formation, imine formation, oxidation, Wittig); to relate carbonyl reactivity to electronic and steric effects; and to identify unknown carbonyl compounds using diagnostic microscale tests (2,4-DNP, Tollens', Fehling's, iodoform, Schiff's, chromic acid).

Naming summary (Section I results)

Structure shownIUPAC nameClassHybridisation at C=O
CH₃CH₂CHOPropanalAldehydesp²
CH₃COCH₂CH₃Butan-2-oneKetonesp²
C₆H₅CHOBenzaldehydeAromatic aldehydesp²
C₆H₅COCH₃1-Phenylethan-1-one (acetophenone)Aromatic ketonesp²
CyclohexanoneCyclohexanoneCyclic ketonesp²
(E)-PhCH=CHCHO(2E)-3-Phenylprop-2-enal (cinnamaldehyde)α,β-Unsaturated aldehydesp² (both C=C and C=O)
(E)-CH₃CH=CHCHO(2E)-But-2-enal (crotonaldehyde)α,β-Unsaturated aldehydesp²
HCHOMethanal (formaldehyde)Aldehydesp²

Reaction bench observations (Section II results)

CarbonylReagentProduct (predicted)Class of product
PropanalNaBH₄ / MeOHPropan-1-ol1° alcohol
Butan-2-oneCH₃MgBr, then H₂O / H⁺2-Methylbutan-2-ol3° alcohol
Propanal2 equiv MeOH, H⁺ catalystPropanal dimethyl acetalAcetal
CyclohexanoneMethylamine (MeNH₂)N-MethylcyclohexanimineImine
ButanalTollens' reagent (Ag(NH₃)₂⁺/OH⁻)Butanoic acid + Ag mirrorCarboxylic acid
CyclohexanonePh₃P=CH₂ (Wittig ylide)MethylenecyclohexaneAlkene

Microscale test results (Section IV, Round 2)

UnknownTest appliedObservationIdentified as
Sample 12,4-DNPYellow-orange precipitateAldehyde or ketone (general carbonyl)
Sample 2Tollens'Silver mirror on flask wallAldehyde (specific)
Sample 3Fehling'sBrick-red precipitateAliphatic aldehyde
Sample 4IodoformYellow CHI₃ precipitateMethyl ketone (R-CO-CH₃) or CH₃CHO
Sample 5Schiff'sMagenta colour returnsAldehyde
Sample 62,4-DNP only (Tollens' negative)Orange precipitate; no silver mirrorKetone

Discussion

The simulation reinforced the centrality of the carbonyl group to organic synthesis: from a single C=O, six different functional-group transformations were possible (alcohol, larger alcohol, acetal, imine, carboxylic acid, alkene). The choice of reagent determines the outcome — reduction (NaBH₄, LiAlH₄, H₂/Pt) gives an alcohol; carbon nucleophiles (Grignard, organolithium, cyanide, Wittig) form new C–C bonds; nitrogen nucleophiles (amines, hydrazines) form C=N bonds; oxygen nucleophiles (alcohols) under acid catalysis form acetals.

The reactivity order RCHO > R₂C=O made physical sense: more alkyl substitution at the carbonyl C provides more steric hindrance for the incoming nucleophile and more electron density to the carbonyl C (reducing its δ+ character through hyperconjugation). Aromatic carbonyls were less reactive than aliphatic because the aryl ring conjugates with the C=O, delocalising the partial positive charge. This explained why benzaldehyde does not react with Fehling's solution but propanal does — the rate is too slow under the test conditions.

The diagnostic tests in Section IV were complementary, not redundant. 2,4-DNP detects all carbonyls (aldehyde or ketone); Tollens' and Fehling's distinguish aldehydes from ketones; iodoform identifies the methyl ketone substructure (or CH₃CHO); Schiff's confirms an aldehyde; chromic acid distinguishes 1° and 2° alcohols (positive) from tertiary alcohols (negative). A typical unknown identification therefore requires running multiple tests and interpreting the pattern of positives and negatives.

Section IV's SDS round emphasised that LiAlH₄ is among the most hazardous reagents commonly used in undergraduate labs — pyrophoric in moist air, violently reactive with water and protic solvents, and capable of igniting spontaneously. NaBH₄, while still a hydride reducing agent, is much milder and tolerates protic solvents. Formaldehyde (methanal) is an IARC Group 1 carcinogen with mandatory fume hood handling. Methyl vinyl ketone is highly toxic and a strong Michael acceptor. None of these reagents should be used without first reading the SDS.

Conclusion

Aldehydes and ketones share the C=O functional group but differ subtly in reactivity (RCHO > R₂CO) and dramatically in their response to oxidising agents (aldehydes oxidise to carboxylic acids; ketones do not). The combination of structure-determining information (IUPAC name, classification, hybridisation), reaction predictions (six major reactions), and diagnostic tests (six microscale tests) gives a complete framework for identifying and transforming carbonyl compounds in the laboratory.

References

1. Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry, 2nd ed., Oxford University Press, 2012, Chs 6, 11, 12, 26–28.
2. McMurry, J. Organic Chemistry, 9th ed., Cengage, 2016, Chs 19–22.
3. IUPAC. Recommendations on Organic Nomenclature, 2013.
4. Sigma-Aldrich SDS for NaBH₄ (CAS 16940-66-2), LiAlH₄ (CAS 16853-85-3), formaldehyde 37% w/v aq. (CAS 50-00-0), and methyl vinyl ketone (CAS 78-94-4), accessed online March 2026.

Practice Questions

Work through each before peeking at the hint.

Practice 1 — Naming
Name (CH₃)₂CHCH₂CHO using IUPAC rules.
Hint: 3-methylbutanal. The CHO is C1; numbering goes from CHO outward; methyl branch is on C3.
Practice 2 — Reactivity
Rank the following from most to least reactive toward NaBH₄: acetone, formaldehyde, benzaldehyde, propanal.
Hint: formaldehyde > propanal > benzaldehyde > acetone. Aldehydes more reactive than ketones (steric + electronic); aliphatic more reactive than aromatic (no conjugation to delocalise δ+).
Practice 3 — Reduction
What product results from butan-2-one + NaBH₄ followed by aqueous workup?
Hint: butan-2-ol (2° alcohol, racemic). NaBH₄ delivers H⁻ to the carbonyl C; aqueous workup protonates the alkoxide. Product is racemic because attack from either face of the planar carbonyl is equally likely.
Practice 4 — Grignard
Predict the major product of: propan-2-one (acetone) + ethylmagnesium bromide, then H₂O/H⁺.
Hint: 2-methylbutan-2-ol — a 3° alcohol. The ethyl group attacks the carbonyl C; aqueous workup protonates the resulting alkoxide. Three R groups on the carbinol C → 3°.
Practice 5 — Acetal as protecting group
You have a molecule containing a ketone and an ester. You want to reduce the ester to a primary alcohol with LiAlH₄ without touching the ketone. Outline a 3-step sequence.
Hint: (1) Protect the ketone as an acetal (e.g., HOCH₂CH₂OH + cat. H⁺, remove water with Dean–Stark). (2) Reduce the ester with LiAlH₄ in dry THF — the acetal is stable to base/hydride. (3) Hydrolyse the acetal back to the ketone with aqueous H⁺. Result: primary alcohol from ester, ketone unchanged.
Practice 6 — Imine vs enamine
Cyclohexanone is treated with (a) methylamine, (b) dimethylamine. What product class results in each case?
Hint: (a) Methylamine (RNH₂, primary) → imine: N-methylcyclohexanimine. (b) Dimethylamine (R₂NH, secondary) → enamine: 1-(dimethylamino)cyclohex-1-ene. Primary amines have an N–H to lose (giving C=N); secondary amines do not, so the system loses an α-H from C instead (giving C=C–N).
Practice 7 — Wittig
What aldehyde and what phosphonium ylide are needed to make (E)-1-phenylprop-1-ene (PhCH=CHCH₃) by Wittig?
Hint: Two retrosynthesis options: (i) benzaldehyde + Ph₃P=CHCH₃ (ethylidene ylide) — stabilised? No, this is unstabilised so favours Z, not what we want. (ii) acetaldehyde + Ph₃P=CHPh (benzyl ylide) — stabilised by phenyl, gives E. Better choice: option (ii) for E selectivity.
Practice 8 — Diagnostic tests
An unknown gives: 2,4-DNP positive (orange precipitate); Tollens' negative (no silver mirror); iodoform positive (yellow CHI₃ precipitate). Identify the functional group present.
Hint: Methyl ketone (R-CO-CH₃). 2,4-DNP+ confirms carbonyl; Tollens'− rules out aldehyde; iodoform+ confirms the CH₃ next to C=O. Could also be acetaldehyde (CH₃CHO), but that would give Tollens'+. Likely candidates: acetone, butan-2-one, acetophenone, etc.
Practice 9 — α-acidity
Of these three compounds, which has the most acidic α-hydrogen and why? (a) acetone (pKa ≈ 20), (b) ethanol (pKa ≈ 16), (c) pentane-2,4-dione (pKa ≈ 9).
Hint: Pentane-2,4-dione (c). Its α-carbon sits between TWO C=O groups, so the resulting enolate is stabilised by delocalisation onto both oxygens, making the conjugate base much more stable than the enolate of acetone (one O) or the alkoxide of ethanol (no resonance stabilisation).
Practice 10 — Mechanism
Draw the mechanism for the acid-catalysed reaction of cyclohexanone with methanol (excess) to form the dimethyl acetal. How many steps?
Hint: Six key steps: (1) protonation of C=O by H⁺ → oxocarbenium-like activation. (2) Nucleophilic addition of MeOH to C+. (3) Deprotonation gives hemiacetal. (4) Protonation of hemiacetal -OH. (5) Loss of water → second oxocarbenium. (6) Second MeOH adds; deprotonation gives the dimethyl acetal. Note: water must be removed to drive equilibrium forward (Dean–Stark or molecular sieves).