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questions answered out of 20
Q1
The carbonyl carbon in an aldehyde is:
A. sp² hybridised
B. sp³ hybridised
C. sp hybridised
D. Unhybridised
✓ Answer: A
The carbonyl carbon forms three σ-bonds and one π-bond, requiring sp² hybridisation. This gives a trigonal planar geometry with bond angles of approximately 120°. The unhybridised p orbital overlaps with the oxygen p orbital to form the C=O π bond. McMurry 8e
Common trap: Do not confuse with sp³ (tetrahedral). The carbonyl carbon loses one degree of tetrahedral character because the π bond keeps it planar.
Q2
Which reagent produces a silver mirror only with aldehydes and not with ketones?
A. Fehling's solution
B. Tollens' reagent
C. Benedict's solution
D. Jones reagent
✓ Answer: B
Tollens' reagent contains the diamminesilver(I) complex [Ag(NH₃)₂]⁺. Aldehydes reduce Ag⁺ to metallic silver (Ag⁰), depositing the characteristic mirror on the inner wall of the test tube. Ketones cannot be oxidised under these mild conditions. McMurry 8e
Common trap: Fehling's and Benedict's also test for aldehydes but produce a brick-red precipitate (Cu₂O), not a silver mirror. The silver mirror is unique to Tollens'.
Q3
Fehling's / Benedict's test produces a __________ precipitate when positive with an aldehyde:
A. White
B. Yellow
C. Brick-red
D. Black
✓ Answer: C
Both Fehling's and Benedict's reagents contain Cu²⁺ ions (blue solution). An aldehyde reduces Cu²⁺ to Cu⁺, which precipitates as copper(I) oxide Cu₂O — a brick-red (or orange-red) solid. The aldehyde itself is oxidised to the corresponding carboxylate. McMurry 8e
Common trap: White precipitate = AgCl (irrelevant here). Yellow-green colour change is not the diagnostic endpoint; the brick-red solid is.
Q4
The IUPAC suffix for naming an aldehyde is:
A. -one
B. -oic acid
C. -yl
D. -al
✓ Answer: D
Aldehydes are named by replacing the terminal "-e" of the parent alkane with "-al" (e.g., methanal, ethanal, propanal). The carbonyl carbon is always C-1 and is implicit — no locant is needed. McMurry 8e
Common trap: "-one" is the ketone suffix. "-oic acid" belongs to carboxylic acids. "-yl" denotes an alkyl group.
Q5
The IUPAC suffix for naming a ketone is:
A. -al
B. -one
C. -oate
D. -yl
✓ Answer: B
Ketones use the suffix "-one" (e.g., propan-2-one = acetone, butan-2-one = methyl ethyl ketone). A locant is required to specify which carbon bears the carbonyl group unless symmetry makes it unnecessary. McMurry 8e
Common trap: "-oate" is the ester suffix. Do not confuse "-one" (ketone) with "-anol" (alcohol).
Q6
NaBH₄ reduces an aldehyde to a:
A. Carboxylic acid
B. Alkane
C. Primary alcohol
D. Secondary alcohol
✓ Answer: C
NaBH₄ delivers a hydride ion (H⁻) to the electrophilic carbonyl carbon of the aldehyde. After aqueous workup, a primary alcohol is formed: R–CHO → R–CH₂OH. The carbonyl carbon gains one C–H bond and one O–H bond. McMurry 8e
Common trap: Secondary alcohols come from reduction of ketones (R–CO–R' → R–CHOH–R'). Alkane would require complete deoxygenation, which requires different conditions (e.g., Wolff–Kishner).
Q7
Which reagent selectively reduces aldehydes but does NOT reduce esters?
A. LiAlH₄
B. H₂ / Pd
C. NaBH₄
D. None — all reduce both equally
✓ Answer: C
NaBH₄ is a mild, selective hydride donor. It readily reduces aldehydes and ketones but is not nucleophilic enough to reduce the less electrophilic esters or carboxylic acids. LiAlH₄, by contrast, is a powerful reagent that reduces both esters AND aldehydes (and most other carbonyl-containing functional groups). McMurry 8e
Common trap: Students assume LiAlH₄ and NaBH₄ have the same selectivity — they do not. NaBH₄ is selective; LiAlH₄ is not.
Q8
Acetal formation from an aldehyde requires:
A. Base catalyst and water
B. An oxidising agent
C. NaBH₄ and methanol
D. Acid catalyst and excess alcohol (ROH)
✓ Answer: D
Acetal formation is an acid-catalysed reversible reaction. The aldehyde first reacts with one equivalent of ROH to give a hemiacetal R–CH(OH)(OR'); a second equivalent of ROH then displaces –OH (under acid) to give the full acetal R–CH(OR')₂. Water must be removed (e.g., molecular sieves or Dean–Stark) to drive equilibrium. McMurry 8e
Common trap: Base does not catalyse acetal formation — the mechanism requires protonation of the hemiacetal –OH to generate a good leaving group (water).
Q9
The aldol condensation requires:
A. At least one carbonyl compound bearing an α-hydrogen, in the presence of base
B. Two ketone molecules only, no α-hydrogen needed
C. Any aldehyde with a strong acid catalyst
D. Formaldehyde and dilute NaOH only
✓ Answer: A
In base, the α-hydrogen (pKₐ ~20) is abstracted to form an enolate ion. This enolate attacks the carbonyl carbon of a second molecule in a nucleophilic addition, forming a β-hydroxy carbonyl compound (the aldol product). Without an α-hydrogen, no enolate can form and no aldol occurs. McMurry 8e
Common trap: Benzaldehyde (no α-H) cannot undergo a simple aldol with itself; it undergoes the Cannizzaro reaction instead.
Q10
The iodoform test is positive for compounds containing the structural feature:
A. –CHO group
B. –CH₂OH group
C. –COCH₃ group (methyl ketone) or acetaldehyde
D. –COOH group
✓ Answer: C
The iodoform test specifically detects the CH₃C(=O)– methyl ketone motif, or acetaldehyde (CH₃CHO). I₂ / NaOH progressively halogenates the methyl group, then base cleaves the triiodomethyl group to yield iodoform (CHI₃). Ethanol (which oxidises to acetaldehyde in situ) also tests positive. McMurry 8e
Common trap: A general –CHO aldehyde is NOT sufficient — only CH₃CHO (acetaldehyde) tests positive among aldehydes, because the methyl group is required.
Q11
The diagnostic product that precipitates in the iodoform test is:
A. CH₂I₂ (diiodomethane)
B. CCl₄
C. CHI₃ — a pale yellow solid
D. CHCl₃ (chloroform)
✓ Answer: C
Iodoform CHI₃ precipitates as a pale yellow solid with a distinctive antiseptic odour. It is insoluble in the aqueous alkaline reaction medium. The combination of pale yellow colour + characteristic smell constitutes a positive iodoform test. McMurry 8e
Common trap: CHCl₃ is the product of the chloroform (haloform) reaction using Cl₂/NaOH. The iodoform test specifically uses I₂/NaOH and yields CHI₃.
Q12
The Cannizzaro reaction requires:
A. Dilute base and any aldehyde
B. Acid catalyst with formaldehyde only
C. H₂ / Ni catalyst
D. Concentrated NaOH and an aldehyde lacking α-hydrogen
✓ Answer: D
The Cannizzaro reaction is a disproportionation: one molecule of the aldehyde is oxidised to a carboxylate while another is reduced to an alcohol. It requires concentrated (strong) NaOH and an aldehyde without α-hydrogens (e.g., formaldehyde, benzaldehyde), because otherwise the base would simply form an enolate and give an aldol product instead. McMurry 8e
Common trap: Aldehydes with α-hydrogens (e.g., ethanal) give aldol products under base, NOT the Cannizzaro reaction.
Q13
In nucleophilic addition to a carbonyl group, the nucleophile attacks the:
A. Carbonyl carbon (δ⁺)
B. Oxygen atom (δ⁻)
C. α-Carbon
D. β-Carbon
✓ Answer: A
The C=O bond is polarised with δ⁺ on carbon and δ⁻ on oxygen, because oxygen is more electronegative. Nucleophiles (electron-rich species) are attracted to the electrophilic carbonyl carbon. The π bond breaks heterolytically, the electrons go to oxygen, and the carbon rehybridises from sp² to sp³. McMurry 8e
Common trap: Although oxygen bears the δ⁻ charge, it is the electrophile that receives the nucleophile, not the nucleophile that attacks the δ⁻ end.
Q14
Regarding reactivity in nucleophilic addition to the carbonyl group:
A. Ketones > Aldehydes
B. Aldehydes > Ketones
C. Both react equally
D. Depends entirely on solvent polarity
✓ Answer: B
Aldehydes (R–CHO) are more reactive than ketones (R–CO–R') in nucleophilic addition for two reasons: (1) Steric factor — aldehydes have only one R group blocking approach of the nucleophile; ketones have two. (2) Electronic factor — two alkyl groups in ketones donate electron density to the carbonyl carbon, reducing its electrophilicity (δ⁺). McMurry 8e
Common trap: "More stable carbonyl" does not mean "more reactive." Greater stability of ketones correlates with lower reactivity.
Q15
Formaldehyde (methanal, HCHO) at room temperature (25 °C) is:
A. A colourless liquid
B. A white solid
C. A colourless gas
D. Available only as an aqueous solution
✓ Answer: C
Methanal has a boiling point of –19 °C, so it is a gas at room temperature. It is commonly encountered as formalin (a 37–40% aqueous solution) in laboratories, but the pure compound is a gas. It has a sharp, pungent odour and is a known carcinogen. McMurry 8e
Common trap: Formalin (the aqueous solution) is a liquid — but this describes the solution, not pure formaldehyde. Pure methanal is a gas at room temperature.
Q16
Acetaldehyde (ethanal, CH₃CHO) is treated with NaBH₄ followed by aqueous workup. The product is:
A. Ethanoic acid (acetic acid)
B. Ethanediol
C. Acetone (propan-2-one)
D. Ethanol (CH₃CH₂OH)
✓ Answer: D
NaBH₄ reduces the aldehyde carbonyl of ethanal by delivering H⁻ to the carbonyl carbon: CH₃CHO → CH₃CH₂OH. This is a primary alcohol (ethanol). The oxygen receives a proton during aqueous workup. McMurry 8e
Common trap: Ethanoic acid would be the oxidation product, not the reduction product. Reduction adds hydrogen; oxidation adds oxygen or removes hydrogen.
Q17
Propanal (CH₃CH₂CHO) treated with excess methanol (CH₃OH) and an acid catalyst gives:
A. A full acetal (1,1-dimethoxypropane)
B. An ester (methyl propanoate)
C. A hemiacetal only
D. An enol ether
✓ Answer: A
With excess alcohol and an acid catalyst, the aldehyde reacts with two equivalents of methanol. The hemiacetal intermediate (one –OCH₃, one –OH) loses water under acid to form the full acetal: CH₃CH₂CH(OCH₃)₂. Both oxygens are now ethers. McMurry 8e
Common trap: Using only one equivalent of alcohol gives the hemiacetal. "Excess" alcohol + acid catalyst is the key to driving the reaction to the full acetal.
Q18
In an aldol condensation of two molecules of ethanal (CH₃CHO), the β-hydroxy carbonyl product (before dehydration) contains how many carbon atoms?
A. 2
B. 3
C. 4
D. 5
✓ Answer: C
Two ethanal molecules (each with 2 carbons) combine via C–C bond formation at the α-carbon of one and the carbonyl of the other. The product is 3-hydroxybutanal CH₃CH(OH)CH₂CHO, which has 4 carbons. This is the key result of aldol: doubling the carbon chain. McMurry 8e
Common trap: Students subtract 1 for water loss — but the question specifies "before dehydration," so all 4 carbons are retained. The dehydration product (crotonaldehyde) still has 4 carbons anyway.
Q19
Acetone (propan-2-one) gives a positive result in which of the following tests?
A. Tollens' test
B. Fehling's test
C. Iodoform test
D. Benedict's test
✓ Answer: C
Acetone is a methyl ketone CH₃–CO–CH₃. The iodoform test (I₂ / NaOH) is positive for any compound with the CH₃CO– motif, producing the pale yellow CHI₃ precipitate. Acetone gives neither a silver mirror (Tollens') nor a brick-red precipitate (Fehling's/Benedict's) because it is a ketone and cannot be oxidised by these mild reagents. McMurry 8e
Common trap: Students sometimes think all ketones fail all tests. Acetone specifically passes the iodoform test because of its methyl ketone structure.
Q20
Which of the following compounds is an aldehyde?
A. Acetone (propan-2-one)
B. Cyclohexanone
C. Benzaldehyde (phenylmethanal)
D. Butanone (butan-2-one)
✓ Answer: C
Benzaldehyde (C₆H₅–CHO) has its carbonyl group at a terminal carbon bonded to one hydrogen — the defining feature of an aldehyde. Acetone, cyclohexanone, and butanone are all ketones: their carbonyl carbons are flanked by two carbon-containing groups with no hydrogen on the carbonyl carbon. McMurry 8e
Common trap: The "benz-" prefix might make benzaldehyde seem like a different class. The "-aldehyde" in its common name is the definitive clue: it ends in –CHO.
D-1
Carbonyl Group
+
A carbon–oxygen double bond (C=O). The carbonyl carbon is sp² hybridised, trigonal planar, with bond angles of approximately 120°. The oxygen is more electronegative, creating a permanent dipole: the carbon bears a partial positive charge (δ⁺) and is therefore electrophilic, while the oxygen bears δ⁻. This polarisation drives nucleophilic addition reactions. All aldehydes and ketones contain a carbonyl group as their functional group.
McMurry & Ballantine, Fundamentals of GOB Chemistry 8e
D-2
Aldehyde
+
An organic compound in which the carbonyl group is located at the terminal carbon of the carbon chain, giving the structural unit R–CHO (or H–CHO for formaldehyde). The carbonyl carbon bears one hydrogen and one R group (or two hydrogens in methanal). Aldehydes are readily oxidised to carboxylic acids by mild oxidising agents. They give positive results with Tollens', Fehling's, and Benedict's tests. IUPAC names end in "-al."
McMurry & Ballantine, Fundamentals of GOB Chemistry 8e
D-3
Ketone
+
An organic compound in which the carbonyl group is flanked by two carbon-containing groups: R–CO–R'. The carbonyl carbon carries no hydrogen atom. Ketones are more thermodynamically stable than structurally similar aldehydes because of electron donation from two alkyl groups. They are resistant to oxidation by mild reagents (Tollens', Fehling's, Benedict's give negative results) but yield secondary alcohols on reduction. IUPAC names end in "-one" with a locant for the carbonyl position.
McMurry & Ballantine, Fundamentals of GOB Chemistry 8e
D-4
Nucleophilic Addition
+
The characteristic reaction mechanism of aldehydes and ketones. A nucleophile (electron-rich species: H⁻, CN⁻, RMgX, ROH, H₂O) attacks the electrophilic δ⁺ carbonyl carbon. The π bond of C=O breaks heterolytically — both electrons migrate to oxygen — and the carbon rehybridises from sp² (planar) to sp³ (tetrahedral). The oxygen then picks up a proton during workup, giving the final product (alcohol, hemiacetal, cyanohydrin, etc.).
McMurry & Ballantine, Fundamentals of GOB Chemistry 8e
D-5
Acetal
+
The product formed when an aldehyde or ketone reacts with two equivalents of an alcohol under acid catalysis, with removal of water: R–CH(OR')₂. The reaction proceeds via a hemiacetal intermediate (one –OH, one –OR'). The full acetal has two ether linkages on the same carbon. Acetals are stable to base and nucleophiles, making them valuable as protecting groups in multi-step synthesis — they shield the carbonyl during reactions that would otherwise attack it, and are removed later under mild acid hydrolysis.
McMurry & Ballantine, Fundamentals of GOB Chemistry 8e
D-6
Aldol Condensation
+
A base-catalysed reaction in which an enolate ion (formed by deprotonation of the α-hydrogen of one carbonyl compound) attacks the carbonyl carbon of a second carbonyl compound, forming a new C–C bond. The immediate product is a β-hydroxy carbonyl compound (the "aldol" product — from aldehyde + alcohol). On heating, this undergoes dehydration (condensation) to yield an α,β-unsaturated carbonyl compound (e.g., an enone). Aldol condensations are among the most important carbon–carbon bond forming reactions in synthetic and biological chemistry (e.g., citrate synthase in the TCA cycle).
McMurry & Ballantine, Fundamentals of GOB Chemistry 8e
E1
Compare the Tollens', Fehling's, and Benedict's tests for aldehydes. What is oxidised, what is reduced, what is the diagnostic observation, and why do ketones generally fail?
8 marks
Overview
All three are oxidative tests. The aldehyde is oxidised to a carboxylate (or carboxylic acid), while the oxidising agent in the reagent is reduced to produce a visible precipitate or deposit.
Tollens' Test
- Reagent: Diamminesilver(I) complex, [Ag(NH₃)₂]⁺, in dilute ammonia.
- What is oxidised: Aldehyde → carboxylate (R–CHO → R–COO⁻).
- What is reduced: Ag⁺ → Ag⁰ (metallic silver).
- Observation: A bright silver mirror deposits on the inner wall of the test tube.
- Condition: Warm gently; do not overheat (risk of explosive Ag₃N).
Fehling's Test
- Reagent: Fehling's A (CuSO₄) + Fehling's B (NaOH + sodium potassium tartrate); mixed just before use. The tartrate complexes Cu²⁺ to keep it in solution at alkaline pH.
- What is oxidised: Aldehyde → carboxylate.
- What is reduced: Cu²⁺ (blue) → Cu⁺ precipitated as Cu₂O (brick-red).
- Observation: Brick-red precipitate; solution changes from blue → green → brick-red.
Benedict's Test
- Reagent: CuSO₄ + Na₂CO₃ + sodium citrate (single stable solution — advantage over Fehling's).
- Chemistry: Same as Fehling's — Cu²⁺ → Cu₂O brick-red precipitate.
- Clinical use: Historically used for urine glucose detection.
Why Ketones Fail
Ketones cannot be oxidised by these mild, aqueous oxidising agents. To oxidise a ketone, a C–C bond must be broken, requiring very harsh conditions (e.g., hot concentrated KMnO₄). Under the mild conditions of Tollens'/Fehling's/Benedict's, no reaction occurs and no precipitate forms.
Marking (8): Tollens' — reagent + observation (2) · Fehling's — reagent + observation (2) · Benedict's — reagent + note on stability (1) · Oxidised species named correctly for all three (1) · Reduced species named correctly for all three (1) · Explanation of ketone failure (1)
E2
Draw the mechanism for nucleophilic addition of HCN to ethanal, and explain why aldehydes are more reactive than ketones in such reactions.
6 marks
Reaction Overview
HCN adds across the C=O bond of ethanal (CH₃CHO) to form a cyanohydrin. The reaction is typically carried out with NaCN (providing CN⁻) and a trace of base or acid, because HCN itself is a weak acid (pKₐ ≈ 9.2).
Mechanism (Two Steps)
- Step 1 — Nucleophilic attack: The cyanide ion (CN⁻) acts as the nucleophile, attacking the electrophilic δ⁺ carbonyl carbon of ethanal from above or below the plane (the trigonal planar carbonyl). The π bond breaks heterolytically; both electrons move to oxygen, forming an alkoxide intermediate: CH₃CH(CN)O⁻.
- Step 2 — Proton transfer: The alkoxide ion picks up a proton (from solvent H₂O or HCN) to give the cyanohydrin product: CH₃CH(OH)CN (2-hydroxypropanenitrile, also called acetaldehyde cyanohydrin).
Why Aldehydes React Faster Than Ketones
- Steric factor: In an aldehyde (R–CHO), only one R group flanks the carbonyl carbon; the other substituent is a small hydrogen. Nucleophilic approach to the carbonyl carbon faces minimal steric congestion. In a ketone (R–CO–R'), two R groups create greater steric bulk, hindering nucleophilic attack.
- Electronic factor: Alkyl groups are electron-donating (+I effect). In a ketone, two alkyl groups donate electron density to the carbonyl carbon, reducing its partial positive charge (δ⁺) and making it less electrophilic. An aldehyde has only one alkyl donor, so the carbonyl carbon retains greater δ⁺ character and is more susceptible to nucleophilic attack.
Significance
Cyanohydrin formation is biologically important — the enzyme oxynitrilase catalyses the addition of HCN to aldehydes in some plants as a defence mechanism, generating toxic HCN on demand.
Marking (6): CN⁻ as nucleophile identified (1) · Attack at carbonyl carbon drawn/described correctly (1) · Alkoxide intermediate (1) · Proton transfer to give cyanohydrin (1) · Steric explanation for aldehyde > ketone (1) · Electronic explanation (1)
E3
Explain acetal formation from propanal and methanol. Include the hemiacetal intermediate, conditions required, and how acetals serve as protecting groups in synthesis.
7 marks
Overall Reaction
Propanal CH₃CH₂CHO + 2 CH₃OH ⇌ CH₃CH₂CH(OCH₃)₂ + H₂O
Conditions: dry HCl (or H₂SO₄) as acid catalyst; excess methanol; remove water to drive equilibrium forward (e.g., molecular sieves or Dean–Stark trap).
Step 1 — Hemiacetal Formation
- Acid catalyst protonates the carbonyl oxygen of propanal, enhancing the electrophilicity of the carbonyl carbon.
- Methanol (nucleophile) attacks the activated carbonyl carbon.
- Deprotonation gives the hemiacetal: CH₃CH₂CH(OH)(OCH₃). This intermediate has one –OCH₃ (ether) and one –OH group on the same carbon.
Step 2 — Acetal Formation
- Acid protonates the –OH group of the hemiacetal, converting it to a good leaving group (water).
- Water departs, generating an oxocarbenium ion (resonance-stabilised carbocation).
- A second methanol molecule attacks the oxocarbenium ion; deprotonation gives the full acetal: CH₃CH₂CH(OCH₃)₂ (1,1-dimethoxypropane).
Use as a Protecting Group
- Acetals are stable to basic conditions and to nucleophiles (e.g., RMgX, LiAlH₄, NaBH₄, base-mediated reactions).
- In multi-step synthesis, a sensitive aldehyde can be masked as its acetal before performing a reaction incompatible with a free carbonyl. After the desired transformation elsewhere in the molecule, the acetal is unmasked by mild acid hydrolysis (H₃O⁺/H₂O, 25–60 °C) to regenerate the original aldehyde.
- Example scenario: protecting an aldehyde during a Grignard reaction on a distant ketone, preventing the Grignard reagent from attacking both carbonyls.
Marking (7): Overall equation with correct product and conditions (1) · Acid catalysis + activation mechanism (1) · Hemiacetal intermediate correctly described (2) · Oxocarbenium ion / second alcohol attack to full acetal (1) · Stability to base/nucleophiles stated (1) · Acid hydrolysis to regenerate aldehyde (1)
E4
Describe the iodoform test. What structural feature does it detect? Give the reaction with acetone and state the clinical/biochemical significance (e.g. diabetic ketoacidosis).
6 marks
The Iodoform Test
Reagent: I₂ dissolved in dilute NaOH (alkaline iodine solution). A positive result is indicated by the formation of iodoform CHI₃, a pale yellow solid with a characteristic antiseptic odour, that precipitates from the aqueous reaction mixture.
Structural Feature Detected
The test is positive for any compound containing the CH₃–C(=O)– methyl ketone motif. This includes:
- All methyl ketones: acetone, methyl ethyl ketone (butanone), acetophenone, etc.
- Acetaldehyde (CH₃CHO) — the only aldehyde that is positive.
- Ethanol (CH₃CH₂OH) and secondary alcohols of the form CH₃CH(OH)R — these are first oxidised in situ to the corresponding methyl ketone/acetaldehyde by I₂/NaOH before the triiodomethyl cleavage occurs.
Reaction with Acetone
- Step 1 — Trihalogenation: I₂/OH⁻ progressively replaces all three α-hydrogens of the CH₃ group: CH₃COCH₃ → CI₃COCH₃.
- Step 2 — Cleavage: OH⁻ attacks the carbonyl, and the electron-withdrawing CI₃ group (now a good leaving group) departs as CHI₃⁻, which is immediately protonated to iodoform.
- Overall: CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃↓ + CH₃COONa + 3NaI + 3H₂O
Clinical / Biochemical Significance
In diabetic ketoacidosis (DKA), the body cannot utilise glucose for energy (due to insulin deficiency). Fatty acid β-oxidation generates large quantities of acetyl-CoA, which is converted in the liver to ketone bodies — primarily acetone, acetoacetate, and β-hydroxybutyrate. Acetone, a methyl ketone, is volatile and exhaled, causing the characteristic "fruity" or "acetone" breath of DKA. A positive iodoform test on urine or breath condensate can indicate elevated ketone body levels. Historically, the iodoform test was used as a bedside screening for acetonuria.
Marking (6): Reagent (I₂/NaOH) and positive observation (CHI₃ pale yellow ppt) (1) · Structural feature identified: CH₃CO– (1) · Balanced reaction with acetone or correct stepwise description (2) · DKA mechanism — insulin deficiency → ketone bodies (1) · Acetone in DKA breath / iodoform as clinical indicator (1)
E5
Explain the aldol condensation using two molecules of ethanal as substrate. Show the product, the role of the base, and state one biological analogue.
7 marks
Overview
The aldol condensation is a base-catalysed reaction in which the α-carbon of one carbonyl compound forms a new C–C bond with the carbonyl carbon of another. The word "aldol" reflects the dual functionality of the product: an aldehyde and an alcohol.
Step 1 — Enolate Formation
- A base (e.g., dilute NaOH) abstracts one of the α-hydrogens from the methyl group of ethanal (CH₃CHO). The pKₐ of the α-H is approximately 17–20, making it weakly acidic but deprotonatable by a strong enough base.
- This generates the enolate ion: ⁻CH₂CHO (resonance-stabilised; charge delocalised onto oxygen).
Step 2 — Nucleophilic Addition (Aldol Step)
- The nucleophilic α-carbon of the enolate attacks the electrophilic carbonyl carbon of a second ethanal molecule.
- The π bond of the second ethanal breaks; electrons shift to oxygen, forming an alkoxide intermediate.
- Protonation (from water) gives the β-hydroxy aldehyde product: CH₃CH(OH)CH₂CHO — 3-hydroxybutanal (4 carbons).
Step 3 — Dehydration (Condensation Step)
- On warming, the β-hydroxy aldehyde loses water (dehydration/elimination) to give the α,β-unsaturated aldehyde: CH₃CH=CHCHO — but-2-enal (crotonaldehyde). This step is called "condensation" (loss of a small molecule).
Role of the Base
The base serves a catalytic role: it abstracts the α-hydrogen to generate the enolate nucleophile, but is regenerated in the protonation step. The base does not oxidise or reduce anything; it simply deprotonates to activate the α-carbon for C–C bond formation.
Biological Analogue
The aldol reaction has a direct equivalent in cellular metabolism: the enzyme aldolase (in glycolysis) catalyses the reversible aldol cleavage of fructose-1,6-bisphosphate into dihydroxyacetone phosphate (DHAP) and glyceraldehyde-3-phosphate (G3P). This is formally the reverse of an aldol condensation — the enzyme cleaves a β-hydroxy ketone into two smaller carbonyl fragments. The citrate synthase reaction in the TCA cycle is another biological aldol-type condensation, forming a C–C bond between acetyl-CoA and oxaloacetate.
Marking (7): Enolate formation — base abstracts α-H (1) · Enolate structure or description (1) · Nucleophilic attack on second ethanal — C–C bond forms (1) · Correct β-hydroxy aldehyde product (3-hydroxybutanal, 4C) (1) · Dehydration to α,β-unsaturated product named (1) · Role of base as catalytic deprotonator (1) · One valid biological analogue named with brief context (1)