MCAT Chemical and Physical Foundations of Biological Systems Quiz: 5d Carbonyl Chemistry Reactivity
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5d Carbonyl Chemistry ReactivityQuestion 1 of 20

A peptide chemist attempts to protect a ketone-containing side chain during a multi-step aqueous synthesis. The compound contains a simple ketone (R–CO–R') and is treated with ethylene glycol (excess) and catalytic p-toluenesulfonic acid in toluene under reflux with water removal. The central concept is acetal (ketal) formation via nucleophilic addition to carbonyls. Which product is most likely formed from the reaction described?

A cyclic ketal, in which the ketone carbonyl is converted to an acetal carbon bonded to two oxygens from ethylene glycol.
A cyclic epoxide formed by intramolecular substitution of the ketone oxygen by ethylene glycol.
An amide formed by reaction of the ketone with p-toluenesulfonic acid as the nucleophile.
A carboxylic acid formed by oxidation of the ketone during reflux in toluene.
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MCAT Chemical and Physical Foundations of Biological Systems Quiz

MCAT Chemical and Physical Foundations of Biological Systems Quiz: 5d Carbonyl Chemistry Reactivity

Practice 5d Carbonyl Chemistry Reactivity in MCAT Chemical and Physical Foundations of Biological Systems with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on 5d Carbonyl Chemistry Reactivity, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A peptide chemist attempts to protect a ketone-containing side chain during a multi-step aqueous synthesis. The compound contains a simple ketone (R–CO–R') and is treated with ethylene glycol (excess) and catalytic p-toluenesulfonic acid in toluene under reflux with water removal. The central concept is acetal (ketal) formation via nucleophilic addition to carbonyls. Which product is most likely formed from the reaction described?

  1. A cyclic ketal, in which the ketone carbonyl is converted to an acetal carbon bonded to two oxygens from ethylene glycol. (correct answer)
  2. A cyclic epoxide formed by intramolecular substitution of the ketone oxygen by ethylene glycol.
  3. An amide formed by reaction of the ketone with p-toluenesulfonic acid as the nucleophile.
  4. A carboxylic acid formed by oxidation of the ketone during reflux in toluene.

Explanation: This question tests Carbonyl Chemistry and Reactivity (5D), focusing on ketal formation as a carbonyl protection strategy. Carbonyl groups in ketones react with diols under acid catalysis to form cyclic ketals through nucleophilic addition followed by intramolecular cyclization with water removal. In the given reaction, the ketone is treated with ethylene glycol and acid catalyst under dehydrating conditions. The correct choice, A, is expected because the diol attacks the protonated carbonyl to form a hemiketal intermediate, which then cyclizes to the stable five-membered cyclic ketal with loss of water. Choice B is incorrect as it suggests epoxide formation, which requires different reagents and doesn't involve carbonyl chemistry. To apply this concept, recognize that ketals serve as protecting groups for carbonyls, forming under acid catalysis with water removal and cleaving under aqueous acidic conditions.

Question 2

A formulation scientist tests reversible covalent inhibition via carbonyl chemistry. A protease is incubated with an electrophilic inhibitor containing either a nitrile (R–C≡N) or an aldehyde (R–CHO). Under identical aqueous conditions (pH 7.4, 37°C), the aldehyde inhibitor shows time-dependent loss of enzyme activity that is partially reversed upon dilution, consistent with a reversible covalent adduct at the active-site serine. The central concept is nucleophilic addition to carbonyls. Which adduct is most consistent with serine attacking the aldehyde?

  1. A tetrahedral hemiacetal (enzyme–O–CH(OH)–R) formed by addition of serine oxygen to the aldehyde carbonyl. (correct answer)
  2. An amide (enzyme–NH–C(=O)–R) formed by acyl substitution on the aldehyde.
  3. An acetal (enzyme–O–CH(OR)–R) requiring two equivalents of serine oxygen to add to the aldehyde.
  4. A carboxylate salt (R–COO−^-) formed by base-promoted oxidation of the aldehyde in buffer.

Explanation: This question tests Carbonyl Chemistry and Reactivity (5D), focusing on reversible nucleophilic addition to aldehydes by serine residues. Carbonyl groups in aldehydes are electrophilic and undergo nucleophilic addition with alcohols to form hemiacetals, which are reversible under physiological conditions. In the given reaction, the aldehyde inhibitor forms a time-dependent, reversible adduct with the active-site serine. The correct choice, A, is expected because serine's hydroxyl group attacks the aldehyde carbonyl to form a tetrahedral hemiacetal, which can dissociate upon dilution. Choice C is incorrect as it suggests acetal formation, which requires two alcohol equivalents and acid catalysis not present under these conditions. To apply this concept, remember that hemiacetal formation is reversible and occurs readily between aldehydes and alcohols at neutral pH, making it useful for reversible covalent inhibition.

Question 3

A biochemistry lab monitors keto–enol tautomerization of pyruvate (CH3–CO–COO−) in D2O at pD 7.0. Over time, they observe deuterium incorporation at the methyl group adjacent to the ketone carbonyl (by 1H NMR signal loss), without net change in the carboxylate. The interpretation uses the reactivity concept of carbonyl enolization at the α-carbon.

Which statement best supports the observed labeling pattern?

  1. Deuterium incorporates at the α-carbon because reversible enolization allows exchange of α-hydrogens with solvent (correct answer)
  2. Deuterium incorporates at the carboxylate carbon because nucleophilic addition of D2O occurs at the carbonyl carbon of the carboxylate
  3. No incorporation should occur because carboxylates prevent any resonance stabilization of an enolate
  4. Deuterium incorporates at the carbonyl carbon because tautomerization proceeds through a carbocation intermediate

Explanation: This question tests Carbonyl Chemistry and Reactivity (5D), focusing on α-hydrogen exchange through enolization. Carbonyl groups undergo reversible enolization, where α-hydrogens become acidic and can exchange with solvent through enol or enolate intermediates. In the given reaction, pyruvate's methyl group exchanges H for D in D₂O, indicating enolization at the α-position. The correct choice, A, is expected because the α-hydrogens of pyruvate are acidic due to stabilization by both the ketone and carboxylate groups, allowing reversible deprotonation and reprotonation with D₂O. Choice B is incorrect as it suggests deuterium adds to the carboxylate carbon, but carboxylates are not electrophilic and don't undergo nucleophilic addition. To apply this concept, remember that α-hydrogen exchange is diagnostic for enolization and occurs readily for carbons between two electron-withdrawing groups.

Question 4

To interpret stability of a drug candidate in plasma, a team compares hydrolysis of two carbonyl-containing functional groups at pH 7.4: an amide linkage vs an ester linkage, both otherwise similar in size and substitution. They note the ester hydrolyzes measurably over hours, while the amide is largely unchanged. The analysis focuses on carbonyl reactivity toward nucleophilic acyl substitution.

Which explanation is most consistent with the observed difference?

  1. Amides are more reactive because nitrogen is more electronegative than oxygen, increasing carbonyl electrophilicity
  2. Esters are more reactive because the amide carbonyl is resonance-stabilized by the nitrogen lone pair, reducing electrophilicity (correct answer)
  3. Amides hydrolyze faster because their leaving group (NH2−) is more stable than alkoxide at neutral pH
  4. Both should hydrolyze at the same rate; the difference must arise from unequal buffer concentrations

Explanation: This question tests Carbonyl Chemistry and Reactivity (5D), focusing on relative reactivity of esters versus amides in nucleophilic acyl substitution. Carbonyl groups in carboxylic acid derivatives undergo nucleophilic acyl substitution, with reactivity determined by resonance stabilization and leaving group ability. In the given comparison, esters hydrolyze faster than amides at physiological pH due to differences in carbonyl electrophilicity. The correct choice, B, is expected because the nitrogen lone pair in amides donates electron density to the carbonyl through resonance more effectively than oxygen in esters, reducing the amide carbonyl's electrophilicity. Choice A is incorrect as it claims nitrogen's electronegativity increases reactivity, but nitrogen's lone pair donation actually decreases carbonyl electrophilicity. To apply this concept, remember that amide resonance stabilization makes them the least reactive carboxylic acid derivatives, explaining their biological stability.

Question 5

A lab evaluates base-catalyzed enolization of carbonyl compounds as a predictor of racemization risk in a chiral drug candidate. Compound X is 2-butanone, and Compound Y is tert-butyl methyl ketone (pinacolone). Each (10 mM) is placed in D2O\mathrm{D_2O} with 10 mM NaOD at 25°C. After 5 minutes, 1^1H NMR shows substantial loss of the α\alpha-C–H signal for X but minimal change for Y. Which interpretation is most consistent with carbonyl enolization?

  1. X exchanges faster because it has more acidic α\alpha-hydrogens and less steric hindrance to deprotonation than Y. (correct answer)
  2. Y exchanges faster because tert-butyl groups donate electron density, increasing α\alpha-hydrogen acidity.
  3. Neither exchanges because ketones cannot form enolates in water.
  4. X exchanges faster because it undergoes nucleophilic acyl substitution at the carbonyl carbon by OD−\mathrm{OD^-}.

Explanation: This question tests Carbonyl Chemistry and Reactivity (5D), focusing on α-hydrogen acidity and enolization rates. Carbonyl compounds with α-hydrogens can form enolates under basic conditions, with the rate depending on both α-hydrogen acidity and steric accessibility. In the given reaction, 2-butanone (X) shows rapid H/D exchange while pinacolone (Y) shows minimal exchange under identical conditions. The correct choice, A, is expected because 2-butanone has less sterically hindered α-hydrogens that are more easily deprotonated by base compared to pinacolone's tert-butyl-adjacent position. Choice B is incorrect as it misunderstands electronic effects - alkyl groups are electron-donating through induction, which decreases α-hydrogen acidity. To apply this concept, evaluate both electronic effects (which influence acidity) and steric effects (which influence accessibility) when predicting enolization rates.

Question 6

An organic synthesis step in a radiotracer preparation uses nucleophilic addition of hydride to a carbonyl. A solution of cyclohexanone is treated with NaBH4_4 in methanol at 0°C, then quenched with water. The chemist expects reduction of the carbonyl without changing the carbon skeleton. Which molecule is most likely formed from the reaction described?

  1. Cyclohexanol, formed by hydride addition to the ketone followed by protonation. (correct answer)
  2. Cyclohexene, formed by dehydration of the ketone under basic conditions.
  3. Cyclohexanecarboxylic acid, formed by oxidation of the ketone during quench.
  4. A methyl ketal, formed by acetalization of cyclohexanone in methanol without acid.

Explanation: This question tests Carbonyl Chemistry and Reactivity (5D), focusing on hydride reduction of ketones. Carbonyl groups undergo nucleophilic addition with hydride donors like NaBH4, converting ketones to secondary alcohols through irreversible hydride transfer. In the given reaction, cyclohexanone is treated with NaBH4 in methanol, followed by aqueous quench. The correct choice, A, is expected because NaBH4 selectively reduces the ketone carbonyl to an alcohol without affecting the carbon skeleton or causing elimination. Choice B is incorrect as it suggests dehydration under basic reducing conditions, which would require acid and heat. To apply this concept, recognize that NaBH4 is a mild, selective reducing agent for aldehydes and ketones that preserves other functional groups and doesn't cause skeletal rearrangements.

Question 7

To assess carbonyl activation, an analyst measures initial rates for nucleophilic addition of cyanide to two substrates at 25°C in aqueous buffer (pH 9.5). Substrate P is propanal (CH3_3CH2_2CHO) and substrate Q is 2-propanone (acetone). With [CN−^-] held constant, the initial rate for P is higher than for Q. No other reagents are present. Which explanation is most consistent with carbonyl reactivity?

  1. Propanal reacts faster because aldehydes are generally more electrophilic and less sterically hindered than ketones. (correct answer)
  2. Acetone reacts faster because its two alkyl groups withdraw electron density, increasing electrophilicity.
  3. Propanal reacts faster because cyanide performs nucleophilic acyl substitution on aldehydes but not ketones.
  4. The rate difference implies the reaction is under equilibrium control and P forms the more stable enol tautomer.

Explanation: This question tests Carbonyl Chemistry and Reactivity (5D), focusing on the relative electrophilicity of aldehydes versus ketones. Carbonyl groups undergo nucleophilic addition with rates determined by both electronic and steric factors - aldehydes are generally more reactive than ketones. In the given reaction, propanal shows a higher initial rate with cyanide than acetone under identical conditions. The correct choice, A, is expected because aldehydes have only one electron-donating alkyl group (versus two in ketones) and less steric hindrance, making them more electrophilic and accessible to nucleophiles. Choice B is incorrect as it mischaracterizes alkyl groups as electron-withdrawing when they are actually electron-donating through induction. To apply this concept, remember the reactivity order for nucleophilic addition: aldehydes > ketones due to both electronic and steric effects.

Question 8

A researcher compares two carbonyl compounds for propensity to undergo aldol condensation under basic conditions. Sample 1 is benzaldehyde (Ph–CHO), and Sample 2 is acetaldehyde (CH3_3–CHO). Each (0.10 M) is treated separately with 0.10 M NaOH in water at room temperature for 15 minutes. A new C–C bond product is readily observed for Sample 2, while Sample 1 shows no analogous self-aldol product. Which conclusion is most consistent with the observations?

  1. Benzaldehyde fails to self-aldol because it lacks α\alpha-hydrogens needed to form an enolate nucleophile. (correct answer)
  2. Benzaldehyde fails to self-aldol because aromatic rings cannot stabilize carbonyls in water.
  3. Acetaldehyde reacts because aldehydes undergo nucleophilic acyl substitution with hydroxide to form esters.
  4. Acetaldehyde reacts because hydroxide converts it into an acetal that then couples to another acetal.

Explanation: This question tests Carbonyl Chemistry and Reactivity (5D), focusing on the requirement for α-hydrogens in aldol condensation reactions. Carbonyl compounds undergo aldol reactions only if they possess α-hydrogens that can be deprotonated to form enolate nucleophiles under basic conditions. In the given reaction, acetaldehyde undergoes self-aldol while benzaldehyde shows no reaction under identical conditions. The correct choice, A, is expected because benzaldehyde lacks α-hydrogens entirely (the carbon adjacent to the carbonyl is part of the aromatic ring), preventing enolate formation and subsequent aldol reaction. Choice B is incorrect as it misunderstands the role of aromatic rings - they actually stabilize carbonyls through conjugation. To apply this concept, check for α-hydrogens when predicting aldol reactivity, as compounds without them can only serve as electrophiles, not nucleophiles.

Question 9

To probe carbonyl nucleophilic addition in a drug–protein adduct model, researchers incubate an aldehyde-containing ligand (R–CHO) with a lysine side-chain mimic (n-butylamine, 20 mM) at pH 7.4. After 1 hour, IR spectroscopy shows decreased C=O stretch intensity and appearance of a new C=N stretch. No external reducing agent is present. Which product is most likely formed from the reaction described?

  1. A hemiacetal (R–CH(OH)–OR) formed from addition of buffer alcohols to the aldehyde.
  2. An imine (Schiff base), R–CH=N–Bu, formed by addition of amine followed by dehydration. (correct answer)
  3. An amide, R–C(=O)–NH–Bu, formed by nucleophilic acyl substitution on the aldehyde.
  4. A carboxylate, R–COO−^-, formed by oxidation of the aldehyde by dissolved oxygen under neutral conditions.

Explanation: This question tests Carbonyl Chemistry and Reactivity (5D), focusing on imine (Schiff base) formation from aldehydes and amines. Carbonyl groups undergo nucleophilic addition with primary amines, followed by dehydration to form C=N double bonds under mild conditions. In the given reaction, the aldehyde reacts with n-butylamine at physiological pH, showing loss of C=O stretch and appearance of C=N stretch. The correct choice, B, is expected because aldehydes readily form imines with primary amines through addition-elimination, with water loss occurring spontaneously at neutral pH. Choice C is incorrect as it suggests acyl substitution, which requires a leaving group that aldehydes lack - they undergo addition reactions instead. To apply this concept, remember that aldehydes and ketones form imines with primary amines without requiring additional reagents, distinguishing them from carboxylic acid derivatives.

Question 10

A biochemical model system is used to mimic an early step in glycolysis: formation of a carbon–carbon bond via an aldol reaction. In vitro, dihydroxyacetone phosphate (DHAP, a ketone) is mixed with glyceraldehyde-3-phosphate (G3P, an aldehyde) in aqueous solution at pH 8.0 with a catalytic amount of a lysine-containing peptide that transiently forms an enamine with DHAP. After 10 minutes, the major product is a phosphorylated hexose (aldol addition product) rather than a dehydration product. Which statement is most consistent with the carbonyl reactivity described?

  1. The peptide most likely increases the electrophilicity of DHAP by converting it into an enolate that is attacked by G3P.
  2. The peptide most likely converts DHAP into a nucleophilic enamine that attacks the electrophilic carbonyl carbon of G3P. (correct answer)
  3. The peptide most likely oxidizes G3P to a carboxylic acid, enabling acyl substitution to form the C–C bond.
  4. The peptide most likely protonates G3P to form an acetal directly with DHAP, bypassing carbon–carbon bond formation.

Explanation: This question tests Carbonyl Chemistry and Reactivity (5D), focusing on enamine-catalyzed aldol reactions. Carbonyl groups can be converted to nucleophilic enamines through condensation with amines, enabling them to attack electrophilic carbonyls in aldol reactions. In the given reaction, DHAP forms an enamine with the lysine-containing peptide, which then attacks the electrophilic aldehyde carbonyl of G3P. The correct choice, B, is expected because enamine formation converts the normally electrophilic ketone (DHAP) into a nucleophile that can attack the aldehyde. Choice A is incorrect as it confuses enolates (anionic) with enamines (neutral) and reverses the nucleophile-electrophile roles. To apply this concept, recognize that amine catalysts enable aldol reactions by forming nucleophilic enamines from ketones, which then attack aldehydes or other ketones.

Question 11

In a medicinal chemistry optimization, a team compares nucleophilic addition to two carbonyl-containing fragments under identical conditions (25°C, pH 7.4 buffer). Fragment 1 is acetone, and Fragment 2 is acetamide. Each is incubated separately with 50 mM methanol and a catalytic amount of HCl (final [HCl] = 1 mM) for 30 minutes, then quenched. LC-MS detects a new species only in the acetone sample consistent with addition of methanol. The central reactivity concept is nucleophilic addition to carbonyls. Which conclusion is most consistent with the observed difference in reactivity?

  1. Acetamide undergoes faster nucleophilic addition because resonance donation from nitrogen increases carbonyl electrophilicity.
  2. Acetone forms a methanol addition product more readily because amide carbonyls are less electrophilic due to n→π∗n\rightarrow\pi^* resonance stabilization. (correct answer)
  3. Acetone fails to react because ketones cannot undergo nucleophilic addition without strong base.
  4. Acetamide forms the methanol addition product, but it is not detected because it spontaneously oxidizes under acidic conditions.

Explanation: This question tests Carbonyl Chemistry and Reactivity (5D), focusing on the relative electrophilicity of ketones versus amides. Carbonyl groups undergo nucleophilic addition, with reactivity determined by the electron density at the carbonyl carbon - ketones are more electrophilic than amides due to nitrogen's resonance donation in amides. In the given reaction, acetone (ketone) and acetamide are each treated with methanol under acidic conditions, but only acetone shows addition product formation. The correct choice, B, is expected because the nitrogen in acetamide donates electron density through n→π* resonance, making the carbonyl less electrophilic and less reactive toward nucleophiles. Choice A is incorrect as it suggests resonance increases electrophilicity when it actually decreases it. To apply this concept, remember that amides are the least reactive carbonyl derivatives due to resonance stabilization, while ketones readily undergo nucleophilic addition under mild conditions.

Question 12

A pharmacology group studies formation of acetals as prodrugs. They treat an aldehyde-containing compound R–CHO with excess ethylene glycol (HO–CH2_2CH2_2–OH) and catalytic acid, removing water as it forms. The product is stable to base but hydrolyzes back to the aldehyde in dilute acid. The key concept is acetal formation from carbonyls. Which functional group is most consistent with the protected product?

  1. Cyclic acetal (1,3-dioxolane) at the former carbonyl carbon (correct answer)
  2. Carboxylate salt formed by deprotonation of the aldehyde
  3. Amide formed by reaction of aldehyde with glycol
  4. Epoxide formed by intramolecular substitution

Explanation: This question tests Carbonyl Chemistry and Reactivity (5D), focusing on acetal formation for carbonyl protection. Carbonyl groups react with diols under acidic conditions to form cyclic acetals, stable to base but hydrolysable in acid. In the given reaction, the aldehyde interacts with ethylene glycol and catalytic acid, forming a product that protects the carbonyl. The correct choice, A, is expected due to the 1,3-dioxolane cyclic acetal, matching stability and hydrolysis behavior. Choice B is incorrect as it assumes deprotonation to carboxylate, but aldehydes lack acidic protons. To apply this concept, ensure that reaction conditions align with expected reactivity, and distinguish between similar functional groups like acetals and hemiacetals, noting acetals require water removal.

Question 13

A kinetics study examines the rate-limiting step of acid-catalyzed acetal formation from an aldehyde and methanol. Under conditions of excess methanol and catalytic H+^+, the initial rate increases with increasing aldehyde concentration but is relatively insensitive to methanol concentration. The reactivity concept is nucleophilic addition to a protonated carbonyl. Which step is most likely rate-limiting?

  1. Protonation of the carbonyl oxygen by H+^+
  2. Nucleophilic attack of methanol on the protonated carbonyl carbon (correct answer)
  3. Rapid deprotonation of the oxonium intermediate by solvent
  4. Final neutralization during aqueous workup

Explanation: This question tests Carbonyl Chemistry and Reactivity (5D), focusing on mechanism of acid-catalyzed acetal formation. Carbonyl groups are protonated in acid, increasing electrophilicity for nucleophilic attack by alcohols in the rate-limiting step. In the given reaction, the rate depends on aldehyde concentration but not methanol, under excess alcohol. The correct choice, B, is expected due to nucleophilic attack on protonated carbonyl being rate-limiting. Choice A is incorrect as it assumes protonation is rate-limiting, but it's fast and pre-equilibrium. To apply this concept, ensure that reaction conditions align with expected reactivity, and distinguish between similar functional groups like aldehydes and ketones, noting similar mechanisms but slower for ketones.

Question 14

An analytical chemist compares relative electrophilicity of carbonyl derivatives by measuring initial rates of reaction with methanol (0.50 M) in the presence of catalytic HCl (same conditions for each). Substrates: acetyl chloride (CH3_3COCl), methyl acetate (CH3_3COOCH3_3), and acetamide (CH3_3CONH2_2). The observed rate order is acetyl chloride  methyl acetate  acetamide. Based on carbonyl reactivity, which conclusion is most consistent with these data?

  1. Leaving-group basicity controls acyl substitution rates (Cl−^- best, NH2−_2^- worst) (correct answer)
  2. Amides react fastest because nitrogen donates electron density to activate the carbonyl
  3. Esters react faster than acid chlorides due to stronger C–O bonds
  4. All three react at identical rates because they share a C=O group

Explanation: This question tests Carbonyl Chemistry and Reactivity (5D), focusing on relative reactivity in nucleophilic acyl substitution. Carbonyl derivatives react via nucleophilic addition-elimination, with rates depending on leaving group ability and carbonyl electrophilicity. In the given reaction, acetyl chloride, methyl acetate, and acetamide interact with methanol under acidic conditions, showing rate order based on leaving group basicity. The correct choice, A, is expected due to Cl⁻ being the best leaving group and NH₂⁻ the worst, explaining the observed rates. Choice B is incorrect as it assumes amides are fastest, ignoring resonance stabilization that decreases reactivity. To apply this concept, ensure that reaction conditions align with expected reactivity, and distinguish between similar functional groups like acid chlorides and esters, noting acid chlorides are most reactive.

Question 15

An analytical chemist monitors hydration (gem-diol formation) of carbonyls in water by 1H^1\mathrm{H} NMR. Two samples at 25°C contain either acetaldehyde (\ceCH3CHO\ce{CH3CHO}) or acetone (\ce(CH3)2CO\ce{(CH3)2CO}), each 50 mM in pure water. The central concept is equilibrium nucleophilic addition of water to a carbonyl; aldehydes typically hydrate more than ketones due to sterics and electronic effects.

Which observation is most consistent with carbonyl reactivity in these samples?

  1. Acetone shows a larger fraction of gem-diol because ketones stabilize tetrahedral hydrates better than aldehydes
  2. Acetaldehyde shows a larger fraction of gem-diol because it is less hindered and more electrophilic (correct answer)
  3. Both show complete conversion to gem-diols because water irreversibly adds to carbonyls
  4. Neither shows gem-diol because hydration requires a hydride donor such as \ceNaBH4\ce{NaBH4}

Explanation: This question tests Carbonyl Chemistry and Reactivity (5D), focusing on carbonyl hydration equilibria. Carbonyl groups undergo reversible nucleophilic addition of water to form gem-diols (hydrates), with aldehydes typically showing greater hydration than ketones. In the given reaction, water adds to either acetaldehyde or acetone to form tetrahedral hydrates. The correct choice, B, is expected because acetaldehyde, being less sterically hindered and more electrophilic than acetone, shifts the hydration equilibrium further toward the gem-diol form. Choice A is incorrect as it reverses the trend - ketones actually destabilize tetrahedral intermediates due to steric crowding. To apply this concept, remember that carbonyl hydration is an equilibrium process, and that electron-withdrawing groups or less steric hindrance favor hydrate formation, explaining why formaldehyde exists predominantly as its hydrate in aqueous solution.

Question 16

A researcher is designing a selective protection strategy for a carbohydrate-derived aldehyde that also contains an alcohol. They treat the substrate with ethylene glycol and catalytic p-toluenesulfonic acid in toluene, removing water as it forms. The key transformation is described by the reactivity concept of acetal formation from a carbonyl.

Which outcome is most consistent with these conditions?

  1. Formation of a cyclic acetal at the aldehyde carbonyl, increasing resistance to nucleophilic addition at that carbon (correct answer)
  2. Conversion of the aldehyde to an amide via nucleophilic acyl substitution by ethylene glycol
  3. Direct dehydration of the alcohol to form an alkene because acid catalysis favors elimination over carbonyl chemistry
  4. Reduction of the aldehyde to a primary alcohol because toluene acts as a hydride donor under acid

Explanation: This question tests Carbonyl Chemistry and Reactivity (5D), focusing on acetal formation as a carbonyl protection strategy. Carbonyl groups undergo nucleophilic addition with diols under acid catalysis, forming cyclic acetals that are stable to nucleophiles and bases. In the given reaction, the aldehyde reacts with ethylene glycol in the presence of acid catalyst, with water removal driving the equilibrium toward acetal formation. The correct choice, A, is expected because acetals protect carbonyls by converting the sp² carbon to a tetrahedral sp³ carbon bearing two alkoxy groups, eliminating electrophilicity. Choice D is incorrect as it suggests toluene acts as a hydride donor, but aromatic solvents are inert under these conditions. To apply this concept, remember that acetal formation is reversible and requires acid catalysis plus water removal, making it an ideal protecting group for carbonyls.

Question 17

A peptide sample containing lysine residues is incubated with an aldehyde-containing drug fragment to form a reversible imine (Schiff base). The reaction is run at 25°C in aqueous buffer at pH 6.0. In a follow-up experiment, the same incubation is performed but with 10 mM sodium borohydride (\ceNaBH4\ce{NaBH4}) added at the start (pH maintained at 6.0). The central concept is nucleophilic addition of amines to carbonyls and how reduction can trap intermediates.

Which outcome is most consistent with carbonyl reactivity when \ceNaBH4\ce{NaBH4} is present from the start?

  1. More stable secondary amine linkage forms because the imine (or iminium) is reduced to a C–N single bond (correct answer)
  2. No covalent modification occurs because \ceNaBH4\ce{NaBH4} converts the aldehyde into a carboxylic acid
  3. Less modification occurs because \ceNaBH4\ce{NaBH4} oxidizes lysine to an aldehyde, preventing imine formation
  4. Only an ester linkage forms because aldehydes undergo acyl substitution with amines in water

Explanation: This question tests Carbonyl Chemistry and Reactivity (5D), focusing on reductive trapping of imine intermediates. Carbonyl groups react with primary amines to form imines (Schiff bases), which can be reduced to stable secondary amines by mild reducing agents like NaBH4. In the given reaction, lysine residues form reversible imines with the aldehyde drug fragment, but when NaBH4 is present, it reduces the C=N double bond to a C-N single bond. The correct choice, A, is expected because NaBH4 converts the reversible imine linkage into an irreversible secondary amine, creating a more stable covalent modification. Choice B is incorrect as NaBH4 reduces aldehydes to alcohols, not carboxylic acids. To apply this concept, understand that reductive amination combines imine formation with immediate reduction, converting reversible modifications into stable ones, which is useful for bioconjugation and crosslinking applications.

Question 18

An organic synthesis team performs a crossed aldol reaction to build a single new C–C bond. They combine cyclohexanone with 1.1 equiv LDA at −78°C in THF to form an enolate, then add benzaldehyde. After aqueous workup, the major product is a β-hydroxy ketone. This relies on the reactivity concept of enolate nucleophilic addition to an aldehyde carbonyl.

Which outcome is most expected after workup?

  1. A β-hydroxy ketone formed by addition of the cyclohexanone enolate to benzaldehyde, followed by protonation (correct answer)
  2. An α,β-unsaturated aldehyde formed immediately because dehydration is unavoidable at −78°C
  3. A diol formed by double hydride addition to benzaldehyde because LDA acts as a reducing agent
  4. A benzyl ester formed by nucleophilic acyl substitution of benzaldehyde by the enolate oxygen

Explanation: This question tests Carbonyl Chemistry and Reactivity (5D), focusing on directed aldol addition using preformed enolates. Carbonyl groups undergo nucleophilic addition when attacked by enolates, with LDA generating kinetic enolates that add to aldehydes preferentially over ketones. In the given reaction, cyclohexanone is deprotonated by LDA at -78°C to form an enolate, which then attacks benzaldehyde. The correct choice, A, is expected because the enolate adds to the aldehyde carbonyl forming a new C-C bond, and aqueous workup protonates the alkoxide to yield a β-hydroxy ketone. Choice B is incorrect as it suggests immediate dehydration at -78°C, but such low temperatures prevent elimination reactions. To apply this concept, recognize that LDA forms enolates quantitatively, and aldehydes are more reactive than ketones toward enolate addition, allowing selective crossed aldol reactions.

Question 19

A formulation scientist evaluates whether a ketone-containing fragrance molecule will form a bisulfite adduct during purification. The ketone (2-butanone) is shaken with aqueous NaHSO3 at room temperature. After equilibration, a new, more water-soluble species is detected. The transformation is described by the reactivity concept of nucleophilic addition to a carbonyl to form an addition product.

Which structure is most consistent with the major new species?

  1. An enamine formed by reaction of 2-butanone with bisulfite followed by dehydration
  2. A bisulfite addition product in which HSO3− has added to the carbonyl carbon, yielding a sulfonate-bearing alcohol (correct answer)
  3. A sulfate ester formed by substitution of the ketone oxygen with HSO3− to give an O–S bond and loss of CO
  4. A carboxylic acid formed by oxidation of 2-butanone by bisulfite under neutral conditions

Explanation: This question tests Carbonyl Chemistry and Reactivity (5D), focusing on bisulfite addition to ketones. Carbonyl groups undergo nucleophilic addition with bisulfite ion (HSO₃⁻), forming water-soluble addition products useful for purification. In the given reaction, 2-butanone reacts with sodium bisulfite to form a more water-soluble species. The correct choice, B, is expected because HSO₃⁻ acts as a nucleophile, attacking the carbonyl carbon to form a tetrahedral intermediate bearing both hydroxyl and sulfonate groups on the same carbon. Choice C is incorrect as it suggests substitution of the carbonyl oxygen with loss of CO, which would require breaking strong C-C bonds under mild conditions. To apply this concept, remember that bisulfite addition is reversible and specific for aldehydes and methyl ketones, making it useful for separation and purification.

Question 20

A lab investigates why carboxylic acids do not typically undergo direct nucleophilic addition like aldehydes/ketones. Benzoic acid (Ph–COOH) is treated with NaBH4_4 in methanol under conditions that readily reduce benzaldehyde. No significant reduction is observed. The key concept is carbonyl reactivity differences among functional groups. Which explanation is most consistent with the result?

  1. The carboxylic acid exists largely as a resonance-stabilized carboxylate/acid, reducing electrophilicity and favoring acid–base reactions (correct answer)
  2. NaBH4_4 can only reduce alkenes, not carbonyls
  3. Benzoic acid is reduced faster than benzaldehyde but the product is volatile and lost
  4. Reduction fails because benzoic acid lacks a carbonyl group

Explanation: This question tests Carbonyl Chemistry and Reactivity (5D), focusing on differences in nucleophilic addition reactivity between carboxylic acids and aldehydes or ketones. Carbonyl groups in aldehydes and ketones are electrophilic and readily undergo nucleophilic addition with reducing agents like NaBH₄, but carboxylic acids are less reactive due to resonance stabilization and a tendency to favor acid-base reactions over addition. In the given reaction, benzoic acid is treated with NaBH₄ in methanol, conditions that reduce benzaldehyde to benzyl alcohol, but no significant reduction of the carboxylic acid occurs. The correct choice, A, is expected due to the carboxylic acid existing in equilibrium as a resonance-stabilized carboxylate ion in the presence of basic NaBH₄, which reduces the carbonyl's electrophilicity and promotes deprotonation rather than hydride addition. Choice B is incorrect as it assumes NaBH₄ reduces alkenes instead of carbonyls, which is the opposite of its actual selectivity. To apply this concept, ensure that reaction conditions align with expected reactivity, and distinguish between similar functional groups by considering their electrophilicity and competing reaction pathways like acid-base equilibria.