Carbonyl Chemistry and Reactivity (5D)
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MCAT Chemical and Physical Foundations of Biological Systems › Carbonyl Chemistry and Reactivity (5D)
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 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.
A cyclic ketal, in which the ketone carbonyl is converted to an acetal carbon bonded to two oxygens from ethylene glycol.
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.
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?
A tetrahedral hemiacetal (enzyme–O–CH(OH)–R) formed by addition of serine oxygen to the aldehyde carbonyl.
An amide (enzyme–NH–C(=O)–R) formed by acyl substitution on the aldehyde.
An acetal (enzyme–O–CH(OR)–R) requiring two equivalents of serine oxygen to add to the aldehyde.
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.
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?
Deuterium incorporates at the α-carbon because reversible enolization allows exchange of α-hydrogens with solvent
Deuterium incorporates at the carbonyl carbon because tautomerization proceeds through a carbocation intermediate
No incorporation should occur because carboxylates prevent any resonance stabilization of an enolate
Deuterium incorporates at the carboxylate carbon because nucleophilic addition of D2O occurs at the carbonyl carbon of the carboxylate
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.
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?
Both should hydrolyze at the same rate; the difference must arise from unequal buffer concentrations
Esters are more reactive because the amide carbonyl is resonance-stabilized by the nitrogen lone pair, reducing electrophilicity
Amides hydrolyze faster because their leaving group (NH2−) is more stable than alkoxide at neutral pH
Amides are more reactive because nitrogen is more electronegative than oxygen, increasing carbonyl electrophilicity
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.
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 $\mathrm{D_2O}$ with 10 mM NaOD at 25°C. After 5 minutes, $^1$H NMR shows substantial loss of the $\alpha$-C–H signal for X but minimal change for Y. Which interpretation is most consistent with carbonyl enolization?
X exchanges faster because it undergoes nucleophilic acyl substitution at the carbonyl carbon by $\mathrm{OD^-}$.
X exchanges faster because it has more acidic $\alpha$-hydrogens and less steric hindrance to deprotonation than Y.
Neither exchanges because ketones cannot form enolates in water.
Y exchanges faster because tert-butyl groups donate electron density, increasing $\alpha$-hydrogen acidity.
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.
An organic synthesis step in a radiotracer preparation uses nucleophilic addition of hydride to a carbonyl. A solution of cyclohexanone is treated with NaBH$_4$ 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?
Cyclohexanol, formed by hydride addition to the ketone followed by protonation.
Cyclohexene, formed by dehydration of the ketone under basic conditions.
A methyl ketal, formed by acetalization of cyclohexanone in methanol without acid.
Cyclohexanecarboxylic acid, formed by oxidation of the ketone during quench.
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.
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 (CH$_3$CH$_2$CHO) 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?
The rate difference implies the reaction is under equilibrium control and P forms the more stable enol tautomer.
Propanal reacts faster because cyanide performs nucleophilic acyl substitution on aldehydes but not ketones.
Acetone reacts faster because its two alkyl groups withdraw electron density, increasing electrophilicity.
Propanal reacts faster because aldehydes are generally more electrophilic and less sterically hindered than ketones.
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.
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 (CH$_3$–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?
Benzaldehyde fails to self-aldol because aromatic rings cannot stabilize carbonyls in water.
Acetaldehyde reacts because hydroxide converts it into an acetal that then couples to another acetal.
Acetaldehyde reacts because aldehydes undergo nucleophilic acyl substitution with hydroxide to form esters.
Benzaldehyde fails to self-aldol because it lacks $\alpha$-hydrogens needed to form an enolate nucleophile.
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.
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?
An imine (Schiff base), R–CH=N–Bu, formed by addition of amine followed by dehydration.
A hemiacetal (R–CH(OH)–OR) formed from addition of buffer alcohols to the aldehyde.
An amide, R–C(=O)–NH–Bu, formed by nucleophilic acyl substitution on the aldehyde.
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.
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?
The peptide most likely oxidizes G3P to a carboxylic acid, enabling acyl substitution to form the C–C bond.
The peptide most likely increases the electrophilicity of DHAP by converting it into an enolate that is attacked by G3P.
The peptide most likely protonates G3P to form an acetal directly with DHAP, bypassing carbon–carbon bond formation.
The peptide most likely converts DHAP into a nucleophilic enamine that attacks the electrophilic carbonyl carbon of G3P.
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.