Carbonyl Chemistry and Reactivity (5D)
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MCAT Chemical and Physical Foundations of Biological Systems › Carbonyl Chemistry and Reactivity (5D)
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?
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.
Cyclohexanol, formed by hydride addition to the ketone followed by protonation.
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.
In a stability study of a carbonyl-containing prodrug, hydrolysis is proposed to occur through formation of a tetrahedral intermediate. The prodrug is an ester (R–C(=O)–OR') and is incubated at pH 1.0 (0.10 M HCl) and separately at pH 13.0 (0.10 M NaOH) at 25°C. Degradation is faster at pH 13. The central concept is carbonyl reactivity and acyl substitution. Which step is most likely rate-limiting under the basic condition for ester hydrolysis?
Nucleophilic attack of OH$^-$ on the ester carbonyl to form the tetrahedral intermediate.
Protonation of the carbonyl oxygen by H$_3$O$^+$ to activate the ester.
Reduction of the ester carbonyl by hydride transfer from water to form an aldehyde.
Formation of an enolate at the $\alpha$-carbon followed by C–C bond formation.
Explanation
This question tests Carbonyl Chemistry and Reactivity (5D), focusing on the mechanism of base-catalyzed ester hydrolysis. Carbonyl groups in esters undergo nucleophilic acyl substitution, with hydroxide attacking the carbonyl carbon to form a tetrahedral intermediate in the rate-determining step. In the given reaction, ester hydrolysis is faster at pH 13 than pH 1, indicating base catalysis dominates. The correct choice, A, is expected because in basic conditions, hydroxide acts as a strong nucleophile attacking the electrophilic carbonyl carbon, forming the tetrahedral intermediate in the slowest step. Choice B is incorrect as it describes acid-catalyzed hydrolysis, which would be favored at pH 1, not pH 13. To apply this concept, remember that base-catalyzed ester hydrolysis proceeds through nucleophilic attack as the rate-limiting step, while acid-catalyzed hydrolysis involves carbonyl protonation.
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?
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.
A cyclic epoxide formed by intramolecular substitution of the ketone oxygen by 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.
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?
Propanal reacts faster because aldehydes are generally more electrophilic and less sterically hindered than ketones.
Acetone reacts faster because its two alkyl groups withdraw electron density, increasing electrophilicity.
Propanal reacts faster because cyanide performs nucleophilic acyl substitution on aldehydes but not ketones.
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.
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.
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?
An acetal (enzyme–O–CH(OR)–R) requiring two equivalents of serine oxygen to add to the aldehyde.
A tetrahedral hemiacetal (enzyme–O–CH(OH)–R) formed by addition of serine oxygen to the aldehyde carbonyl.
A carboxylate salt (R–COO$^-$) formed by base-promoted oxidation of the aldehyde in buffer.
An amide (enzyme–NH–C(=O)–R) formed by acyl substitution on the aldehyde.
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.
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?
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.
An imine (Schiff base), R–CH=N–Bu, formed by addition of amine followed by dehydration.
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 converts DHAP into a nucleophilic enamine that attacks the electrophilic carbonyl carbon of G3P.
The peptide most likely protonates G3P to form an acetal directly with DHAP, bypassing carbon–carbon bond formation.
The peptide most likely increases the electrophilicity of DHAP by converting it into an enolate that is attacked by 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.
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?
Acetamide undergoes faster nucleophilic addition because resonance donation from nitrogen increases carbonyl electrophilicity.
Acetone forms a methanol addition product more readily because amide carbonyls are less electrophilic due to $n\rightarrow\pi^*$ resonance stabilization.
Acetone fails to react because ketones cannot undergo nucleophilic addition without strong base.
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.
A pharmacology group studies formation of acetals as prodrugs. They treat an aldehyde-containing compound R–CHO with excess ethylene glycol (HO–CH$_2$CH$_2$–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?
Cyclic acetal (1,3-dioxolane) at the former carbonyl carbon
Carboxylate salt formed by deprotonation of the aldehyde
Amide formed by reaction of aldehyde with glycol
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.