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Carbonyl Chemistry and Reactivity (5D) - Chem & Physics

  1. 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?
    Answer: A bisulfite addition product in which HSO3− has added to the carbonyl carbon, yielding a sulfonate-bearing alcohol 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.
  2. An enzyme-mimic catalyst is proposed to accelerate nucleophilic addition to a carbonyl by stabilizing negative charge on oxygen in the tetrahedral intermediate. In a model reaction, a thiolate nucleophile (\ceRS−\ce{RS^-}, 5 mM) adds to an aldehyde (\ceR′CHO\ce{R'CHO}, 5 mM) in water at pH 8.0 and 25°C to form a thiohemiacetal. Kinetic data show the rate increases when a Lewis acid (1 mM \ceZn2+\ce{Zn^{2+}}) is added, while reactant concentrations are unchanged. The central concept is carbonyl activation by Lewis acids increasing electrophilicity and stabilizing the oxyanion intermediate. Which step is most likely rate-limiting in the uncatalyzed reaction (before adding \ceZn2+\ce{Zn^{2+}})?
    Answer: Nucleophilic attack of \ceRS−\ce{RS^-} on the carbonyl carbon to form the tetrahedral intermediate This question tests Carbonyl Chemistry and Reactivity (5D), focusing on the mechanism and rate-limiting step of nucleophilic addition to carbonyls. Carbonyl groups undergo nucleophilic addition through attack at the electrophilic carbon, forming a tetrahedral intermediate with negative charge on oxygen. In the given reaction, thiolate attacks the aldehyde carbonyl, and the question asks which step is rate-limiting before Lewis acid catalysis. The correct choice, B, is expected because nucleophilic attack to form the tetrahedral intermediate is typically rate-limiting, and Lewis acids like Zn2+ accelerate this step by coordinating to the carbonyl oxygen and increasing electrophilicity. Choice A is incorrect as enolate formation is not part of the nucleophilic addition mechanism for forming thiohemiacetals. To apply this concept, recognize that Lewis acids catalyze carbonyl reactions by activating the C=O bond toward nucleophilic attack, explaining why many carbonyl-processing enzymes use metal cofactors.
  3. 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?
    Answer: Propanal reacts faster because aldehydes are generally more electrophilic and less sterically hindered than ketones. 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.
  4. 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?
    Answer: Nucleophilic attack of OH−^- on the ester carbonyl to form the tetrahedral intermediate. 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.
  5. To compare nucleophilic addition to aldehydes vs ketones in a biomimetic reduction, equal concentrations of formaldehyde (HCHO) and acetone ((CH3_3)2_2CO) are separately treated with the same amount of NaBH4_4 in methanol for a short time. Assuming similar mixing and temperature, which substrate is expected to react faster, based on carbonyl reactivity?
    Answer: Formaldehyde, because it is less sterically hindered and more electrophilic This question tests Carbonyl Chemistry and Reactivity (5D), focusing on relative electrophilicity of aldehydes vs ketones in reductions. Carbonyl reactivity increases with less steric hindrance and higher electrophilicity, making aldehydes more reactive than ketones to nucleophiles like hydride. In the given reaction, formaldehyde and acetone are treated separately with NaBH₄ in methanol, comparing rates. The correct choice, B, is expected due to formaldehyde's higher reactivity from minimal steric bulk and greater electrophilicity. Choice A is incorrect as it assumes acetone is more electrophilic, but alkyl groups decrease carbonyl electrophilicity. To apply this concept, ensure that reaction conditions align with expected reactivity, and distinguish between similar functional groups like formaldehyde and other aldehydes, noting its unique high reactivity.
  6. 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?
    Answer: Nucleophilic attack of methanol on the protonated carbonyl carbon 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.
  7. To mimic a key step in glycolysis involving carbonyl chemistry, a researcher mixes dihydroxyacetone phosphate (DHAP, a ketose phosphate) with a catalytic amount of base (0.5 mM \ceOH−\ce{OH^-}) in water at 37°C. The reaction is quenched after 30 seconds. The central concept is enolization to form an enediolate intermediate, enabling aldose–ketose isomerization. Assume phosphate remains attached and no bond cleavage occurs. Which transformation is most consistent with base-catalyzed carbonyl reactivity in this system?
    Answer: Conversion of DHAP to glyceraldehyde-3-phosphate via enediol/enediolate-mediated isomerization This question tests Carbonyl Chemistry and Reactivity (5D), focusing on base-catalyzed keto-enol tautomerization and aldose-ketose isomerization. Carbonyl groups with α-hydrogens undergo enolization in basic conditions, forming enolate intermediates that enable isomerization reactions. In the given reaction, DHAP (a ketose) forms an enediolate intermediate when deprotonated at the α-position by hydroxide. The correct choice, A, is expected because this enediolate can rearrange to form glyceraldehyde-3-phosphate (an aldose) through proton transfer, mimicking the triose phosphate isomerase reaction in glycolysis. Choice B is incorrect as it assumes oxidation without an oxidizing agent present. To apply this concept, recognize that base-catalyzed carbonyl chemistry often involves enolate intermediates, and that aldose-ketose isomerization proceeds through enediol/enediolate species without breaking C-C bonds.
  8. A biochemistry lab compares carbonyl electrophilicity by measuring the initial rate of cyanohydrin formation (nucleophilic addition of \ceCN−\ce{CN^-}) for two substrates at 25°C in water with 0.10 M NaCN and pH held constant at 9.5 (so \ceHCN\ce{HCN} formation is minimized). Substrate 1 is acetaldehyde (\ceCH3CHO\ce{CH3CHO}). Substrate 2 is acetone (\ce(CH3)2CO\ce{(CH3)2CO}). The only difference between trials is the carbonyl compound (each 10 mM). The central concept is that aldehydes are generally more reactive toward nucleophilic addition than ketones due to sterics and inductive effects. Which conclusion is most consistent with carbonyl reactivity under these conditions?
    Answer: Acetaldehyde reacts faster because its carbonyl carbon is less sterically hindered and more electrophilic This question tests Carbonyl Chemistry and Reactivity (5D), focusing on relative electrophilicity of aldehydes versus ketones. Carbonyl groups undergo nucleophilic addition, with aldehydes being more reactive than ketones due to less steric hindrance and greater electrophilicity from fewer electron-donating alkyl groups. In the given reaction, cyanide ion attacks the carbonyl carbon of either acetaldehyde or acetone to form cyanohydrins. The correct choice, B, is expected because acetaldehyde has only one methyl group compared to acetone's two, making its carbonyl carbon more accessible and more electron-deficient. Choice A is incorrect as it reverses the reactivity trend - ketones actually form less stable tetrahedral intermediates due to steric crowding. To apply this concept, remember that aldehyde carbonyls are generally more reactive than ketone carbonyls in nucleophilic addition reactions, and that steric effects become more pronounced with bulkier nucleophiles.
  9. To model thiohemiacetal formation in cysteine proteases, a chemist mixes acetone with excess ethanethiol (EtSH) under mildly acidic conditions (catalytic H+^+) in anhydrous solvent. A new product forms that retains the carbonyl carbon but now shows an –OH and –SEt on the same carbon (no oxidation/reduction reagents present). Which species is most consistent with carbonyl nucleophilic addition under these conditions?
    Answer: Thiohemiacetal: (CH3_3)2_2C(OH)(SEt) This question tests Carbonyl Chemistry and Reactivity (5D), focusing on thiohemiacetal formation via nucleophilic addition. Carbonyl groups react with thiols under acidic conditions to form thiohemiacetals, analogous to hemiacetals but with sulfur. In the given reaction, acetone interacts with ethanethiol under catalytic acid, forming a product with –OH and –SEt on the same carbon. The correct choice, A, is expected due to the thiohemiacetal structure, consistent with addition without oxidation. Choice B is incorrect as it assumes thioester formation, requiring different conditions like oxidation. To apply this concept, ensure that reaction conditions align with expected reactivity, and distinguish between similar functional groups like hemiacetals and acetals, noting thio analogs are often more stable.
  10. 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?
    Answer: An imine (Schiff base), R–CH=N–Bu, formed by addition of amine followed by dehydration. 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.
  11. In a model for nonenzymatic glycation, researchers incubate D-glucose (predominantly cyclic hemiacetal in water) with lysine side-chain analog n-butylamine in phosphate buffer at pH 7.4, 37∘^\circC. Over hours, LC-MS detects a new species consistent with loss of H2_2O relative to the initial adduct. The key carbonyl-reactivity concept is nucleophilic addition of an amine to an aldehyde followed by dehydration. Which functional group is most consistent with the newly formed linkage in the major product at early time points?
    Answer: Imine/Schiff base (R–CH=NR') This question tests Carbonyl Chemistry and Reactivity (5D), focusing on imine formation from aldehydes and amines. Carbonyl groups undergo nucleophilic addition with primary amines, forming carbinolamines that dehydrate to imines, especially under mild conditions. In the given reaction, D-glucose (opening to aldehyde) interacts with n-butylamine in neutral buffer, leading to initial adduct followed by water loss detected by LC-MS. The correct choice, B, is expected due to the formation of a Schiff base linkage, consistent with the dehydration product. Choice A is incorrect as it assumes amide formation, which requires oxidation or different conditions not present. To apply this concept, ensure that reaction conditions align with expected reactivity, and distinguish between similar functional groups like aldehydes and ketones, noting aldehydes form imines more readily.
  12. To probe aldol reactivity relevant to polyketide biosynthesis, a chemist mixes acetaldehyde (CH3_3CHO) in 0.10 M NaOH at 0∘^\circC for 5 min, then neutralizes. GC-MS indicates formation of a C4_4H8_8O2_2 compound. The central concept is base-catalyzed aldol addition (not dehydration). Which molecule is most likely formed as the major product under these cold, aqueous conditions?
    Answer: 3-hydroxybutanal (CH3_3CH(OH)CH2_2CHO) This question tests Carbonyl Chemistry and Reactivity (5D), focusing on base-catalyzed aldol addition without dehydration. Carbonyl groups with alpha-hydrogens form enolates that add to other carbonyls, yielding beta-hydroxy carbonyls under cold, aqueous conditions. In the given reaction, acetaldehyde interacts with itself under basic conditions at 0°C, forming a C₄H₈O₂ adduct detected by GC-MS. The correct choice, B, is expected due to the aldol addition product 3-hydroxybutanal, matching the molecular formula and conditions preventing dehydration. Choice C is incorrect as it assumes dehydration to crotonaldehyde, which requires heat or acid not used here. To apply this concept, ensure that reaction conditions align with expected reactivity, and distinguish between similar functional groups like aldehydes and ketones, noting aldehydes are more electrophilic in aldol reactions.
  13. 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?
    Answer: A cyclic ketal, in which the ketone carbonyl is converted to an acetal carbon bonded to two oxygens from ethylene glycol. 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.
  14. An enzyme-mimic catalyst accelerates aldol condensation between acetone and benzaldehyde in ethanolic NaOH at room temperature. After 2 h, the isolated major product is an α,β\alpha,\beta-unsaturated ketone (confirmed by IR: C=O at ~1665 cm−1^{-1} and C=C at ~1600 cm−1^{-1}). The key reactivity concept is aldol addition followed by dehydration. Which product is most likely formed from acetone + benzaldehyde under these conditions?
    Answer: Chalcone analog: Ph–CH=CH–C(=O)CH3_3 This question tests Carbonyl Chemistry and Reactivity (5D), focusing on crossed aldol condensation with dehydration. Carbonyl groups undergo aldol addition followed by dehydration to alpha,beta-unsaturated carbonyls under basic conditions, especially with aromatic aldehydes. In the given reaction, benzaldehyde interacts with acetone in ethanolic NaOH, forming an unsaturated ketone confirmed by IR. The correct choice, B, is expected due to the chalcone analog from acetone enolate adding to benzaldehyde, followed by dehydration. Choice C is incorrect as it assumes benzoin formation, which requires cyanide catalysis not present. To apply this concept, ensure that reaction conditions align with expected reactivity, and distinguish between similar functional groups like aldehydes without alpha-H and ketones, noting the former act as electrophiles.
  15. 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?
    Answer: A β-hydroxy ketone formed by addition of the cyclohexanone enolate to benzaldehyde, followed by protonation 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.
  16. A stability study examines why amides are less reactive toward nucleophilic acyl substitution than esters in aqueous solution. Two compounds are placed in 0.10 M NaOH at 25∘^\circC: ethyl acetate and acetamide. After 1 h, ethyl acetate shows significant hydrolysis, while acetamide shows minimal change. Based on carbonyl reactivity, which explanation is most consistent?
    Answer: Amides are resonance-stabilized by N donation, decreasing carbonyl electrophilicity This question tests Carbonyl Chemistry and Reactivity (5D), focusing on relative reactivity in nucleophilic acyl substitution. Carbonyl groups in amides are less electrophilic due to nitrogen resonance donation, making them slower to hydrolyze than esters. In the given reaction, ethyl acetate hydrolyzes significantly in NaOH, while acetamide does not. The correct choice, A, is expected due to amide stabilization decreasing carbonyl electrophilicity. Choice B is incorrect as it assumes amides hydrolyze faster, but NH₂⁻ is a poorer leaving group. To apply this concept, ensure that reaction conditions align with expected reactivity, and distinguish between similar functional groups like amides and esters, noting amides require harsher conditions for hydrolysis.
  17. A researcher studies selective reactivity of carbonyl-containing biomolecules with hydrazine derivatives. Pyruvate (CH3_3COCOO−^-) is treated with 2,4-dinitrophenylhydrazine (DNPH) under acidic conditions, forming a precipitate (hydrazone). The key concept is nucleophilic addition of hydrazine to a carbonyl followed by dehydration. Which carbonyl in pyruvate is most likely to form the hydrazone most readily?
    Answer: The ketone carbonyl (C=O adjacent to CH3_3) This question tests Carbonyl Chemistry and Reactivity (5D), focusing on selective hydrazone formation. Carbonyl groups in ketones react with hydrazines to form hydrazones, while carboxylates are less reactive to nucleophiles. In the given reaction, pyruvate interacts with DNPH under acid, forming a hydrazone precipitate. The correct choice, A, is expected due to the ketone carbonyl being more electrophilic and reactive. Choice B is incorrect as it assumes carboxylate reactivity, but it's deactivating at low pH. To apply this concept, ensure that reaction conditions align with expected reactivity, and distinguish between similar functional groups like keto-acids and simple ketones, noting selectivity for the ketone.
  18. 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?
    Answer: The peptide most likely converts DHAP into a nucleophilic enamine that attacks the electrophilic carbonyl carbon of G3P. 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.
  19. In a medicinal chemistry study modeling lysine side-chain modification, a buffered aqueous solution (pH 5.0, 25°C) contains acetone (\ce(CH3)2C=O\ce{(CH3)2C=O}, 10 mM) and methylamine (\ceCH3NH2\ce{CH3NH2}, 10 mM). The mixture is treated with a mild reducing agent, sodium cyanoborohydride (\ceNaBH3CN\ce{NaBH3CN}, 20 mM), and stirred for 1 hour. No strong acid is added; pH is maintained at 5.0 throughout. The central carbonyl reactivity concept is nucleophilic addition to form an iminium intermediate followed by selective reduction (reductive amination). Which product is most likely formed in highest yield under these conditions? Assume \ceNaBH3CN\ce{NaBH3CN} selectively reduces iminium ions over unprotonated ketones at pH 5, and acetone does not undergo self-aldol under these conditions.
    Answer: N-methyl-2-propanamine (\ce(CH3)2CHNHCH3\ce{(CH3)2CHNHCH3}) from reductive amination of acetone This question tests Carbonyl Chemistry and Reactivity (5D), focusing on reductive amination of ketones. Reductive amination involves nucleophilic addition of an amine to a carbonyl to form an iminium ion, followed by selective reduction with NaBH3CN. In the given reaction, acetone reacts with methylamine at pH 5.0, where the amine is partially protonated but still nucleophilic enough to attack the carbonyl carbon. The correct choice, C (N-methyl-2-propanamine), is expected because NaBH3CN selectively reduces the iminium intermediate formed between acetone and methylamine while leaving unreacted ketone intact. Choice A is incorrect as it assumes direct reduction of the ketone, which NaBH3CN does not efficiently perform at pH 5. To apply this concept, remember that reductive amination requires both amine and carbonyl to form an iminium intermediate, and that pH affects both amine nucleophilicity and the selectivity of the reducing agent.
  20. A formulation scientist evaluates imine formation between a ketone drug (R–CO–CH3_3) and a primary amine impurity (R'NH2_2). In a dry solvent with a trace of acid catalyst, the reaction proceeds, but when 5% water is added, the imine signal in IR decreases and the carbonyl signal returns. The reactivity concept is reversible nucleophilic addition/elimination at carbonyls. Which interpretation is most consistent with the observation?
    Answer: Water shifts equilibrium toward carbonyl + amine by hydrolyzing the imine This question tests Carbonyl Chemistry and Reactivity (5D), focusing on reversible imine formation and hydrolysis. Carbonyl groups react with amines to form imines via nucleophilic addition-elimination, but water can shift equilibrium back to reactants by hydrolysis. In the given reaction, the ketone and amine form an imine in dry solvent, but added water reverses it, restoring the carbonyl IR signal. The correct choice, A, is expected due to water hydrolyzing the imine, favoring the carbonyl and amine. Choice B is incorrect as it assumes irreversible reduction, but no reducing agent is present. To apply this concept, ensure that reaction conditions align with expected reactivity, and distinguish between similar functional groups like imines and enamines, noting imines are more susceptible to hydrolysis.

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