Organic Chemistry 2 Quiz: Nucleophilic Acyl Substitution Mechanism
14 questions · exam conditions
0:00
Nucleophilic Acyl Substitution MechanismQuestion 1 of 14

In the hydrolysis of ethyl acetate, an oxygen-18 isotope is incorporated into the acetic acid product when the reaction is run in H₂¹⁸O. Which statement provides the best mechanistic explanation for this observation?

The carbonyl oxygen of the ester directly exchanges with the solvent in a rapid equilibrium before any hydrolysis occurs.
The reaction proceeds through a symmetrical tetrahedral intermediate where both single-bonded oxygens can be protonated and become equivalent.
The reaction proceeds via an SN2 mechanism where H₂¹⁸O attacks the ethyl group, releasing ethanol that contains the ¹⁸O label.
The ester undergoes enolization under the reaction conditions, and the ¹⁸O from water is incorporated at the alpha-carbon position.
← Back to quizzes

Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: Nucleophilic Acyl Substitution Mechanism

Practice Nucleophilic Acyl Substitution Mechanism in Organic Chemistry 2 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 Nucleophilic Acyl Substitution Mechanism, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry 2.

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.

All questions

Question 1

In the hydrolysis of ethyl acetate, an oxygen-18 isotope is incorporated into the acetic acid product when the reaction is run in H₂¹⁸O. Which statement provides the best mechanistic explanation for this observation?

  1. The carbonyl oxygen of the ester directly exchanges with the solvent in a rapid equilibrium before any hydrolysis occurs.
  2. The reaction proceeds through a symmetrical tetrahedral intermediate where both single-bonded oxygens can be protonated and become equivalent. (correct answer)
  3. The reaction proceeds via an SN2 mechanism where H₂¹⁸O attacks the ethyl group, releasing ethanol that contains the ¹⁸O label.
  4. The ester undergoes enolization under the reaction conditions, and the ¹⁸O from water is incorporated at the alpha-carbon position.
Explanation: The mechanism of ester hydrolysis involves the nucleophilic addition of water to the carbonyl carbon, forming a tetrahedral intermediate. This intermediate has two -OH groups (one from the original carbonyl oxygen, one from the attacking water) and one -OEt group. Proton transfers can make the two -OH groups chemically equivalent. When the intermediate collapses, either the original carbonyl oxygen or the newly added oxygen from H₂¹⁸O can be retained in the final carboxylic acid product. This scrambling is direct evidence for the formation of a tetrahedral intermediate. Choice A is not the mechanism. Choice C describes an SN2 reaction at the ethyl group, which is incorrect; attack is at the acyl carbon and the acetic acid is labeled, not the ethanol. Choice D describes an irrelevant process.

Question 2

A student proposes that the reaction of acetyl chloride with sodium methoxide could proceed via a direct SN2-type displacement of chloride by methoxide at the carbonyl carbon. Why is the accepted addition-elimination mechanism strongly favored over this direct displacement pathway?

  1. The sp² hybridization and trigonal planar geometry of the carbonyl carbon prevent the required backside attack for an SN2 transition state. (correct answer)
  2. The chloride ion is a poor leaving group in this context and requires the formation of a tetrahedral intermediate to facilitate its departure.
  3. The addition-elimination pathway is catalyzed by the sodium ion, whereas the SN2 pathway is inhibited by it.
  4. Methoxide is a strong base that would exclusively cause E2 elimination rather than substitution at the carbonyl carbon.
Explanation: The SN2 mechanism requires the nucleophile to attack the carbon atom from the side opposite to the leaving group (backside attack), leading to a trigonal bipyramidal transition state. An sp²-hybridized carbon, like a carbonyl carbon, has a trigonal planar geometry. The lobes of the pi bond obstruct the trajectory for backside attack, and the geometry is wrong for this type of transition state. The energetically favorable pathway is for the nucleophile to attack the empty π* orbital of the C=O bond, perpendicular to the plane of the molecule, leading to the tetrahedral intermediate of the addition-elimination mechanism.

Question 3

The reaction of a carboxylic acid with an amine to form an amide often requires high temperatures to proceed, despite being thermodynamically favorable overall. What is the best mechanistic explanation for this high activation energy barrier?

  1. The initial, rapid acid-base reaction between the two starting materials forms a stable and unreactive ammonium carboxylate salt. (correct answer)
  2. The tetrahedral intermediate is particularly unstable in this reaction and requires significant thermal energy to form.
  3. The water molecule that must be eliminated is an exceptionally poor leaving group even after protonation, requiring heat to be expelled.
  4. The C-N bond of the final amide product is very high in energy, and reaching this product state requires a large input of thermal energy.
Explanation: Before any nucleophilic acyl substitution can occur, the basic amine and the acidic carboxylic acid undergo a very fast and favorable acid-base reaction. This forms an ammonium carboxylate salt (RCOO⁻ H₃N⁺R'). In this salt form, the nucleophile has been protonated (and thus deactivated), and the electrophile has been deprotonated (and thus deactivated). The species are unreactive toward each other. High temperatures are required to overcome this thermodynamic sink and establish a small equilibrium concentration of the neutral starting materials, which can then react via the nucleophilic acyl substitution pathway, ultimately losing water.

Question 4

In the acid-catalyzed hydrolysis of an ester, several proton transfer steps are essential parts of the mechanism. Which of the following proposed proton transfers is NOT a productive step in the accepted pathway?

  1. Initial protonation of the ester's carbonyl oxygen by the acid catalyst to activate the electrophile.
  2. Protonation of the attacking water molecule by the acid catalyst before it performs the nucleophilic attack. (correct answer)
  3. Deprotonation of the oxygen atom from the attacking water molecule after the tetrahedral intermediate has formed.
  4. Protonation of the ester's alkoxy (-OR) group within the tetrahedral intermediate to make it a better leaving group.
Explanation: Protonating the nucleophile (water) before it attacks would form the hydronium ion (H₃O⁺). H₃O⁺ is no longer a nucleophile because it has no lone pairs available for donation; in fact, it is an electrophile. This step would deactivate the nucleophile and prevent the reaction from proceeding. All other options are essential proton transfers in the accepted mechanism: (A) activates the carbonyl, (C) neutralizes the intermediate after attack, and (D) prepares the leaving group for elimination.

Question 5

Based on the principles of the nucleophilic acyl substitution mechanism, particularly the stability of the leaving group, which of the following reactions is the most thermodynamically favorable and likely to proceed as written?

  1. Reacting ethyl acetate with NaCl to form acetyl chloride and sodium ethoxide.
  2. Reacting acetic anhydride with ethanol to form ethyl acetate and acetic acid. (correct answer)
  3. Reacting acetamide with sodium ethoxide to form ethyl acetate and sodium amide (NaNH₂).
  4. Reacting acetic acid with NaCl to form acetyl chloride and sodium hydroxide.
Explanation: Nucleophilic acyl substitution reactions generally proceed in the direction that converts a more reactive acyl derivative into a less reactive one. This corresponds to ejecting a leaving group that is more stable (a weaker base) than the nucleophile's conjugate base. In choice B, the starting material is a reactive anhydride, and the leaving group is a stable carboxylate. The product is a less reactive ester. In choice A, the leaving group (ethoxide) is much more basic than the nucleophile (chloride), so the reaction is unfavorable. In choice C, the leaving group (amide anion) is extremely basic. In choice D, the leaving group (hydroxide) is much more basic than the nucleophile (chloride). Therefore, only reaction B is thermodynamically downhill.

Question 6

The hydrolysis of an amide requires harsh conditions (strong acid or base and heat). What is the primary mechanistic barrier that makes the amide so unreactive compared to other carboxylic acid derivatives?

  1. The planarity of the amide bond creates significant steric hindrance that prevents the approach of a nucleophile.
  2. The initial acid-base reaction between the amide and water is highly endothermic, creating a large activation barrier.
  3. The nitrogen atom is sp² hybridized, which prevents it from being an effective leaving group under any conditions.
  4. The C-N bond has substantial double bond character due to resonance, making the carbonyl carbon a poor electrophile. (correct answer)
Explanation: When you encounter questions about amide reactivity, think about how electron delocalization affects the electrophilicity of the carbonyl carbon. Amides are uniquely unreactive among carboxylic acid derivatives because of their special resonance stabilization. The correct answer is D because amides exhibit significant resonance between the nitrogen's lone pair and the carbonyl group. This creates a resonance structure where the C-N bond has partial double bond character (approximately 40% based on bond length measurements). This electron donation from nitrogen makes the carbonyl carbon much less electrophilic and less susceptible to nucleophilic attack. Additionally, this resonance stabilization must be disrupted during hydrolysis, creating a high energy barrier. Answer A is incorrect because while the amide bond is planar due to resonance, this planarity actually reduces steric hindrance rather than creating it. The planar geometry allows for better orbital overlap. Answer B misidentifies the rate-limiting step. The initial protonation or deprotonation isn't the primary barrier - it's the subsequent nucleophilic attack and C-N bond breaking that requires harsh conditions. Answer C incorrectly focuses on hybridization. While nitrogen is sp² hybridized in amides, this isn't what prevents it from being a leaving group. The issue is the resonance stabilization, not the hybridization state itself. Remember this key principle: among carboxylic acid derivatives, reactivity toward nucleophiles decreases as the leaving group's ability to donate electrons through resonance increases. Amides have the most electron-donating leaving group (−NR2-NR_2), making them the least reactive.

Question 7

During the uncatalyzed reaction of an ester with a primary amine (aminolysis), the rate is often observed to increase in the presence of excess amine. What is the most likely mechanistic role of the second amine molecule?

  1. It acts as a Lewis acid, coordinating to the carbonyl oxygen and activating the ester.
  2. It acts as a proton shuttle, assisting in proton transfers within the tetrahedral intermediate to facilitate its collapse. (correct answer)
  3. It deprotonates the first amine molecule, generating a highly reactive amide anion (RNH⁻) that is the true nucleophile.
  4. It forms a hydrogen bond with the ester's alkoxy group, making it a better leaving group.
Explanation: The tetrahedral intermediate formed from the initial attack of the amine is zwitterionic (R-C(O⁻)(OR')-N⁺H₂R''). For this intermediate to collapse and expel the alkoxide (⁻OR'), two things must happen: the alkoxide needs to become a better leaving group, and the nitrogen needs to be deprotonated. A second molecule of the amine can act as a base to deprotonate the N⁺H₂ group and then as an acid to protonate the ⁻OR' group as it leaves (as HOR'). This 'proton shuttle' role facilitates the breakdown of the tetrahedral intermediate, which can be a rate-limiting step, thereby accelerating the overall reaction.

Question 8

The base-promoted hydrolysis (saponification) of an amide is significantly slower than that of an ester with similar steric bulk. What is the primary mechanistic reason for this substantial difference in reaction rate?

  1. The C=O bond in an amide has less double-bond character than in an ester, making it a weaker electrophile.
  2. The nitrogen atom in an amide is much more electronegative than the oxygen atom in an ester, which destabilizes the transition state.
  3. The amide anion (R₂N⁻) is a much poorer leaving group than an alkoxide anion (RO⁻), making the collapse of the tetrahedral intermediate the difficult, rate-determining step. (correct answer)
  4. The initial nucleophilic attack by hydroxide is reversible for the ester but irreversible for the amide, slowing the overall reaction.
Explanation: The key difference lies in the leaving group ability. In the tetrahedral intermediate's collapse, the amide would need to expel an amide anion (e.g., R₂N⁻), while the ester expels an alkoxide anion (RO⁻). The pKa of the conjugate acid of the amide anion (an amine, pKa ≈ 38) is much higher than that of the alkoxide (an alcohol, pKa ≈ 16-18). This means the amide anion is a much stronger base and a vastly inferior leaving group. This makes the elimination step (collapse of the intermediate) for the amide very slow and energetically costly, thus controlling the overall rate.

Question 9

Saponification, the hydrolysis of an ester with stoichiometric NaOH, is effectively irreversible. In contrast, acid-catalyzed hydrolysis is a reversible equilibrium. What is the key mechanistic step that accounts for the irreversibility under basic conditions?

  1. The initial attack of the hydroxide ion is much faster than the attack of water, making the first step irreversible.
  2. The tetrahedral intermediate formed under basic conditions is resonance-stabilized and cannot collapse back to the starting materials.
  3. The final step is an acid-base reaction where the carboxylic acid produced is deprotonated by base to form a carboxylate anion. (correct answer)
  4. The alkoxide leaving group is immediately protonated by the hydroxide catalyst, trapping it and preventing the reverse reaction.
Explanation: While the initial attack of hydroxide is reversible, the overall reaction is driven to completion by a final, irreversible acid-base step. Once the tetrahedral intermediate collapses to form a carboxylic acid and an alkoxide, the alkoxide (or another hydroxide ion) immediately deprotonates the carboxylic acid. This forms a resonance-stabilized carboxylate anion. The carboxylate is negatively charged and a very poor electrophile, so it will not be attacked by the alcohol (the conjugate acid of the alkoxide) to reform the ester. This final, thermodynamically downhill deprotonation step makes the entire sequence irreversible.

Question 10

Transesterification is the conversion of one ester to another by reacting it with an alcohol, often in the presence of an acid or base catalyst. In the acid-catalyzed mechanism, which species is the direct leaving group from the tetrahedral intermediate?

  1. An alkoxide anion (RO⁻), which is then protonated by the catalyst.
  2. A protonated ether (R-O⁺H-R), which rapidly fragments.
  3. A hydronium ion (H₃O⁺), formed after the addition of the new alcohol.
  4. An alcohol molecule (ROH), which was the alkoxy portion of the original ester. (correct answer)
Explanation: When analyzing transesterification mechanisms, focus on what happens at the tetrahedral intermediate stage and remember that acid catalysis means all leaving groups must be neutral or positively charged species. In acid-catalyzed transesterification, the original ester first gets protonated, then the new alcohol attacks the carbonyl carbon, forming a tetrahedral intermediate. This intermediate has both the original alkoxy group and the new alcohol attached. For the reaction to proceed, one of these must leave. Since we're under acidic conditions, the original alkoxy group gets protonated, converting it from a potential alkoxide anion (which would be a terrible leaving group) into a neutral alcohol molecule (ROH). This protonated alcohol then leaves directly as ROH, making answer D correct. Answer A is wrong because alkoxide anions (RO⁻) are extremely basic and would be terrible leaving groups under acidic conditions. The protonation happens before departure, not after. Answer B describes a protonated ether, but the tetrahedral intermediate doesn't contain an ether linkage—it has two separate alcohol-type groups attached to the same carbon. Answer C incorrectly identifies hydronium ion as the leaving group, but H₃O⁺ isn't part of the tetrahedral intermediate structure; it's just the catalytic species floating around in solution. Remember this key principle: in acid-catalyzed reactions, poor leaving groups get protonated first to become better (neutral) leaving groups. Always look for the neutral molecule that departs, not charged species.

Question 11

An ester reacts with two equivalents of a Grignard reagent (RMgBr) followed by aqueous workup to yield a tertiary alcohol. Which statement provides the correct mechanistic description for the formation of the intermediate ketone?

  1. The first equivalent of Grignard reagent attacks the carbonyl carbon, and the resulting tetrahedral intermediate collapses by eliminating an alkoxide group. (correct answer)
  2. The Grignard reagent first deprotonates the alpha-carbon to form an enolate, which then rearranges to form the ketone.
  3. Two molecules of the Grignard reagent coordinate to the carbonyl oxygen, weakening the C-O bond and allowing the alkoxide to leave, forming an acylium ion which is then trapped.
  4. The ester undergoes an SN2 reaction where the R-group of the Grignard reagent displaces the ester's -OR' group in a single step.
Explanation: The reaction proceeds in two stages. The first stage is a nucleophilic acyl substitution. The carbanion-like R-group from the Grignard reagent acts as a strong nucleophile, attacking the ester's carbonyl carbon to form a tetrahedral intermediate. This intermediate is unstable and collapses, eliminating the alkoxide (-OR') group, which is a better leaving group than the R-group. This process forms a ketone. The newly formed ketone is more reactive than the starting ester and is immediately attacked by the second equivalent of Grignard reagent in a nucleophilic addition reaction, which after workup yields the tertiary alcohol.

Question 12

The conversion of an acid anhydride to an ester by reaction with an alcohol is a thermodynamically favorable process. Which statement best explains the energetic driving force for this nucleophilic acyl substitution?

  1. The ester product has stronger resonance stabilization than the acid anhydride starting material, making it much lower in energy.
  2. The carboxylate anion that serves as the leaving group is a weaker base and thus more stable than the alkoxide that would be the leaving group in the reverse reaction. (correct answer)
  3. The tetrahedral intermediate formed from the acid anhydride is unusually stable compared to other acyl derivatives, lowering the reaction barrier.
  4. The alcohol nucleophile is significantly more reactive than the carboxylate leaving group, making the initial addition step irreversible.
Explanation: In nucleophilic acyl substitution, the equilibrium favors the formation of the more stable, less reactive acyl derivative. This stability is directly related to the basicity (and thus stability) of the leaving group. In this reaction, the leaving group is a carboxylate anion. In the reverse reaction (ester to anhydride), the leaving group would be an alkoxide anion. A carboxylate anion is significantly stabilized by resonance and is a much weaker base (pKa of conjugate acid ≈ 5) than an alkoxide anion (pKa of conjugate acid ≈ 16-18). Therefore, the equilibrium strongly favors the products containing the more stable carboxylate anion.

Question 13

In the acid-catalyzed Fischer esterification of a carboxylic acid with an alcohol, what is the principal role of the acid catalyst in the forward reaction mechanism?

  1. To protonate the alcohol, converting it into a more potent nucleophile for attacking the carboxylic acid.
  2. To deprotonate the carboxylic acid, forming a highly reactive carboxylate that is readily attacked by the alcohol.
  3. To protonate the carbonyl oxygen of the carboxylic acid, which significantly increases the electrophilicity of the carbonyl carbon. (correct answer)
  4. To form a covalent bond with the carboxylic acid's hydroxyl group, transforming it into a better leaving group before the alcohol attacks.
Explanation: The first and most crucial role of the acid catalyst is to activate the carbonyl group. By protonating the carbonyl oxygen, a positive charge is placed on the oxygen, which can be delocalized to the carbonyl carbon through resonance. This makes the carbonyl carbon much more electrophilic and thus more susceptible to attack by the weak nucleophile (the alcohol). Protonating the nucleophile (A) deactivates it. Acidic conditions prevent deprotonation of the carboxylic acid (B). While the hydroxyl group is eventually protonated to become a good leaving group (H₂O), this happens within the tetrahedral intermediate, not as the first step (D). The activation of the electrophile (C) is the primary initial role.

Question 14

Why is it mechanistically infeasible to convert an ester directly into an acid chloride by reacting it with a salt like NaCl?

  1. The chloride ion is not nucleophilic enough to attack the relatively unreactive ester carbonyl group under neutral conditions.
  2. The tetrahedral intermediate in this reaction would be unstable and immediately revert to starting materials without expelling chloride.
  3. The sodium ion forms a stable complex with the ester's carbonyl oxygen, preventing any reaction from occurring.
  4. The alkoxide ion (RO⁻) that would be the leaving group is a much stronger base than the chloride ion (Cl⁻) nucleophile. (correct answer)
Explanation: This question tests your understanding of nucleophilic acyl substitution mechanisms and the critical role of leaving group ability in determining reaction feasibility. In nucleophilic acyl substitution reactions, the mechanism proceeds through a tetrahedral intermediate where a nucleophile attacks the carbonyl carbon, followed by elimination of a leaving group. For the reaction to proceed forward, the leaving group must be more stable (weaker base) than the incoming nucleophile. Answer D correctly identifies the fundamental thermodynamic problem: alkoxide ions (RO⁻) are much stronger bases than chloride ions (Cl⁻). Since stronger bases are poorer leaving groups, the equilibrium heavily favors the starting materials. Even if chloride could attack the ester carbonyl, the tetrahedral intermediate would collapse by expelling the weaker base (Cl⁻) rather than the stronger base (RO⁻), regenerating the original ester. Answer A is incorrect because chloride ion is actually quite nucleophilic and can attack carbonyl groups under appropriate conditions. Answer B misidentifies the problem—the tetrahedral intermediate itself isn't inherently unstable, but rather it preferentially eliminates the wrong group. Answer C is wrong because sodium ion coordination to the carbonyl oxygen would actually make the carbon more electrophilic and reactive, not prevent reaction. Remember this key principle: in nucleophilic substitution reactions, you need a leaving group that's a weaker base than your incoming nucleophile. Always compare basicity when evaluating whether acyl substitutions are thermodynamically favorable.