Organic Chemistry 2 Quiz: Amide Formation And Hydrolysis
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Amide Formation And HydrolysisQuestion 1 of 15

A direct Friedel-Crafts acylation of aniline with an acyl chloride and AlCl3 fails. A successful multi-step procedure involves first reacting aniline with excess acetyl chloride to form acetanilide, performing the Friedel-Crafts acylation, and finally hydrolyzing the amide. What is the essential function of converting aniline to acetanilide in this sequence?

To decrease the steric hindrance of the substituent, allowing easier access to the ortho positions of the aromatic ring.
To moderate the directing effect of the nitrogen, switching it from a meta-director to an ortho,para-director.
To increase the reactivity of the aromatic ring, making it a better nucleophile for the Friedel-Crafts reaction.
To prevent the basic nitrogen atom from coordinating with the Lewis acid catalyst (AlCl3), which would deactivate the ring.
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Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: Amide Formation And Hydrolysis

Practice Amide Formation And Hydrolysis in Organic Chemistry 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Amide Formation And Hydrolysis, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry 2.

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Question 1

A direct Friedel-Crafts acylation of aniline with an acyl chloride and AlCl3 fails. A successful multi-step procedure involves first reacting aniline with excess acetyl chloride to form acetanilide, performing the Friedel-Crafts acylation, and finally hydrolyzing the amide. What is the essential function of converting aniline to acetanilide in this sequence?

  1. To decrease the steric hindrance of the substituent, allowing easier access to the ortho positions of the aromatic ring.
  2. To moderate the directing effect of the nitrogen, switching it from a meta-director to an ortho,para-director.
  3. To increase the reactivity of the aromatic ring, making it a better nucleophile for the Friedel-Crafts reaction.
  4. To prevent the basic nitrogen atom from coordinating with the Lewis acid catalyst (AlCl3), which would deactivate the ring. (correct answer)
Explanation: The primary amino group of aniline is a Lewis base. The Friedel-Crafts catalyst, AlCl3, is a strong Lewis acid. When mixed directly, they undergo a Lewis acid-base reaction where the nitrogen's lone pair coordinates to the aluminum. This forms an -NH2-AlCl3 complex, which places a positive charge on or near the nitrogen. This complex is a very strong deactivating group and a meta-director, shutting down the desired electrophilic aromatic substitution. Converting the amine to an amide (acetanilide) makes the nitrogen lone pair much less basic due to resonance with the acetyl carbonyl. This prevents the destructive coordination with AlCl3, allowing the Friedel-Crafts reaction to proceed with the (now moderately activating) ortho,para-directing acetamido group.

Question 2

A compound with the molecular formula C4H9NO is subjected to vigorous acid-catalyzed hydrolysis. The two organic products are isolated and identified as propanoic acid and a primary amine. The ¹H NMR spectrum of the original C4H9NO compound shows signals at approximately δ 2.8 (doublet, 3H), δ 2.2 (quartet, 2H), and δ 1.1 (triplet, 3H). What is the structure of the C4H9NO compound?

  1. N-methylpropanamide (correct answer)
  2. N-ethylacetamide
  3. Butanamide
  4. N,N-dimethylacetamide
Explanation: First, analyze the hydrolysis products. The reaction yields propanoic acid (a 3-carbon carboxylic acid) and a primary amine. Since the starting material has 4 carbons, the primary amine must be methylamine (1 carbon). An amide formed from propanoic acid and methylamine is N-methylpropanamide. Now, we verify this structure with the NMR data. For CH3CH2C(=O)NHCH3: The ethyl group's -CH2- is adjacent to a -CH3, so it appears as a quartet (at ~2.2 ppm). The ethyl group's -CH3 is adjacent to a -CH2-, so it appears as a triplet (at ~1.1 ppm). The N-methyl group's protons would be coupled to the single N-H proton, appearing as a doublet (at ~2.8 ppm). The data perfectly matches the structure of N-methylpropanamide.

Question 3

A proposed synthesis of N-phenylacetamide involves heating phenyl acetate with a concentrated solution of ammonia in ethanol. This reaction, known as aminolysis of an ester, is expected to be very inefficient. Which statement provides the best explanation for the low yield?

  1. Ammonia is a weaker nucleophile than the phenoxide leaving group is a base, making the reaction thermodynamically unfavorable.
  2. The phenol produced as a byproduct is acidic enough (pKa ≈ 10) to protonate the ammonia nucleophile (pKa of NH4+ ≈ 9.3), quenching the reaction. (correct answer)
  3. The reaction requires a strong acid catalyst, which is absent in the proposed conditions, to activate the ester carbonyl.
  4. Phenyl acetate is sterically hindered, preventing effective nucleophilic attack by ammonia at the carbonyl carbon.
Explanation: The reaction produces N-phenylacetamide and phenol. Phenol is an acid (pKa ≈ 10), and the nucleophile, ammonia, is a base (pKa of conjugate acid NH4+ ≈ 9.3). Since the pKa of phenol is lower than the pKa of the ammonium ion, phenol is a stronger acid than the ammonium ion. Consequently, the phenol byproduct will readily protonate the ammonia nucleophile. This acid-base reaction removes the nucleophile from the reaction mixture, effectively stopping or severely inhibiting the desired aminolysis reaction.

Question 4

In the mechanism of nucleophilic acyl substitution, the quality of the leaving group is critical. Which statement best explains why an amine (RNH2) can function as a leaving group during acid-catalyzed amide hydrolysis?

  1. Under acidic conditions, the departing amino group is protonated, allowing it to leave as a neutral, weakly basic amine molecule. (correct answer)
  2. The tetrahedral intermediate formed in acid catalysis is exceptionally stable, which lowers the activation energy for amine departure.
  3. The amine is a strong nucleophile, and by the principle of microscopic reversibility, it must also be a good leaving group.
  4. The C-N amide bond is inherently weaker than other sigma bonds, such as C-O or C-C, making it easy to cleave.
Explanation: The rule for leaving groups is that good leaving groups are weak bases. An amide anion (RNH⁻) is an extremely strong base and therefore a terrible leaving group. This is why amides are stable under basic/neutral conditions. However, in acid-catalyzed hydrolysis, the mechanism involves proton transfers within the tetrahedral intermediate. The nitrogen atom becomes protonated before or as it departs. This allows it to leave as a neutral amine molecule (RNH2), which is a weak base and thus a reasonably good leaving group. This protonation of the leaving group is a key feature that makes the reaction feasible.

Question 5

A student wants to synthesize the dipeptide Ala-Gly (alanylglycine) by reacting the carboxylic acid of alanine with the amino group of glycine using DCC as a coupling agent. However, they neglect to use protecting groups. What is the most likely outcome of this reaction?

  1. The desired dipeptide Ala-Gly will be formed cleanly and in high yield.
  2. A mixture of four dipeptides (Ala-Gly, Gly-Ala, Ala-Ala, Gly-Gly) and longer polypeptides will be formed. (correct answer)
  3. No reaction will occur because the zwitterionic forms of the amino acids are unreactive.
  4. Only the cyclic dipeptide (a diketopiperazine) of alanine will be formed.
Explanation: This question applies the principles of amide formation to peptide synthesis. Both alanine and glycine have a nucleophilic amino group and an electrophilic carboxylic acid group (after activation by DCC). Without protecting groups, the alanine carboxylic acid can react with the glycine amine (desired) or another alanine amine (forming Ala-Ala). Similarly, the glycine carboxylic acid can react with the alanine amine (forming Gly-Ala) or another glycine amine (forming Gly-Gly). This lack of control leads to a complex mixture of all possible dipeptides, as well as longer chain polypeptides, resulting in a very low yield of the desired product. This illustrates the critical need for protecting groups in peptide synthesis.

Question 6

N-methylbutanamide is subjected to hydrolysis using excess aqueous NaOH and heat. After the reaction is complete, the solution is cooled and extracted with diethyl ether (no acidic workup is performed). Which pair of species will be the major organic products present in the two separate phases (aqueous and organic)?

  1. Aqueous phase: Butanoic acid; Organic phase: Methylamine
  2. Aqueous phase: Sodium butanoate; Organic phase: Sodium methylamide
  3. Aqueous phase: Sodium butanoate; Organic phase: Methylamine (correct answer)
  4. Aqueous phase: Butanoic acid; Organic phase: Methylammonium hydroxide
Explanation: Base-catalyzed hydrolysis (saponification) of an amide cleaves the C-N bond. The initial products are a carboxylate anion and a neutral amine. Since the reaction is run in excess NaOH, the carboxylic acid product (butanoic acid) will be deprotonated to form sodium butanoate, which is a salt and is water-soluble. The other product, methylamine, is a small, neutral organic molecule that will preferentially partition into the organic diethyl ether layer. An acidic workup would be required to convert sodium butanoate back to butanoic acid.

Question 7

The base-promoted hydrolysis of an amide is mechanistically irreversible, whereas acid-catalyzed hydrolysis is a reversible equilibrium. What is the fundamental reason for the irreversibility of the base-promoted pathway?

  1. Hydroxide is a significantly stronger nucleophile than water, making the initial attack step irreversible under basic conditions.
  2. The tetrahedral intermediate formed under basic conditions is resonance-stabilized and cannot collapse back to the starting materials.
  3. The final step in the basic mechanism is the deprotonation of the carboxylic acid product by a base, which is a highly favorable acid-base reaction. (correct answer)
  4. The amine leaving group formed under basic conditions is immediately protonated by water, preventing it from re-attacking the carbonyl.
Explanation: While hydroxide is a better nucleophile than water (affecting the rate), the key to irreversibility lies at the end of the reaction. In base-promoted hydrolysis, the products are an amine and a carboxylic acid. In the basic solution, the carboxylic acid (pKa ~5) is immediately and irreversibly deprotonated by the base (e.g., OH-, conjugate acid H2O has pKa ~15.7) to form a resonance-stabilized carboxylate anion. This final, highly exergonic acid-base step effectively removes the carboxylic acid product from the equilibrium, driving the entire process to completion according to Le Châtelier's principle.

Question 8

The peptide bond that links amino acids in proteins is a specific type of amide bond. At neutral pH and body temperature, these bonds are exceptionally stable and hydrolyze extremely slowly. Which factor is the most important contributor to the kinetic stability of the peptide bond?

  1. The C-N bond has significant double bond character due to resonance, which lowers the energy of the ground state. (correct answer)
  2. The zwitterionic nature of amino acids at neutral pH prevents the necessary protonation steps for hydrolysis.
  3. Significant steric hindrance from the amino acid side chains (R groups) physically blocks the approach of water.
  4. The leaving group, an amine, is a strong base and therefore cannot be displaced by a weak nucleophile like water.
Explanation: The primary reason for the kinetic stability of amides, including peptide bonds, is resonance. The lone pair of electrons on the nitrogen atom is delocalized into the carbonyl group. This resonance has two major effects: 1) It imparts significant double bond character to the C-N bond, and 2) It makes the carbonyl carbon less electrophilic. This delocalization stabilizes the amide ground state, increasing the activation energy required for nucleophilic attack and subsequent hydrolysis. While other factors contribute, this electronic stabilization is the most fundamental reason for their stability.

Question 9

The synthesis of N-propylbenzamide from benzoyl chloride and propylamine requires either two equivalents of propylamine or one equivalent of propylamine plus one equivalent of a tertiary amine like pyridine. What is the chemical justification for this stoichiometry?

  1. The second equivalent of amine is required to deprotonate the tetrahedral intermediate, which facilitates collapse to the final product.
  2. The reaction is an equilibrium that lies far to the left, and a large excess of the amine nucleophile is needed to drive the reaction forward.
  3. The reaction generates one equivalent of hydrochloric acid (HCl), which must be neutralized to prevent it from protonating and deactivating the propylamine nucleophile. (correct answer)
  4. The second equivalent of the amine acts as a Lewis base catalyst, coordinating to the carbonyl carbon and increasing its electrophilicity.
Explanation: The reaction between an acyl chloride and an amine produces an amide and one equivalent of HCl. Amines are basic. If the HCl is not neutralized, it will protonate the starting amine (propylamine), converting it into an ammonium salt. The propylammonium ion has no lone pair and is no longer nucleophilic, which would stop the reaction after a maximum of 50% conversion. Therefore, a second equivalent of the amine (or another non-nucleophilic base like pyridine or triethylamine) must be added to act as a base to neutralize the HCl as it is formed.

Question 10

A student attempts to synthesize N-ethylpropanamide by mixing propanoic acid and ethylamine in diethyl ether at room temperature. After several hours, TLC analysis indicates that very little of the desired amide has formed. What is the primary reason for this low yield under these conditions?

  1. The tetrahedral intermediate required for the reaction is too unstable at room temperature and rapidly collapses back to the starting materials.
  2. A rapid acid-base reaction occurs first, forming ethylammonium propanoate, a salt which is unreactive toward nucleophilic acyl substitution. (correct answer)
  3. Ethylamine is not a strong enough nucleophile to attack the unactivated carbonyl of a carboxylic acid at room temperature.
  4. The hydroxide ion that would serve as the leaving group is too strong a base, preventing the reaction from proceeding to completion.
Explanation: The primary reason for the failure of this reaction under neutral, room-temperature conditions is that the carboxylic acid (propanoic acid) is acidic and the amine (ethylamine) is basic. They undergo a rapid acid-base reaction to form an ammonium carboxylate salt. The resulting carboxylate anion is resonance-stabilized and its carbonyl carbon is no longer electrophilic, preventing the nucleophilic amine from attacking. To form the amide, one must typically use high heat to drive off water from the salt or use a coupling agent like DCC.

Question 11

Which of the following amino acids would most readily undergo intramolecular cyclization to form a stable lactam (a cyclic amide) when gently heated?

  1. 3-aminopropanoic acid
  2. 6-aminohexanoic acid
  3. 2-aminobutanoic acid
  4. 5-aminopentanoic acid (correct answer)
Explanation: The formation of cyclic amides (lactams) is most favorable for the formation of 5-membered (gamma-lactams) and 6-membered (delta-lactams) rings due to low ring strain and favorable entropy. 5-aminopentanoic acid has the amine on carbon 5 and the carboxylic acid on carbon 1. Cyclization will form a 6-membered ring, which is highly stable. Choice A would form a strained 4-membered ring. Choice B would form a 7-membered ring, which is entropically less favorable to form than a 6-membered ring. Choice C is an alpha-amino acid and would not readily form a stable monomeric lactam under these conditions.

Question 12

Dicyclohexylcarbodiimide (DCC) is a common reagent for forming amide bonds from carboxylic acids and amines without requiring high heat. What is the primary mechanistic role of DCC?

  1. It functions as a powerful dehydrating agent, directly removing water from the tetrahedral intermediate.
  2. It activates the carboxylic acid by reacting with the carboxyl group to form an O-acylisourea intermediate, which has a good leaving group. (correct answer)
  3. It acts as a phase-transfer catalyst, bringing the nonpolar amine and the polar carboxylic acid salt into the same phase.
  4. It functions as a strong base, deprotonating the amine to form a highly nucleophilic amide anion.
Explanation: The main challenge in forming an amide from a carboxylic acid and an amine is that the hydroxyl group (-OH) of the acid is a poor leaving group. DCC overcomes this by activating the carboxylic acid. The carboxylic acid adds to one of the C=N double bonds of DCC, forming a highly reactive O-acylisourea intermediate. In this intermediate, the dicyclohexylurea portion is an excellent leaving group. The amine can then attack the carbonyl carbon, displacing the activated leaving group and forming the amide bond. The byproduct is dicyclohexylurea (DCU).

Question 13

In the accepted mechanism for acid-catalyzed hydrolysis of N-methylacetamide, which of the following steps is generally considered to be rate-determining?

  1. Initial protonation of the carbonyl oxygen by the acid catalyst.
  2. Nucleophilic attack of a water molecule on the carbonyl carbon of the protonated amide. (correct answer)
  3. Proton transfer from the oxygen of the tetrahedral intermediate to the nitrogen atom.
  4. Collapse of the protonated tetrahedral intermediate to expel the neutral methylamine leaving group.
Explanation: Acid-catalyzed hydrolysis of amides proceeds via a nucleophilic acyl substitution mechanism. The first step, protonation of the carbonyl oxygen, is a fast acid-base reaction that activates the carbonyl. The key step that determines the overall rate is the subsequent attack by the weak nucleophile (water) on the now highly electrophilic carbonyl carbon to form the tetrahedral intermediate. The following proton transfers and collapse of the intermediate are typically faster steps.

Question 14

The molecule N-(4-nitrophenyl)acetamide contains an amide linkage. How would its rate of acid-catalyzed hydrolysis compare to that of N-phenylacetamide (acetanilide), and why?

  1. It would hydrolyze slower because the nitro group deactivates the ring, making protonation of the amide more difficult.
  2. It would hydrolyze faster because the strongly electron-withdrawing nitro group pulls electron density from the nitrogen, making the amide carbonyl more electrophilic. (correct answer)
  3. The rates would be nearly identical because the nitro group is para to the amide and has no significant electronic effect on the amide bond itself.
  4. It would hydrolyze slower because the nitro group adds steric bulk to the molecule, hindering the approach of water.
Explanation: The stability of an amide is largely due to the resonance donation from the nitrogen lone pair into the carbonyl. In N-(4-nitrophenyl)acetamide, the para-nitro group is a powerful electron-withdrawing group. It withdraws electron density from the aromatic ring and, in turn, from the amide nitrogen atom. This reduces the ability of the nitrogen lone pair to donate into the carbonyl group. As a result, the carbonyl carbon becomes more electron-deficient (more electrophilic) and thus more susceptible to nucleophilic attack by water, leading to a faster rate of hydrolysis compared to the unsubstituted acetanilide.

Question 15

A chemist wishes to convert N-benzylpropanamide into benzylamine. Which of the following reagents is most suitable for this specific transformation?

    1. H3O+, heat; 2. NaOH
  2. H2 (1 atm), Pd/C, ethanol
    1. LiAlH4, THF; 2. H2O
     (correct answer)
  4. NaBH4, CH3OH
Explanation: The desired transformation is the conversion of an amide to an amine, which is a reduction of the carbonyl group to a methylene (-C=O to -CH2-). Lithium aluminum hydride (LiAlH4) is a powerful reducing agent capable of reducing amides to amines. Sodium borohydride (NaBH4) is not strong enough for this reaction. Acidic or basic hydrolysis (Choice A) would cleave the amide bond to produce propanoic acid and benzylamine, not just benzylamine. Catalytic hydrogenation (H2/Pd-C) can cleave benzyl groups but would not reduce the amide carbonyl under these mild conditions.