Organic Chemistry 2 Quiz: Carboxylic Acids And Derivatives Reactivity Order
16 questions · exam conditions
0:00
Carboxylic Acids And Derivatives Reactivity OrderQuestion 1 of 16

An equimolar mixture of ethyl acetate and acetic anhydride is treated with one equivalent of methanol under mild acid catalysis. Assuming the reaction goes to completion, what is the major organic product formed from the methanol nucleophile?

Methyl acetate, formed from the reaction with acetic anhydride.
Methyl acetate, formed from transesterification of ethyl acetate.
An equimolar mixture of methyl acetate and ethyl acetate.
No significant reaction occurs under these mild conditions.
← Back to quizzes

Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: Carboxylic Acids And Derivatives Reactivity Order

Practice Carboxylic Acids And Derivatives Reactivity Order 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 Carboxylic Acids And Derivatives Reactivity Order, 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

An equimolar mixture of ethyl acetate and acetic anhydride is treated with one equivalent of methanol under mild acid catalysis. Assuming the reaction goes to completion, what is the major organic product formed from the methanol nucleophile?

  1. Methyl acetate, formed from the reaction with acetic anhydride. (correct answer)
  2. Methyl acetate, formed from transesterification of ethyl acetate.
  3. An equimolar mixture of methyl acetate and ethyl acetate.
  4. No significant reaction occurs under these mild conditions.
Explanation: The reactivity of carboxylic acid derivatives towards nucleophilic acyl substitution follows the order: acid chloride > acid anhydride > ester > amide. In a competitive reaction with a limited amount of nucleophile, the more reactive derivative will react preferentially. Acetic anhydride is significantly more reactive than ethyl acetate, so the methanol will selectively attack the anhydride to form methyl acetate and acetic acid.

Question 2

The carbonyl (C=O) stretching frequency in an IR spectrum is a good indicator of the bond's character and, consequently, the functional group's reactivity. Given that acid chlorides are the most reactive and amides are among the least reactive common derivatives, which set of typical C=O frequencies correctly corresponds to the order: Acid Chloride, Ester, Amide?

  1. ~1680 cm⁻¹, ~1740 cm⁻¹, ~1810 cm⁻¹
  2. ~1810 cm⁻¹, ~1740 cm⁻¹, ~1680 cm⁻¹ (correct answer)
  3. ~1710 cm⁻¹, ~1810 cm⁻¹, ~1680 cm⁻¹
  4. ~1680 cm⁻¹, ~1810 cm⁻¹, ~1740 cm⁻¹
Explanation: A higher C=O stretching frequency corresponds to a stronger, more double-bond-like carbonyl. In amides, strong resonance donation from nitrogen gives the C-O bond more single-bond character, lowering its frequency (~1680 cm⁻¹). In acid chlorides, the highly electronegative chlorine atom's inductive effect pulls electron density away, strengthening the C=O bond and increasing its frequency (~1810 cm⁻¹). Esters fall in between, with some resonance stabilization but less than in amides, giving a frequency around 1740 cm⁻¹. Thus, the order of decreasing frequency (Acid Chloride > Ester > Amide) matches the order of decreasing reactivity.

Question 3

Which of the following transformations represents a thermodynamically favorable interconversion of carboxylic acid derivatives that can be achieved in a single, high-yielding step under standard conditions?

  1. Ethyl acetate reacting with sodium chloride to form acetyl chloride.
  2. N-methylacetamide reacting with methanol and heat to form methyl acetate.
  3. Acetic acid reacting with ammonia at room temperature to form acetamide.
  4. Acetyl chloride reacting with sodium acetate to form acetic anhydride. (correct answer)
Explanation: Carboxylic acid derivatives can be readily converted from a more reactive form to a less reactive form. The reactivity order is acid chloride > anhydride > ester > amide. In choice D, a highly reactive acid chloride is attacked by a carboxylate nucleophile (from sodium acetate) to form a less reactive acid anhydride. This is a standard and favorable synthesis. Choices A and B represent converting a less reactive derivative to a more reactive one, which is unfavorable. Choice C is an acid-base reaction; forming the amide from the resulting ammonium carboxylate salt requires high heat to drive off water.

Question 4

The rate of acid-catalyzed hydrolysis depends on the electrophilicity of the carbonyl carbon. Which of the following correctly predicts the order of initial reaction rates for the hydrolysis of propionyl chloride (I), ethyl propionate (II), and N-ethylpropionamide (III)?

  1. I > II > III (correct answer)
  2. III > II > I
  3. II > I > III
  4. I > III > II
Explanation: The rate of hydrolysis is directly related to the reactivity of the carboxylic acid derivative. The general reactivity order is acid chloride > ester > amide. Propionyl chloride (I) is an acid chloride and is the most reactive due to the highly electrophilic carbonyl carbon. Ethyl propionate (II) is an ester, which is moderately reactive. N-ethylpropionamide (III) is an amide, which is the least reactive due to strong resonance stabilization of the carbonyl group. Therefore, the order of hydrolysis rates will be I > II > III.

Question 5

A chemist plans to synthesize N-ethylbenzamide from benzoic acid and ethylamine. The direct reaction requires high temperatures and gives low yields. A better, two-step route involves first converting benzoic acid to an intermediate, then reacting the intermediate with ethylamine. Why is converting benzoic acid to benzoyl chloride with SOCl₂ a critical first step in this improved synthesis?

  1. It prevents an unproductive acid-base reaction between the carboxylic acid and the amine.
  2. It converts the moderately reactive carboxylic acid into a highly reactive acid chloride, which readily undergoes substitution. (correct answer)
  3. It removes water from the reaction mixture, which would otherwise hydrolyze the product amide.
  4. It protects the carboxylic acid group while another reaction is performed elsewhere in the molecule.
Explanation: The core principle of synthesis involving carboxylic acid derivatives is to convert a less reactive derivative into a more reactive one to facilitate a desired reaction. Benzoic acid is only moderately reactive. The direct reaction with ethylamine is slow primarily because it's an acid-base reaction first, forming a salt, and the subsequent nucleophilic acyl substitution requires harsh conditions. By converting benzoic acid to benzoyl chloride (an acid chloride) with thionyl chloride (SOCl₂), we create a much more electrophilic species. This highly reactive acid chloride can then be easily attacked by the ethylamine nucleophile to form the desired amide in high yield under mild conditions.

Question 6

A solution contains equimolar amounts of p-toluic acid (pKa ≈ 4.4) and methyl p-toluate. One equivalent of sodium methoxide is added to this solution at room temperature. What is the most likely initial outcome?

  1. Methoxide attacks the ester's carbonyl carbon in a transesterification reaction.
  2. Methoxide deprotonates the p-toluic acid in an acid-base reaction. (correct answer)
  3. Methoxide attacks the carboxylic acid's carbonyl carbon.
  4. No reaction occurs as both substrates are relatively unreactive.
Explanation: This question tests the relative rates of different reaction types. Acid-base reactions are extremely fast, generally much faster than nucleophilic substitution reactions. Sodium methoxide is a strong base. P-toluic acid is a carboxylic acid and is acidic. Before any nucleophilic attack on either carbonyl can occur, the methoxide will rapidly and irreversibly deprotonate the most acidic species present, the carboxylic acid. This forms methanol and sodium p-toluate. The resulting carboxylate is negatively charged and highly unreactive towards further nucleophilic attack.

Question 7

Thioesters (R-CO-SR') are biochemically important acylating agents and are generally more reactive than their oxygen ester analogues (R-CO-OR'). Which statement provides the most accurate electronic basis for this difference?

  1. Sulfur is much more electronegative than oxygen, making the carbonyl carbon more electrophilic by induction.
  2. The C-S bond is significantly weaker than the C-O bond, making bond cleavage easier in the tetrahedral intermediate.
  3. Poor orbital overlap between the sulfur 3p and carbon 2p orbitals reduces resonance stabilization of the thioester. (correct answer)
  4. A thiolate anion (RS⁻) is a much stronger base than an alkoxide anion (RO⁻), making it a better leaving group.
Explanation: The primary reason for the enhanced reactivity of thioesters is electronic. The lone pairs on the sulfur atom are in 3p orbitals, which have poor size and energy match with the 2p orbital of the carbonyl carbon. This results in less effective orbital overlap for resonance compared to the 2p-2p overlap between oxygen and carbon in an ester. Consequently, the thioester carbonyl is less stabilized by resonance, has more double-bond character, and its carbon is more electrophilic and reactive. While the leaving group ability of thiolate is also a factor, the effect on the ground state electrophilicity is the key distinction.

Question 8

The reactivity of a carboxylic acid derivative R-CO-L is inversely related to the basicity of its leaving group L⁻. Given the following pKa values for the conjugate acids (HL): H₂S (pKa=7.0), H₂O (pKa=15.7), CH₃COOH (pKa=4.8), NH₃ (pKa=38). Which derivative would be the LEAST reactive toward nucleophilic acyl substitution?

  1. R-CO-SH (a thioacid, if it acted as an acylating agent)
  2. R-CO-OCOCH₃ (an anhydride)
  3. R-CO-NH₂ (an amide) (correct answer)
  4. R-CO-OH (a carboxylic acid)
Explanation: A better leaving group is a weaker base. The strength of a base is inversely related to the pKa of its conjugate acid; a higher pKa for HL means L⁻ is a stronger base and a poorer leaving group. Comparing the pKa values: NH₃ (38) >> H₂O (15.7) > H₂S (7.0) > CH₃COOH (4.8). The leaving group for an amide is NH₂⁻, whose conjugate acid NH₃ has the highest pKa (38). This means NH₂⁻ is the strongest base and the worst leaving group among the options, making the amide (R-CO-NH₂) the least reactive derivative.

Question 9

β-Lactams (cyclic amides in a four-membered ring) are key structural features in antibiotics like penicillin and are known to be much more reactive towards hydrolysis than typical acyclic amides. What is the best explanation for this enhanced reactivity?

  1. The planarity of the ring system enhances the resonance stabilization of the amide group.
  2. The amide nitrogen in a β-lactam is more basic than in an acyclic amide, making it a better leaving group.
  3. Significant angle strain in the four-membered ring is released when the ring is opened by a nucleophile. (correct answer)
  4. The carbonyl carbon is less sterically hindered in a β-lactam than in a typical acyclic amide.
Explanation: The high reactivity of β-lactams is primarily due to ring strain. The ideal bond angles for an sp² carbon (carbonyl) are 120° and for an sp³ nitrogen are ~109.5°, but they are constrained to ~90° in a four-membered ring. This strain makes the β-lactam unstable and high in energy. Nucleophilic attack opens the ring, forming a stable, acyclic tetrahedral intermediate and releasing this strain. This large, favorable change in enthalpy makes the activation energy for the reaction much lower than for a strain-free acyclic amide. Furthermore, ring strain inhibits the planarity required for effective resonance stabilization, also contributing to higher reactivity.

Question 10

Imagine a hypothetical leaving group, Z, where the conjugate acid HZ has a pKa of 12. How would the reactivity of an acyl derivative R-CO-Z compare to that of a typical ester (R-CO-OR') and a typical thioester (R-CO-SR')?

  1. More reactive than both the ester and the thioester.
  2. Less reactive than the ester but more reactive than the thioester.
  3. More reactive than the ester but less reactive than the thioester. (correct answer)
  4. Less reactive than both the ester and the thioester.
Explanation: Reactivity is related to leaving group ability, which can be estimated by the pKa of the conjugate acid (a lower pKa for HL means L- is a better leaving group). A typical alcohol (R'OH), the conjugate acid for an ester's leaving group, has a pKa ≈ 16-18. A typical thiol (R'SH), the conjugate acid for a thioester's leaving group, has a pKa ≈ 10. The hypothetical HZ has a pKa of 12. Since 10 < 12 < 16, the leaving group ability will be Thiolate > Z⁻ > Alkoxide. Therefore, the reactivity of the acyl derivatives will follow the same order: Thioester > R-CO-Z > Ester.

Question 11

Consider the nucleophilic acyl substitution mechanism. When comparing the reaction of an acid chloride versus an amide with a given nucleophile, which statement best describes the relative stability of their tetrahedral intermediates and the consequence for reactivity?

  1. The intermediate from the acid chloride is more stable due to the inductive effect of chlorine, leading to a faster reaction.
  2. The intermediate from the amide is less stable because nitrogen cannot support a negative charge as well as oxygen.
  3. The intermediates have similar stabilities, but the ground state energy of the amide is much lower, leading to a slower reaction.
  4. The intermediate from the acid chloride more readily collapses to products because chloride is a much better leaving group than the amide ion. (correct answer)
Explanation: The rate of a nucleophilic acyl substitution depends on both the formation of the tetrahedral intermediate and its collapse to products. A key factor in the collapse is the ability of the leaving group to depart. Chloride (Cl⁻) is the conjugate base of a strong acid (HCl) and is therefore a very weak base and an excellent leaving group. The amide ion (NH₂⁻) is the conjugate base of a very weak acid (NH₃) and is a very strong base and a terrible leaving group. Therefore, even after it is formed, the tetrahedral intermediate from the acid chloride collapses rapidly and irreversibly to products, while the intermediate from the amide is much more likely to revert to starting materials.

Question 12

Rank the following esters in order of decreasing reactivity towards saponification (base-catalyzed hydrolysis): Methyl trifluoroacetate (I), Methyl acetate (II), and Methyl p-methoxybenzoate (III).

  1. I > II > III (correct answer)
  2. III > II > I
  3. II > I > III
  4. I > III > II
Explanation: Reactivity in nucleophilic acyl substitution is enhanced by electron-withdrawing groups (EWGs) and diminished by electron-donating groups (EDGs) attached to the acyl group. The trifluoromethyl group (-CF₃) in (I) is a very strong EWG, making the carbonyl carbon highly electrophilic. The methyl group in (II) is neutral/weakly donating. The p-methoxy group in (III) is a strong EDG via resonance, which stabilizes the starting material and makes the carbonyl carbon less electrophilic. Therefore, the order of decreasing reactivity is I > II > III.

Question 13

A student attempts to synthesize methyl butanoate by reacting butanoic acid with sodium methoxide. The reaction fails to produce a significant yield. What is the fundamental reason for this failure?

  1. Sodium methoxide is a weak nucleophile and cannot attack the carboxylic acid carbonyl.
  2. The reaction generates water, which hydrolyzes the ester product back to the starting material.
  3. Methoxide, being a strong base, deprotonates the carboxylic acid to form a carboxylate anion that is resistant to nucleophilic attack. (correct answer)
  4. This transformation requires an acid catalyst (Fischer esterification), and the basic conditions inhibit the reaction.
Explanation: This is a common pitfall. While methoxide is a good nucleophile, it is also a strong base. A carboxylic acid has an acidic proton. The fastest reaction is always the acid-base reaction. The methoxide will immediately deprotonate the butanoic acid to form sodium butanoate and methanol. The resulting butanoate anion has a negative charge, which makes the carbonyl carbon electron-rich and extremely unreactive towards nucleophiles. To form an ester from a carboxylic acid, one must use acid catalysis (Fischer esterification) with an alcohol, not a strong base.

Question 14

In the presence of pyridine, acetic anhydride reacts with ethanol to form ethyl acetate. Which statement best describes the role of pyridine in this reaction?

  1. Pyridine protonates the anhydride's carbonyl oxygen, activating it for nucleophilic attack.
  2. Pyridine coordinates to the carbonyl carbon, making it more electrophilic toward nucleophilic attack.
  3. Pyridine deprotonates ethanol to form the more nucleophilic ethoxide ion.
  4. Pyridine acts as a non-nucleophilic base to neutralize the acetic acid byproduct, driving the equilibrium forward. (correct answer)
Explanation: The reaction of acetic anhydride with ethanol produces ethyl acetate and one equivalent of acetic acid. This acid can protonate the ethanol or other species, potentially slowing the reaction. Pyridine is a base that scavenges this acid byproduct, neutralizing it to form pyridinium acetate. By removing a product, it helps drive the reaction to completion according to Le Châtelier's principle. This is its primary role in this context. Choice C is incorrect as pyridine is not strong enough to deprotonate ethanol significantly. Choice B is incorrect as pyridine coordination would not significantly increase electrophilicity.

Question 15

Which of the following correctly ranks the given compounds from MOST reactive to LEAST reactive towards hydrolysis?

  1. Phenyl acetate > Acetyl chloride > Acetamide > Ethyl acetate
  2. Acetyl chloride > Phenyl acetate > Ethyl acetate > Acetamide (correct answer)
  3. Acetamide > Ethyl acetate > Phenyl acetate > Acetyl chloride
  4. Acetyl chloride > Ethyl acetate > Phenyl acetate > Acetamide
Explanation: This question requires ranking four common derivatives. The general order is Acid Chloride > Anhydride > Ester > Amide. Acetyl chloride is an acid chloride, so it is the most reactive. Acetamide is an amide, so it is the least reactive. We must then compare the two esters: phenyl acetate and ethyl acetate. The reactivity of an ester is influenced by the leaving group. Phenoxide is a better leaving group than ethoxide because its conjugate acid, phenol (pKa ≈ 10), is more acidic than ethanol (pKa ≈ 16). Therefore, phenyl acetate is more reactive than ethyl acetate. The correct overall order is Acetyl chloride > Phenyl acetate > Ethyl acetate > Acetamide.

Question 16

Amides are substantially less reactive towards nucleophilic acyl substitution than esters. What is the primary electronic reason for this significant difference in reactivity?

  1. Nitrogen is less electronegative than oxygen, making it a poorer inductive electron-withdrawer.
  2. The nitrogen atom's lone pair provides stronger resonance stabilization to the carbonyl group than the oxygen's lone pair. (correct answer)
  3. The N-H bonds in primary and secondary amides create steric hindrance around the carbonyl carbon.
  4. The amide anion (NH₂⁻) is a much poorer leaving group than an alkoxide anion (OR⁻).
Explanation: The primary reason for the low reactivity of amides is the strong resonance stabilization of the starting material. Nitrogen is less electronegative than oxygen, making it a better electron-pair donor. This results in a more significant resonance contributor with a C=N double bond and a negative charge on the oxygen. This delocalization stabilizes the ground state of the amide, increasing the activation energy for nucleophilic attack. While leaving group ability (D) is also a factor, the ground state stabilization (B) is considered the most fundamental reason for the large reactivity gap.