Organic Chemistry 2 Quiz: Acid Chlorides And Anhydrides Synthesis Reactions
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Acid Chlorides And Anhydrides Synthesis ReactionsQuestion 1 of 15

In a Friedel-Crafts acylation reaction, benzene is treated with phthalic anhydride and a Lewis acid catalyst (AlCl₃), followed by an aqueous workup. What is the structure of the resulting major product?

A spirocyclic compound where one carbonyl of the anhydride has been replaced by the benzene ring.
Diphenyl phthalate
2-Benzoylbenzoic acid
Anthraquinone
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Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: Acid Chlorides And Anhydrides Synthesis Reactions

Practice Acid Chlorides And Anhydrides Synthesis Reactions 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 Acid Chlorides And Anhydrides Synthesis Reactions, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry 2.

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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.

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Question 1

In a Friedel-Crafts acylation reaction, benzene is treated with phthalic anhydride and a Lewis acid catalyst (AlCl₃), followed by an aqueous workup. What is the structure of the resulting major product?

  1. A spirocyclic compound where one carbonyl of the anhydride has been replaced by the benzene ring.
  2. Diphenyl phthalate
  3. 2-Benzoylbenzoic acid (correct answer)
  4. Anthraquinone
Explanation: This is an intramolecular Friedel-Crafts acylation. The Lewis acid activates one of the carbonyl groups of the phthalic anhydride. The benzene ring acts as a nucleophile and attacks this activated carbonyl, leading to the opening of the anhydride ring. The intermediate is an acylium ion attached to a carboxylate. After aqueous workup, the final product is 2-benzoylbenzoic acid, which contains both a ketone and a carboxylic acid functional group.

Question 2

Which combination of reagents is most suitable for the synthesis of acetophenone (phenyl methyl ketone) with minimal side products?

  1. Benzoyl chloride and methylmagnesium bromide
  2. Benzoyl chloride and lithium dimethylcuprate
  3. Acetyl chloride and benzene with AlCl₃ (correct answer)
  4. Acetyl chloride and phenylmagnesium bromide
Explanation: Friedel-Crafts acylation (option C) is the most common and practical method for synthesizing acetophenone. Acetyl chloride with benzene and AlCl₃ directly gives acetophenone in good yield. Option B (cuprate with acid chloride) would also work but is less commonly used for this specific synthesis. Options A and D would give tertiary alcohols because Grignard reagents add twice to acid chlorides - first to give the ketone, then immediately to the ketone product to form a tertiary alcohol.

Question 3

What is the final major organic product when butanoyl chloride is treated with two equivalents of phenylmagnesium bromide (PhMgBr), followed by an aqueous acid workup (H₃O⁺)?

  1. Butyrophenone (1-phenyl-1-butanone)
  2. 1,1-Diphenyl-1-butanol (correct answer)
  3. 4-Phenyl-1-butanol
  4. 1-Phenylbutane
Explanation: Grignard reagents react twice with acid chlorides. The first equivalent of PhMgBr adds to the carbonyl of butanoyl chloride to form an unstable tetrahedral intermediate that collapses to form butyrophenone (a ketone). This ketone is more reactive towards the Grignard reagent than the starting acid chloride. A second equivalent of PhMgBr immediately attacks the ketone, forming a new tetrahedral intermediate. Aqueous acid workup protonates the resulting alkoxide to yield the tertiary alcohol, 1,1-diphenyl-1-butanol.

Question 4

A student plans to synthesize N-ethylpropanamide from propanoic acid and ethylamine. Why is the direct reaction between propanoic acid and ethylamine at room temperature generally ineffective for forming the amide?

  1. The carboxylic acid is not electrophilic enough to react with the amine without heat.
  2. The reaction produces water, which immediately hydrolyzes the amide product back to the starting materials.
  3. The ethylamine is not nucleophilic enough to attack the carboxylic acid.
  4. An acid-base reaction occurs, forming an unreactive ammonium carboxylate salt. (correct answer)
Explanation: When you encounter amide synthesis problems, consider the fundamental chemical properties of the reactants involved. Carboxylic acids and amines are both capable of donating and accepting protons, which significantly affects their reactivity. The direct reaction fails because propanoic acid (pKa ≈ 4.9) readily donates a proton to ethylamine (pKb ≈ 3.3), forming an acid-base adduct: CH₃CH₂COOH + CH₃CH₂NH₂ → CH₃CH₂COO⁻ ⁺NH₃CH₂CH₃. This ammonium carboxylate salt is much less reactive than the original starting materials. The carboxylate anion is stabilized by resonance and won't act as an electrophile, while the protonated ammonium ion has lost its nucleophilic lone pair. This makes answer D correct. Answer A is incorrect because carboxylic acids are sufficiently electrophilic—the issue isn't electrophilicity but rather the acid-base chemistry that prevents the desired nucleophilic acyl substitution. Answer B misunderstands the kinetics; while water could theoretically hydrolyze amides, this isn't why the initial formation fails, and amide hydrolysis is typically slow under neutral conditions. Answer C is wrong because ethylamine is quite nucleophilic as a primary amine—again, the problem isn't nucleophilicity but the competing acid-base reaction. For successful amide formation, you need activated carboxylic acid derivatives (like acid chlorides) or coupling reagents that prevent the acid-base reaction. Remember: whenever you see carboxylic acid + amine reactions, always consider whether acid-base chemistry will compete with the desired substitution mechanism.

Question 5

An equimolar mixture of propanoyl chloride and propanoic anhydride is treated with 0.5 equivalents of a sterically hindered amine, such as di-isopropylamine. Which statement accurately predicts the outcome?

  1. The amine will react exclusively with propanoyl chloride due to its higher intrinsic reactivity. (correct answer)
  2. The amine will react exclusively with propanoic anhydride because it is less sterically demanding.
  3. The amine will react with both compounds at roughly equal rates, yielding a mixture of amides.
  4. No reaction will occur because the amine is too sterically hindered to act as a nucleophile.
Explanation: Acid chlorides are significantly more reactive electrophiles than anhydrides because the chloride ion is a better leaving group than the carboxylate ion. Even with a sterically hindered nucleophile, the large difference in electrophilicity of the carbonyls will dominate. The amine will preferentially attack the more 'activated' carbonyl of the propanoyl chloride, leading to the formation of N,N-di-isopropylpropanamide.

Question 6

When synthesizing an amide, it is common to add the acid chloride slowly to a solution containing excess amine, rather than adding the amine to the acid chloride. What is the primary chemical reason for this specific order of addition?

  1. To prevent the highly reactive acid chloride from reacting with the amide product.
  2. To ensure an excess of base is always present to neutralize the HCl byproduct immediately. (correct answer)
  3. To better control the exothermic nature of the reaction by having the amine act as a heat sink.
  4. To avoid the formation of a stable, unreactive tetrahedral intermediate.
Explanation: The reaction produces one equivalent of HCl for every equivalent of amide formed. The amine serves as both the nucleophile and the base. If the amine is added slowly to the acid chloride, the initially formed HCl will protonate the incoming amine, rendering it non-nucleophilic and halting the reaction. By adding the acid chloride slowly to an excess of the amine, there is always a sufficient amount of free amine base available to neutralize the HCl as it is formed, allowing the reaction to proceed to completion.

Question 7

Anisole (methoxybenzene) is subjected to Friedel-Crafts acylation with propanoyl chloride and AlCl₃. What is the predicted major organic product?

  1. 2-Methoxypropiophenone
  2. 3-Methoxypropiophenone
  3. 4-Methoxypropiophenone (correct answer)
  4. Phenyl propanoate
Explanation: This is an electrophilic aromatic substitution (EAS) reaction. The methoxy group (-OCH₃) on anisole is a strong activating group and an ortho, para-director. The incoming electrophile is the propanoyl cation (or its complex with AlCl₃). Substitution will occur at the positions most activated by the methoxy group, which are the ortho and para positions. Due to the steric bulk of the propanoyl group, substitution at the less hindered para position is strongly favored over the ortho position. Thus, the major product is 4-methoxypropiophenone.

Question 8

During the preparation of an ester from an acid chloride and an alcohol, a student uses reagents that are contaminated with a small amount of water. Besides the expected ester, what high-boiling point impurity is most likely to form as a result of this contamination?

  1. A carboxylic acid
  2. An ether
  3. A symmetrical anhydride (correct answer)
  4. An aldehyde
Explanation: The presence of water will cause some of the highly reactive acid chloride to hydrolyze into the corresponding carboxylic acid. This carboxylic acid (or its carboxylate form) can then act as a nucleophile, attacking another molecule of the unreacted acid chloride. This nucleophilic acyl substitution reaction results in the formation of a symmetrical anhydride. Anhydrides have significantly higher boiling points than the corresponding esters or carboxylic acids due to their larger size and polarity, making them a likely high-boiling impurity.

Question 9

4-Aminobutanol is treated with one equivalent of acetic anhydride at a low temperature (0 °C). Which functional group is expected to react preferentially, and what is the major product?

  1. The alcohol reacts to form 4-aminobutyl acetate.
  2. The amine reacts to form N-(4-hydroxybutyl)acetamide. (correct answer)
  3. Both groups react to form N-(4-acetoxybutyl)acetamide.
  4. An intramolecular cyclization occurs to form a seven-membered ring.
Explanation: In competitive reactions with an acylating agent like acetic anhydride, the relative nucleophilicity of the functional groups determines the outcome. Primary amines are significantly more nucleophilic than primary alcohols. Therefore, the amino group will selectively attack the anhydride to form an amide, leaving the alcohol group untouched, especially when using only one equivalent of the anhydride at low temperature. The product is N-(4-hydroxybutyl)acetamide.

Question 10

In the synthesis of an ester from an acid chloride and an alcohol, pyridine is often used as a solvent or additive. What is the consequence if pyridine is omitted from the reaction of acetyl chloride and ethanol?

  1. The reaction will not proceed at all because pyridine is a required catalyst.
  2. The reaction will proceed faster because pyridine is an inhibitor for this transformation.
  3. The reaction will yield acetic anhydride as the major product instead of the ester.
  4. The reaction rate will decrease significantly as the HCl byproduct protonates the ethanol. (correct answer)
Explanation: When you encounter questions about acid chloride reactions, focus on the role of bases in managing acidic byproducts. Acid chlorides react readily with alcohols to form esters, but this reaction produces HCl as a byproduct that can significantly impact the reaction. In the reaction between acetyl chloride and ethanol, HCl forms immediately when the ester bond creates. Without pyridine present, this HCl will protonate the ethanol nucleophile, converting it from ROH to ROH₂⁺. Protonated alcohols are much weaker nucleophiles because the positive charge makes them less likely to donate their electron pairs for bond formation. This dramatically slows the reaction rate as fewer alcohol molecules are available in their reactive, unprotonated form. Looking at the incorrect answers: (A) is wrong because pyridine isn't a catalyst for the actual ester formation - it's a base that neutralizes HCl. The reaction can still proceed without it. (B) incorrectly suggests pyridine inhibits the reaction, when actually it facilitates it by maintaining favorable conditions. (C) confuses this reaction with anhydride formation, which requires different conditions and reactants - you won't get acetic anhydride from acetyl chloride and ethanol. Pyridine's role is to act as an HCl scavenger, forming pyridinium chloride and keeping the alcohol in its nucleophilic form. Study tip: Remember that in organic reactions producing acidic byproducts, bases like pyridine or triethylamine are added not as catalysts, but to neutralize acids that would otherwise protonate and deactivate your nucleophiles.

Question 11

Benzoyl chloride is treated with exactly one equivalent of ammonia (NH₃). Assuming the reaction goes to completion based on the limiting reagent, what is the theoretical yield of benzamide?

  1. 0%
  2. 25%
  3. 50% (correct answer)
  4. 100%
Explanation: The reaction of an acid chloride with ammonia produces an amide and HCl. Ammonia is a base and will react with the HCl produced to form ammonium chloride (NH₄Cl). Therefore, two equivalents of ammonia are required for the reaction to go to completion: one as the nucleophile and one as the base. Since only one equivalent of ammonia is used, half will act as a nucleophile to produce benzamide and HCl, and the other half will be immediately protonated by that HCl to form non-nucleophilic NH₄Cl. Thus, only 50% of the benzoyl chloride can react, leading to a theoretical yield of 50%.

Question 12

Which of the following pairs of reactants is most suitable for the controlled synthesis of the mixed anhydride, acetic benzoic anhydride?

  1. Acetic acid and benzoic acid, with P₂O₅ and heat
  2. Acetic anhydride and sodium benzoate
  3. Acetyl chloride and benzoic acid
  4. Benzoyl chloride and sodium acetate (correct answer)
Explanation: When synthesizing mixed anhydrides, you need a controlled, selective reaction that prevents the formation of symmetric anhydrides or other unwanted products. Mixed anhydrides form when two different carboxylic acid derivatives combine, but the challenge lies in achieving selectivity. The most effective approach uses an acid chloride (highly reactive) with a carboxylate salt (nucleophilic but controlled reactivity). In option D, benzoyl chloride reacts with sodium acetate through nucleophilic acyl substitution. The acetate ion attacks the carbonyl carbon of benzoyl chloride, displacing chloride and forming the desired mixed anhydride while producing NaCl as a harmless byproduct. This reaction is clean, selective, and occurs under mild conditions. Option A fails because heating acetic acid and benzoic acid with P₂O₅ creates a mixture of products: both symmetric anhydrides (acetic anhydride and benzoic anhydride) plus some mixed anhydride. This lacks the selectivity needed for controlled synthesis. Option B presents a reactivity mismatch. Acetic anhydride is less electrophilic than acid chlorides, making the reaction with sodium benzoate sluggish and potentially incomplete. Option C suffers from competing reactions. When acetyl chloride (highly reactive) meets benzoic acid, you risk forming multiple products including symmetric anhydrides, especially since benzoic acid can act as both a nucleophile and undergo side reactions. Remember this pattern: for selective mixed anhydride synthesis, pair an acid chloride with a carboxylate salt rather than using two acids or mismatched reactivity partners. The acid chloride provides the driving force while the salt ensures clean substitution.

Question 13

Which of the following reaction conditions is most effective for the high-yield synthesis of pure, symmetrical isobutyric anhydride?

  1. Heating isobutyric acid with a catalytic amount of sulfuric acid.
  2. Reacting isobutyryl chloride with one equivalent of sodium isobutyrate. (correct answer)
  3. Heating an equimolar mixture of isobutyric acid and acetic anhydride.
  4. Reacting isobutyric acid with one equivalent of sodium hydroxide, followed by heating.
Explanation: The most effective method to prepare a pure, symmetrical anhydride is the reaction of an acid chloride with a carboxylate salt. This is an irreversible nucleophilic acyl substitution that gives the anhydride and a salt byproduct (NaCl), which is easily removed. Option A is a reversible equilibrium that requires removal of water to achieve high yield. Option C involves anhydride exchange and will result in a mixture of three different anhydrides (isobutyric, acetic, and the mixed anhydride). Option D simply forms the sodium salt of the acid, which will not form an anhydride upon heating alone.

Question 14

Which of the following dicarboxylic acids will most readily form a stable, cyclic anhydride upon gentle heating?

  1. Oxalic acid (ethanedioic acid)
  2. Malonic acid (propanedioic acid)
  3. Glutaric acid (pentanedioic acid) (correct answer)
  4. Adipic acid (hexanedioic acid)
Explanation: Formation of cyclic anhydrides is most favorable for dicarboxylic acids that can form stable five- or six-membered rings. Glutaric acid (a 5-carbon diacid) will form a six-membered cyclic anhydride, which is thermodynamically stable. Succinic acid (a 4-carbon diacid) forms a stable five-membered ring. Oxalic acid (2C) and malonic acid (3C) would form highly strained four- and five-membered rings with two carbonyls, but malonic acid readily decarboxylates upon heating. Adipic acid (6C) would form a seven-membered ring, which is less entropically and enthalpically favored than a six-membered ring.

Question 15

Acetic isobutyric anhydride is treated with one equivalent of sodium methoxide (NaOCH₃). Which of the following best describes the major products after an acidic workup?

  1. Methyl acetate and isobutyric acid (correct answer)
  2. Methyl isobutyrate and acetic acid
  3. Equal amounts of methyl acetate, methyl isobutyrate, acetic acid, and isobutyric acid
  4. Methanol, acetic acid, and isobutyric acid
Explanation: In a mixed anhydride, a nucleophile will preferentially attack the more electrophilic (less sterically hindered) carbonyl carbon. The carbonyl of the acetyl group is less sterically hindered than the carbonyl of the isobutyryl group. Therefore, the methoxide ion will selectively attack the acetyl carbonyl, leading to the formation of methyl acetate and sodium isobutyrate. Subsequent acidic workup protonates the isobutyrate to form isobutyric acid.