Organic Chemistry 2 Quiz: Claisen Condensation
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Claisen CondensationQuestion 1 of 10

Consider the Claisen condensation of ethyl acetate with one equivalent of sodium ethoxide in ethanol. What is the major organic species present in the reaction mixture immediately before the addition of aqueous acid (H₃O⁺)?

Ethyl acetoacetate
The tetrahedral addition intermediate
The sodium enolate of ethyl acetoacetate
The sodium enolate of ethyl acetate
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Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: Claisen Condensation

Practice Claisen Condensation 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 Claisen Condensation, 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

Consider the Claisen condensation of ethyl acetate with one equivalent of sodium ethoxide in ethanol. What is the major organic species present in the reaction mixture immediately before the addition of aqueous acid (H₃O⁺)?

  1. Ethyl acetoacetate
  2. The tetrahedral addition intermediate
  3. The sodium enolate of ethyl acetoacetate (correct answer)
  4. The sodium enolate of ethyl acetate
Explanation: The reaction proceeds by forming ethyl acetoacetate. However, this β-keto ester is acidic (pKa ~11) and is immediately deprotonated by the sodium ethoxide base (conjugate acid pKa ~16) in a thermodynamically favorable step. This acid-base reaction is the driving force. Therefore, before the acidic workup, the product exists as its resonance-stabilized sodium enolate salt. The acidic workup is required to protonate this enolate and isolate the neutral β-keto ester.

Question 2

Which of the following compounds can act as the electrophilic partner in a crossed Claisen condensation but CANNOT act as the nucleophilic (enolate) partner?

  1. Diethyl carbonate (correct answer)
  2. Diethyl malonate
  3. Ethyl acetate
  4. Ethyl acetoacetate
Explanation: The crossed Claisen condensation involves two different ester compounds: one that provides an enolate nucleophile and another that serves as the electrophile. For a compound to act as the nucleophilic partner, it must have acidic α-hydrogens that can be deprotonated to form an enolate. For the electrophilic partner, the compound should lack α-hydrogens (preventing it from forming competing enolates) but still be reactive toward nucleophilic attack at the carbonyl carbon. Diethyl carbonate (A) is the perfect electrophilic partner because it has no α-hydrogens—there's no carbon adjacent to the carbonyl since it's a carbonate ester (\ce(EtO)2CO\ce{(EtO)2CO}). This means it cannot form an enolate but readily accepts nucleophilic attack. This makes A the correct answer. Diethyl malonate (B) has very acidic α-hydrogens between two electron-withdrawing ester groups, making it an excellent enolate source. It can definitely act as the nucleophilic partner. Ethyl acetate (C) has α-hydrogens on its methyl group, so it can form enolates and serve as the nucleophilic partner in Claisen condensations. Ethyl acetoacetate (D) contains highly acidic α-hydrogens between the ketone and ester carbonyls, making it another excellent nucleophile source. When approaching crossed Claisen problems, look for compounds lacking α-hydrogens as electrophiles (like formate esters, carbonate esters, or benzoate esters) and compounds with acidic α-hydrogens as nucleophiles. The key is identifying which partner can only play one role in the reaction.

Question 3

A crossed Claisen condensation is performed by slowly adding ethyl acetate to a solution containing sodium ethoxide and ethyl benzoate. Why is this specific procedure (adding the enolizable ester to the non-enolizable one) likely to result in a high yield of a single major product?

  1. Ethyl benzoate is a much stronger electrophile than ethyl acetate, so it reacts exclusively regardless of the enolate concentration.
  2. This procedure ensures that the concentration of the ethyl acetate enolate is always low, minimizing its self-condensation while it can readily react with the abundant ethyl benzoate. (correct answer)
  3. Ethyl acetate cannot self-condense under these conditions, forcing it to react only with ethyl benzoate.
  4. The sodium ethoxide selectively forms the enolate of ethyl benzoate, which then reacts with ethyl acetate.
Explanation: In a successful crossed Claisen, we want to prevent the self-condensation of the enolizable partner (ethyl acetate). By adding ethyl acetate slowly to the mixture of base and the non-enolizable partner (ethyl benzoate), any ethyl acetate enolate that forms is in a low concentration but is surrounded by a high concentration of the electrophile (ethyl benzoate). This maximizes the probability of the desired cross-reaction and minimizes the probability of the enolate finding another molecule of ethyl acetate to react with, thus preventing self-condensation.

Question 4

A student attempts a Claisen condensation using ethyl benzoate and ethyl isobutyrate with sodium ethoxide. Despite ethyl benzoate being unable to form an enolate, the reaction yields almost no desired crossed product. What is the most likely reason for this failure?

  1. The reaction forms the desired β-keto ester, but this product lacks an α-hydrogen, so the equilibrium cannot be driven forward by deprotonation. (correct answer)
  2. The enolate of ethyl isobutyrate is too stable to react with the electrophilic ethyl benzoate.
  3. Ethyl benzoate is sterically hindered by the benzene ring, preventing attack by any enolate.
  4. Sodium ethoxide is not a strong enough base to deprotonate ethyl isobutyrate at its sterically hindered α-carbon.
Explanation: The Claisen condensation relies on a key thermodynamic principle: the reaction equilibrium is driven forward by deprotonation of the β-keto ester product, which has a highly acidic α-hydrogen (pKa ~11). This removes the product from equilibrium, making the reaction irreversible. In this crossed Claisen attempt, ethyl isobutyrate would form an enolate that attacks ethyl benzoate. However, the resulting β-keto ester product has its carbonyl groups separated by a carbon bearing two methyl groups from the isobutyrate. This carbon has no hydrogens—it's a quaternary carbon. Without an acidic α-hydrogen to deprotonate, the equilibrium cannot be driven forward, so the reaction remains reversible and yields minimal product. Looking at the wrong answers: B) incorrectly suggests the enolate of ethyl isobutyrate is too stable—enolates are actually quite reactive nucleophiles, and steric hindrance at the α-carbon doesn't prevent enolate formation significantly. C) misidentifies steric hindrance from the benzene ring as the issue, but aromatic esters are commonly used electrophiles in Claisen condensations. D) wrongly claims sodium ethoxide isn't strong enough—it's the standard base for Claisen condensations and readily deprotonates α-hydrogens on esters. The correct answer is A because the lack of α-hydrogens in the product prevents the crucial deprotonation step that drives Claisen condensations to completion. Remember: successful Claisen condensations require the product to have at least one α-hydrogen for deprotonation. Always check if your expected β-keto ester product will have this feature before predicting reaction success.

Question 5

The Claisen condensation of ethyl acetate requires a full equivalent of sodium ethoxide, not a catalytic amount. Which statement provides the most accurate mechanistic reason for this requirement?

  1. The alkoxide base is consumed by reacting with the ethanol solvent, requiring a stoichiometric amount to drive the reaction forward.
  2. The formation of the initial enolate is an irreversible step that consumes one equivalent of the base.
  3. The final deprotonation of the β-keto ester product is thermodynamically favorable and consumes the base, driving the overall equilibrium toward the product. (correct answer)
  4. The tetrahedral intermediate is exceptionally stable and requires a full equivalent of base to facilitate the elimination of the leaving group.
Explanation: The overall equilibrium for the condensation steps of the Claisen reaction is not highly favorable. However, the β-keto ester product is significantly more acidic (pKa ≈ 11) than the alcohol solvent (pKa ≈ 16). Therefore, the alkoxide base irreversibly deprotonates the product to form a resonance-stabilized enolate. This final, essentially irreversible acid-base reaction consumes the base and pulls the entire equilibrium toward the final product, making a stoichiometric amount of base necessary.

Question 6

What is the primary reason that a crossed Claisen condensation between ethyl propanoate and ethyl butanoate is synthetically inefficient?

  1. The two esters have different boiling points, making temperature control of the reaction impossible.
  2. One ester is significantly more acidic than the other, preventing the formation of one of the enolates.
  3. Both esters can act as both the nucleophile (enolate) and the electrophile, leading to a complex mixture of at least four different products. (correct answer)
  4. Steric hindrance between the ethyl groups of the different esters prevents the condensation reaction from occurring at a reasonable rate.
Explanation: Both ethyl propanoate and ethyl butanoate have α-hydrogens and can be deprotonated to form enolates. Consequently, four possible reactions can occur: (1) self-condensation of ethyl propanoate, (2) self-condensation of ethyl butanoate, (3) ethyl propanoate enolate attacking ethyl butanoate, and (4) ethyl butanoate enolate attacking ethyl propanoate. This results in a complex, difficult-to-separate mixture of products, making the reaction synthetically useless without special techniques.

Question 7

If ethyl propanoate is treated with sodium methoxide in methanol, a standard Claisen condensation is complicated by a significant side reaction. What is this side reaction?

  1. Saponification, forming propanoate and methanol.
  2. Transesterification, forming methyl propanoate and ethanol. (correct answer)
  3. Decarboxylation, losing CO₂ to form ethane.
  4. E2 elimination, forming ethyl acrylate.
Explanation: When the alkoxide base does not match the alcohol portion of the ester, transesterification can occur. The methoxide ion can act as a nucleophile, attacking the carbonyl of ethyl propanoate. The resulting tetrahedral intermediate can eject ethoxide, forming methyl propanoate. This leads to a mixture of starting materials and products (four different esters), which can all potentially participate in subsequent Claisen condensations, leading to a complex product mixture.

Question 8

During the mechanism of the Claisen condensation between two molecules of ethyl acetate, which step is considered the rate-determining step?

  1. The initial deprotonation of ethyl acetate by ethoxide to form the enolate.
  2. The nucleophilic attack of the enolate on the carbonyl carbon of a second ethyl acetate molecule. (correct answer)
  3. The collapse of the tetrahedral intermediate to eject the ethoxide leaving group.
  4. The final deprotonation of the β-keto ester product by ethoxide.
Explanation: The formation of the new carbon-carbon bond, which involves the nucleophilic attack of the enolate on the electrophilic carbonyl carbon, is the slowest and therefore rate-determining step of the condensation phase. The initial proton transfer (A) is typically fast. The collapse of the tetrahedral intermediate (C) is also relatively fast. The final deprotonation (D) is a fast acid-base reaction that serves as the thermodynamic sink but does not determine the rate of C-C bond formation.

Question 9

Which of the following reaction conditions would most likely result in a retro-Claisen condensation of ethyl acetoacetate?

  1. Treatment with dilute HCl in water at room temperature.
  2. Treatment with LiAlH₄ followed by H₃O⁺ workup.
  3. Treatment with H₂ and a Palladium catalyst.
  4. Treatment with excess sodium ethoxide in ethanol with heating. (correct answer)
Explanation: When you encounter questions about retro-Claisen condensations, focus on the reaction conditions that favor the reverse of the original condensation process. A retro-Claisen breaks a β-keto ester back into its component carbonyl compounds and requires specific conditions to drive the equilibrium backward. The correct answer is D because excess sodium ethoxide in ethanol with heating provides the perfect conditions for retro-Claisen condensation. The strong base (ethoxide) can deprotonate the active methylene group in ethyl acetoacetate, and heating provides the energy needed to break the C-C bond. The excess ethoxide shifts the equilibrium toward the retro-Claisen products by removing one of the products (acetate) from the equilibrium. Option A is incorrect because dilute HCl at room temperature is too mild and acidic - retro-Claisen reactions require strong basic conditions, not acidic ones. Option B represents a reduction reaction where LiAlH₄ would reduce the carbonyl groups to alcohols rather than cleaving the molecule via retro-Claisen mechanism. The H₃O⁺ workup confirms this is a reduction pathway. Option C describes catalytic hydrogenation, which would reduce C=C bonds or potentially carbonyl groups, but wouldn't cause the specific C-C bond cleavage characteristic of retro-Claisen reactions. Remember this key pattern: retro-Claisen condensations require strong base and heat to reverse the original condensation. Look for alkoxide bases (like sodium ethoxide) combined with heating as your primary clue for these reactions.

Question 10

The final step in a Claisen condensation procedure is the addition of a dilute acid, such as H₃O⁺, to the reaction mixture. This step is necessary primarily because:

  1. The acid catalyzes the elimination of the alkoxide leaving group from the tetrahedral intermediate.
  2. The reaction is reversible, and adding acid quenches the basic catalyst, preventing the product from reverting to the starting materials.
  3. The acid is required to hydrolyze any unreacted starting ester, which simplifies the purification of the desired product.
  4. The product, a β-keto ester, exists as its conjugate base (enolate) under the basic reaction conditions and must be protonated to yield the neutral final product. (correct answer)
Explanation: The Claisen condensation is a base-catalyzed reaction that creates C-C bonds between ester molecules, producing β-keto esters. Understanding what happens to your product under the reaction conditions is crucial for grasping why acid workup is essential. During the Claisen condensation, you're working under strongly basic conditions (typically using alkoxide bases like sodium ethoxide). The β-keto ester product that forms has acidic hydrogen atoms between the two carbonyls - this methylene group has a pKa around 10-11, making it quite acidic. Under the basic reaction conditions, this acidic hydrogen is immediately deprotonated, converting your desired neutral β-keto ester product into its enolate anion (conjugate base). To isolate the actual β-keto ester product, you must protonate this enolate back to the neutral form, which requires adding dilute acid like H₃O⁺. This is exactly what answer D describes. Let's examine why the other options miss the mark: A incorrectly suggests the acid helps with elimination from a tetrahedral intermediate - but that elimination happens during the condensation step itself, not during workup. B misunderstands the reaction mechanism; while Claisen condensations can be reversible, the primary reason for acid addition isn't to prevent reversal. C describes an unnecessary step - hydrolyzing unreacted starting material isn't the main purpose of the acid workup. Remember this pattern: whenever you see base-catalyzed reactions that produce compounds with acidic protons (like β-keto esters), expect an acid workup step to protonate the product back to its neutral form.