Content check — Organic Chemistry, Chapter 23: Carbonyl Condensation Reactions

Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 23: Carbonyl Condensation Reactions

22 problems · 47 figures

23-1 §23.1 source ↗

Predict the aldol reaction product of the following compounds:

(a)

(b)

(c)

Answer (3 parts)

(a)

(b)

(c)

23-2 §23.1 source ↗

Using curved arrows to indicate the electron flow in each step, show how the base-catalyzed reverse-aldol reaction of 4-hydroxy-4-methyl-2-pentanone takes place to yield 2 equivalents of acetone.

Answer

The reverse reaction is the exact opposite of the forward reaction shown in Figure 23.2.

The three-step reaction is involved in the formation of beta-hydroxy carbonyl compound on the reaction of a carbonyl compound with base. — Figure 23.2 in the book ↗

23-3 §23.3 source ↗

What enone product would you expect from aldol condensation of each of the following compounds?

(a)

(b)

(c)

Answer (3 parts)

(a)

(b)

(c)

23-4 §23.3 source ↗

Aldol condensation of 3-methylcyclohexanone leads to a mixture of two enone products, not counting double-bond isomers. Draw them.

Answer

23-5 §23.4 source ↗

Which of the following compounds are aldol condensation products? What is the aldehyde or ketone precursor of each?

(a)

2-Hydroxy-2-methylpentanal

(b)

5-Ethyl-4-methyl-4-hepten-3-one

Answer (2 parts)

(a)

Not an aldol product

(b)

3-Pentanone

23-6 §23.4 source ↗

1-Butanol is prepared commercially by a route that begins with an aldol reaction. Show the steps that are likely to be involved.

Answer

1. NaOH; 2. LiAlH₄; 3. H₂/Pd

23-7 §23.4 source ↗

Show how you would synthesize the following compound using an aldol reaction:

Answer

23-8 §23.5 source ↗

Which of the following compounds can probably be prepared by a mixed aldol reaction? Show the reactants you would use in each case.

(a)

(b)

(c)

Answer (3 parts)

(a)

C₆H₅CHO + CH₃COCH₃

(b)

Not easily prepared

(c)

Not easily prepared

23-9 §23.6 source ↗

Treatment of a 1,3-diketone such as 2,4-pentanedione with base does not give an aldol condensation product. Explain.

Answer

The CH₂ position between the two carbonyl groups is so acidic that it is completely deprotonated to give a stable enolate ion.

23-10 §23.6 source ↗

What product would you expect to obtain from base treatment of 1,6-cyclodecanedione?

Answer

23-11 §23.7 source ↗

Show the products you would expect to obtain by Claisen condensation of the following esters:

(a)

(CH₃)₂CHCH₂CO₂Et

(b)

Ethyl phenylacetate

(c)

Ethyl cyclohexylacetate

Answer (3 parts)

(a)

(b)

(c)

23-12 §23.7 source ↗

As shown in Figure 23.5, the Claisen reaction is reversible. That is, a β-keto ester can be cleaved by base into two fragments. Using curved arrows to indicate electron flow, show the mechanism by which this cleavage occurs.

Reversible mechanism of Claisen condensation of ethyl acetate with arrows showing electron flow: alpha proton abstraction, nucleophilic addition of enolate, ethoxide expelled, deprotonation between carbonyls, reprotonation to form ethyl acetoacetate. — Figure 23.5 in the book ↗

Answer

The cleavage reaction is the exact reverse of the forward reaction.

23-13 §23.8 source ↗

What product would you expect from the following mixed Claisen-like reaction?

Answer

23-14 §23.9 source ↗

What product would you expect from the following reaction?

Answer

23-15 §23.9 source ↗

Dieckmann cyclization of diethyl 3-methylheptanedioate gives a mixture of two β-keto ester products. What are their structures, and why is a mixture formed?

Answer

23-16 §23.10 source ↗

What product would you obtain from a base-catalyzed Michael reaction of 2,4-pentanedione with each of the following α,β-unsaturated acceptors?

(a)

2-Cyclohexenone

(b)

Propenenitrile

(c)

Ethyl 2-butenoate

Answer (3 parts)

(a)

(b)

(c)

23-17 §23.10 source ↗

What product would you obtain from a base-catalyzed Michael reaction of 3-buten-2-one with each of the following nucleophilic donors?

(a)

(b)

Answer (2 parts)

(a)

(b)

23-18 §23.10 source ↗

How would you prepare the following compound using a Michael reaction (gray = H, black = C, red = O, blue = N)?

Answer

CH₃CH₂COCH ═ CH₂ + CH₃CH₂NO₂

23-19 §23.11 source ↗

What products would result after hydrolysis from reaction of the enamine prepared from cyclopentanone and pyrrolidine with the following α,β-unsaturated acceptors?

(a)

H₂C ═ CHCO₂Et

(b)

H₂C ═ CHCHO

(c)

CH₃CH ═ CHCOCH₃

Answer (3 parts)

(a)

(b)

(c)

23-20 §23.11 source ↗

Show how you might use an enamine reaction to prepare each of the following compounds:

(a)

(b)

Answer (2 parts)

(a)

Cyclopentanone enamine + propenenitrile

(b)

Cyclohexanone enamine + methyl propenoate

23-21 §23.12 source ↗

What product would you expect from a Robinson annulation reaction of 2-methyl-1,3-cyclopentanedione with 3-buten-2-one?

Answer

23-22 §23.12 source ↗

How would you prepare the following compound using a Robinson annulation reaction between a β-diketone and an α,β-unsaturated ketone? Draw the structures of both reactants and the intermediate Michael addition product.

Answer

2,5,5-Trimethyl-1,3-cyclohexanedione + 1-penten-3-one

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Problems and answers are adapted from Organic Chemistry: A Tenth Edition by John McMurry, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/organic-chemistry/pages/1-why-this-chapter. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.