Organic Chemistry CH22 Practice: Enolate and Carbonyl Chemistry

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Q1. (2a) Provide the product for the following sequence: 1. LDA, 2. CH3I

Background

Topic: Enolate Alkylation

This question tests your understanding of enolate formation using a strong base (LDA) and subsequent alkylation with an alkyl halide (CH3I).

Key Terms and Formulas

LDA (Lithium Diisopropylamide): A strong, non-nucleophilic base used to generate enolates from carbonyl compounds.
Enolate: The resonance-stabilized anion formed by deprotonation of the alpha hydrogen of a carbonyl compound.
Alkylation: The process of adding an alkyl group to a molecule, here via SN2 reaction with an alkyl halide.

Step-by-Step Guidance

Identify the alpha carbon(s) in the starting ketone that can be deprotonated by LDA.
Draw the enolate ion formed after treatment with LDA.
Show how the enolate acts as a nucleophile and attacks the methyl iodide (CH3I) in an SN2 reaction.
Draw the structure of the product after the alkylation step, but do not finalize the answer yet.

Try solving on your own before revealing the answer!

Final Answer:

The product is the methylated ketone at the alpha position:

LDA forms the enolate at the less hindered alpha carbon, which then attacks methyl iodide to give the alkylated product.

Q2. (3a) Provide the product for the following sequence: 1. CH3I, 2. I-

Background

Topic: Enolate Alkylation and Iodine-Mediated Cleavage

This question tests your knowledge of enolate chemistry and the use of iodide as a nucleophile or leaving group.

Key Terms and Formulas

Enolate Alkylation: Introduction of an alkyl group at the alpha position of a carbonyl compound.
Iodide (I-): Can act as a nucleophile or facilitate elimination/substitution reactions.

Step-by-Step Guidance

Identify the enolate-forming position in the starting molecule.
Show the alkylation step with CH3I, adding a methyl group to the alpha position.
Consider the role of I- in the second step—does it act as a nucleophile or promote elimination?
Draw the intermediate after methylation, and set up the next step for the student to complete.

Try solving on your own before revealing the answer!

Final Answer:

The product is the methylated compound at the alpha position; I- may facilitate elimination or substitution if a good leaving group is present.

For the specific structure, see the worked solution in your notes.

Q3. (4a) Provide the product for the following sequence: 1. CH2CH2Cl, 2. NaOEt

Background

Topic: Alkylation and Intramolecular Aldol Condensation

This question tests your understanding of carbon-carbon bond formation via alkylation and subsequent cyclization (aldol condensation).

Key Terms and Formulas

Alkylation: Addition of an alkyl group to a molecule.
Aldol Condensation: Formation of a β-hydroxy carbonyl compound followed by dehydration to yield an α,β-unsaturated carbonyl.

Step-by-Step Guidance

Identify the nucleophilic site for alkylation with CH2CH2Cl.
Draw the intermediate after the alkylation step.
Show how NaOEt can deprotonate to form an enolate, which can then attack the carbonyl group intramolecularly.
Set up the cyclization step, but do not complete the final ring closure or dehydration yet.

Try solving on your own before revealing the answer!

Final Answer:

The product is a cyclized compound formed via intramolecular aldol condensation after alkylation.

Refer to your notes for the exact structure; the key is recognizing the formation of a new ring.

Q4. (5a) Provide the product for the following reaction sequence involving an enolate and an α,β-unsaturated carbonyl compound.

Background

Topic: Michael Addition and Aldol Cyclization

This question tests your understanding of conjugate addition (Michael addition) followed by intramolecular aldol condensation.

Key Terms and Formulas

Michael Addition: 1,4-conjugate addition of an enolate to an α,β-unsaturated carbonyl compound.
Aldol Cyclization: Formation of a ring via intramolecular aldol reaction.

Step-by-Step Guidance

Identify the nucleophile (enolate) and the electrophile (α,β-unsaturated carbonyl).
Show the 1,4-addition (Michael addition) to form a new carbon-carbon bond.
Draw the intermediate and identify the positions for possible cyclization (aldol reaction).
Set up the cyclization step, but do not complete the final ring closure or dehydration yet.

Try solving on your own before revealing the answer!

Final Answer:

The product is a bicyclic compound formed via Michael addition followed by intramolecular aldol condensation.

See your notes for the detailed structure; the key is recognizing the sequence of addition and cyclization.

Q5. (6a) 2 equivalents of phenylacetaldehyde react with NaOH. What is the product?

Background

Topic: Aldol Condensation

This question tests your understanding of the base-catalyzed aldol condensation between two equivalents of an aldehyde.

Key Terms and Formulas

Aldol Addition: Formation of a β-hydroxy aldehyde or ketone from two carbonyl compounds.
Condensation: Dehydration of the aldol product to form an α,β-unsaturated carbonyl compound.

Step-by-Step Guidance

Draw the enolate formed from one equivalent of phenylacetaldehyde under basic conditions.
Show the nucleophilic attack of the enolate on the carbonyl carbon of the second equivalent.
Draw the resulting β-hydroxy aldehyde intermediate.
Set up the dehydration (elimination of water) to form the α,β-unsaturated product, but do not complete the final structure yet.

Try solving on your own before revealing the answer!

Final Answer:

The product is an α,β-unsaturated aldehyde formed by aldol condensation of two phenylacetaldehyde molecules.

Refer to your notes for the exact structure; the key is the formation of a double bond between the α and β carbons.

Q6. (6S) 2 pentanal react with NaOH. What is the product?

Background

Topic: Aldol Condensation

This question tests your understanding of the self-aldol condensation of an aldehyde under basic conditions.

Key Terms and Formulas

Pentanal: An aldehyde with five carbons.
Aldol Condensation: Formation of a β-hydroxy aldehyde followed by dehydration to an α,β-unsaturated aldehyde.

Step-by-Step Guidance

Draw the enolate formed from pentanal under basic conditions.
Show the nucleophilic attack of the enolate on another pentanal molecule.
Draw the β-hydroxy aldehyde intermediate.
Set up the elimination of water to form the α,β-unsaturated aldehyde, but do not complete the final structure yet.

Try solving on your own before revealing the answer!

Final Answer:

The product is 2-pentenal, an α,β-unsaturated aldehyde formed by aldol condensation of pentanal.

The key is recognizing the formation of a double bond between the α and β carbons after dehydration.

Q7. (6g) Benzophenone and pentanal react under basic conditions. What is the product?

Background

Topic: Mixed Aldol Condensation (Crossed Aldol)

This question tests your understanding of the crossed aldol reaction between a ketone (benzophenone) and an aldehyde (pentanal).

Key Terms and Formulas

Crossed Aldol Reaction: Aldol condensation between two different carbonyl compounds.
Enolate Formation: The more acidic α-hydrogen is abstracted to form the enolate, which attacks the other carbonyl compound.

Step-by-Step Guidance

Identify which compound forms the enolate (usually the aldehyde, pentanal, due to more acidic α-hydrogens).
Show the nucleophilic attack of the enolate on the carbonyl carbon of benzophenone.
Draw the β-hydroxy ketone intermediate.
Set up the dehydration step to form the α,β-unsaturated ketone, but do not complete the final structure yet.

Try solving on your own before revealing the answer!

Final Answer:

The product is a β,β-diphenyl-α,β-unsaturated ketone formed by crossed aldol condensation.

See your notes for the exact structure; the key is the formation of a double bond adjacent to the carbonyl group.

Q8. (7g) 2,6-heptanedione reacts with NaOH. What is the product?

Background

Topic: Intramolecular Aldol Condensation (Dieckmann Condensation)

This question tests your understanding of the intramolecular aldol condensation of a 1,6-diketone to form a cyclic compound.

Key Terms and Formulas

Dieckmann Condensation: Intramolecular version of the Claisen condensation, forming a cyclic β-keto ester or ketone.
Enolate Formation: The enolate attacks the other carbonyl group within the same molecule.

Step-by-Step Guidance

Draw the enolate formed from one of the carbonyl groups of 2,6-heptanedione.
Show the intramolecular attack on the other carbonyl group to form a five- or six-membered ring.
Draw the cyclic β-keto ketone intermediate.
Set up the final step (possible tautomerization or dehydration), but do not complete the final structure yet.

Try solving on your own before revealing the answer!

Final Answer:

The product is 3-methylcyclohex-2-en-1-one, formed by intramolecular aldol condensation (Dieckmann condensation) of 2,6-heptanedione.

The key is recognizing the formation of a six-membered ring with a double bond adjacent to the carbonyl group.