EBK ORGANIC CHEMISTRY: PRINCIPLES AND M
EBK ORGANIC CHEMISTRY: PRINCIPLES AND M
2nd Edition
ISBN: 9780393543971
Author: KARTY
Publisher: VST
Question
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Chapter 18, Problem 18.65P
Interpretation Introduction

(a)

Interpretation:

The complete, detailed mechanism and the major organic product for the given reaction is to be drawn.

Concept introduction:

When an aldehyde is treated with sodium hydroxide, the product β -hydroxycarbonyl compound (an aldol) is formed. The product contains both a CH=O group (characteristic of an aldehyde) and an O-H group (characteristic of an alcohol). The carbonyl and the hydroxyl groups in the product are separated by two carbon atoms. In Step 1, HO- deprotonates the α carbon to yield an enolate anion. In Step 2, the newly formed enolate anion, which is nucleophilic at the α carbon, attacks a second molecule of the aldehyde that still has its proton (and hence is uncharged). The immediate product is an alkoxide anion (RO-), which is subsequently protonated in Step 3. If the reaction is heated, dehydration occurs, and water is eliminated from the β -hydroxycarbonyl compound, giving the overall product a C=C double bond. This overall reaction, in which an aldol addition is followed by dehydration, is called an aldol condensation. The product of an aldol condensation is an α,β -unsaturated carbonyl compound. Under basic conditions, dehydration of β -hydroxycarbonyl compound takes place via an E1cb mechanism, which stands for elimination, unimolecular, conjugate base. Step 1 is a proton transfer, generating a resonance-stabilized enolate anion. In Step 2, the HO- leaving group departs to yield the overall product.

Expert Solution
Check Mark

Answer to Problem 18.65P

The complete, detailed mechanism and the major organic product for the given reaction are shown below:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 18, Problem 18.65P , additional homework tip  1

Explanation of Solution

The given reaction is as shown below:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 18, Problem 18.65P , additional homework tip  2

The given reaction is aldol condensation as the sodium hydroxide and heat is given. Firstly, the given aldehyde is treated with sodium hydroxide to produce β -hydroxycarbonyl compound (an aldol ). In Step 1, HO- deprotonates the α carbon to yield an enolate anion. In Step 2, the newly formed enolate anion, which is nucleophilic at the α carbon, attacks a second molecule of the given aldehyde. The immediate product is an alkoxide anion (RO-), which is subsequently protonated in Step 3. As the reaction is heated, dehydration occurs, and water is eliminated from the β -hydroxycarbonyl compound, giving the overall product a C=C double bond. The dehydration of β -hydroxycarbonyl compound takes place via an E1cb mechanism. This step 1 is a proton transfer, generating a resonance-stabilized enolate anion. In Step 2, the HO- leaving group departs to yield the overall product.

The complete, detailed mechanism and the major organic product for the given reaction are shown below:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 18, Problem 18.65P , additional homework tip  3

Conclusion

The complete, detailed mechanism and the major organic product for the given reaction are drawn by aldol addition reaction, followed by E1cb mechanism.

Interpretation Introduction

(b)

Interpretation:

The complete, detailed mechanism and the major organic product for the given reaction are to be drawn.

Concept introduction:

When an aldehyde is treated with sodium hydroxide, the product β -hydroxycarbonyl compound (an aldol) is formed. The product contains both a CH=O group (characteristic of an aldehyde) and an O-H group (characteristic of an alcohol). The carbonyl and the hydroxyl groups in the product are separated by two carbon atoms. In Step 1, HO- deprotonates the α carbon to yield an enolate anion. In Step 2, the newly formed enolate anion, which is nucleophilic at the α carbon, attacks a second molecule of the aldehyde that still has its proton (and hence is uncharged). The immediate product is an alkoxide anion (RO-), which is subsequently protonated in Step 3. If the reaction is heated, dehydration occurs, and water is eliminated from the β -hydroxycarbonyl compound, giving the overall product a C=C double bond. This overall reaction, in which an aldol addition is followed by dehydration, is called an aldol condensation. The product of an aldol condensation is an α,β -unsaturated carbonyl compound. Under basic conditions, dehydration of β -hydroxycarbonyl compound takes place via an E1cb mechanism, which stands for elimination, unimolecular, conjugate base. Step 1 is a proton transfer, generating a resonance-stabilized enolate anion. In Step 2, the HO- leaving group departs to yield the overall product.

Expert Solution
Check Mark

Answer to Problem 18.65P

The complete, detailed mechanism and the major organic product for the given reaction are as shown below:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 18, Problem 18.65P , additional homework tip  4

Explanation of Solution

The given reaction is shown below:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 18, Problem 18.65P , additional homework tip  5

The given reaction is aldol condensation as the sodium hydroxide and heat is given. Firstly, the given aldehyde is treated with sodium hydroxide to produce β -hydroxycarbonyl compound (an aldol). In Step 1, HO- deprotonates the α carbon to yield an enolate anion. In Step 2, the newly formed enolate anion, which is nucleophilic at the α carbon, attacks a second molecule of the given aldehyde. The immediate product is an alkoxide anion (RO-), which is subsequently protonated in Step 3. As the reaction is heated, dehydration occurs, and water is eliminated from the β -hydroxycarbonyl compound, giving the overall product a C=C double bond. The dehydration of β -hydroxycarbonyl compound takes place via an E1cb mechanism. This step 1 is a proton transfer, generating a resonance-stabilized enolate anion. In Step 2, the HO- leaving group departs to yield the overall product.

The complete, detailed mechanism and the major organic product for the given reaction are shown below:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 18, Problem 18.65P , additional homework tip  6

Conclusion

The complete, detailed mechanism and the major organic product for the given reaction are drawn by aldol addition reaction followed by E1cb mechanism.

Interpretation Introduction

(c)

Interpretation:

The complete, detailed mechanism and the major organic product for the given reaction is to be drawn.

Concept introduction:

When an aldehyde is treated with sodium hydroxide, the product β -hydroxycarbonyl compound (an aldol) is formed. The product contains both a CH=O group (characteristic of an aldehyde) and an O-H group (characteristic of an alcohol). The carbonyl and the hydroxyl groups in the product are separated by two carbon atoms. In Step 1, HO- deprotonates the α carbon to yield an enolate anion. In Step 2, the newly formed enolate anion, which is nucleophilic at the α carbon, attacks a second molecule of the carbonyl carbon present in the given molecule. The immediate product is an alkoxide anion (RO-), which is subsequently protonated in Step 3.

Expert Solution
Check Mark

Answer to Problem 18.65P

The complete, detailed mechanism and the major organic product for the given reaction are as shown below:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 18, Problem 18.65P , additional homework tip  7

Explanation of Solution

The given reaction is shown below:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 18, Problem 18.65P , additional homework tip  8

The given reaction is aldol addition as the sodium hydroxide is given. The given aldehyde is treated with sodium hydroxide to produce β -hydroxycarbonyl compound (an aldol). In Step 1, HO- deprotonates the α carbon of one carbonyl group to yield an enolate anion. In Step 2, the newly formed enolate anion, which is nucleophilic at the α carbon, attacks the carbon of the second carbonyl group of the given aldehyde. The intramolecular addition takes place. The immediate product is an alkoxide anion (RO-), which is subsequently protonated in Step 3.

The complete, detailed mechanism and the major organic product for the given reaction are shown below:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 18, Problem 18.65P , additional homework tip  9

Conclusion

The complete, detailed mechanism and the major organic product for the given reaction are drawn by intramolecular aldol addition reaction mechanism.

Interpretation Introduction

(d)

Interpretation:

The complete, detailed mechanism and the major organic product for the given reaction are to be drawn.

Concept introduction:

When aldehyde is treated with sodium hydroxide, the product β -hydroxycarbonyl compound (an aldol ) is formed. The product contains both a CH=O group (characteristic of an aldehyde) and an O-H group (characteristic of an alcohol). The carbonyl and the hydroxyl groups in the product are separated by two carbon atoms. In Step 1, HO- deprotonates the α carbon to yield an enolate anion. In Step 2, the newly formed enolate anion, which is nucleophilic at the α carbon, attacks a second molecule of the aldehyde that still has its proton (and hence is uncharged). The immediate product is an alkoxide anion (RO-), which is subsequently protonated in Step 3. If the reaction is heated, dehydration occurs, and water is eliminated from the β -hydroxycarbonyl compound, giving the overall product a C=C double bond. Under basic conditions, dehydration of β -hydroxycarbonyl compound takes place via an E1cb mechanism. However, if the α carbon of β -hydroxycarbonyl compound does not have the hydrogen atom, the E1cb mechanism would not take place.

Expert Solution
Check Mark

Answer to Problem 18.65P

The complete, detailed mechanism and the major organic product for the given reaction are as shown below:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 18, Problem 18.65P , additional homework tip  10

Explanation of Solution

The given reaction is shown below:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 18, Problem 18.65P , additional homework tip  11

The given reaction is aldol condensation as the sodium hydroxide and heat is given. Firstly, the given aldehyde is treated with sodium hydroxide, to produce β -hydroxycarbonyl compound (an aldol ). In Step 1, HO- deprotonates the α carbon to yield an enolate anion. In Step 2, the newly formed enolate anion, which is nucleophilic at the α carbon, attacks a second molecule of the given aldehyde. The immediate product is an alkoxide anion (RO-), which is subsequently protonated in Step 3. As the reaction is heated, dehydration occurs, and water is eliminated from the β -hydroxycarbonyl compound, giving the overall product a C=C double bond. However, as the α carbon of β -hydroxycarbonyl compound does not have any H atom, the E1cb mechanism does not takes place. Therefore, only the aldol addition takes place.

The complete, detailed mechanism and the major organic product for given reaction are shown below:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 18, Problem 18.65P , additional homework tip  12

Conclusion

The complete, detailed mechanism and the major organic product for given reaction are drawn on the basis of aldol addition reaction mechanism.

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Chapter 18 Solutions

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M

Ch. 18 - Prob. 18.11PCh. 18 - Prob. 18.12PCh. 18 - Prob. 18.13PCh. 18 - Prob. 18.14PCh. 18 - Prob. 18.15PCh. 18 - Prob. 18.16PCh. 18 - Prob. 18.17PCh. 18 - Prob. 18.18PCh. 18 - Prob. 18.19PCh. 18 - Prob. 18.20PCh. 18 - Prob. 18.21PCh. 18 - Prob. 18.22PCh. 18 - Prob. 18.23PCh. 18 - Prob. 18.24PCh. 18 - Prob. 18.25PCh. 18 - Prob. 18.26PCh. 18 - Prob. 18.27PCh. 18 - Prob. 18.28PCh. 18 - Prob. 18.29PCh. 18 - Prob. 18.30PCh. 18 - Prob. 18.31PCh. 18 - Prob. 18.32PCh. 18 - Prob. 18.33PCh. 18 - Prob. 18.34PCh. 18 - Prob. 18.35PCh. 18 - Prob. 18.36PCh. 18 - Prob. 18.37PCh. 18 - Prob. 18.38PCh. 18 - Prob. 18.39PCh. 18 - Prob. 18.40PCh. 18 - Prob. 18.41PCh. 18 - Prob. 18.42PCh. 18 - Prob. 18.43PCh. 18 - Prob. 18.44PCh. 18 - Prob. 18.45PCh. 18 - Prob. 18.46PCh. 18 - Prob. 18.47PCh. 18 - Prob. 18.48PCh. 18 - Prob. 18.49PCh. 18 - Prob. 18.50PCh. 18 - Prob. 18.51PCh. 18 - Prob. 18.52PCh. 18 - Prob. 18.53PCh. 18 - Prob. 18.54PCh. 18 - Prob. 18.55PCh. 18 - Prob. 18.56PCh. 18 - Prob. 18.57PCh. 18 - Prob. 18.58PCh. 18 - Prob. 18.59PCh. 18 - Prob. 18.60PCh. 18 - Prob. 18.61PCh. 18 - Prob. 18.62PCh. 18 - Prob. 18.63PCh. 18 - Prob. 18.64PCh. 18 - Prob. 18.65PCh. 18 - Prob. 18.66PCh. 18 - Prob. 18.67PCh. 18 - Prob. 18.68PCh. 18 - Prob. 18.69PCh. 18 - Prob. 18.70PCh. 18 - Prob. 18.71PCh. 18 - Prob. 18.72PCh. 18 - Prob. 18.73PCh. 18 - Prob. 18.74PCh. 18 - Prob. 18.75PCh. 18 - Prob. 18.76PCh. 18 - Prob. 18.77PCh. 18 - Prob. 18.78PCh. 18 - Prob. 18.79PCh. 18 - Prob. 18.80PCh. 18 - Prob. 18.81PCh. 18 - Prob. 18.82PCh. 18 - Prob. 18.83PCh. 18 - Prob. 18.84PCh. 18 - Prob. 18.85PCh. 18 - Prob. 18.86PCh. 18 - Prob. 18.87PCh. 18 - Prob. 18.88PCh. 18 - Prob. 18.89PCh. 18 - Prob. 18.90PCh. 18 - Prob. 18.91PCh. 18 - Prob. 18.92PCh. 18 - Prob. 18.93PCh. 18 - Prob. 18.94PCh. 18 - Prob. 18.95PCh. 18 - Prob. 18.96PCh. 18 - Prob. 18.97PCh. 18 - Prob. 18.98PCh. 18 - Prob. 18.99PCh. 18 - Prob. 18.100PCh. 18 - Prob. 18.101PCh. 18 - Prob. 18.102PCh. 18 - Prob. 18.103PCh. 18 - Prob. 18.1YTCh. 18 - Prob. 18.2YTCh. 18 - Prob. 18.3YTCh. 18 - Prob. 18.4YTCh. 18 - Prob. 18.5YTCh. 18 - Prob. 18.6YTCh. 18 - Prob. 18.7YTCh. 18 - Prob. 18.8YTCh. 18 - Prob. 18.9YTCh. 18 - Prob. 18.10YTCh. 18 - Prob. 18.11YTCh. 18 - Prob. 18.12YTCh. 18 - Prob. 18.13YTCh. 18 - Prob. 18.14YTCh. 18 - Prob. 18.15YT
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