ORGANIC CHEM +SG +SAPLING >IP<
ORGANIC CHEM +SG +SAPLING >IP<
6th Edition
ISBN: 9781319171179
Author: LOUDON
Publisher: MAC HIGHER
Question
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Chapter 22, Problem 22.7P
Interpretation Introduction

(a)

Interpretation:

The enol structure of the compound 2-methylcyclohexanone is to be drawn. The explanation as to why the compound does not form enol structure is to be stated.

Concept introduction:

Ketone also exists in two forms which are commonly known as keto-enol tautomerism. Tautomers are isomers which differ only in the position of the protons and electrons of the double bond of the electronegative atom in the compound. There is no change in the carbon skeleton of the compound. This phenomenon which involves simple proton transfer in an intramolecular fashion is known as tautomerism.

Expert Solution
Check Mark

Answer to Problem 22.7P

The enol structures of the compound 2-methylcyclohexanone are shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 22, Problem 22.7P , additional homework tip  1

Explanation of Solution

The structure of 2-methylcyclohexanone is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 22, Problem 22.7P , additional homework tip  2

Figure 1

In the compound, 2-methylcyclohexanone, there are α-hydrogens present on both sides of the carbonyl group. The compound 2-methylcyclohexanone can undergo keto-enoltautomerism by the transfer of α hydrogen. Therefore, the keto-enoltautomerism in 2-methylcyclohexanone is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 22, Problem 22.7P , additional homework tip  3

Figure 2

The compound 2-methylcyclohexanone has two enol forms.

Conclusion

The enol structures of the compound 2-methylcyclohexanone are shown in Figure 2.

Interpretation Introduction

(b)

Interpretation:

The enol structure of the compound 2-methylpentanoicacid is to be drawn. The explanation as to why the compound does not form enol structure.

Concept introduction:

Ketone also exists in two forms which are commonly known as keto-enol tautomerism. Tautomers are isomers which differ only in the position of the protons and electrons of the double bond of the electronegative atom in the compound. There is no change in the carbon skeleton of the compound. This phenomenon which involves simple proton transfer in an intramolecular fashion is known as tautomerism.

Expert Solution
Check Mark

Answer to Problem 22.7P

The enol structure of the compound 2-methylpentanoicacid is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 22, Problem 22.7P , additional homework tip  4

Explanation of Solution

The structure of given compound 2-methylpentanoicacid is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 22, Problem 22.7P , additional homework tip  5

Figure 3

In the compound, 2-methylpentanoicacid, there is α hydrogen present on one side of the carbonyl group. The compound 2-methylpentanoicacid can undergo keto-enoltautomerism by the transfer of α hydrogen. Therefore, the keto-enoltautomerism of 2-methylpentanoicacid is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 22, Problem 22.7P , additional homework tip  6

Figure 4

Conclusion

The enol structure of the compound 2-methylpentanoicacid is shown in Figure 4.

Interpretation Introduction

(c)

Interpretation:

The enol structure of the compound benzaldehyde is to be drawn. The explanation as to why the compound does not form enol structure.

Concept introduction:

The carbonyl compound contains a >C=O group. They are of two types aldehyde and ketone. Ketone also exists in two forms which are commonly known as keto-enoltautomerism. Tautomers are isomers which differ only in the position of the protons and electrons of the double bond of the electronegative atom in the compound. There is no change in the carbon skeleton of the compound. This phenomenon which involves simple proton transfer in an intramolecular fashion is known as tautomerism.

Expert Solution
Check Mark

Answer to Problem 22.7P

There is no α hydrogen in benzaldehyde. Therefore, it does not have any enol structure.

Explanation of Solution

The given compound benzaldehydeis shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 22, Problem 22.7P , additional homework tip  7

Figure 5

In the compoundbenzaldehydethere is no α hydrogens present on either side of the carbonyl group. It is not possible that benzaldehyde will undergo keto-enoltautomerism. Therefore, no enol forms will be observed.

Conclusion

In benzaldehyde, due to the absence of α hydrogen no enol structures will be obtained.

Interpretation Introduction

(d)

Interpretation:

The enol structure of the compound N,N-dimethylacetamide is to be drawn. The explanation as to why the compound does not form enol structure.

Concept introduction:

Ketone also exists in two forms which are commonly known as keto-enol tautomerism. Tautomers are isomers which differ only in the position of the protons and electrons of the double bond of the electronegative atom in the compound. There is no change in the carbon skeleton of the compound. This phenomenon which involves simple proton transfer in an intramolecular fashion is known as tautomerism.

Expert Solution
Check Mark

Answer to Problem 22.7P

The enol structure of the compound N,N-dimethylacetamide is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 22, Problem 22.7P , additional homework tip  8

Explanation of Solution

The structure of the given compound N,N-dimethylacetamide is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 22, Problem 22.7P , additional homework tip  9

Figure 6

In the compound, N,N-dimethylacetamide there are α hydrogens present on one side of the carbonyl group. The compound N,N-dimethylacetamide can undergo keto-enoltautomerism by the transfer of α hydrogen. Therefore, the keto-enoltautomerism in N,N-dimethylacetamide is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 22, Problem 22.7P , additional homework tip  10

Figure 7

Conclusion

The enol structure of the compound N,N-dimethylacetamide is shown in Figure 7.

Interpretation Introduction

(e)

Interpretation:

The enol structure of the compound N,N-dimethylformamide is to be drawn. The explanation as to why the compound does not form enol structure.

Concept introduction:

Ketone also exists in two forms which are commonly known as keto-enol tautomerism. Tautomers are isomers which differ only in the position of the protons and electrons of the double bond of the electronegative atom in the compound. There is no change in the carbon skeleton of the compound. This phenomenon which involves simple proton transfer in an intramolecular fashion is known as tautomerism.

Expert Solution
Check Mark

Answer to Problem 22.7P

There is no α- hydrogen in N,N-dimethylformamide. Therefore, it does not have any enol structure.

Explanation of Solution

The given compound N,N-dimethylformamide is shown below.

ORGANIC CHEM +SG +SAPLING >IP<, Chapter 22, Problem 22.7P , additional homework tip  11

Figure 8

In the compound N,N-dimethylformamide, there are no α- hydrogens present on either side of the carbonyl group. It is not possible that N,N-dimethylformamide will undergo keto-enoltautomerism. Therefore, no enol forms will be observed.

Conclusion

In N,N-dimethylformamide, due to the absence of α- hydrogen, no enol structures will be obtained.

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Chapter 22 Solutions

ORGANIC CHEM +SG +SAPLING >IP<

Ch. 22 - Prob. 22.11PCh. 22 - Prob. 22.12PCh. 22 - Prob. 22.13PCh. 22 - Prob. 22.14PCh. 22 - Prob. 22.15PCh. 22 - Prob. 22.16PCh. 22 - Prob. 22.17PCh. 22 - Prob. 22.18PCh. 22 - Prob. 22.19PCh. 22 - Prob. 22.20PCh. 22 - Prob. 22.21PCh. 22 - Prob. 22.22PCh. 22 - Prob. 22.23PCh. 22 - Prob. 22.24PCh. 22 - Prob. 22.25PCh. 22 - Prob. 22.26PCh. 22 - Prob. 22.27PCh. 22 - Prob. 22.28PCh. 22 - Prob. 22.29PCh. 22 - Prob. 22.30PCh. 22 - Prob. 22.31PCh. 22 - Prob. 22.32PCh. 22 - Prob. 22.33PCh. 22 - Prob. 22.34PCh. 22 - Prob. 22.35PCh. 22 - Prob. 22.36PCh. 22 - Prob. 22.37PCh. 22 - Prob. 22.38PCh. 22 - Prob. 22.39PCh. 22 - Prob. 22.40PCh. 22 - Prob. 22.41PCh. 22 - Prob. 22.42PCh. 22 - Prob. 22.43PCh. 22 - Prob. 22.44PCh. 22 - Prob. 22.45PCh. 22 - Prob. 22.46PCh. 22 - Prob. 22.47PCh. 22 - Prob. 22.48PCh. 22 - Prob. 22.49PCh. 22 - Prob. 22.50PCh. 22 - Prob. 22.51PCh. 22 - Prob. 22.52PCh. 22 - Prob. 22.53PCh. 22 - Prob. 22.54PCh. 22 - Prob. 22.55APCh. 22 - Prob. 22.56APCh. 22 - Prob. 22.57APCh. 22 - Prob. 22.58APCh. 22 - Prob. 22.59APCh. 22 - Prob. 22.60APCh. 22 - Prob. 22.61APCh. 22 - Prob. 22.62APCh. 22 - Prob. 22.63APCh. 22 - Prob. 22.64APCh. 22 - Prob. 22.65APCh. 22 - Prob. 22.66APCh. 22 - Prob. 22.67APCh. 22 - Prob. 22.68APCh. 22 - Prob. 22.69APCh. 22 - Prob. 22.70APCh. 22 - Prob. 22.71APCh. 22 - Prob. 22.72APCh. 22 - Prob. 22.73APCh. 22 - Prob. 22.74APCh. 22 - Prob. 22.75APCh. 22 - Prob. 22.76APCh. 22 - Prob. 22.77APCh. 22 - Prob. 22.78APCh. 22 - Prob. 22.79APCh. 22 - Prob. 22.80APCh. 22 - Prob. 22.81APCh. 22 - Prob. 22.82APCh. 22 - Prob. 22.83APCh. 22 - Prob. 22.84APCh. 22 - Prob. 22.85APCh. 22 - Prob. 22.86APCh. 22 - Prob. 22.87APCh. 22 - Prob. 22.88APCh. 22 - Prob. 22.89APCh. 22 - Prob. 22.90APCh. 22 - Prob. 22.91APCh. 22 - Prob. 22.92APCh. 22 - Prob. 22.93APCh. 22 - Prob. 22.94APCh. 22 - Prob. 22.95APCh. 22 - Prob. 22.96APCh. 22 - Prob. 22.97APCh. 22 - Prob. 22.98AP
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