Chapter 22, Problem 22.7P

Interpretation Introduction

(a)

Interpretation:

The enol structure of the compound $\text{2-methylcyclohexanone}$ 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.

Answer to Problem 22.7P

The enol structures of the compound $\text{2-methylcyclohexanone}$ are shown below.

Explanation of Solution

The structure of $\text{2-methylcyclohexanone}$ is shown below.

Figure 1

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

Figure 2

The compound $\text{2-methylcyclohexanone}$ has two enol forms.

Conclusion

The enol structures of the compound $\text{2-methylcyclohexanone}$ are shown in Figure 2.

Interpretation Introduction

(b)

Interpretation:

The enol structure of the compound $\text{2-methylpentanoic}\text{acid}$ 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.

Answer to Problem 22.7P

The enol structure of the compound $\text{2-methylpentanoic}\text{acid}$ is shown below.

Explanation of Solution

The structure of given compound $\text{2-methylpentanoic}\text{acid}$ is shown below.

Figure 3

In the compound, $\text{2-methylpentanoic}\text{acid}$, there is $\alpha -$ hydrogen present on one side of the carbonyl group. The compound $\text{2-methylpentanoic}\text{acid}$ can undergo keto-enoltautomerism by the transfer of $\alpha -$ hydrogen. Therefore, the keto-enoltautomerism of $\text{2-methylpentanoic}\text{acid}$ is shown below.

Figure 4

Conclusion

The enol structure of the compound $\text{2-methylpentanoic}\text{acid}$ 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 $>\text{C=O}$ 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.

Answer to Problem 22.7P

There is no $\alpha -$ hydrogen in benzaldehyde. Therefore, it does not have any enol structure.

Explanation of Solution

The given compound benzaldehydeis shown below.

Figure 5

In the compoundbenzaldehydethere is no $\alpha -$ 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 $\alpha -$ hydrogen no enol structures will be obtained.

Interpretation Introduction

(d)

Interpretation:

The enol structure of the compound $N,N\text{-dimethylacetamide}$ 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.

Answer to Problem 22.7P

The enol structure of the compound $N,N\text{-dimethylacetamide}$ is shown below.

Explanation of Solution

The structure of the given compound $N,N\text{-dimethylacetamide}$ is shown below.

Figure 6

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

Figure 7

Conclusion

The enol structure of the compound $N,N\text{-dimethylacetamide}$ is shown in Figure 7.

Interpretation Introduction

(e)

Interpretation:

The enol structure of the compound $N,N\text{-dimethylformamide}$ 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.

Answer to Problem 22.7P

There is no $\alpha \text{-}$ hydrogen in $N,N\text{-dimethylformamide}$. Therefore, it does not have any enol structure.

Explanation of Solution

The given compound $N,N\text{-dimethylformamide}$ is shown below.

Figure 8

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

Conclusion

In $N,N\text{-dimethylformamide}$, due to the absence of $\alpha \text{-}$ hydrogen, no enol structures will be obtained.