Chapter 10, Problem 10.31P

Interpretation Introduction

(a)

Interpretation:

The alcohol from which the compound $\text{5-methylhexanoic acid}$ is prepared by the oxidation process is to be stated.

Concept introduction:

Primary and secondary alcohols can be oxidized into aldehydes and ketones using $\text{Cr(VI)}$ compounds as they are very good oxidizing agents. Few forms of chromium used to oxidize are chromate $\left({\text{CrO}}_{\text{4}}^{2-}\right)$ and dichromate $\left({\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\right)$ and chromium trioxide $\left({\text{CrO}}_{\text{3}}\right)$. Primary alcohols are oxidized into aldehydes and secondary alcohols are oxidized into ketones.

Answer to Problem 10.31P

The $\text{5-methylhexanol}$ is the alcohol from which the compound $\text{5-methylhexanoic acid}$ is prepared by the oxidation process.

Explanation of Solution

Primary and secondary alcohols can be oxidized into aldehydes and ketones using $\text{Cr(VI)}$ compounds as they are very good oxidizing agents. They are also further oxidized into carboxylic acids in the presence of water with chromium compounds therefore in order to stop this process, generally anhydrous conditions are taken. Some examples of the $\text{Cr(VI)}$ compounds used are PCC, ${\text{CrO}}_{\text{3}}$, and ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$.

The compound $\text{5-methylhexanoic acid}$ can be best prepared from the alcohol given below by the oxidation method.

$\begin{array}{l}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_2\text{OH}\underset{\text{aq}{\text{. H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\overset{{\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}{\to }}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{COOH}\\ \text{ 5-methylhexanol 5-methylhexanoic acid}\end{array}$

The carboxylic acid, $\text{5-methylhexanoic acid}$ can only be obtained when the primary alcohols are oxidized into aldehydes first which are easily oxidized into carboxylic acids. Therefore, the primary alcohol that can give $\text{5-methylhexanoic acid}$ as a product on oxidation is $\text{5-methylhexanol}$.

Conclusion

The alcohol from which the compound $\text{5-methylhexanoic acid}$ can be best prepared by the oxidation process is $\text{5-methylhexanol}$.

Interpretation Introduction

(b)

Interpretation:

The alcohol from which the compound $\text{pentan-3-one}$ can be best prepared by the oxidation process is to be stated.

Concept introduction:

Primary and secondary alcohols can be oxidized into aldehydes and ketones using $\text{Cr(VI)}$ compounds as they are very good oxidizing agents. Few forms of chromium used to oxidize are chromate $\left({\text{CrO}}_{\text{4}}^{2-}\right)$ and dichromate $\left({\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\right)$ and chromium trioxide $\left({\text{CrO}}_{\text{3}}\right)$. Primary alcohols are oxidized into aldehydes and secondary alcohols are oxidized into ketones.

Answer to Problem 10.31P

The $\text{pentan-3-ol}$ is the alcohol from which the compound $\text{pentan-3-one}$ can be best prepared by the oxidation process which is shown below.

Explanation of Solution

Primary and secondary alcohols are oxidized into aldehydes and ketones using $\text{Cr(VI)}$ compounds as they are very good oxidizing agents. They are also further oxidized into carboxylic acids in the presence of water with chromium compounds. Therefore, in order to stop this process, generally anhydrous conditions are taken. Some examples of the $\text{Cr(VI)}$ compounds used are PCC, ${\text{CrO}}_{\text{3}}$, and ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$.

The secondary alcohols are converted into ketones on oxidation.

The compound $\text{pentan-3-one}$ is prepared from the alcohol given below by the oxidation method.

Figure 1

The ketone $\text{pentan-3-one}$ is obtained from a secondary alcohol as only secondary alcohols give ketones on oxidation. Therefore, the best alcohol for the oxidation process to give $\text{pentan-3-one}$ is $\text{pentan-3-ol}$ as shown in Figure 1.

Conclusion

The alcohol from which the compound $\text{pentan-3-one}$ can be best prepared by the oxidation process is shown in Figure 1.

Interpretation Introduction

(c)

Interpretation:

The alcohol from which the compound $\text{5-hydroxy-5-methylhexanal}$ can be best prepared by the oxidation process is to be stated.

Concept introduction:

Primary and secondary alcohols can be oxidized into aldehydes and ketones using $\text{Cr(VI)}$ compounds as they are very good oxidizing agents. Few forms of chromium used to oxidize are chromate $\left({\text{CrO}}_{\text{4}}^{2-}\right)$ and dichromate $\left({\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\right)$ and chromium trioxide $\left({\text{CrO}}_{\text{3}}\right)$. Primary alcohols are oxidized into aldehydes and secondary alcohols are oxidized into ketones.

Answer to Problem 10.31P

The $\text{5-methylhexane-1, 5-diol}$ is the alcohol from which the compound $\text{5-hydroxy-5-methylhexanal}$ can be best prepared by the oxidation process which is shown below.

Explanation of Solution

Primary and secondary alcohols can be oxidized into aldehydes and ketones using $\text{Cr(VI)}$ compounds as they are very good oxidizing agents. They are also further oxidized into carboxylic acids in the presence of water with chromium compounds therefore in order to stop this process, generally anhydrous conditions are taken. Some examples of the $\text{Cr(VI)}$ compounds used are PCC, ${\text{CrO}}_{\text{3}}$, and ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$.

The primary alcohols are converted into aldehydes under anhydrous conditions.

The compound $\text{5-hydroxy-5-methylhexanal}$ is prepared from the alcohol given below by the oxidation method.

Figure 2

The aldehyde, $\text{5-hydroxy-5-methylhexanal}$ is obtained from the alcohol $\text{5-methylhexan-1, 5-diol}$ under the anhydrous conditions of oxidation using PCC otherwise it would be further oxidized to the carboxylic acid form.

Conclusion

The alcohol from which the compound $\text{5-hydroxy-5-methylhexanal}$ can be best prepared by the oxidation process is shown in Figure 2.

Interpretation Introduction

(d)

Interpretation:

The alcohol from which the compound $\text{cyclopent-2-enecarbaldehyde}$ can be best prepared by the oxidation process is to be stated.

Concept introduction:

Primary and secondary alcohols can be oxidized into aldehydes and ketones using $\text{Cr(VI)}$ compounds as they are very good oxidizing agents. Few forms of chromium used to oxidize are chromate $\left({\text{CrO}}_{\text{4}}^{2-}\right)$ and dichromate $\left({\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\right)$ and chromium trioxide $\left({\text{CrO}}_{\text{3}}\right)$. Primary alcohols are oxidized into aldehydes and secondary alcohols are oxidized into ketones.

Answer to Problem 10.31P

The $\text{cyclopent-2-en-1-ylmethanol}$ is the alcohol from which the compound $\text{cyclopent-2-enecarbaldehyde}$ can be best prepared by the oxidation process which is shown below.

Explanation of Solution

Primary and secondary alcohols can be oxidized into aldehydes and ketones using $\text{Cr(VI)}$ compounds as they are very good oxidizing agents. They are also further oxidized into carboxylic acids in the presence of water with chromium compounds. Therefore, in order to stop this process, generally anhydrous conditions are taken. Some examples of the $\text{Cr(VI)}$ compounds used are PCC, ${\text{CrO}}_{\text{3}}$, and ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$.

The primary alcohols are oxidized into aldehydes under anhydrous conditions.

The compound $\text{cyclopent-2-enecarbaldehyde}$ is prepared from the alcohol given below by the oxidation method.

Figure 3

The compound $\text{cyclopent-2-enecarbaldehyde}$ is obtained from the alcohol $\text{cyclopent-2-en-1-ylmethanol}$ under the anhydrous conditions of oxidation using PCC otherwise it would be further oxidized to the carboxylic acid form.

Conclusion

The alcohol from which the compound $\text{cyclopent-2-enecarbaldehyde}$ can be best prepared by the oxidation process is shown in Figure 3.