Chapter 12, Problem 28P

Draw the products formed when A is treated with each reagent: (a) ${\text{H}}_{\text{2}}+\text{Pd-C}$; (b) m-CPBA; (c) $\text{PCC}$; (d)

${\text{CrO}}_{\text{3}},{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}},{\text{H}}_{\text{2}}\text{O}$; (e) Sharpless reagent with (+)-DET.

A

Interpretation Introduction

(a)

Interpretation: The product formed when A is treated with ${\text{H}}_{\text{2}}+\text{Pd-C}$ is to be predicted.

Concept introduction: The addition of ${\text{H}}_{\text{2}}$ in presence of catalyst is known as catalytic hydrogenation reaction. This is a syn addition. Alkyne is reduced to alkane. In this reaction two weak pi bond of alkyne and sigma bond between ${\text{H}}_{\text{2}}$ is broken and four $\text{C}-\text{H}$ bonds are formed.

Answer to Problem 28P

The product formed when A is treated with ${\text{H}}_{\text{2}}+\text{Pd-C}$ is shown in Figure 1.

Explanation of Solution

The given molecule is,

Figure 1

The red coloured balls have two bonds. So, these are the oxygen atoms. Black coloured atoms have four bonds. So, these are the carbon atoms. The grey coloured balls have one bond. So, these are the hydrogen atoms. The molecular structure of A is,

Figure 2

When A is treated with ${\text{H}}_{\text{2}}+\text{Pd-C}$, the $\text{C}-\text{C}$ double bond is reduced to $\text{C}-\text{C}$ single bond. The reaction is as follows:

Figure 3

Interpretation Introduction

(b)

Interpretation: The product formed when A is treated with $\text{mCPBA}$ is to be predicted.

Concept introduction: In presence of peroxide alkene is oxidized to epoxide this is known as epoxidation. The weak pi bond of alkene and weak $\text{O}-\text{O}$ bond of peroxide are broken and two new $\text{C}-\text{O}$ bonds are formed. Breaking of bonds and forming of bonds takes place in single step. $\text{mCPBA}$ is $\text{meta-chloroperoxybenzoic}\text{acid}$.

Answer to Problem 28P

The product formed when A is treated with $\text{mCPBA}$ is shown in Figure 2.

Explanation of Solution

The given molecule is,

Figure 1

The red coloured balls have two bonds. So, these are the oxygen atoms. Black coloured atoms have four bonds. So, these are the carbon atoms. The grey coloured balls have one bond. So, these are the hydrogen atoms. The molecular structure of the A is,

Figure 2

In presence of $\text{mCPBA}$ ($\text{meta-chloroperoxybenzoic}\text{acid}$), alkene is oxidized to epoxide. The reaction is as follows:

Figure 4

Interpretation Introduction

(c)

Interpretation: The product formed when A is treated with $\text{PCC}$ is to be predicted.

Concept introduction: Alcohols are oxidized to different carbonyl compounds depending upon the reagents and alcohol used. In presence of strong oxidizing reagents such as ${\text{KMnO}}_{\text{4}},{\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ and ${\text{CrO}}_{\text{3}}$, primary alcohols are oxidized to carboxylic acids and secondary alcohols to ketones. In presence of mild oxidizing agents such as $\text{PCC}$, primary alcohols are oxidized to aldehydes and secondary alcohols to ketones.

Answer to Problem 28P

The product formed when A is treated with $\text{PCC}$ is shown in Figure 3.

Explanation of Solution

The given molecule is,

Figure 1

The red coloured balls have two bonds. So, these are the oxygen atoms. Black coloured atoms have four bonds. So, these are the carbon atoms. The grey coloured balls have one bond. So, these are the hydrogen atoms. The molecular structure of the A is,

Figure 2

The chemical reaction of A with $\text{PCC}$ is,

Figure 5

The primary alcohol is oxidized to aldehyde in presence of mild oxidizing reagent $\text{PCC}$.

Interpretation Introduction

(d)

Interpretation: The product formed when A is treated with ${\text{CrO}}_{\text{3}},{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}},{\text{H}}_{\text{2}}\text{O}$ is to be predicted.

Concept introduction: Alcohols are oxidized to different carbonyl compounds depending upon the reagents and alcohol used. In presence of strong oxidizing reagents such as ${\text{KMnO}}_{\text{4}},{\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ and ${\text{CrO}}_{\text{3}}$, primary alcohols are oxidized to carboxylic acids and secondary alcohols to ketones. In presence of mild oxidizing agents such as $\text{PCC}$, primary alcohols are oxidized to aldehydes and secondary alcohols to ketones.

Answer to Problem 28P

The product formed when A is treated with ${\text{CrO}}_{\text{3}},{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}},{\text{H}}_{\text{2}}\text{O}$ is shown in Figure 4.

Explanation of Solution

The given molecule is

Figure 1

The red coloured balls have two bonds. So, these are the oxygen atoms. Black coloured atoms have four bonds. So, these are the carbon atoms. The grey coloured balls have one bond. So, these are the hydrogen atoms. The molecular structure of the A is,

Figure 2

When A is treated with ${\text{CrO}}_{\text{3}},{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}},{\text{H}}_{\text{2}}\text{O}$, the reaction is as follows,

Figure 6

The primary alcohol is oxidized to carboxylic acid in presence of strong oxidizing reagent ${\text{CrO}}_{\text{3}},{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}},{\text{H}}_{\text{2}}\text{O}$.

Interpretation Introduction

(e)

Interpretation: The product of formed when A is treated with Sharpless reagent $\left(\text{+}\right)\text{-DET}$ is to be predicted.

Concept introduction: Sharpless epoxidation involves the oxidation of double bond between carbon atoms to epoxide. This oxidation occurs only in allylic alcohol. This is an enantioselective oxidation, which means predominantly one enantiomer is formed. Sharpless reagents are $tert\text{-butyl}\text{hydroperoxide,}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{COOH}$; titanium $\left(\text{IV}\right)$ isopropoxide, $\text{Ti}{\left[\text{OCH}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\right]}_{\text{4}}$; and diethyl tartrate. There are two different chiral diethyl tartrate isomers, $\left(\text{+}\right)\text{-DET}$ and $(-)\text{-DET}$ to indicate the direction in which they rotate the plane polarized light. The identity of isomer determines which enantiomer is the major product.

Answer to Problem 28P

The product of formed when A is treated with Sharpless reagent $\left(\text{+}\right)\text{-DET}$ is shown in Figure 5.

Explanation of Solution

The given molecule is

Figure 1

The red coloured balls have two bonds. So, these are the oxygen atoms. Black coloured atoms have four bonds. So, these are the carbon atoms. The grey coloured balls have one bond. So, these are the hydrogen atoms. The molecular structure of the A is,

Figure 2

When epoxidation is done using $\left(\text{+}\right)\text{-DET}$, epoxide is formed from below the plane as shown below.

Figure 7

Conclusion

(a) The product formed when A is treated with ${\text{H}}_{\text{2}}+\text{Pd-C}$ is shown in Figure 3.

(b) The product formed when A is treated with $\text{mCPBA}$ is shown in Figure 4.

(c) The product formed when A is treated with $\text{PCC}$ is shown in Figure 5.

(d) The product formed when A is treated with ${\text{CrO}}_{\text{3}},{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}},{\text{H}}_{\text{2}}\text{O}$ is shown in Figure 6.

(e) The product of formed when A is treated with Sharpless reagent $\left(\text{+}\right)\text{-DET}$ is shown in Figure 7.