Chapter 12, Problem 12.40P

Draw the organic products formed when allylic alcoholA is treated with each reagent.

a. ${\text{H}}_{\text{2}}+\text{Pd-C}$

b. $\text{mCPBA}$

c. $\text{PCC}$

d. ${\text{CrO}}_{\text{3}},{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}},{\text{H}}_{\text{2}}\text{O}$

e. ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{COOH}$, $\text{Ti}{\left[\text{OCH}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\right]}_{\text{4}}$, $(+)\text{-DET}$

f. ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{COOH}$, $\text{Ti}{\left[\text{OCH}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\right]}_{\text{4}}$, $(-)\text{-DET}$

g. [1] ${\text{PBr}}_{\text{3}}$ ; [2] ${\text{LiAlH}}_{\text{4}}$;[3] ${\text{H}}_{\text{2}}\text{O}$

h. ${\text{HCrO}}_4^{-}\text{-Amberlyst}\text{A-}26$ resin

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 a 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 12.40P

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

Explanation of Solution

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 product formed is $\text{3-methyl-1-pentanol}$. The reaction is as follows,

Figure 1

Conclusion

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

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 formation of bonds takes place in single step. $\text{mCPBA}$ is $\text{meta-chloroperoxybenzoic}\text{acid}$.

Answer to Problem 12.40P

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

Explanation of Solution

In the given reaction, when A is treated with $\text{mCPBA}$ ($\text{meta-chloroperoxybenzoic}\text{acid}$), alkene is oxidized to epoxide. The product formed is $\text{3-ethyl-3-methyloxiran-2-yl methanol}$. The reaction is as follows,

Figure 2

Conclusion

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

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 are oxidized 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 12.40P

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

Explanation of Solution

In the given reaction, when A is treated with $\text{PCC}$, the primary alcohol is oxidized to aldehyde. The corresponding chemical reaction is given below. The product formed is $\text{3-methylpent-2-enal}$.

Figure 3

Conclusion

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

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 the 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 are oxidised to ketones. In the presence of mild oxidizing agents such as $\text{PCC}$, primary alcohols are oxidized to aldehydes and secondary alcohols to ketones.

Answer to Problem 12.40P

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

In the given reaction, when A is treated with ${\text{CrO}}_{\text{3}},{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}},{\text{H}}_{\text{2}}\text{O}$, the primary alcohol is oxidized to carboxylic acid. The corresponding chemical reaction is given below. The product formed is $\text{3-methylpent-2-enoic acid}$.

Figure 4

Conclusion

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.

Interpretation Introduction

(e)

Interpretation: The product 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.

Answer to Problem 12.40P

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

Explanation of Solution

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 configuration of isomer determines which enantiomer is the major product.

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

Figure 5

Conclusion

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

Interpretation Introduction

(f)

Interpretation: The product formed when A is treated with Sharpless reagent $(-)\text{-DET}$ 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.

Answer to Problem 12.40P

The product of formed when A is treated with Sharpless reagent $(-)\text{-DET}$ is shown in Figure 6.

Explanation of Solution

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 configuration of isomer determines which enantiomer is the major product.

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

Figure 6

Conclusion

The product formed when A is treated with Sharpless reagent $(-)\text{-DET}$ is shown in Figure 6

Interpretation Introduction

(g)

Interpretation: The product formed when A is treated with given reagent is shown in Figure 7.

Concept introduction: Alcohols on treatment with phosphorus tribromide gives alkyl bromide. The $\text{C}-\text{X}$ bond in alkyl halides can be reduced using metal hydride reagents. The alkyl halides are reduced to alkanes with ${\text{X}}^{-}$ acting as leaving group.

Answer to Problem 12.40P

The product formed in the given reaction is shown in Figure 7.

Explanation of Solution

The given reaction is,

Figure 7

When the given alcohol is treated with ${\text{PBr}}_{\text{3}}$, it forms alkyl bromide. The alkyl bromide on further reaction with ${\text{LiAlH}}_{\text{4}},{\text{H}}_{\text{2}}\text{O}$ produces alkane. Hence, The final product is $\text{3-methylpent-2-ene}$.

Conclusion

The product formed when A is treated with given reagent is shown in Figure 7.

Interpretation Introduction

(h)

Interpretation: The product formed when A is treated with ${\text{HCrO}}_4^{-}\text{-Amberlyst}\text{A-}26$ resin is to be predicted.

Concept introduction: ${\text{HCrO}}_4^{-}\text{-Amberlyst}\text{A-}26$ is an insoluble polymeric reagent. It consists of complex hydrocarbon network with cationic ammonium ion which acts a counter ion to anionic chromium ion. This is a green chemistry method for oxidation of alcohols as it avoids the usage of strong acids and also it forms by-product with carcinogenic ${\text{Cr}}^{3+}$ which can be removed by filtration.

Answer to Problem 12.40P

The product formed when A is treated with ${\text{HCrO}}_4^{-}\text{-Amberlyst}\text{A-}26$ resin is shown in Figure 8.

Explanation of Solution

In the given reaction, when A is treated with ${\text{HCrO}}_4^{-}\text{-Amberlyst}\text{A-}26$, the primary alcohol is oxidized to aldehyde. The corresponding chemical reaction is given below. The product formed is $\text{3-methylpent-2-enal}$.

Figure 8

Conclusion

The product formed when A is treated with ${\text{HCrO}}_4^{-}\text{-Amberlyst}\text{A-}26$ resin is shown in Figure 8.