Chapter 9, Problem 9.37P

Interpretation Introduction

(a)

Interpretation: The product formed by the treatment of given alcohol with $\text{HBr}$ is to be drawn and the stereochemistry of product is to be indicated.

Concept introduction: The reaction of alcohols with halogen acids $\left(\text{HX}\right)$ is a general method to obtain $\text{1}\text{°}\text{,}\text{2}\text{°}\text{or}\text{3}\text{°}$ alkyl halides. This method is preferred due to the formation of ${\text{H}}_{\text{2}}\text{O}$ as leaving group. The mechanism of the reaction is ${\text{S}}_{\text{N}}\text{1}$ for $\text{2}\text{°}\text{or}\text{3}\text{°}$ alcohols, and ${\text{S}}_{\text{N}}2$ for $\text{1}\text{°}$ alcohols.

Answer to Problem 9.37P

The products formed by the treatment of given alcohol with $\text{HBr}$ (including stereochemistry) are drawn in Figure 3.

Explanation of Solution

The given structure of alcohol is in the form of ball-and-stick model. It is converted into skeletal structure by replacing black ball with $\text{C}$, red ball with $\text{O}$, and gray ball with $\text{H}$.

Figure 1

The above skeletal structure indicates that the given alcohol is secondary. It contains only one stereogenic center at $\text{C3}$ where it possesses $\left(S\right)$ configuration as follows.

Figure 2

The reaction of alcohols with halogen acids $\left(\text{HX}\right)$ is a general method to obtain $\text{1}\text{°}\text{,}\text{2}\text{°}\text{or}\text{3}\text{°}$ alkyl halides. This method is preferred due to the formation of ${\text{H}}_{\text{2}}\text{O}$ as leaving group. The mechanism of the reaction is ${\text{S}}_{\text{N}}\text{1}$ for $\text{2}\text{°}\text{or}\text{3}\text{°}$ alcohols, and ${\text{S}}_{\text{N}}2$ for $\text{1}\text{°}$ alcohols.

Since the given alcohol is secondary, it follows ${\text{S}}_{\text{N}}\text{1}$ path.

The ${\text{S}}_{\text{N}}\text{1}$ reaction results racemic of mixture in the final product. Thus, the products formed by the treatment of given alcohol with $\text{HBr}$ (including stereochemistry) are,

Figure 3

Conclusion

The products formed by the treatment of given alcohol with $\text{HBr}$ (including stereochemistry) are drawn in Figure 3.

Interpretation Introduction

(b)

Interpretation: The product formed by the treatment of given alcohol with ${\text{PBr}}_{\text{3}}$ is to be drawn and the stereochemistry of product is to be indicated.

Concept introduction: Alkyl bromides are obtained by the reaction of $\text{1}\text{°}\text{or}\text{2}\text{°}$ alcohols with phosphorous tribromide $\left({\text{PBr}}_{\text{3}}\right)$. The reaction yields ${\text{HOPBr}}_{\text{2}}$ as by-product. The mechanism of the reaction involves two steps. The first step is conversion of the $\text{OH}$ group into a good leaving group. The second step is nucleophilic attack by ${\text{Br}}^{-}$ through an ${\text{S}}_{\text{N}}2$ reaction.

Answer to Problem 9.37P

The product formed by the treatment of given alcohol with ${\text{PBr}}_{\text{3}}$ (including stereochemistry) is drawn in Figure 4.

Explanation of Solution

Alkyl bromides are obtained by the reaction of $\text{1}\text{°}\text{or}\text{2}\text{°}$ alcohols with phosphorous tribromide $\left({\text{PBr}}_{\text{3}}\right)$. The reaction yields ${\text{HOPBr}}_{\text{2}}$ as by-product. The mechanism of the reaction involves two steps. The first step is conversion of the $\text{OH}$ group into a good leaving group. The second step is nucleophilic attack by ${\text{Br}}^{-}$ through an ${\text{S}}_{\text{N}}2$ reaction.

Since the given alcohol is secondary, it follows ${\text{S}}_{\text{N}}2$ path.

The ${\text{S}}_{\text{N}}2$ reaction results final product with inversion of configuration. Thus, the product formed by the treatment of given alcohol with ${\text{PBr}}_{\text{3}}$ (including stereochemistry) is,

Figure 4

Conclusion

The product formed by the treatment of given alcohol with ${\text{PBr}}_{\text{3}}$ (including stereochemistry) is drawn in Figure 4.

Interpretation Introduction

(c)

Interpretation: The product formed by the treatment of given alcohol with $\text{HCl}$ is to be drawn and the stereochemistry of product is to be indicated.

Concept introduction: The reaction of alcohols with halogen acids $\left(\text{HX}\right)$ is a general method to obtain $\text{1}\text{°}\text{,}\text{2}\text{°}\text{or}\text{3}\text{°}$ alkyl halides. This method is preferred due to the formation of ${\text{H}}_{\text{2}}\text{O}$ as leaving group. The mechanism of the reaction is ${\text{S}}_{\text{N}}\text{1}$ for $\text{2}\text{°}\text{or}\text{3}\text{°}$ alcohols, and ${\text{S}}_{\text{N}}2$ for $\text{1}\text{°}$ alcohols.

Answer to Problem 9.37P

The products formed by the treatment of given alcohol with ${\text{PBr}}_{\text{3}}$ (including stereochemistry) are drawn in Figure 5.

Explanation of Solution

The reaction of alcohols with halogen acids $\left(\text{HX}\right)$ is a general method to obtain $\text{1}\text{°}\text{,}\text{2}\text{°}\text{or}\text{3}\text{°}$ alkyl halides. This method is preferred due to the formation of ${\text{H}}_{\text{2}}\text{O}$ as leaving group. The mechanism of the reaction is ${\text{S}}_{\text{N}}\text{1}$ for $\text{2}\text{°}\text{or}\text{3}\text{°}$ alcohols, and ${\text{S}}_{\text{N}}2$ for $\text{1}\text{°}$ alcohols.

Since the given alcohol is secondary, it follows ${\text{S}}_{\text{N}}\text{1}$ path.

The ${\text{S}}_{\text{N}}\text{1}$ reaction results racemic of mixture in the final product. Thus, the products formed by the treatment of given alcohol with ${\text{PBr}}_{\text{3}}$ (including stereochemistry) are,

Figure 5

Conclusion

The products formed by the treatment of given alcohol with ${\text{PBr}}_{\text{3}}$ (including stereochemistry) are drawn in Figure 5.

Interpretation Introduction

(d)

Interpretation: The product formed by the treatment of given alcohol with ${\text{SOCl}}_2$ and pyridine is to be drawn and the stereochemistry of product is to be indicated.

Concept introduction: Alkyl chlorides are obtained by the reaction of $\text{1}\text{°}\text{or}\text{2}\text{°}$ alcohols with thionyl chloride $\left({\text{SOCl}}_2\right)$, and pyridine. The reaction yields ${\text{SO}}_{\text{2}}$, and $\text{HCl}$ as by-products. The mechanism of the reaction involves two steps. The first step is conversion of the $\text{OH}$ group into a good leaving group. The second step is nucleophilic attack by ${\text{Cl}}^{-}$ through an ${\text{S}}_{\text{N}}2$ reaction.

Answer to Problem 9.37P

The product formed by the treatment of given alcohol with ${\text{SOCl}}_2$ and pyridine (including stereochemistry) is drawn in Figure 6.

Explanation of Solution

Alkyl chlorides are obtained by the reaction of $\text{1}\text{°}\text{or}\text{2}\text{°}$ alcohols with thionyl chloride $\left({\text{SOCl}}_2\right)$, and pyridine. The reaction yields ${\text{SO}}_{\text{2}}$, and $\text{HCl}$ as by-products. The mechanism of the reaction involves two steps. The first step is conversion of the $\text{OH}$ group into a good leaving group. The second step is nucleophilic attack by ${\text{Cl}}^{-}$ through an ${\text{S}}_{\text{N}}2$ reaction.

Since the given alcohol is secondary, it follows ${\text{S}}_{\text{N}}2$ path.

The ${\text{S}}_{\text{N}}2$ reaction results final product with inversion of configuration. Thus, the product formed by the treatment of given alcohol with ${\text{SOCl}}_2$ and pyridine (including stereochemistry) is,

Figure 6

Conclusion

The product formed by the treatment of given alcohol with ${\text{SOCl}}_2$ and pyridine (including stereochemistry) is drawn in Figure 6.