Chapter 9, Problem 9.46P

Draw the organic product(s) formed when ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ is treated with each reagent.

a. ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ d. $\text{HBr}$ g. $\text{TsCl},\text{pyridine}$

b. $\text{NaH}$ e. ${\text{SOCl}}_2,\text{pyridine}$ h. $\left[1\right]\text{NaH;}\left[2\right]{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{Br}$

c. $\text{HCl}+{\text{ZnCl}}_{\text{2}}$ f. ${\text{PBr}}_3$ i. $\left[1\right]\text{TsCl},\text{pyridine;}\left[2\right]\text{NaSH}$

Interpretation Introduction

(a)

Interpretation: The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is to be drawn.

Concept introduction: Alcohols undergo dehydration reaction in the presence of strong acids like ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ or $\text{p-toluenesulfonic}\text{acid}\left(\text{TsOH}\right)$ and yield alkenes. The dehydration of primary alcohols occurs in two steps and it is $\text{E2}$ reaction. The first step is protonation of $\text{OH}$ to form good leaving group. The second step is loss of the leaving group and removal of the $\text{β}$ proton.

Answer to Problem 9.46P

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is shown below.

Explanation of Solution

The given reagent is ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$.

Alcohols undergo dehydration reaction in the presence of strong acids like ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ or $\text{p-toluenesulfonic}\text{acid}\left(\text{TsOH}\right)$ and yield alkenes. The dehydration of primary alcohols occurs in two steps and it is $\text{E2}$ reaction. The first step is protonation of $\text{OH}$ to form good leaving group. The second step is loss of the leaving group and removal of the $\text{β}$ proton.

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is drawn in Figure 1.

Figure 1

Conclusion

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is drawn in Figure 1.

Interpretation Introduction

(b)

Interpretation: The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is to be drawn.

Concept introduction: An alkoxide salt is required to prepare ether. The alkoxide salts are prepared from alcohols through the Bronsted-Lowry acid-base reaction. In this reaction, $\text{NaH}$ or ${\text{NaNH}}_{\text{2}}$ are especially utilized, due to the by-product of the reaction $\left({\text{H}}_{\text{2}}\text{or}{\text{NH}}_{\text{3}}\right)$. Both ${\text{H}}_{\text{2}}$ and ${\text{NH}}_{\text{3}}$ are gases that bubbles out rapidly from the reaction mixture.

Answer to Problem 9.46P

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is shown below.

.

Explanation of Solution

The given reagent is $\text{NaH}$.

An alkoxide salt is required to prepare ether. The alkoxide salts are prepared from alcohols through the Bronsted-Lowry acid-base reaction. In this reaction, $\text{NaH}$ or ${\text{NaNH}}_{\text{2}}$ are especially utilized, due to the by-product of the reaction $\left({\text{H}}_{\text{2}}\text{or}{\text{NH}}_{\text{3}}\right)$. Both ${\text{H}}_{\text{2}}$ and ${\text{NH}}_{\text{3}}$ are gases that bubbles out rapidly from the reaction mixture.

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is drawn in Figure 2.

Figure 2

Conclusion

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is drawn in Figure 2.

Interpretation Introduction

(c)

Interpretation: The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is to be drawn.

Concept introduction: The reactivity of $\text{HCl}$ is low towards the primary alcohols. It is due to low nucleophilicity of ${\text{Cl}}^{-}$. Therefore, an additional catalyst ${\text{ZnCl}}_{\text{2}}$ is added along with it, that makes a Lewis acid-base reaction with primary alcohols. The mechanism of the reaction is ${\text{S}}_{\text{N}}\text{2}$.

Answer to Problem 9.46P

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is shown below.

Explanation of Solution

The given reagent is $\text{HCl}+{\text{ZnCl}}_{\text{2}}$.

The reactivity of $\text{HCl}$ is low towards the primary alcohols. It is due to low nucleophilicity of ${\text{Cl}}^{-}$. Therefore, an additional catalyst ${\text{ZnCl}}_{\text{2}}$ is added along with it, that makes a Lewis acid-base reaction with primary alcohols. The mechanism of the reaction is ${\text{S}}_{\text{N}}\text{2}$.

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is shown below.

Figure 3

Conclusion

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is drawn in Figure 3.

Interpretation Introduction

(d)

Interpretation: The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is to be drawn.

Concept introduction: The reaction of alcohols with halogen acids $\left(\text{HX}\right)$ is a general method to obtain ${\text{1}}^{\text{ο}}\text{,}{\text{2}}^{\text{ο}}\text{or}{\text{3}}^{\text{ο}}$ alkyl halides. This method is preferred due to the formation of ${\text{H}}_{\text{2}}\text{O}$ as leaving group. The more substituted alcohols generally react rapidly with halogen acids.

Answer to Problem 9.46P

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent are shown below.

Explanation of Solution

The given reagent is $\text{HBr}$.

The reaction of alcohols with halogen acids $\left(\text{HX}\right)$ is a general method to obtain ${\text{1}}^{\text{ο}}\text{,}{\text{2}}^{\text{ο}}\text{or}{\text{3}}^{\text{ο}}$ alkyl halides. This method is preferred due to the formation of ${\text{H}}_{\text{2}}\text{O}$ as leaving group. The more substituted alcohols generally react rapidly with halogen acids.

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is drawn in Figure 4.

Figure 4

Conclusion

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is drawn in Figure 4.

Interpretation Introduction

(e)

Interpretation: The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is to be drawn.

Concept introduction: Alkyl chlorides are obtained by the reaction of ${\text{1}}^{\text{ο}}\text{or}{\text{2}}^{\text{ο}}$ 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 9.46P

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent are shown below.

Explanation of Solution

The given reagent is ${\text{SOCl}}_2,\text{pyridine}$.

Alkyl chlorides are obtained by the reaction of ${\text{1}}^{\text{ο}}\text{or}{\text{2}}^{\text{ο}}$ 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.

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is drawn in Figure 5.

Figure 5

Conclusion

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is drawn in Figure 5.

Interpretation Introduction

(f)

Interpretation: The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is to be drawn.

Concept introduction: Alkyl bromides are obtained by the reaction of ${\text{1}}^{\text{ο}}\text{or}{\text{2}}^{\text{ο}}$ 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 9.46P

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is shown below.

Explanation of Solution

The given reagent is ${\text{PBr}}_3$.

Alkyl bromides are obtained by the reaction of ${\text{1}}^{\text{ο}}\text{or}{\text{2}}^{\text{ο}}$ 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. The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is drawn in Figure 6.

Figure 6

Conclusion

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is drawn in Figure 6.

Interpretation Introduction

(g)

Interpretation: The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is to be drawn.

Concept introduction: Alcohols are converted into alkyl tosylates by treatment with $p\text{-toluenesulfonyl}\text{chloride}\left(\text{TsCl}\right)$ in the presence of pyridine. The overall reaction converts ${}^{-}\text{OH}$ (a poor leaving group) into ${}^{-}\text{OTs}$ (a good leaving group).

Answer to Problem 9.46P

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is shown below.

Explanation of Solution

The given reagent is $\text{TsCl},\text{pyridine}$.

Alcohols are converted into alkyl tosylates by treatment with $p\text{-toluenesulfonyl}\text{chloride}\left(\text{TsCl}\right)$ in the presence of pyridine. The overall reaction converts ${}^{-}\text{OH}$ (a poor leaving group) into ${}^{-}\text{OTs}$ (a good leaving group).

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is drawn in Figure 7.

Figure 7

Conclusion

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is drawn in Figure 7.

Interpretation Introduction

(h)

Interpretation: The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is to be drawn.

Concept introduction: An alkoxide salt is required to prepare ether. The alkoxide salts are prepared from alcohols through the Bronsted-Lowry acid-base reaction. In this reaction, $\text{NaH}$ or ${\text{NaNH}}_{\text{2}}$ are especially utilized, due to the by-product of the reaction $\left({\text{H}}_{\text{2}}\text{or}{\text{NH}}_{\text{3}}\right)$.

The formed alkoxide is allowed to react with an alkyl halide to obtain ether. The mechanism of the reaction is ${\text{S}}_{\text{N}}\text{2}$.

Answer to Problem 9.46P

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is shown below.

Explanation of Solution

The given reagents are $\left[1\right]\text{NaH;}\left[2\right]{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{Br}$.

An alkoxide salt is required to prepare ether. The alkoxide salts are prepared from alcohols through the Bronsted-Lowry acid-base reaction. In this reaction, $\text{NaH}$ or ${\text{NaNH}}_{\text{2}}$ are especially utilized, due to the by-product of the reaction $\left({\text{H}}_{\text{2}}\text{or}{\text{NH}}_{\text{3}}\right)$.

The formed alkoxide is allowed to react with an alkyl halide to obtain ether. The mechanism of the reaction is ${\text{S}}_{\text{N}}\text{2}$.

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is drawn in Figure 8.

Figure 8

Conclusion

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is drawn in Figure 8.

Interpretation Introduction

(i)

Interpretation: The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is to be drawn.

Concept introduction: Alcohols are converted into alkyl tosylates by treatment with $p\text{-toluenesulfonyl}\text{chloride}\left(\text{TsCl}\right)$ in the presence of pyridine. The overall reaction converts ${}^{-}\text{OH}$ (a poor leaving group) into ${}^{-}\text{OTs}$ (a good leaving group).

The formed alkyl tosylate reacts with strong nucleophile $\left({}^{-}\text{SH}\right)$ and yields thiol. The mechanism of the reaction is ${\text{S}}_{\text{N}}\text{2}$.

Answer to Problem 9.46P

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is shown below.

Explanation of Solution

The given reagents are $\left[1\right]\text{TsCl},\text{pyridine;}\left[2\right]\text{NaSH}$.

Alcohols are converted into alkyl tosylates by treatment with $p\text{-toluenesulfonyl}\text{chloride}\left(\text{TsCl}\right)$ in the presence of pyridine. The overall reaction converts ${}^{-}\text{OH}$ (a poor leaving group) into ${}^{-}\text{OTs}$ (a good leaving group).

The formed alkyl tosylate react with strong nucleophile $\left({}^{-}\text{SH}\right)$ and yield thiol. The mechanism of the reaction is ${\text{S}}_{\text{N}}\text{2}$.

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is drawn in Figure 9.

Figure 9

Conclusion

The organic product(s) formed by the treatment of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ with given reagent is drawn in Figure 9.