Chapter 25, Problem 25.54P

How would you prepare $\text{3-phenyl-1-propanamine}$ $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\right)$ from

each compound?

a. ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Br}$ c. ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NO}}_{\text{2}}$ e. ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CHO}$

b. ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Br}$ d. ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CONH}}_2$

Interpretation Introduction

(a)

Interpretation:

The preparation of $\text{3-phenyl-1-propanamine}$ $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\right)$ from ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Br}$ is to be stated.

Concept introduction:

Synthesis is one of the major areas in the field of organic chemistry. It can be a simple one-step reaction or it may involve many steps. The reaction of alkyl halide with ammonia results in the formation of primary amine. Such reactions are commonly known as alkylation of ammonia.

Answer to Problem 25.54P

The preparation of $\text{3-phenyl-1-propanamine}$ $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\right)$ from ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Br}$ is shown below.

$\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Br}+{\text{NH}}_{\text{3}}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}+{\text{NH}}_{\text{4}}^{\text{+}}{\text{Br}}^{-}\\ \text{ excess}\end{array}$

Explanation of Solution

The given compound is $\left(\text{3-bromopropyl}\right)\text{benzene}$. The conversion of this to $\text{3-phenyl-1-propanamine}$ occurs by using excess of ammonia. The reaction for the preparation of $\text{3-phenyl-1-propanamine}$ from $\left(\text{3-bromopropyl}\right)\text{benzene}$ is shown below.

$\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Br}+{\text{NH}}_{\text{3}}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}+{\text{NH}}_{\text{4}}^{\text{+}}{\text{Br}}^{-}\\ \text{ excess}\end{array}$

Conclusion

The compound $\text{3-phenyl-1-propanamine}$ is prepared from ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Br}$ by using excess of ammonia.

Interpretation Introduction

(b)

Interpretation:

The preparation of $\text{3-phenyl-1-propanamine}$ $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\right)$ from ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Br}$ is to be stated.

Concept introduction:

Synthesis is one of the major areas in the field of organic chemistry. It can be a simple one-step reaction or it may involve many steps. Lithium aluminum hydride is a strong reducing agent. It is an inorganic compound which is used as a reducing agent in organic synthesis.

Answer to Problem 25.54P

The preparation of $\text{3-phenyl-1-propanamine}$ $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\right)$ from ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Br}$ is shown below.

$\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Br}+\text{NaCN}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CN}\\ {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CN}\underset{\left(2\right){\text{H}}_{\text{2}}\text{O}}{\overset{\left(1\right){\text{LiAlH}}_{\text{4}}}{\to }}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\end{array}$

Explanation of Solution

The given compound is $\left(\text{2-bromoethyl}\right)\text{benzene}$. The conversion of this to $\text{3-phenyl-1-propanamine}$ occurs by using reagents like sodium cyanide and Lithium aluminum hydride. The reaction for the preparation of $\text{3-phenyl-1-propanamine}$ from $\left(\text{2-bromoethyl}\right)\text{benzene}$ is shown below.

$\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Br}+\text{NaCN}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CN}\\ {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CN}\underset{\left(2\right){\text{H}}_{\text{2}}\text{O}}{\overset{\left(1\right){\text{LiAlH}}_{\text{4}}}{\to }}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\end{array}$

In the first step of reaction, $\left(\text{2-bromoethyl}\right)\text{benzene}$ reacts with sodium cyanide to form $\text{3-}$ phenylpropanenitrile. In second step, $\text{3-}$ phenylpropanenitrile reacts with Lithium aluminum hydride in the presence of ${\text{H}}_{\text{2}}\text{O}$ to form the final product $\text{3-phenyl-1-propanamine}$.

Conclusion

The conversion of $\left(\text{2-bromoethyl}\right)\text{benzene}$ into $\text{3-phenyl-1-propanamine}$ occurs by using reagents like sodium cyanide and Lithium aluminum hydride.

Interpretation Introduction

(c)

Interpretation:

The preparation of $\text{3-phenyl-1-propanamine}$ $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\right)$ from ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NO}}_{\text{2}}$ is to be stated.

Concept introduction:

Synthesis is one of the major areas in the field of organic chemistry. It can be a simple one-step reaction or it may involve many steps. The conversion of nitroalkanes to primary amines occur by using the reducing agent $\text{Sn/HCl}$.

Answer to Problem 25.54P

The preparation of $\text{3-phenyl-1-propanamine}$ $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\right)$ from ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NO}}_{\text{2}}$ is shown below.

${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NO}}_{\text{2}}\underset{\text{HCl}}{\overset{\text{Sn}}{\to }}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}$

Explanation of Solution

The given compound is $\left(\text{3-nitropropyl}\right)\text{benzene}$. The conversion of this to $\text{3-phenyl-1-propanamine}$ occurs by using the reducing agent $\text{Sn/HCl}$. The reaction for the preparation of $\text{3-phenyl-1-propanamine}$ from $\left(\text{3-nitropropyl}\right)\text{benzene}$ is shown below.

${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NO}}_{\text{2}}\underset{\text{HCl}}{\overset{\text{Sn}}{\to }}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}$

Conclusion

The conversion of $\left(\text{3-nitropropyl}\right)\text{benzene}$ into $\text{3-phenyl-1-propanamine}$ occurs by using the reducing agent $\text{Sn/HCl}$.

Interpretation Introduction

(d)

Interpretation:

The preparation of $\text{3-phenyl-1-propanamine}$ $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\right)$ from ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CONH}}_2$ is to be stated.

Concept introduction:

Synthesis is one of the major areas in the field of organic chemistry. It can be a simple one-step reaction or it may involve many steps. Lithium aluminum hydride is a strong reducing agent. It is an inorganic compound which is used as a reducing agent in organic synthesis.

Answer to Problem 25.54P

The preparation of $\text{3-phenyl-1-propanamine}$ $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\right)$ from ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CONH}}_2$ is shown below.

${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CONH}}_2\underset{\left(2\right){\text{H}}_{\text{2}}\text{O}}{\overset{\left(1\right){\text{LiAlH}}_{\text{4}}}{\to }}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}$

Explanation of Solution

The given compound is $\text{3-phenylpropanamide}$. The conversion of this to $\text{3-phenyl-1-propanamine}$ occurs by using reagent Lithium aluminum hydride. The reaction for the preparation of $\text{3-phenyl-1-propanamine}$ from $\text{3-phenylpropanamide}$ is shown below.

${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CONH}}_2\underset{\left(2\right){\text{H}}_{\text{2}}\text{O}}{\overset{\left(1\right){\text{LiAlH}}_{\text{4}}}{\to }}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}$

Conclusion

The conversion of this to $\text{3-phenyl-1-propanamine}$ occurs by using reagent Lithium aluminum hydride.

Interpretation Introduction

(e)

Interpretation:

The preparation of $\text{3-phenyl-1-propanamine}$ $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\right)$ from ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CHO}$ is to be stated.

Concept introduction:

Synthesis is one of the major areas in the field of organic chemistry. It can be a simple one-step reaction or it may involve many steps. When an aldehyde or a ketone is converted to an amine, the process is known as reductive amination. The conversion takes place through the formation of an intermediate known as imine. The excess ammonia is used when a primary amine is synthesized by reductive amination.

Answer to Problem 25.54P

The preparation of $\text{3-phenyl-1-propanamine}$ $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\right)$ from ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CHO}$ is shown below.

$\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CHO}\stackrel{{\text{NH}}_{\text{3}}}{\to }{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH=NH}\\ {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH=NH}\stackrel{{\text{NaBH}}_{\text{3}}\text{CN}}{\to }{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\end{array}$

Explanation of Solution

The given compound is $\text{3-phenylpropanal}$. The conversion of this to $\text{3-phenyl-1-propanamine}$ occurs by reductive animation. The reaction for the preparation of $\text{3-phenyl-1-propanamine}$ from $\text{3-phenylpropanal}$ is shown below.

$\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CHO}\stackrel{{\text{NH}}_{\text{3}}}{\to }{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH=NH}\\ {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH=NH}\stackrel{{\text{NaBH}}_{\text{3}}\text{CN}}{\to }{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\end{array}$

Conclusion

The conversion of $\text{3-phenylpropanal}$ into $\text{3-phenyl-1-propanamine}$ occurs by reductive animation using reagents such as ammonia and sodium cyanoborohydride.