Chapter 23, Problem 23.47AP

Interpretation Introduction

(a)

Interpretation:

The structure of the primary chiral amine with formula ${\text{C}}_{\text{4}}{\text{H}}_{\text{7}}\text{N}$

without any triple bond is to be stated.

Concept introduction:

Amines are the organic compounds that are formed by replacement of hydrogen from ammonia with a substituent. It may be alkyl or aryl group. Amines are basic in nature because the nitrogen can donate its lone pairs and also the ability of the nitrogen to accept the proton in water. Chiral amines are compound which are not superimposable to each other.

Answer to Problem 23.47AP

The structure of the chiral primary amine ${\text{C}}_{\text{4}}{\text{H}}_{\text{7}}\text{N}$ with no triple bonds is shown below.

Explanation of Solution

The given compound has a molecular formula ${\text{C}}_{\text{4}}{\text{H}}_{\text{7}}\text{N}$ with no triple bond. It may have a double bond or maybe a completely saturated compound. In order to find, if there is any double bond present or not. The double bond equivalent formula is applied. The double bond equivalent (DBE) formula is shown below.

$DBE=\frac{2C+2+N-H-X}{2}$ …(1)

Where

• $C$ is the number of the carbon atom.
• $N$ is the number of the nitrogen atom.
• $H$ is the number of the hydrogen atom.
• $X$ is the number of the halogen atom.

Substitute the value of each atom in equation (1). Then the unsaturation number is calculated as shown below.

$\begin{array}{rll}DBE& =& \frac{2C+2+N-H-X}{2}\\ & =& \frac{2\times 4+2+1-7-0}{2}\\ & =& \frac{8+2+1-7}{2}\\ & =& 2\end{array}$

Its unsaturation number is $2$ which means two double bonds or one cyclic ring with one double bond is present. Since it is a chiral primary amine. Then the possible structure of a chiral primary amine ${\text{C}}_{\text{4}}{\text{H}}_{\text{7}}\text{N}$ with no triple bonds is shown below.

Figure 1

Therefore, the primary amine ${\text{C}}_{\text{4}}{\text{H}}_{\text{7}}\text{N}$ with no triple bond is identified as $2-\text{methylcycloprop}-2-\text{enamine}$.

Conclusion

The chiral primary amine ${\text{C}}_{\text{4}}{\text{H}}_{\text{7}}\text{N}$ with no triple bond is $2-\text{methylcycloprop}-2-\text{enamine}$ and its structure is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

The structure of the primary chiral amine with formula ${\text{C}}_{\text{4}}{\text{H}}_{11}\text{N}$ is to be stated.

Concept introduction:

Amines are the organic compounds that are formed by replacement of hydrogen from ammonia with a substituent. It may be alkyl or aryl group. Amines are basic in nature because the nitrogen can donate its lone pairs and also the ability of the nitrogen to accept the proton in water. Chiral amines are compound which are not superimposable to each other.

Answer to Problem 23.47AP

The structure of the chiral primary amine ${\text{C}}_{\text{4}}{\text{H}}_{11}\text{N}$ is shown below.

Explanation of Solution

The given compound has a molecular formula ${\text{C}}_{\text{4}}{\text{H}}_{11}\text{N}$. It may have a double bond or it maybe a completely saturated compound. In order to find, if there is any double bond present or not. The double bond equivalent formula is applied. The formula is shown below.

$DBE=\frac{2C+2+N-H-X}{2}$ …(1)

Where

• $C$ is the number of the carbon atom.
• $N$ is the number of the nitrogen atom.
• $H$ is the number of the hydrogen atom.
• $X$ is the number of the halogen atom.

Substitute the value of each atom in equation (1). Then the unsaturation number is calculated as shown below.

$\begin{array}{rll}DBE& =& \frac{2C+2+N-H-X}{2}\\ & =& \frac{2\times 4+2+1-11-0}{2}\\ & =& \frac{8+2+1-11}{2}\\ & =& 0\end{array}$

Its unsaturation number is $0$ which means the given compound is a saturated compound. Then the possible structure of the chiral primary amine compound with the molecular formula ${\text{C}}_{\text{4}}{\text{H}}_{11}\text{N}$ is shown below.

Figure 2

Therefore, the chiral primary amine ${\text{C}}_{\text{4}}{\text{H}}_{11}\text{N}$ is identified as $\text{butan}-2-\text{amine}$.

Conclusion

The chiral primary amine ${\text{C}}_{\text{4}}{\text{H}}_{11}\text{N}$ is $\text{butan}-2-\text{amine}$ and its structure is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation:

The structure of two secondary amines which gives propene and $N,N-\text{dimethylamine}$ on reaction with ${\text{CH}}_{\text{3}}\text{I}$, then ${\text{Ag}}_{\text{2}}\text{O}$ and heat is to be stated.

Concept introduction:

Amines are the organic compounds that are formed by replacement of hydrogen from ammonia with a substituent. It may be alkyl or aryl group. Hofmann elimination reaction occurs as an anti-elimination reaction. In this reaction, the starting material is quaternary ammonium hydroxide. When quaternary ammonium hydroxide is heated, $\text{β}$ elimination occurs to form an alkene, amines act as a leaving group.

Answer to Problem 23.47AP

The compound $N-\text{isopropylaniline}$ and $N-\text{propylaniline}$ are the two secondary amines which on Hofmann elimination reaction gives propene and $N,N-\text{dimethylamine}$. Their reactions and structure are shown below.

Explanation of Solution

The two secondary amines which under given reaction conditions may undergo Hofmann elimination reaction. In the Hofmann elimination reaction, the starting material is quaternary ammonium hydroxide which is formed by reaction with ${\text{CH}}_{\text{3}}\text{I}$ then ${\text{Ag}}_{\text{2}}\text{O}$. When quaternary ammonium hydroxide is heated, $\text{β}$ elimination occurs to form an alkene, amines act as a leaving group. Then the reaction with two possible secondary amines structures which gives propene and $N,N-\text{dimethylamine}$ is shown below.

Figure 3

Therefore, the two secondary amines which undergo Hofmann elimination reaction are $N-\text{isopropylaniline}$ and $N-\text{propylaniline}$.

Conclusion

The two secondary amines which on Hofmann elimination reaction gives propene and $N,N-\text{dimethylamine}$ are $N-\text{isopropylaniline}$ and $N-\text{propylaniline}$ their structure and reaction are shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

The structure of ${\text{C}}_{\text{4}}{\text{H}}_{\text{9}}\text{N}$ which on reaction with $\text{NaBH}{\left(\text{OAc}\right)}_{\text{3}}$ and $1$ equivalent $\text{HOAc}$, then $\text{KOH}$, to give $N-\text{methyl}-2-\text{propanamine}$ is to be stated.

Concept introduction:

Amines are the organic compounds that are formed by replacement of hydrogen from ammonia with a substituent. It may be alkyl or aryl group. Amines are basic in nature because the nitrogen can donate its lone pairs and also the ability of the nitrogen to accept the proton in water.

Answer to Problem 23.47AP

The given compound is identified as $N-\text{methylenepropan}-2-\text{amine}$ which undergo reductive amination reaction. Its reaction and structure is shown below.

Explanation of Solution

The given compound has a molecular formula ${\text{C}}_{\text{4}}{\text{H}}_{\text{9}}\text{N}$. It may have a double bond or it maybe a completely saturated compound. In order to find, if there is any double bond is present or not. The double bond equivalent formula is applied. The formula is shown below.

$DBE=\frac{2C+2+N-H-X}{2}$ …(1)

Where

• $C$ is the number of the carbon atom.
• $N$ is the number of the nitrogen atom.
• $H$ is the number of the hydrogen atom.
• $X$ is the number of the halogen atom.

Substitute the value of each atom in equation (1). Then the unsaturation number is calculated as shown below.

$\begin{array}{rll}DBE& =& \frac{2C+2+N-H-X}{2}\\ & =& \frac{2\times 4+2+1-9-0}{2}\\ & =& \frac{8+2+1-9}{2}\\ & =& 1\end{array}$

Its unsaturation number is $1$ which means it contains one acyclic double bond. The given compound, ${\text{C}}_{\text{4}}{\text{H}}_{\text{9}}\text{N}$ on reaction with $\text{NaBH}{\left(\text{OAc}\right)}_{\text{3}}$ and $1$ equivalent $\text{HOAc}$, then $\text{KOH}$, to give $N-\text{methyl}-2-\text{propanamine}$. It depicts that the given compound undergoes reductive amination reaction. Reductive amination reaction is the conversion of the carbonyl group to the amine or it converts one amine to newer amine. The reaction and structure of the given compound which undergo reductive amination reaction is shown below.

Figure 4

Therefore, the given compound ${\text{C}}_{\text{4}}{\text{H}}_{\text{9}}\text{N}$ is identified as $N-\text{methylenepropan}-2-\text{amine}$.

Conclusion

The given compound is $N-\text{methylenepropan}-2-\text{amine}$ which undergo reductive amination reaction and gives $N-\text{methylpropan}-2-\text{amine}$ is shown in Figure 4.