Chapter 20, Problem 20.31AP

Interpretation Introduction

(a)

Interpretation:

The product which is formed by the reaction of $0.01\text{ mole}$ of ${\text{Na}}^{+}{\text{CH}}_{\text{3}}{\text{O}}^{-}$ with $0.01\text{ mole}$ of acetic acid and the corresponding curved-arrow notation are to be stated.

Concept introduction:

The curved-arrow notation is used to show the transfer of electrons from one atom to another. The curved arrow has two barbs (head and tail) which represent the direction of electron flow.

Answer to Problem 20.31AP

The curved arrow notation and product for the reaction of $0.01\text{ mole}$ of ${\text{Na}}^{+}{\text{CH}}_{\text{3}}{\text{O}}^{-}$ with $0.01\text{ mole}$ of acetic acid are shown below.

Explanation of Solution

The treatment of sodium methoxide with a carboxylic acid produces an ester. The steps for the formation of the product of the reaction between $0.01\text{ mole}$ of ${\text{Na}}^{+}{\text{CH}}_{\text{3}}{\text{O}}^{-}$ and $0.01\text{ mole}$ of acetic acid are as follows.

In the first step, the electron pairs of oxygen present in methoxide ion attacks the acyl carbon atom of acetic acid and bonding pair of electrons present in the pi bond of acyl group shifts towards oxygen atom simultaneously. Due this an intermediate is formed as shown below.

Figure 1

In the second step, the regeneration of the acyl group takes place by the movement of electron pair of negative charged oxygen atom and also the removal of hydroxide ion takes place from the intermediate simultaneously. Hydroxide ion gets removed by grabbing the electron pair of $\text{C}-\text{O}$ bond which results in the formation of an ester, methyl acetate as shown below.

Figure 2

Therefore, methyl acetate is desired product formed by the reaction of $0.01\text{ mole}$ of $$ ${\text{Na}}^{+}{\text{CH}}_{\text{3}}{\text{O}}^{-}$ with $0.01\text{ mole}$ of acetic acid.

Conclusion

The curved arrow notation and product for the reaction of $0.01\text{ mole}$ of ${\text{Na}}^{+}{\text{CH}}_{\text{3}}{\text{O}}^{-}$ with $0.01\text{ mole}$ of acetic acid are shown in Figure 1 and Figure 2.

Interpretation Introduction

(b)

Interpretation:

The product which is formed by the reaction of $0.01\text{ mole}$ of ${\text{Cs}}^{+}{}^{-}\text{OH}$ with $0.01\text{ mole}$ of acetic acid and the corresponding curved-arrow notation are to be stated.

Concept introduction:

The curved-arrow notation is used to show the transfer of electrons from one atom to another. The curved arrow has two barbs (head and tail) which represent the direction of electron flow.

Answer to Problem 20.31AP

The curved arrow notation and product for the reaction of $0.01\text{ mole}$ of ${\text{Cs}}^{+}{}^{-}\text{OH}$ with of acetic acid are shown below.

Explanation of Solution

The treatment of a base like ${\text{Cs}}^{+}{}^{-}\text{OH}$ with acetic acid produces a cesium salt of acetic acid. It is a neutralization acid-base reaction. In the first step of the reaction between the $0.01\text{ mole}$ of ${\text{Cs}}^{+}{}^{-}\text{OH}$ and $0.01\text{ mole}$ of acetic acid, hydroxide ion abstracts the proton of acetic acid and bonding pair of electrons present in the pi bond of acyl group shifts towards oxygen atom simultaneously. This reaction results in the formation of cesium acetate and water molecule as shown below.

Figure 3

Therefore, cesium acetate is desired product formed by the reaction of of ${\text{Cs}}^{+}{}^{-}\text{OH}$ with $0.01\text{ mole}$ of acetic acid.

Conclusion

The curved arrow notation and product for the reaction of $0.01\text{ mole}$ of ${\text{Cs}}^{+}{}^{-}\text{OH}$ with $0.01\text{ mole}$ of acetic acid are shown in Figure 3.

Interpretation Introduction

(c)

Interpretation:

The product which is formed by the reaction of $0.01\text{ mole}$ of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}-\text{MgBr}$ with of acetic acid and the corresponding curved-arrow notation are to be stated.

Concept introduction:

The curved-arrow notation is used to show the transfer of electrons from one atom to another. The curved arrow has two barbs (head and tail) which represent the direction of electron flow.

Answer to Problem 20.31AP

The curved arrow notation and product for the reaction of $0.01\text{ mole}$ of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}-\text{MgBr}$ with $0.01\text{ mole}$ of acetic acid are shown below.

Explanation of Solution

The treatment of Grignard reagent with a carboxylic acid results into a nucleophilic substitution reaction. The steps for the formation of the product of the reaction between $0.01\text{ mole}$ of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}-\text{MgBr}$ and $0.01\text{ mole}$ of acetic acid are as follows.

In the first step, the bonding electron pairs of ${\text{CH}}_{\text{3}}{\text{-CH}}_{\text{2}}$ present in ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}-\text{MgBr}$ attacks the acyl carbon atom of acetic acid and bonding pair of electrons present in the pi bond of acyl group shifts towards oxygen atom simultaneously. Due this an intermediate is formed as shown below.

Figure 4

In the second step, the regeneration of the acyl group takes place by the movement of electron pair of negative charged oxygen atom and also the removal of hydroxide ion takes place from the intermediate simultaneously. Hydroxide ion gets removed by grabbing the electron pair of $\text{C}-\text{O}$ bond which results in the formation of a ketone, $\text{2-butanone}$ as shown below.

Figure 5

Therefore, $\text{2-butanone}$ is desired product formed by the reaction of of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}-\text{MgBr}$ with $0.01\text{ mole}$ of acetic acid.

Conclusion

The curved arrow notation and product for the reaction of $0.01\text{ mole}$ of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}-\text{MgBr}$ with $0.01\text{ mole}$ of acetic acid are shown in Figure 4 and Figure 5.

Interpretation Introduction

(d)

Interpretation:

The product which is formed by the reaction of $0.01\text{ mole}$ of ${\text{H}}_{\text{3}}\text{C}-\text{Li}$ with $0.01\text{ mole}$ of acetic acid and the corresponding curved-arrow notation are to be stated.

Concept introduction:

The curved-arrow notation is used to show the transfer of electrons from one atom to another. The curved arrow has two barbs (head and tail) which represent the direction of electron flow.

Answer to Problem 20.31AP

The curved arrow notation and product for the reaction of $0.01\text{ mole}$ of ${\text{H}}_{\text{3}}\text{C}-\text{Li}$ with $0.01\text{ mole}$ of acetic acid are shown below.

Explanation of Solution

The treatment of a base like ${\text{H}}_{\text{3}}\text{C}-\text{Li}$ with acetic acid produces a lithium salt of acetic acid. In the first step of the reaction between $0.01\text{ mole}$ of ${\text{H}}_{\text{3}}\text{C}-\text{Li}$ and $0.01\text{ mole}$ of acetic acid, bonding electron pairs of ${\text{H}}_{\text{3}}\text{C}-\text{Li}$ abstracts the proton of acetic acid. This reaction results in the formation of lithium acetate and methane as shown below.

Figure 6

Therefore, lithium acetate is desired product formed by the reaction of $0.01\text{ mole}$ of ${\text{H}}_{\text{3}}\text{C}-\text{Li}$ with $0.01\text{ mole}$ of acetic acid.

Conclusion

The curved arrow notation and product for the reaction of $0.01\text{ mole}$ of ${\text{H}}_{\text{3}}\text{C}-\text{Li}$ with $0.01\text{ mole}$ of acetic acid are shown in Figure 6.

Interpretation Introduction

(e)

Interpretation:

The product which is formed by the reaction of $0.01\text{ mole}$ of $:{\text{NH}}_{\text{3}}$ with $0.01\text{ mole}$ of acetic acid and the corresponding curved-arrow notation are to be stated.

Concept introduction:

The curved-arrow notation is used to show the transfer of electrons from one atom to another. The curved arrow has two barbs (head and tail) which represent the direction of electron flow.

Answer to Problem 20.31AP

The curved arrow notation and product for the reaction of $0.01\text{ mole}$ of $:{\text{NH}}_{\text{3}}$ with $0.01\text{ mole}$ of acetic acid are shown below.

Explanation of Solution

The treatment of ammonia with a carboxylic acid results into a nucleophilic acyl substitution reaction. The steps for the formation of the product of the reaction between $0.01\text{ mole}$ of $:{\text{NH}}_{\text{3}}$ and $0.01\text{ mole}$ of acetic acid are as follows.

In the first step, the lone of pair of electrons of $:{\text{NH}}_{\text{3}}$ attacks the acyl carbon atom of acetic acid and bonding pair of electrons present in the pi bond of acyl group shifts towards oxygen atom simultaneously. Due this an intermediate is formed as shown below.

Figure 7

In the second step, electron pairs of oxygen in $-\text{OH}$ group present in the intermediate attacks the proton present in the positively charged nitrogen atom to form other intermediate as shown below.

Figure 8

In the third step, the regeneration of the acyl group takes place by the movement of electron pair of negative charged oxygen atom and also the removal of hydroxide ion takes place from the intermediate simultaneously. Hydroxide ion gets removed by grabbing the electron pair of $\text{C}-\text{O}$ bond which results in the formation of a ketone, $2-\text{acetamide}$ as shown below.

Figure 9

Therefore, $2-\text{acetamide}$ is desired product formed by the reaction of $0.01\text{ mole}$ of $:{\text{NH}}_{\text{3}}$ with $0.01\text{ mole}$ of acetic acid.

Conclusion

The curved arrow notation and product for the reaction of $0.01\text{ mole}$ of $:{\text{NH}}_{\text{3}}$ with $0.01\text{ mole}$ of acetic acid are shown in Figure 7, Figure 8 and Figure 9.

Interpretation Introduction

(f)

Interpretation:

The product which is formed by the reaction of $0.01\text{ mole}$ of $\text{NaH}$ with $0.01\text{ mole}$ of acetic acid and the corresponding curved-arrow notation are to be stated.

Concept introduction:

The curved-arrow notation is used to show the transfer of electrons from one atom to another. The curved arrow has two barbs (head and tail) which represent the direction of electron flow.

Answer to Problem 20.31AP

The curved arrow notation and product for the reaction of $0.01\text{ mole}$ of $\text{NaH}$ with $0.01\text{ mole}$ of acetic acid are shown below.

Explanation of Solution

The treatment of a base like $\text{NaH}$ with acetic acid produces sodium acetate. In the first step of the reaction between $0.01\text{ mole}$ of $\text{NaH}$ and $0.01\text{ mole}$ of acetic acid, bonding electron pairs of $\text{Na-H}$ abstracts the proton of acetic acid. This reaction results in the formation of sodium acetate and hydrogen gas as shown below.

Figure 10

Therefore, sodium acetate is desired product formed by the reaction of $0.01\text{ mole}$ of $\text{NaH}$ with $0.01\text{ mole}$ of acetic acid.

Conclusion

The curved arrow notation and product for the reaction of $0.01\text{ mole}$ of $\text{NaH}$ with $0.01\text{ mole}$ of acetic acid are shown in Figure 10.

Interpretation Introduction

(g)

Interpretation:

The product which is formed by the reaction of $0.01\text{ mole}$ of sodium hydrogencarbonate with $0.01\text{ mole}$ of acetic acid and the corresponding curved-arrow notation are to be stated.

Concept introduction:

The curved-arrow notation is used to show the transfer of electrons from one atom to another. The curved arrow has two barbs (head and tail) which represent the direction of electron flow.

Answer to Problem 20.31AP

The curved arrow notation and product for the reaction of $0.01\text{ mole}$ of sodium hydrogencarbonate with $0.01\text{ mole}$ of acetic acid are shown below.

Explanation of Solution

The given compound is shown below.

Figure 11

The given compound is sodium hydrogencarbonate. The treatment of a base like sodium hydrogencarbonate with acetic acid produces a salt known as sodium acetate. In the first step of the reaction between $0.01\text{ mole}$ of sodium hydrogencarbonate and $0.01\text{ mole}$ of acetic acid, bonding electron pairs of sodium hydrogencarbonate abstracts the proton of acetic acid and negatively charged oxygen present in sodium hydrogencarbonate shifts to produce a pi bond. This reaction results in the formation of sodium acetate, carbon dioxide and water molecule as shown below.

Figure 12

Therefore, sodium acetate is desired product formed by the reaction of $0.01\text{ mole}$ of sodium hydrogencarbonate with $0.01\text{ mole}$ of acetic acid.

Conclusion

The curved arrow notation and product for the reaction of $0.01\text{ mole}$ of sodium hydrogencarbonate with $0.01\text{ mole}$ of acetic acid are shown in Figure 12.