Chapter 2, Problem 2.67P

Interpretation Introduction

(a)

Interpretation: The product formed by the reaction of Lewis acid ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{C}}^{+}$ with the given Lewis base is to be drawn.

Concept introduction: Lewis acid accepts an electron pair. It acts as an electrophile as it is electron deficient. Lewis base donates an electron pair. It acts as a nucleophile as it is electron loving.

Answer to Problem 2.67P

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{C}}^{+}$ with ${\text{H}}_{\text{2}}\text{O}$ is ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_{\text{3}}\text{C}\stackrel{+}{\text{O}}{\text{H}}_2$.

Explanation of Solution

In the given reaction, Lewis base ${\text{H}}_{\text{2}}\text{O}$ reacts with Lewis acid ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{C}}^{+}$ . Electron pair present on the oxygen atom of Lewis base is donated to the carbon of Lewis acid. The product formed is shown below.

Figure 1

Conclusion

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{C}}^{+}$ with ${\text{H}}_{\text{2}}\text{O}$ is ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_{\text{3}}\text{C}\stackrel{+}{\text{O}}{\text{H}}_2$.

Interpretation Introduction

(b)

Interpretation: The product formed by the reaction of Lewis acid ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{C}}^{+}$ with the given Lewis base is to be drawn.

Concept introduction: Lewis acid accepts an electron pair. It acts as an electrophile as it is electron deficient. Lewis base donates an electron pair. It acts as a nucleophile as it is electron loving.

Answer to Problem 2.67P

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{C}}^{+}$ with ${\text{CH}}_{\text{3}}\text{<munder><mover accent="true"><mo>O</mo><mo>¨</mo></mover><mo>¨</mo></munder> H}$ is ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_{\text{3}}\text{C}\stackrel{+}{\text{O}}{\text{HCH}}_3$.

Explanation of Solution

In the given reaction, Lewis base ${\text{CH}}_{\text{3}}\text{<munder><mover accent="true"><mo>O</mo><mo>¨</mo></mover><mo>¨</mo></munder> H}$ reacts with Lewis acid ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{C}}^{+}$ . Electron pair present on the oxygen atom of Lewis base is donated to the carbon of Lewis acid. The product formed is shown below.

Figure 2

Conclusion

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{C}}^{+}$ with ${\text{CH}}_{\text{3}}\text{<munder><mover accent="true"><mo>O</mo><mo>¨</mo></mover><mo>¨</mo></munder> H}$ is ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_{\text{3}}\text{C}\stackrel{+}{\text{O}}{\text{HCH}}_3$.

Interpretation Introduction

(c)

Interpretation: The product formed by the reaction of Lewis acid ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{C}}^{+}$ with the given Lewis base is to be drawn.

Concept introduction: Lewis acid accepts an electron pair. It acts as an electrophile as it is electron deficient. Lewis base donates an electron pair. It acts as a nucleophile as it is electron loving.

Answer to Problem 2.67P

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{C}}^{+}$ with ${\text{CH}}_{\text{3}}{\underset{¨}{\ddot{O}}\text{CH}}_3$ is ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_{\text{3}}\text{C}\stackrel{+}{\text{O}}{\left({\text{CH}}_3\right)}_2$.

Explanation of Solution

In the given reaction, Lewis base ${\text{CH}}_{\text{3}}{\underset{¨}{\ddot{O}}\text{CH}}_3$ reacts with Lewis acid ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{C}}^{+}$ .Electron pairs present on the oxygen atom of Lewis base is donated to the carbon of Lewis acid. The product formed is shown below.

Figure 3

Conclusion

The product formed by the reaction of ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{C}}^{+}$ with ${\text{CH}}_{\text{3}}{\underset{¨}{\ddot{O}}\text{CH}}_3$ is ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_{\text{3}}\text{C}\stackrel{+}{\text{O}}{\left({\text{CH}}_3\right)}_2$.