Chapter 2, Problem 2.40P

Interpretation Introduction

(a)

Interpretation: The product of the given proton transfers reaction is to be drawn. The acid and base in the starting material, and the conjugate acid and base in the products are to be stated.

Concept introduction: In proton transfer reaction, acid donate protons to from a conjugate base and a base accepts protons to from a conjugate acid. An acid base reaction is represented as,

$\begin{array}{ccccccc}\text{HA}& +& \text{B}& \rightleftharpoons & {\text{BH}}^{+}& +& {\text{A}}^{-}\\ \text{Acid}& & \text{Base}& & \text{Conjufagate acid}& & \text{Conjufagate base}\end{array}$

Answer to Problem 2.40P

The products of the given proton transfer reaction are ${\text{CH}}_{\text{3}}{\text{O}}^{-}$ and ${\text{NH}}_3$. The acid and base in the starting material are ${\text{CH}}_{\text{3}}\text{OH}$ and $\text{N}{\text{H}}_{\text{2}}$ respectively. the conjugate acid and base in the products are ${\text{NH}}_3$ and ${\text{CH}}_{\text{3}}{\text{O}}^{-}$ respectively.

Explanation of Solution

The given starting material ${\text{CH}}_{\text{3}}\text{OH}$ will acts as an acid and gives it proton to the base $\text{N}{\text{H}}_{\text{2}}$ to from ${\text{CH}}_{\text{3}}{\text{O}}^{-}$ and ${\text{NH}}_3$. The reaction that takes place between ${\text{CH}}_{\text{3}}\text{OH}$ and $\text{N}{\text{H}}_{\text{2}}$ is represented as,

$\begin{array}{ccccccc}{\text{CH}}_{\text{3}}\text{OH}& +& \text{N}{\text{H}}_{\text{2}}& \rightleftharpoons & {\text{NH}}_3& +& {\text{CH}}_{\text{3}}{\text{O}}^{-}\\ \text{Acid}& & \text{Base}& & \text{Conjufagate acid}& & \text{Conjufagate base}\end{array}$

Conclusion

The products of the given proton transfer reaction are ${\text{CH}}_{\text{3}}{\text{O}}^{-}$ and ${\text{NH}}_3$. The acid and base in the starting material are ${\text{CH}}_{\text{3}}\text{OH}$ and $\text{N}{\text{H}}_{\text{2}}$ respectively. The conjugate acid and base in the products are ${\text{NH}}_3$ and ${\text{CH}}_{\text{3}}{\text{O}}^{-}$ respectively.

Interpretation Introduction

(b)

Interpretation: The product of the given proton transfers reaction is to be drawn. The acid and base in the starting material, and the conjugate acid and base in the products are to be stated.

Concept introduction: In proton transfer reaction, acid donate protons to from a conjugate base and a base accepts protons to from a conjugate acid. An acid base reaction is represented as,

$\begin{array}{ccccccc}\text{HA}& +& \text{B}& \rightleftharpoons & {\text{BH}}^{+}& +& {\text{A}}^{-}\\ \text{Acid}& & \text{Base}& & \text{Conjufagate acid}& & \text{Conjufagate base}\end{array}$

Answer to Problem 2.40P

The products of the given proton transfer reaction are ${\text{CH}}_{\text{3}}{\text{CH}}_2{\text{COO}}^{-}$ and ${\text{H}}_{\text{3}}\text{COH}$. The acid and base in the starting material are ${\text{CH}}_{\text{3}}{\text{CH}}_2\text{COOH}$ and ${\text{H}}_{\text{3}}{\text{CO}}^{-}$ respectively. The conjugate acid and base in the products are ${\text{H}}_{\text{3}}\text{COH}$ and ${\text{CH}}_{\text{3}}{\text{CH}}_2{\text{COO}}^{-}$ respectively.

Explanation of Solution

The given starting material ${\text{CH}}_{\text{3}}{\text{CH}}_2\text{COOH}$ will acts as an acid and gives it proton to the base ${\text{H}}_{\text{3}}{\text{CO}}^{-}$ to from ${\text{H}}_{\text{3}}\text{COH}$ and ${\text{CH}}_{\text{3}}{\text{CH}}_2{\text{COO}}^{-}$. The reaction that takes place between ${\text{CH}}_{\text{3}}{\text{CH}}_2\text{COOH}$ and ${\text{H}}_{\text{3}}{\text{CO}}^{-}$ is represented as,

Figure 1

Conclusion

The products of the given proton transfer reaction are ${\text{CH}}_{\text{3}}{\text{CH}}_2{\text{COO}}^{-}$ and ${\text{H}}_{\text{3}}\text{COH}$. The acid and base in the starting material are ${\text{CH}}_{\text{3}}{\text{CH}}_2\text{COOH}$ and ${\text{H}}_{\text{3}}{\text{CO}}^{-}$ respectively. The conjugate acid and base in the products are ${\text{H}}_{\text{3}}\text{COH}$ and ${\text{CH}}_{\text{3}}{\text{CH}}_2{\text{COO}}^{-}$ respectively.

Interpretation Introduction

(c)

Interpretation: The product of the given proton transfers reaction is to be drawn. The acid and base in the starting material, and the conjugate acid and base in the products are to be stated.

Concept introduction: In proton transfer reaction, acid donate protons to from a conjugate base and a base accepts protons to from a conjugate acid. An acid base reaction is represented as,

$\begin{array}{ccccccc}\text{HA}& +& \text{B}& \rightleftharpoons & {\text{BH}}^{+}& +& {\text{A}}^{-}\\ \text{Acid}& & \text{Base}& & \text{Conjufagate acid}& & \text{Conjufagate base}\end{array}$

Answer to Problem 2.40P

The products of the given proton transfer reaction are ${\text{CH}}_{\text{3}}{\text{CH}}_2{\text{OH}}_2^{+}$ and ${\text{Br}}^{-}$. The acid and base in the starting material are $\text{HBr}$ and ${\text{CH}}_{\text{3}}{\text{CH}}_2\text{OH}$ respectively. The conjugate acid and base in the products are ${\text{CH}}_{\text{3}}{\text{CH}}_2{\text{OH}}_2^{+}$ and ${\text{Br}}^{-}$ respectively.

Explanation of Solution

The given starting material $\text{HBr}$ will acts as an acid and gives it proton to the base ${\text{CH}}_{\text{3}}{\text{CH}}_2\text{OH}$ to from ${\text{CH}}_{\text{3}}{\text{CH}}_2{\text{OH}}_2^{+}$ and ${\text{Br}}^{-}$. The reaction that takes place between ${\text{CH}}_{\text{3}}{\text{CH}}_2\text{OH}$ and $\text{HBr}$ is represented as,

$\begin{array}{ccccccc}\text{H}-\text{Br}& +& {\text{CH}}_{\text{3}}{\text{CH}}_2-\text{O}-\text{H}& \rightleftharpoons & {\text{CH}}_{\text{3}}{\text{CH}}_2{\text{OH}}_2^{+}& +& {\text{Br}}^{-}\\ \text{Acid}& & \text{Base}& & \text{Conjufagate acid}& & \text{Conjufagate base}\end{array}$

Conclusion

The products of the given proton transfer reaction are ${\text{CH}}_{\text{3}}{\text{CH}}_2{\text{OH}}_2^{+}$ and ${\text{Br}}^{-}$. The acid and base in the starting material are $\text{HBr}$ and ${\text{CH}}_{\text{3}}{\text{CH}}_2\text{OH}$ respectively. The conjugate acid and base in the products are ${\text{CH}}_{\text{3}}{\text{CH}}_2{\text{OH}}_2^{+}$ and ${\text{Br}}^{-}$ respectively.

Interpretation Introduction

(d)

Interpretation: The product of the given proton transfers reaction is to be drawn. The acid and base in the starting material, and the conjugate acid and base in the products are to be stated.

Concept introduction: In proton transfer reaction, acid donate protons to from a conjugate base and a base accepts protons to from a conjugate acid. A acid base reaction is represented as,

$\begin{array}{ccccccc}\text{HA}& +& \text{B}& \rightleftharpoons & {\text{BH}}^{+}& +& {\text{A}}^{-}\\ \text{Acid}& & \text{Base}& & \text{Conjufagate acid}& & \text{Conjufagate base}\end{array}$

Answer to Problem 2.40P

The products of the given proton transfer reaction are ${\text{CH}}_{\text{3}}\text{C}\equiv \text{CH}$ and ${\text{OH}}^{-}$. The acid and base in the starting material are ${\text{H}}_{\text{2}}\text{O}$ and ${\text{CH}}_{\text{3}}\text{C}\equiv {\text{C}}^{-}$ respectively. The conjugate acid and base in the products are ${\text{CH}}_{\text{3}}\text{C}\equiv \text{CH}$ and ${\text{OH}}^{-}$ respectively.

Explanation of Solution

The given starting material ${\text{H}}_{\text{2}}\text{O}$ will acts as an acid and gives it proton to the base ${\text{CH}}_{\text{3}}\text{C}\equiv {\text{C}}^{-}$ to from ${\text{CH}}_{\text{3}}\text{C}\equiv \text{CH}$ and ${\text{OH}}^{-}$. The reaction that takes place between ${\text{H}}_{\text{2}}\text{O}$ and ${\text{CH}}_{\text{3}}\text{C}\equiv {\text{C}}^{-}$ is represented as,

$\begin{array}{ccccccc}{\text{H}}_{\text{2}}\text{O}& +& {\text{CH}}_{\text{3}}\text{C}\equiv {\text{C}}^{-}& \rightleftharpoons & {\text{CH}}_{\text{3}}\text{C}\equiv \text{CH}& +& {\text{OH}}^{-}\\ \text{Acid}& & \text{Base}& & \text{Conjufagate acid}& & \text{Conjufagate base}\end{array}$

Conclusion

The products of the given proton transfer reaction are ${\text{CH}}_{\text{3}}\text{C}\equiv \text{CH}$ and ${\text{OH}}^{-}$. The acid and base in the starting material are ${\text{H}}_{\text{2}}\text{O}$ and ${\text{CH}}_{\text{3}}\text{C}\equiv {\text{C}}^{-}$ respectively. The conjugate acid and base in the products are ${\text{CH}}_{\text{3}}\text{C}\equiv \text{CH}$ and ${\text{OH}}^{-}$ respectively.