Chapter 9, Problem 9.72P

Interpretation Introduction

(a)

Interpretation:

It is to be determined whether the given ${\text{S}}_{\text{N}}\text{2}$ reaction is reversible or irreversible.

Concept introduction:

Competing reactions can take place in kinetic or thermodynamic control. Reactions that tend to take place under the kinetic control are the ones in which the major product is the one that forms the fastest.

Reactions that tend to take place under thermodynamic control are the ones in which a more stable but not necessarily the major product is formed.

Reversible reactions tend to take place under thermodynamic control, while irreversible reactions tend to take place under kinetic control.

Reactions like ${\text{S}}_{\text{N}}{\text{1, S}}_{\text{N}}\text{2, E1, or }\text{}\text{E2}$ usually take place under kinetic control because these reactions are irreversible.

The charge stability decides if the products are more stable than the reactants or vice versa. A reaction is irreversible if it’s ${\text{ΔG}}^{\text{0}}{}_{\text{rxn}}$ is negative i.e. if the products are much more stable than the reactants. Otherwise, the reaction tends to be reversible.

Answer to Problem 9.72P

The given reaction is irreversible.

Explanation of Solution

The given reaction is:

It is mentioned that the reaction would follow the ${\text{S}}_{\text{N}}\text{1}$ mechanism pathway. The products for the above reaction are shown as:

The species ${\text{CH}}_{\text{3}}{\text{S}}^{\text{-}}$ is a better nucleophile than ${\text{Br}}^{-}$, and bromine is a good leaving group than ${\text{CH}}_{\text{3}}\text{S-}$. The charge stability decides if the products are more stable than the reactants or vice versa. The charged species on the left side is ${\text{CH}}_{\text{3}}{\text{S}}^{\text{-}}$ while that on the product side is ${\text{Br}}^{-}$. Out of the two, because Br is a larger atom and the charge is less concentrated. This makes ${\text{ΔG}}^{\text{0}}{}_{\text{rxn}}$ for the forward reaction substantially negative. Therefore, ${\text{ΔG}}^{\text{0}}{}_{\text{reverse}}$ would be very large, so the reverse reaction is extremely slow. Thus, the reaction is irreversible.

As the reaction is irreversible it would take place under kinetic control.

Conclusion

The charge stability decides if the products are more stable than the reactants or vice versa.

Interpretation Introduction

(b)

Interpretation:

It is to be determined whether the given ${\text{S}}_{\text{N}}\text{2}$ reaction is reversible or irreversible.

Concept introduction:

Competing reactions can take place in kinetic or thermodynamic control. Reactions that tend to take place under the kinetic control are the ones in which the major product is the one that forms the fastest.

Reactions that tend to take place under thermodynamic control are the ones in which a more stable but not necessarily the major product is formed.

Reversible reactions tend to take place under thermodynamic control, while irreversible reactions tend to take place under kinetic control.

Reactions like ${\text{S}}_{\text{N}}{\text{1, S}}_{\text{N}}\text{2, E1, or }\text{}\text{E2}$ usually take place under kinetic control because these reactions are irreversible.

The charge stability decides if the products are more stable than the reactants or vice versa. A reaction is irreversible if it’s ${\text{ΔG}}^{\text{0}}{}_{\text{rxn}}$ is substantially negative i.e. if the products are much more stable than the reactants. Otherwise, the reaction tends to be reversible.

Answer to Problem 9.72P

The given reaction is reversible.

Explanation of Solution

The given reaction is:

It is mentioned that the reaction would follow the ${\text{S}}_{\text{N}}\text{1}$ mechanism pathway. The products for the above reaction are shown as:

The charge stability decides if the products are more stable than the reactants or vice versa. The charged species on the left side is ${\text{Br}}^{-}$ while that on the product side is ${\text{Cl}}^{-}$. Out of the two, ${\text{Cl}}^{-}$ is slightly less stable than ${\text{Br}}^{-}$, due to the smaller atomic size of chlorine. Also, ${\text{Br}}^{-}$ is a better leaving group than ${\text{Cl}}^{-}$.

This makes the reaction faster in the reverse direction than in the forward direction under standard conditions. Thus, this reaction is reversible.

Conclusion

The charge stability decides if the products are more stable than the reactants or vice versa.

Interpretation Introduction

(c)

Interpretation:

It is to be determined whether the given ${\text{S}}_{\text{N}}\text{2}$ reaction is reversible or irreversible.

Concept introduction:

Competing reactions can take place in kinetic or thermodynamic control. Reactions that tend to take place under the kinetic control are the ones in which the major product is the one that forms the fastest.

Reactions that tend to take place under thermodynamic control are the ones in which a more stable but not necessarily the major product is formed.

Reversible reactions tend to take place under thermodynamic control, while irreversible reactions tend to take place under kinetic control.

Reactions like ${\text{S}}_{\text{N}}{\text{1, S}}_{\text{N}}\text{2, E1, or }\text{}\text{E2}$ usually take place under kinetic control because these reactions are irreversible.

The charge stability decides if the products are more stable than the reactants or vice versa. A reaction is irreversible if it’s ${\text{ΔG}}^{\text{0}}{}_{\text{rxn}}$ is substantially negative i.e. if the products are much more stable than the reactants. Otherwise, the reaction tends to be reversible.

Answer to Problem 9.72P

The given reaction is irreversible.

Explanation of Solution

The given reaction is:

It is mentioned that the reaction would follow the ${\text{S}}_{\text{N}}\text{1}$ mechanism pathway. The products for the above reaction are shown as:

In the above reaction, ${\text{TsO}}^{-}$ is a good leaving group than ${\text{CN}}^{-}$ because in ${\text{TsO}}^{-}$, the negative charge on the oxygen atom is stabilized via delocalization of electrons more than in ${\text{CN}}^{-}$. Also, ${\text{CN}}^{-}$ is a good nucleophile. Thus, these reaction conditions would highly favor the forward reaction.

The charge stability decides if the products are more stable than the reactants or vice versa. The charged species on the left side is ${\text{CN}}^{-}$ while that on the product side is ${\text{TsO}}^{-}$. Out of the two, ${\text{CN}}^{-}$ is less stable than ${\text{TsO}}^{-}$. This makes the reaction faster in the forward direction than in the reverse direction under standard conditions.

Thus, ${\text{ΔG}}^{\text{0}}{}_{\text{rxn}}$ for the forward reaction substantially negative. Therefore, ${\text{ΔG}}^{\text{0}}{}_{\text{reverse}}$ would be very large, so the reverse reaction is extremely slow. Thus, the reaction is irreversible.

As the reaction is irreversible it would take place under kinetic control.

Conclusion

The charge stability decides if the products are more stable than the reactants or vice versa.