Chapter 7, Problem 7.26P

What happens to the rate of an ${\text{S}}_{\text{N}}\text{1}$ reaction under each of the following conditions?

a. $\left[\text{RX}\right]$ is tripled, and $\left[{\text{Nu}}^{-}\right]$ stays the same.

b. Both $\left[\text{RX}\right]$ and $\left[{\text{Nu}}^{-}\right]$ are tripled.

c. $\left[\text{RX}\right]$ is halved, and $\left[{\text{Nu}}^{-}\right]$ stays the same.

d. $\left[\text{RX}\right]$ is halved, and $\left[{\text{Nu}}^{-}\right]$ is doubled.

Interpretation Introduction

(a)

Interpretation: The change that occurs to the rate of an ${\text{S}}_{\text{N}}\text{1}$ reaction when $\left[\text{RX}\right]$ is tripled and $\left[{\text{Nu}}^{-}\right]$ stays the same is to be predicted.

Concept introduction: The replacement or substitution of one functional group with another different functional group in any chemical reaction is termed as substitution reaction. The electron rich chemical species that contains negative charge or lone pair of electrons are known as a nucleophile. In a nucleophilic substitution reaction, nucleophile takes the position of leaving group by attacking the electron deficient carbon atom.

Answer to Problem 7.26P

The rate of the reaction increases by three times when the concentration of $\left[\text{RX}\right]$ is tripled.

Explanation of Solution

The rate of ${\text{S}}_{\text{N}}\text{1}$ reaction depends upon the concentration of reactant (alkyl halide, $\text{RX}$) only. The concentration of incoming nucleophile $\left({\text{Nu}}^{-}\right)$ has no effect on the rate of ${\text{S}}_{\text{N}}\text{1}$ reaction.

The rate of ${\text{S}}_{\text{N}}\text{1}$ reaction is expressed as,

$\text{Rate}=k\left[\text{RX}\right]$

According the given statement, the concentration of $\left[\text{RX}\right]$ is tripled. Hence, the rate of the reaction is,

$\begin{array}{rll}\text{Rate}& =& k\left[\text{3RX}\right]\\ & =& 3\left(k\left[\text{RX}\right]\right)\end{array}$

Therefore, the rate of the reaction increases by three times when the concentration of $\left[\text{RX}\right]$ is tripled.

Conclusion

The rate of the reaction increases by three times when the concentration of $\left[\text{RX}\right]$ is tripled.

Interpretation Introduction

(b)

Interpretation: The change that occurs to the rate of an ${\text{S}}_{\text{N}}\text{1}$ reaction when both $\left[\text{RX}\right]$ and $\left[{\text{Nu}}^{-}\right]$ are tripled is to be predicted.

Concept introduction: The replacement or substitution of one functional group with another different functional group in any chemical reaction is termed as substitution reaction. The electron rich chemical species that contains negative charge or lone pair of electrons are known as a nucleophile. In a nucleophilic substitution reaction, nucleophile takes the position of leaving group by attacking the electron deficient carbon atom.

Answer to Problem 7.26P

The rate of the reaction increases by three times when the concentration of both $\left[\text{RX}\right]$ and $\left[{\text{Nu}}^{-}\right]$ is tripled.

Explanation of Solution

The rate of ${\text{S}}_{\text{N}}\text{1}$ reaction depends upon the concentration of reactant (alkyl halide, $\text{RX}$) only. The concentration of incoming nucleophile $\left({\text{Nu}}^{-}\right)$ has no effect on the rate of ${\text{S}}_{\text{N}}\text{1}$ reaction.

The rate of ${\text{S}}_{\text{N}}\text{1}$ reaction is expressed as,

$\text{Rate}=k\left[\text{RX}\right]$

According the given statement, the concentration of both $\left[\text{RX}\right]$ and $\left[{\text{Nu}}^{-}\right]$ is tripled. Hence, the rate of the reaction is,

$\begin{array}{rll}\text{Rate}& =& k\left[\text{3RX}\right]\\ & =& 3\left(k\left[\text{RX}\right]\right)\end{array}$

Therefore, the rate of the reaction increases by three times when the concentration of both $\left[\text{RX}\right]$ and $\left[{\text{Nu}}^{-}\right]$ is tripled.

Conclusion

The rate of the reaction increases by three times when the concentration of both $\left[\text{RX}\right]$ and $\left[{\text{Nu}}^{-}\right]$ is tripled.

Interpretation Introduction

(c)

Interpretation: The change that occurs to the rate of an ${\text{S}}_{\text{N}}\text{1}$ reaction when $\left[\text{RX}\right]$ is halved and $\left[{\text{Nu}}^{-}\right]$ stays the same is to be predicted.

Concept introduction: The replacement or substitution of one functional group with another different functional group in any chemical reaction is termed as substitution reaction. The electron rich chemical species that contains negative charge or lone pair of electrons are known as a nucleophile. In a nucleophilic substitution reaction, nucleophile takes the position of leaving group by attacking the electron deficient carbon atom.

Answer to Problem 7.26P

The rate of the reaction decreases by half when the concentration of $\left[\text{RX}\right]$ is halved and $\left[{\text{Nu}}^{-}\right]$ stays the same.

Explanation of Solution

The rate of ${\text{S}}_{\text{N}}\text{1}$ reaction depends upon the concentration of reactant (alkyl halide, $\text{RX}$) only. The concentration of incoming nucleophile $\left({\text{Nu}}^{-}\right)$ has no effect on the rate of ${\text{S}}_{\text{N}}\text{1}$ reaction.

The rate of ${\text{S}}_{\text{N}}\text{1}$ reaction is expressed as,

$\text{Rate}=k\left[\text{RX}\right]$

According the given statement, the concentration of both $\left[\text{RX}\right]$ is halved and $\left[{\text{Nu}}^{-}\right]$ stays the same. Hence, the rate of the reaction is,

$\begin{array}{rll}\text{Rate}& =& k\left[\frac{1}{2}\times \text{RX}\right]\\ & =& \frac{1}{2}\left(k\left[\text{RX}\right]\right)\end{array}$

Therefore, the rate of the reaction decreases by half when the concentration of $\left[\text{RX}\right]$ is halved and $\left[{\text{Nu}}^{-}\right]$ stays the same.

Conclusion

The rate of the reaction decreases by half when the concentration of $\left[\text{RX}\right]$ is halved and $\left[{\text{Nu}}^{-}\right]$ stays the same.

Interpretation Introduction

(d)

Interpretation: The change that occurs to the rate of an ${\text{S}}_{\text{N}}\text{1}$ reaction when $\left[\text{RX}\right]$ is halved and $\left[{\text{Nu}}^{-}\right]$ is doubled is to be predicted.

Concept introduction: The replacement or substitution of one functional group with another different functional group in any chemical reaction is termed as substitution reaction. The electron rich chemical species that contains negative charge or lone pair of electrons are known as a nucleophile. In a nucleophilic substitution reaction, nucleophile takes the position of leaving group by attacking the electron deficient carbon atom.

Answer to Problem 7.26P

The rate of the reaction decreases by half when the concentration of $\left[\text{RX}\right]$ is halved and $\left[{\text{Nu}}^{-}\right]$ is doubled.

Explanation of Solution

The rate of ${\text{S}}_{\text{N}}\text{1}$ reaction depends upon the concentration of reactant (alkyl halide, $\text{RX}$) only. The concentration of incoming nucleophile $\left({\text{Nu}}^{-}\right)$ has no effect on the rate of ${\text{S}}_{\text{N}}\text{1}$ reaction.

The rate of ${\text{S}}_{\text{N}}\text{1}$ reaction is expressed as,

$\text{Rate}=k\left[\text{RX}\right]$

According the given statement, the concentration of both $\left[\text{RX}\right]$ is halved and $\left[{\text{Nu}}^{-}\right]$ is doubled. Hence, the rate of the reaction is,

$\begin{array}{rll}\text{Rate}& =& k\left[\frac{1}{2}\text{RX}\right]\\ & =& \frac{1}{2}k\left[\text{RX}\right]\end{array}$

Therefore, The rate of the reaction decreases by half when the concentration of $\left[\text{RX}\right]$ is halved and $\left[{\text{Nu}}^{-}\right]$ is doubled.

Conclusion

The rate of the reaction decreases by half when the concentration of $\left[\text{RX}\right]$ is halved and $\left[{\text{Nu}}^{-}\right]$ is doubled.