Chapter 7, Problem 7.57P

Consider the following ${\text{S}}_{\text{N}}\text{2}$ reaction.

a. Draw a mechanism using curved arrows.

b. Draw an energy diagram. Label the axes, the reactants, products, $E_{\text{a}}$, and $\Delta H^{\text{ο}}$. Assume that the reaction is exothermic.

c. Draw the structure of the transition state.

d. What is the rate equation?

e. What happens to the reaction rate in each of the following instances? [1] The leaving group is changed from ${\text{Br}}^{-}$ to ${\text{I}}^{-}$; [2] The solvent is changed from acetone to ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$; [3] The alkyl halide is changed from ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}\text{Br}$ to ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}\left(\text{Br}\right){\text{CH}}_{\text{3}}$; [4] The concentration of ${}^{-}\text{CN}$ is increased by a factor of five; and [5] The concentrations of both the alkyl halide and ${}^{-}\text{CN}$ are increased by a factor of five.

Interpretation Introduction

(a)

Interpretation: The mechanism of the given reaction is to be drawn by the use of curved arrows.

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 7.57P

The mechanism of the given reaction is,

Explanation of Solution

The structure of the given alkyl halide shows that carbon atom, on which bromine is present, is bonded to one another carbon atom. Hence, the bromine atom is bonded to primary carbon atom and the given alkyl halide is $1°$. The primary alkyl halide is most likely to undergo nucleophilic substitution reaction through ${\text{S}}_{\text{N}}\text{2}$ mechanism.

In ${\text{S}}_{\text{N}}\text{2}$ mechanism, the removal of halide and the attack of nucleophile take place simultaneously in the transition state. It is a one step reaction. The mechanism of the given reaction is shown in Figure 1.

Figure 1

Conclusion

The mechanism of the given reaction is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation: The energy diagram is to be drawn. The axes, reactants, products, $E_{\text{a}}$ and $\Delta H\text{°}$ is to be labeled.

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 graphical representation of chemical reaction in which x-axis represents energy of the reaction and y-axis represents the reaction coordinate is called energy profile diagram.

Answer to Problem 7.57P

The energy diagram of the given reaction is shown below.

Explanation of Solution

In bimolecular nucleophilic substitution reaction, formation and breakage of bond takes place simultaneously in the transition state. In the given reaction, breakage of $\text{C}-\text{Br}$ bond and formation of $\text{C}-\text{CN}$ bond takes place.

The energy of products is lower than the energy of reactants in exothermic reaction. Therefore, the energy diagram of the given reaction is,

Figure 2

The x-axes of the energy diagram represent the energy of reactants and products and y-axes represent the reaction coordinate.

Conclusion

The energy diagram of the given reaction is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation: The structure of the transition state is to be drawn.

Concept introduction: In bimolecular nucleophilic substitution reaction, formation and breakage of bond takes place simultaneously in the transition state. The energy of transition state is more than the energy of reactants and products.

Answer to Problem 7.57P

The structure of the transition state is shown below.

Explanation of Solution

In bimolecular nucleophilic substitution reaction, formation and breakage of bond takes place simultaneously in the transition state. In the given reaction, breakage of $\text{C}-\text{Br}$ bond and formation of $\text{C}-\text{CN}$ bond takes place.

Therefore, the transition state of the given reaction is,

Figure 3

Conclusion

The structure of the transition state is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation: The rate equation of the given reaction is to be predicted.

Concept introduction: The rate of ${\text{S}}_{\text{N}}\text{2}$ reaction depends upon the concentration of reactant (alkyl halide, $\text{RX}$) and incoming nucleophile $\left({\text{Nu}}^{-}\right)$.

The rate equation of ${\text{S}}_{\text{N}}\text{2}$ reaction is expressed as,

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

Answer to Problem 7.57P

The rate equation of the given reaction is, $\text{Rate}=k\left[{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}\text{Br}\right]\left[{}^{-}\text{CN}\right]$.

Explanation of Solution

The rate of ${\text{S}}_{\text{N}}\text{2}$ reaction depends upon the concentration of reactant (alkyl halide, $\text{RX}$) and incoming nucleophile $\left({\text{Nu}}^{-}\right)$.

The rate equation of ${\text{S}}_{\text{N}}\text{2}$ reaction is expressed as,

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

The alkyl halide of given reaction is ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}\text{Br}$ and the nucleophile is ${}^{-}\text{CN}$.

Therefore, the rate equation of given reaction is,

$\text{Rate}=k\left[{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}\text{Br}\right]\left[{}^{-}\text{CN}\right]$

Conclusion

The rate equation of the given reaction is, $\text{Rate}=k\left[{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}\text{Br}\right]\left[{}^{-}\text{CN}\right]$.

Interpretation Introduction

(e)

Interpretation: The change that occurs to the reaction rate in given instances is to be stated.

Concept introduction: The rate of ${\text{S}}_{\text{N}}\text{2}$ reaction depends upon the concentration of reactant (alkyl halide, $\text{RX}$) and incoming nucleophile $\left({\text{Nu}}^{-}\right)$.

The rate of ${\text{S}}_{\text{N}}\text{2}$ reaction is expressed as,

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

Answer to Problem 7.57P

The change that occurs to the reaction rate in given instances is,

[1] The rate of the reaction will increase.

[2] The rate of the reaction will decrease.

[3] The rate of the reaction will decrease.

[4] The rate of the reaction increases by five times.

[5] The rate of the reaction increases by twenty five times.

Explanation of Solution

[1]

The primary halide undergoes nucleophilic substitution by ${\text{S}}_{\text{N}}\text{2}$ mechanism in which transition state is formed. The removal of halide and the attack of nucleophile take place simultaneously in transition state. Iodine is a good leaving group as compared to bromine. The rate of ${\text{S}}_{\text{N}}\text{2}$ reaction increases in the presence of good leaving group. Therefore, if the leaving group of the reaction is changed from ${\text{Br}}^{-}$ to ${\text{I}}^{-}$, the rate of the reaction will increase.

[2]

The polar protic solvent favors ${\text{S}}_{\text{N}}\text{1}$ reaction whereas polar aprotic solvent favors ${\text{S}}_{\text{N}}\text{2}$ reaction. The mechanism of the given reaction is ${\text{S}}_{\text{N}}\text{2}$. Acetone is a polar aprotic solvent whereas ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is a protic solvent. Therefore, the rate of reaction will decrease if the solvent is changed from acetone to ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$.

[3]

The primary halide undergoes nucleophilic substitution by ${\text{S}}_{\text{N}}\text{2}$ mechanism in which transition state is formed. Bromine atom is bonded to primary carbon atom in ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}\text{Br}$ whereas in ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}\left(\text{Br}\right){\text{CH}}_{\text{3}}$, it is bonded to secondary carbon atom. The order of increasing reactivity toward ${\text{S}}_{\text{N}}\text{2}$ reaction is $3°<2°<1°$. Therefore, the rate of reaction will decrease if the alkyl halide is changed from ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}\text{Br}$ to ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}\left(\text{Br}\right){\text{CH}}_{\text{3}}$.

[4]

The rate of ${\text{S}}_{\text{N}}\text{2}$ reaction depends upon the concentration of reactant (alkyl halide, ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}\text{Br}$) and incoming nucleophile $\left({}^{-}\text{CN}\right)$.

The rate of ${\text{S}}_{\text{N}}\text{2}$ reaction is expressed as,

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

According the given statement, the concentration of $\left[{}^{-}\text{CN}\right]$ is increased by a factor of five. Hence, the rate of the reaction is,

$\begin{array}{rll}\text{Rate}& =& k\left[{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}\text{Br}\right]\left[5{\times }^{-}\text{CN}\right]\\ & =& 5\left(k\left[{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}\text{Br}\right]\left[{}^{-}\text{CN}\right]\right)\end{array}$

Therefore, the rate of the reaction increases by five times when the concentration of $\left[{}^{-}\text{CN}\right]$ is increased by a factor of five.

[5]

The rate of ${\text{S}}_{\text{N}}\text{2}$ reaction depends upon the concentration of reactant (alkyl halide, ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}\text{Br}$) and incoming nucleophile $\left({}^{-}\text{CN}\right)$.

The rate of ${\text{S}}_{\text{N}}\text{2}$ reaction is expressed as,

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

According the given statement, the concentration of $\left[{}^{-}\text{CN}\right]$ and $\left[\text{RX}\right]$ is increased by a factor of five. Hence, the rate of the reaction is,

$\begin{array}{rll}\text{Rate}& =& k\left[\text{5}\times {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}\text{Br}\right]\left[5{\times }^{-}\text{CN}\right]\\ & =& 25\left(k\left[{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}\text{Br}\right]\left[{}^{-}\text{CN}\right]\right)\end{array}$

Therefore, the rate of the reaction increases by twenty five times when the concentration $\left[{}^{-}\text{CN}\right]$ and $\left[\text{RX}\right]$ is increased by a factor of five.

Conclusion

The change that occurs to the reaction rate in given instances is,

[1] The rate of the reaction will increase.

[2] The rate of the reaction will decrease.

[3] The rate of the reaction will decrease.

[4] The rate of the reaction increases by five times.

[5] The rate of the reaction increases by twenty five times.