Chapter 7, Problem 7.64P

Consider the following ${\text{S}}_{\text{N}}\text{1}$ reaction.

a. Draw a mechanism for this reaction 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 starting material and product are equal in energy.

c. Draw the structure of any transition states.

d. What is the rate equation for this reaction?

e. What happens to the reaction rate in each of the following instances? [1] The leaving group is changed from ${\text{I}}^{-}$ to ${\text{Cl}}^{-}$; [2] The solvent is changed from ${\text{H}}_{\text{2}}\text{O}$ to DMF; [3] The alkyl halide is changed from ${\left({\text{CH}}_{\text{3}}\right)}_2\text{C}\left(\text{I}\right){\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ to ${\left({\text{CH}}_{\text{3}}\right)}_2\text{CHCH}\left(\text{I}\right){\text{CH}}_{\text{3}}$; [4] The concentration of ${\text{H}}_{\text{2}}\text{O}$ is increased by a factor of five; and [5] The concentrations of both the alkyl halide and ${\text{H}}_{\text{2}}\text{O}$ 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 7.64P

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

Explanation of Solution

The structure of the given alkyl halide shows that carbon atom, on which bromine is present, is bonded to three other carbon atoms. Hence, the bromine atom is bonded to tertiary carbon atom and the given alkyl halide is $3°$. The tertiary alkyl halide is most likely to undergo nucleophilic substitution reaction through ${\text{S}}_{\text{N}}\text{1}$ mechanism.

In ${\text{S}}_{\text{N}}\text{1}$ mechanism, the removal of halide leads to the formation of carbocation and the incoming nucleophile then attack on the electron deficient carbocation. 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 $\Delta H^{\text{ο}}$ 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 7.64P

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

Explanation of Solution

In ${\text{S}}_{\text{N}}\text{1}$ mechanism, the removal of halide leads to the formation of carbocation and the incoming nucleophile then attack on the electron deficient carbon atom. In the given reaction, first the breakage of $\text{C}-\text{I}$ bond occurs and then the formation of $\text{C}-\text{O}$ bond takes place.

The energy of products is equal to the energy of reactants. 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 ${\text{S}}_{\text{N}}\text{1}$ mechanism, the removal of halide leads to the formation of carbocation and the incoming nucleophile then attack on the electron deficient carbon atom. Carbocation is the intermediate of reaction that proceeds through ${\text{S}}_{\text{N}}\text{1}$ mechanism.

Answer to Problem 7.64P

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

Explanation of Solution

In ${\text{S}}_{\text{N}}\text{1}$ mechanism, the removal of halide leads to the formation of carbocation and the incoming nucleophile then attack on the electron deficient carbon atom. In the given reaction, first the breakage of $\text{C}-\text{I}$ bond occurs and then the formation of $\text{C}-\text{O}$ bond takes place. The intermediate of the given reaction is stable tertiary carbocation.

Therefore, the transition states of the given reaction are,

Figure 3

Conclusion

The structure of the transition states 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 ${\text{S}}_{\text{N}}\text{1}$ 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,

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

Answer to Problem 7.64P

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

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,

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

The alkyl halide of given reaction is ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{I}\right){\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ and the nucleophile is ${\text{H}}_{\text{2}}\text{O}$.

Therefore, the rate equation of given reaction is,

$\text{Rate}=k\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{I}\right){\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}\right]$

Conclusion

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

Interpretation Introduction

(e)

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

Concept introduction: The rate of ${\text{S}}_{\text{N}}\text{1}$ 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,

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

Answer to Problem 7.64P

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

[1] The rate of the reaction will decrease.

[2] The rate of the reaction will decrease.

[3] The rate of the reaction will decrease.

[4] The rate of the reaction remains unchanged.

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

Explanation of Solution

[1]

The tertiary halide undergoes nucleophilic substitution by ${\text{S}}_{\text{N}}1$ mechanism in which carbocation is formed. The removal of halide leads to the formation of carbocation. Iodine is a good leaving group as compared to chlorine. The rate of ${\text{S}}_{\text{N}}\text{1}$ reaction increases in the presence of good leaving group. Therefore, if the leaving group of the reaction is changed from ${\text{I}}^{-}$ to ${\text{Cl}}^{-}$, the rate of the reaction will decrease.

[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 ${\text{S}}_{\text{N}}\text{2}$. Water is a polar protic solvent whereas DMF is a polar aprotic solvent. Therefore, the rate of reaction will decrease if the solvent is changed from ${\text{H}}_{\text{2}}\text{O}$ to DMF.

[3]

The tertiary halide undergoes nucleophilic substitution by ${\text{S}}_{\text{N}}1$ mechanism in which carbocation is formed. The removal of halide leads to the formation of carbocation. Iodine atom is bonded to tertiary carbon atom in ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{I}\right){\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ whereas in ${\left({\text{CH}}_{\text{3}}\right)}_2\text{CHCH}\left(\text{I}\right){\text{CH}}_{\text{3}}$, it is bonded to secondary carbon atom. The order of increasing reactivity of alkyl halides toward ${\text{S}}_{\text{N}}\text{1}$ reaction is $1°<2°<3°$. Therefore, the rate of reaction will decrease if the alkyl halide is changed from ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{I}\right){\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ to ${\left({\text{CH}}_{\text{3}}\right)}_2\text{CHCH}\left(\text{I}\right){\text{CH}}_{\text{3}}$.

[4]

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 $\left[{\text{H}}_2\text{O}\right]$ is increased by a factor of five. Hence, the rate of the reaction is,

$\text{Rate}=k\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{I}\right){\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}\right]$

Therefore, the rate of the reaction remains unchanged when the concentration of $\left[{\text{H}}_2\text{O}\right]$ is increased by a factor of five.

[5]

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 $\left[{\text{H}}_2\text{O}\right]$ and $\left[\text{RX}\right]$ is increased by a factor of five. Hence, the rate of the reaction is,

$\begin{array}{rll}\text{Rate}& =& k\left[\text{5}\times {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{I}\right){\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}\right]\\ & =& 5\left(k\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}\left(\text{I}\right){\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}\right]\right)\end{array}$

Therefore, the rate of the reaction increases by five times when the concentration $\left[{\text{H}}_2\text{O}\right]$ 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 decrease.

[2] The rate of the reaction will decrease.

[3] The rate of the reaction will decrease.

[4] The rate of the reaction remains unchanged.

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