Chapter 9, Problem 9.66AP

Interpretation Introduction

(a)

Interpretation:

The major product of the reaction between methyl iodide and ethanol solution containing nucleophiles $\left({\text{K}}^{\text{+}}{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{S}}^{-}\right)$ and $\left({\text{Na}}^{\text{+}}{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}\right)$ is to be stated.

Concept introduction:

The nucleophilic substitution reactions are the reactions in which a group is substituted by an incoming nucleophile. These rate of reaction depend upon the nucleophilicity and concentration of the incoming nucleophile.

The ${\text{S}}_{\text{N}}\text{2}$ reaction is a nucleophilic substitution bimolecular reaction in which the addition of incoming nucleophile and elimination of a leaving group takes place simultaneously. The rate of reaction depends upon both the substrate and reactant in the rate law equation.

Answer to Problem 9.66AP

The major product of the ${\text{S}}_{\text{N}}\text{2}$ reaction between methyl iodide and ethanol solution containing $\left({\text{K}}^{\text{+}}{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{S}}^{-}\right)$ and $\left({\text{Na}}^{\text{+}}{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}\right)$ comes from the attack of the nucleophile $\left({\text{K}}^{\text{+}}{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{S}}^{-}\right)$ is shown below.

$\begin{array}{c}{\text{H}}_{\text{3}}\text{C}-{\text{SC}}_{\text{2}}{\text{H}}_{\text{5}}\\ \text{Ethyl}\left(\text{methyl}\right)\text{sulfane}\end{array}$

Explanation of Solution

The ${\text{S}}_{\text{N}}\text{2}$ reaction occurs more rapidly in the polar solvent as polar solvents help in the dissolution of both the substrate and reactant.

In polar protic solvents, the reaction occurs due to the presence of polar hydrogen atom. Therefore, the more basic nucleophiles undergo slower reaction giving minor products in competition with less basic nucleophiles and vice-versa for polar aprotic solvents.

The reaction of methyl iodide with both the nucleophiles undergoes via ${\text{S}}_{\text{N}}\text{2}$ reaction pathway but the product which is obtained in major quantity is with the $\left({\text{K}}^{\text{+}}{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{S}}^{-}\right)$ nucleophile This is because of the difference in the nucleophilicity of both the nucleophile in the ethanol solvent.

$\left({\text{K}}^{\text{+}}{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{S}}^{-}\right)$ is more nucleophilic than the $\left({\text{Na}}^{\text{+}}{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}\right)$ because of the stronger hydrogen bonding of the nucleophile containing oxygen with ethanol than the sulfur one. As a result, more energy is needed to break the hydrogen bond for $\left({\text{Na}}^{\text{+}}{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}\right)$ than $\left({\text{K}}^{\text{+}}{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{S}}^{-}\right)$.

The reaction products are shown below.

Figure 1

Conclusion

The major product of the reaction between methyl iodide and $\left({\text{K}}^{\text{+}}{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{S}}^{-}\right)$, $\left({\text{Na}}^{\text{+}}{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}\right)$ in ethanol solvent is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

The major product of the reaction between methyl iodide and the ethanol solution containing $\left({\text{K}}^{\text{+}}{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{S}}^{-}\right)$ and $\left({\text{Na}}^{\text{+}}{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}\right)$ in $\text{DMSO}$ as a solvent is to be stated.

Concept introduction:

The nucleophilic substitution reactions are the reactions in which a group is substituted by an incoming nucleophile. These rate of reaction depend upon the nucleophilicity and concentration of the incoming nucleophile.

The ${\text{S}}_{\text{N}}\text{2}$ reaction is a nucleophilic substitution bimolecular reaction in which the addition of incoming nucleophile and elimination of a leaving group takes place simultaneously. The rate of reaction depends upon both the substrate and reactant in the rate law equation.

Answer to Problem 9.66AP

The major product of the ${\text{S}}_{\text{N}}\text{2}$ reaction between methyl iodide and DMSO solution containing $\left({\text{K}}^{\text{+}}{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{S}}^{-}\right)$ and $\left({\text{Na}}^{\text{+}}{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}\right)$ comes from the attack of the nucleophile $\left({\text{Na}}^{\text{+}}{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}\right)$ is shown below.

$\begin{array}{c}{\text{H}}_{\text{3}}\text{C}-{\text{OC}}_{\text{2}}{\text{H}}_{\text{5}}\\ \text{Methoxyethane}\end{array}$

Explanation of Solution

The ${\text{S}}_{\text{N}}\text{2}$ reaction occurs more rapidly in the polar solvent as polar solvents help in the dissolution of both the substrate and reactant.

In polar protic solvents there undergoes due to the presence of polar hydrogen atom. Therefore, the more basic nucleophiles undergo slower reaction giving minor products in competition with less basic nucleophiles and vice-versa for polar aprotic solvents.

The reaction of methyl iodide with both the nucleophiles undergoes via ${\text{S}}_{\text{N}}\text{2}$ reaction pathway but the product which is obtained in major quantity is with the $\left({\text{Na}}^{\text{+}}{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}\right)$ nucleophile. This is because of the difference in the nucleophilicity of both the nucleophile in the $\text{DMSO}$ solvent.

DMSO is a polar aprotic solvent. Sodium ethoxide $\left({\text{Na}}^{\text{+}}{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}\right)$ is more nucleophilic than $\left({\text{K}}^{\text{+}}{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{S}}^{-}\right)$ in a polar aprotic solvent. There is no such hydrogen bonding undergoes which lowers down the nucleophilicity of $\left({\text{Na}}^{\text{+}}{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}\right)$ than $\left({\text{K}}^{\text{+}}{\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{S}}^{-}\right)$. In general, more the basicity more is the nucleophilicity of the atom or group.

The reaction products are shown below.

Figure 2

Conclusion

The major product obtained when $\text{DMSO}$ is taken instead of ethanol as solvent is shown in the Figure 2.