Chapter 8, Problem 8.11P

Interpretation Introduction

(a)

Interpretation: The change in the rate of an E2 reaction is to be stated when the concentration of alkyl halide is tripled.

Concept introduction: E2 type of reaction follows second order kinetics in which the rate depends on both the reactants. The rate law equation for E2 reaction is expressed as,

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

Answer to Problem 8.11P

The rate of an E2 reaction is tripled when the concentration of alkyl halide is tripled.

Explanation of Solution

E2type of reaction is second order elimination reaction in which the rate of reaction depends on both the reactants i.e. the alkyl halide and the base.

The rate law equation for E2 reaction is expressed as,

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

According to the given statement, concentration of alkyl halide is tripled. Hence, the rate of reaction is

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

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

Conclusion

The rate of an E2 reaction is tripled when the concentration of alkyl halide is tripled.

Interpretation Introduction

(b)

Interpretation: The change in the rate of an E2 reaction is to be stated when the concentration of base is halved.

Concept introduction: E2 type of reaction follows second order kinetics in which the rate depends on both the reactants. The rate law equation for E2 reaction is expressed as,

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

Answer to Problem 8.11P

The rate of an E2 reaction is halved when the concentration of base is halved.

Explanation of Solution

E2 type of reaction is second order elimination reaction in which the rate of reaction depends on both the reactants i.e. the alkyl halide and the base.

The rate law equation for E2 reaction is expressed as,

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

According to the given statement, concentration of base is halved. Hence, the rate of reaction is,

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

Therefore, the rate of the reaction decreases by half when the concentration of $\left[\text{B:}\right]$ is halved.

Conclusion

The rate of an E2 reaction is halved when the concentration of base is halved.

Interpretation Introduction

(c)

Interpretation: The change in the rate of an E2 reaction is to be stated when the solvent is changed from ${\text{CH}}_{\text{3}}\text{OH}$ to DMSO

Concept introduction: E2 reactions are usually preferred in polar aprotic solvents as the transition state involves charge dispersion which is stabilized by polar aprotic solvents.

Answer to Problem 8.11P

The rate of an E2 reaction increases when the solvent is changed from ${\text{CH}}_{\text{3}}\text{OH}$ to DMSO.

Explanation of Solution

E2 type of reaction is second order elimination reaction in which the rate of reaction depends on both the reactants i.e. the alkyl halide and the base. The choice of solvent for this type of reaction is polar aprotic. DMSO is a polar aprotic solvent and ${\text{CH}}_{\text{3}}\text{OH}$ is a polar protic solvent. So changing the latter with the former increases the rate of the reaction.

Conclusion

The rate of an E2 reaction increases when the solvent is changed from ${\text{CH}}_{\text{3}}\text{OH}$ to DMSO.

Interpretation Introduction

(d)

Interpretation: The change in the rate of an E2 reaction is to be stated when the leaving is changed from ${\text{I}}^{\text{-}}$ to ${\text{Br}}^{\text{-}}$

Concept introduction: A leaving group bigger in size is preferred as it leaves easily and fast while a smaller leaving group leaves with difficulty and slow.

Answer to Problem 8.11P

The rate of an E2 reaction is decreased when the leaving group is changed from ${\text{I}}^{\text{-}}$ to ${\text{Br}}^{\text{-}}$.

Explanation of Solution

E2 type of reaction is second order elimination reaction in which the rate of reaction depends on both the reactants i.e. the alkyl halide and the base. Iodine is bigger in size as compared to bromine. An atom bigger in size is a good leaving group. Thus, iodine is a better leaving group then bromine. Thus, when ${\text{I}}^{\text{-}}$ is replaced by a poor leaving group ${\text{Br}}^{\text{-}}$, the rate of elimination decreases.

Conclusion

The rate of an E2 reaction is decreased when the leaving is changed from ${\text{I}}^{\text{-}}$ to ${\text{Br}}^{\text{-}}$.

Interpretation Introduction

(e)

Interpretation: The change in the rate of an E2 reaction is to be stated when the base is changed from ${\text{OH}}^{\text{-}}$ to ${\text{H}}_{\text{2}}\text{O}$

Concept introduction: Strong base is more efficient in proton abstraction and hence, it is more preferred in E2 type of reactions.

Answer to Problem 8.11P

The rate of an E2 reaction is decreased when the base is changed from ${\text{OH}}^{\text{-}}$ to ${\text{H}}_{\text{2}}\text{O}$.

Explanation of Solution

E2 type of reaction is second order elimination reaction in which the rate of reaction depends on both the reactants i.e. the alkyl halide and the base. A stronger base will facilitate the abstraction of proton and hence will increase the rate of E2 reaction. ${\text{OH}}^{\text{-}}$ is a stronger base than ${\text{H}}_{\text{2}}\text{O}$ and replacing it with ${\text{H}}_{\text{2}}\text{O}$ will decrease the rate of the reaction.

Conclusion

The rate of an E2 reaction is decreased when the base is changed from ${\text{OH}}^{\text{-}}$ to ${\text{H}}_{\text{2}}\text{O}$.

Interpretation Introduction

(f)

Interpretation: The change in the rate of an E2 reaction is to be stated when the the alkyl halide is changed from ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{Br}$ to ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHBr}$.

Concept introduction: The rate of E2 reaction depends on the halide being used. A halide in which carbon attached to the leaving group has more number of alkyl groups is usually preferred.

Answer to Problem 8.11P

The rate of E2 reaction increases when the alkyl halide is changed from ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{Br}$ to ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHBr}$

Explanation of Solution

E2 type of reaction is second order elimination reaction in which the rate of reaction depends on both the reactants i.e. the alkyl halide and the base. A highly substituted product is preferred in E2 reaction as it is more stable. This can be obtained when the carbon attached to the leaving group has more number of alkyl groups. Since ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHBr}$ has more number of alkyl groups, substituting it with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{Br}$ will increase the rate of the reaction.

Conclusion

The rate of E2 reaction increases when the alkyl halide is changed from ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{Br}$ to ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHBr}$