Chapter 8, Problem 8.54P

For which reaction mechanisms, ${\text{S}}_{\text{N}}\text{1}$, ${\text{S}}_{\text{N}}2$, $\text{E1}$ or $\text{E2}$-are each of the following statements true? A statement may be true for one or more mechanisms.

a. The mechanism involves carbocation intermediates.

b. The mechanism has two steps.

c. The reaction rate increases with better leaving groups.

d. The reaction rate increases when the solvent is changed from ${\text{CH}}_{\text{3}}\text{OH}$ to ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{SO}$.

e. The reaction rate depends on the concentration of the alkyl halide only.

f. The mechanism is concerted.

g. The reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{Br}$ with $\text{NaOH}$ occurs by this mechanism.

h. Racemization of a stereogenic center occurs.

i. Tertiary $\left({\text{3}}^{\text{o}}\right)$ alkyl halides react faster than ${\text{2}}^{\text{o}}$ or ${\text{1}}^{\text{o}}$ alkyl halides.

j. The reaction follows a second-order rate equation.

Interpretation Introduction

(a)

Interpretation: The validation of the given statement is to be stated according to the mechanism involved in the reaction.

Concept introduction: The two-step unimolecular elimination reaction that favors the removal of a HX substituent and the formation of a carbocation intermediate takes place in its first step. In the second step of the reaction, the carbocation forms a double bond. This type of reaction is termed E1 elimination reaction.

The two-step unimolecular reaction which favors the removal of a HX substituent and the formation of a carbocation intermediate takes place in its first step. Then, in the second step, the carbocation undergoes substitution. This type of reaction is termed as ${\text{S}}_{\text{N}}\text{1}$ reaction.

Answer to Problem 8.54P

The given statement is true for ${\text{S}}_{\text{N}}\text{1}$ and $\text{E1}$ reaction mechanisms.

Explanation of Solution

The two-steps reactions are ${\text{S}}_{\text{N}}\text{1}$ and $\text{E1}$ in which the first step is the formation of carbocation intermediate. In the second step, if the base is strong, negatively charged, then elimination takes place and if the base is weak, then substitution takes place.

Figure 1

Conclusion

${\text{S}}_{\text{N}}1$ and $\text{E1}$ reactions mechanism involves carbocation intermediates.

Interpretation Introduction

(b)

Interpretation: The reaction in which the mechanism has two steps is to be stated.

Concept introduction: The two-step unimolecular elimination reaction that favors the removal of a HX substituent and the formation of a carbocation intermediate takes place in its first step. In the second step of the reaction, the carbocation forms a double bond. This type of reaction is termed $\text{E1}$ elimination reaction.

The two-step unimolecular reaction which favors the removal of a HX substituent and the formation of a carbocation intermediate takes place in its first step. Then, in the second step, the carbocation undergoes substitution. This type of reaction is termed as ${\text{S}}_{\text{N}}\text{1}$ reaction.

Answer to Problem 8.54P

${\text{S}}_{\text{N}}\text{1}$ and $\text{E1}$ reactions have two steps.

Explanation of Solution

The two-steps reactions are ${\text{S}}_{\text{N}}\text{1}$ and $\text{E1}$ in which the first step is the formation of carbocation intermediate. In the second step if the base is strong, negatively charged than elimination takes place and if the base is weak then substitution takes place. If the given base is a good nucleophile then it can undergo nucleophilic substitution reaction or if a strong base then elimination reaction.

Conclusion

${\text{S}}_{\text{N}}1$ and $\text{E1}$ reactions have two steps.

Interpretation Introduction

(c)

Interpretation: The reactions for which the reaction rate increases with better leaving groups is to be stated.

Concept introduction: A good leaving groups is preferred by all substitution and elimination reactions as they increase the rate of the reaction by lowering the transition state energy.

Answer to Problem 8.54P

The rate of ${\text{S}}_{\text{N}}\text{1}$, ${\text{S}}_{\text{N}}\text{2}$, $\text{E1}$ or $\text{E2}$ reactions increases with better leaving groups.

Explanation of Solution

The more the bigger the atom the better it will be as a leaving group and more efficiently it will leave the group to which they are attached and hence increasing the reaction rate. They do so by lowering the transition state energy.

Conclusion

The rate of ${\text{S}}_{\text{N}}1$, ${\text{S}}_{\text{N}}2$, $\text{E1}$ or $\text{E2}$ reactions increases with better leaving groups.

Interpretation Introduction

(d)

Interpretation: The reactions for which the reaction rate increases when the solvent is changed from ${\text{CH}}_{\text{3}}\text{OH}$ to ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{SO}$.is to be stated.

Concept introduction: The choice of the solvent being used determines the mechanism being followed.

Answer to Problem 8.54P

The rate of ${\text{S}}_{\text{N}}\text{2}$ and $\text{E2}$ reactions increases when the solvent is changed from ${\text{CH}}_{\text{3}}\text{OH}$ to ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{SO}$.

Explanation of Solution

${\text{CH}}_{\text{3}}\text{OH}$ is a polar protic solvent while ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{SO}$ is a polar aprotic solvent. The choice of solvent for ${\text{S}}_{\text{N}}\text{2}$ and $\text{E2}$ reactions is a polar aprotic solvent as they stabilize the transition state more efficiently than polar protic solvents.

Conclusion

The rate of ${\text{S}}_{\text{N}}2$ and $\text{E2}$ reactions rate increases when the solvent is changed from ${\text{CH}}_{\text{3}}\text{OH}$ to ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{SO}$.

Interpretation Introduction

(e)

Interpretation: The reactions for which the reaction rate depends on the concentration of the alkyl halide only is to be stated.

Concept introduction: The rate law for ${\text{S}}_{\text{N}}\text{1}$ and $\text{E1}$ reactions is expressed as,

$\text{Rate}=k\left[\text{R-X}\right]$

The rate law depends only on the concentration of alkyl halide.

Answer to Problem 8.54P

For ${\text{S}}_{\text{N}}\text{1}$ and $\text{E1}$ reactions, the reaction rate depends on the concentration of the alkyl halide only.

Explanation of Solution

The two step reactions are ${\text{S}}_{\text{N}}\text{1}$ and $\text{E1}$ in which the first step is the formation of carbocation intermediate. The rate law for both of these reactions is expressed as,

$\text{Rate}=k\left[\text{R-X}\right]$

The rate of both the reactions depends only on the concentration of alkyl halide and is independent of the base concentration.

Conclusion

For ${\text{S}}_{\text{N}}1$ and $\text{E1}$ reactions the reaction rate depends on the concentration of the alkyl halide only.

Interpretation Introduction

(f)

Interpretation: The reaction for which the mechanism is concerted is to be stated.

Concept introduction: The one-step bimolecular elimination reaction that favors the removal of a proton from carbon adjacent to the leaving group by a base that results in the formation of a carbocation. Then, the formation of a double bond takes place simultaneously. This type of reaction is termed $\text{E2}$ elimination reaction.

The nucleophilic reaction that consists of bimolecular as well as bond-making and bond-breaking steps is termed as ${\text{S}}_{\text{N}}\text{2}$ reactions. This is a one-step reaction in which both nucleophile and substrate participate in the rate determining step.

Answer to Problem 8.54P

${\text{S}}_{\text{N}}\text{2}$ and $\text{E2}$ reactions follow concerted mechanism.

Explanation of Solution

In a concerted mechanism, all the bond breaking and bond making takes place simultaneously in a single step. In ${\text{S}}_{\text{N}}\text{2}$ reaction, the cleavage of the carbon-leaving group bond and formation of carbon-nucleophile bond takes place in a single step. In $\text{E2}$ reaction, the cleavage of the carbon-leaving group bond and formation of the double bond takes place in a single step. Hence, both of these reactions follow concerted mechanism.

Conclusion

${\text{S}}_{\text{N}}2$ and $\text{E2}$ reactions follow concerted mechanism.

Interpretation Introduction

(g)

Interpretation: The mechanism followed by the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{Br}$ with $\text{NaOH}$ is to be stated.

Concept introduction: Substitution and elimination depends on the base being employed.

Answer to Problem 8.54P

${\text{S}}_{\text{N}}\text{2}$ and $\text{E2}$ reaction mechanism is followed by the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{Br}$ with $\text{NaOH}$.

Explanation of Solution

In the given question the alkyl halide is primary and the base is a strong base as well as a strong nucleophile. So, substitution as well as elimination is possible. Since, ${\text{S}}_{\text{N}}\text{2}$ reactions favors primary alkyl halide with a strong nucleophile substitution takes place through ${\text{S}}_{\text{N}}\text{2}$ mechanism. Since the alkyl halide is primary, $\text{E2}$ elimination pathway will be followed for elimination reaction by a strong base.

Conclusion

${\text{S}}_{\text{N}}2$ and $\text{E2}$ reaction mechanism is followed by the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{Br}$ with $\text{NaOH}$.

Interpretation Introduction

(h)

Interpretation: The reaction for which racemization of a stereogenic center occurs is to be stated.

Concept introduction: Racemization is possible where a planar carbocation is the intermediate.

Answer to Problem 8.54P

For SN1 reaction, racemization of a stereogenic center occurs.

Explanation of Solution

Racemization can take place where a planar carbocation intermediate is formed and the incoming nucleophile approaches it from both the sides. In a S_N1 reaction, a planar carbocation is formed as the intermediate and attack of the nucleophile is possible from both the sides leading to a mixture of enantiomers being formed.

Conclusion

For ${\text{S}}_{\text{N}}1$ reaction, racemization of a stereogenic center occurs.

Interpretation Introduction

(i)

Interpretation: The reactions for which tertiary $\left({\text{3}}^{\text{o}}\right)$ alkyl halides react faster than ${\text{2}}^{\text{o}}$ or ${\text{1}}^{\text{o}}$ alkyl halides is to be stated.

Concept introduction: A tertiary carbocation is more stable then secondary followed by a primary carbocation.

Answer to Problem 8.54P

For ${\text{S}}_{\text{N}}\text{1}$ $\text{E1}$ and $\text{E2}$ reactions tertiary $\left({\text{3}}^{\text{o}}\right)$ alkyl halides react faster than ${\text{2}}^{\text{o}}$ or ${\text{1}}^{\text{o}}$ alkyl halides.

Explanation of Solution

In S_N1 and E1 reaction, the first step is the formation of carbocation and follows the order $\text{tertiary}>\text{secondary}>\text{primary}$. In an $\text{E2}$ reaction, two $s{p}^3$ carbons are transferred to $s{p}^2$ carbon atoms which leads to substituents moving further apart decreasing any steric interactions. So, substituted compounds undergo $\text{E2}$ reactions rapidly.

Conclusion

For ${\text{S}}_{\text{N}}1$, $\text{E2}$ and $\text{E1}$ reactions tertiary $\left({\text{3}}^{\text{o}}\right)$ alkyl halides react faster than ${\text{2}}^{\text{o}}$ or ${\text{1}}^{\text{o}}$ alkyl halides.

Interpretation Introduction

(j)

Interpretation: The reaction which follows a second-order rate equation is to be stated.

Concept introduction: Second order reaction is one in which the reaction rate depends on the concentration of both the reacting species.

Answer to Problem 8.54P

${\text{S}}_{\text{N}}2$ and $\text{E2}$ reactions follow the second-order rate equation.

Explanation of Solution

In ${\text{S}}_{\text{N}}2$ and $\text{E2}$ reactions, the reaction rate depends on both the alkyl halide and base used. The rate law is expressed as $\mathrm{Rate}=k\left[R-X\right]\left[B\right]$. Second order reaction is one in which the reaction rate depends on the concentration of both the reacting species.

Conclusion

${\text{S}}_{\text{N}}2$ and $\text{E2}$ reactions follow second-order rate equation.