Chapter 9, Problem 9.63P

Interpretation Introduction

(a)

Interpretation:

The complete, detailed mechanism for the reaction is to be provided, and the products are to be predicted including stereochemistry, where appropriate. It is be determined whether the reaction will yield exclusively one product or a mixture of products. If the reaction will yield a mixture of products, then the major product is to be determined.

Concept introduction:

In order to predict the outcome of the given reaction, four factors are considered.

First is to determine if the substrate molecule has a suitable leaving group. If the leaving group is suitable, then evaluate the type of carbon bonded to the leaving group. For an ${\text{S}}_{\text{N}}{\text{1, S}}_{\text{N}}\text{2, E1, or }\text{}\text{E2}$ reaction to take place readily, the carbon atom needs to be ${\text{sp}}^3$ hybridized. If the leaving group is on a primary carbon atom, then ${\text{S}}_{\text{N}}\text{1, and E1}$ reactions are not feasible unless the resulting carbocation is resonance stabilized. If the leaving group is on a tertiary carbon atom, then ${\text{S}}_{\text{N}}\text{2}$ reactions are not feasible. If it is attached to a secondary carbon atom, then all four reactions are equally likely. The next step is to examine the influence of the factors like the strength of the attacking species as a nucleophile and as a base, the concentration of the attacking species, the leaving group ability, and the effect of the solvent. If both substitution and elimination reactions appear to be favored, then the influencing factor is heat.

Answer to Problem 9.63P

The complete, detailed mechanism for the reaction and the products (major) including the stereochemistry are given below:

${\text{S}}_{\text{N}}\text{2}$ pathway:

E2 pathway:

Explanation of Solution

The given reaction is

The structure of $\text{(S)}-\text{3}-\text{chlorooctane}$ is

The substrate has one site for substitution or elimination as chlorine is a moderately good leaving group. The chlorine atom is attached to an ${\text{sp}}^{\text{3}}$ hybridized secondary carbon atom. The reagent and the source of nucleophile is ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OK}$, which dissociates into

the corresponding ions: ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}{\text{ and K}}^{+}$. Here, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$ is a strong nucleophile but a moderately good base, which favors ${\text{S}}_{\text{N}}\text{2, and E2}$ mechanism pathways. However, the high concentration of the nucleophile would favor the ${\text{S}}_{\text{N}}\text{2}$ over the $\text{E2}$ reactions. The solvent used in the above reaction is DMSO, which is a polar aprotic solvent, which favors ${\text{S}}_{\text{N}}\text{2, and E2}$ reactions. Thus, both ${\text{S}}_{\text{N}}\text{2, and E2}$ reactions are equally likely and both of them will occur. The nucleophile will attack the least sterically hindered site from the side opposite the $\text{C-Cl}$ bond. As the nucleophile attacks from the back side of the Cl, which is attached by a wedge bond, the nucleophile would get attached by a dash bond. The stereochemistry is important here, and the ${\text{S}}_{\text{N}}\text{2}$ product that will be formed will have an inversion of configuration, and its stereochemical configuration would be R.

For the E2 product, the base abstracts the alpha hydrogen atom, which results in the formation of a mixture of products. The substrate has two types of alpha hydrogen atoms attached at C2 and C4 carbon atoms. Thus, in all four different products (alkenes) are possible for the $\text{E2}$ reaction. The ${\text{S}}_{\text{N}}\text{2, and E2}$ mechanisms are shown below:

Note the stereochemistry, which is governed by the attack of the nucleophile from the side opposite the leaving group to yield major product in case of ${\text{S}}_{\text{N}}\text{2}$ reaction.

As for the E2 reaction, as there are two different hydrogen atoms attached on either sides of the carbon atom with the leaving group, there could be four stereoisomers for the reaction. All four will be the major products.

Conclusion

The outcome of any given reaction can be predicted by considering the factors like nature of the leaving group, substrate, strength of the reagent used, solvent, and temperature.

Interpretation Introduction

(b)

Interpretation:

The complete, detailed mechanism for the reaction is to be provided and the products are to be predicted including stereochemistry where appropriate. It is be determined whether the reaction will yield exclusively one product or a mixture of products. If the reaction will yield a mixture of products, then the major product is to be determined.

Concept introduction:

In order to predict the outcome of the given reaction, four factors are considered.

First is to determine if the substrate molecule has a suitable leaving group. If the leaving group is suitable, then evaluate the type of carbon bonded to the leaving group. For an ${\text{S}}_{\text{N}}{\text{1, S}}_{\text{N}}\text{2, E1, or }\text{}\text{E2}$ reaction to take place readily, the carbon atom needs to be ${\text{sp}}^3$ hybridized. If the leaving group is on a primary carbon atom, then ${\text{S}}_{\text{N}}\text{1, and E1}$ reactions are not feasible unless the resulting carbocation is resonance stabilized. If the leaving group is on a tertiary carbon atom, then ${\text{S}}_{\text{N}}\text{2}$ reactions are not feasible. If it is attached to a secondary carbon atom, then all four reactions are equally likely. The next step is to examine the influence of the factors like the strength of the attacking species as a nucleophile and as a base, the concentration of the attacking species, the leaving group ability, and the effect of the solvent. If both substitution and elimination reactions appear to be favored, then the influencing factor is heat.

Answer to Problem 9.63P

The complete, detailed mechanism for the reaction and the products (major) including the stereochemistry are given below:

SN1 pathway:

E1 pathway:

Explanation of Solution

The given reaction is

The structure of $\text{(S)}-\text{3}-\text{chlorooctane}$ is

The substrate has one site for substitution or elimination as chlorine is a moderately good leaving group. The chlorine atom is attached to an ${\text{sp}}^{\text{3}}$ hybridized tertiary carbon atom.

As the leaving group is attached to a tertiary carbon atom, ${\text{S}}_{\text{N}}\text{2}$ reactions are not feasible. The reagent and the source of nucleophile is ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OK}$, which dissociates into the corresponding ions: ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}{\text{ and K}}^{+}$ in the solution. Neutral nucleophiles and bases tend to favor the ${\text{S}}_{\text{N}}\text{1, and E1}$ mechanism pathways. The solvent used in the reaction is ethanol, which is a polar protic solvent and would favor ${\text{S}}_{\text{N}}\text{1, and E1}$ mechanism pathways. Considering all the facts, it can be said that the reaction would follow the ${\text{S}}_{\text{N}}\text{1, and E1}$ mechanism pathways.

First step in the ${\text{S}}_{\text{N}}\text{1}$ reaction mechanism is the heterolysis step in which the bond between carbon and chlorine (leaving group) breaks to generate a stable tertiary carbocation.

Step two is the attack of the nucleophile on the carbocation from both the sides to generate mixture of stereoisomers if the substrate has a chiral center and the reaction occurs at the chiral center. The C3 carbon atom where the reaction occurs is a chiral carbon atom. Thus, a mixture of stereoisomers is produced as the racemic mixture; thus, two produces are formed as major products in the ${\text{S}}_{\text{N}}\text{1}$ reaction mechanism pathway. The complete, detailed mechanism is shown below:

For the E1 product, the first step is common, that is, formation of a tertiary carbocation.

In the second step, the hydrogen atom from the adjacent carbon atom of the carbocation is abstracted, resulting into an alkene. The tertiary carbocation formed above has three different adjacent hydrogen atoms.

Elimination usually occurs so as to produce the most stable (the most substituted) alkene as the major product. Thus, abstraction of the green hydrogen atom by the base would result in the least substituted alkene, and therefore, it will be the minor product.

Abstraction of red and blue protons would result in four possible products, all of which are major products.

The complete, detailed mechanism for the E1 reaction is given below:

Note the stereochemistry, there are four major products (trisubstituted alkenes), and one minor product (disubstituted alkene) formed in the above mechanism.

Conclusion

The outcome of any given reaction can be predicted by considering the factors like nature of the leaving group, substrate, strength of the reagent used, solvent, and temperature.

Interpretation Introduction

(c)

Interpretation:

The complete, detailed mechanism for the reaction is to be provided and the products are to be predicted including stereochemistry where appropriate. It is be determined whether the reaction will yield exclusively one product or a mixture of products. If the reaction will yield a mixture of products, then the major product is to be determined.

Concept introduction:

In order to predict the outcome of the given reaction, four factors are considered.

First is to determine if the substrate molecule has a suitable leaving group. If the leaving group is suitable, then evaluate the type of carbon bonded to the leaving group. For an ${\text{S}}_{\text{N}}{\text{1, S}}_{\text{N}}\text{2, E1, or }\text{}\text{E2}$ reaction to take place readily, the carbon atom needs to be ${\text{sp}}^3$ hybridized. If the leaving group is on a primary carbon atom, then ${\text{S}}_{\text{N}}\text{1, and E1}$ reactions are not feasible unless the resulting carbocation is resonance stabilized. If the leaving group is on a tertiary carbon atom, then ${\text{S}}_{\text{N}}\text{2}$ reactions are not feasible. If it is attached to a secondary carbon atom, then all four reactions are equally likely. The next step is to examine the influence of the factors like the strength of the attacking species as a nucleophile and as a base, the concentration of the attacking species, the leaving group ability, and the effect of the solvent. If both substitution and elimination reactions appear to be favored, then the influencing factor is heat.

Answer to Problem 9.63P

The complete, detailed mechanism for the reaction and the products (major) including the stereochemistry are given below:

Explanation of Solution

The given reaction is

The substrate has one site for substitution or elimination; however hydroxyl group is not a good leaving group. The chlorine atom is attached to an ${\text{sp}}^{\text{3}}$ hybridized secondary carbon atom. Thus, all four reaction mechanisms are equally likely.

The reagent used is a phosphoric acid, which is a strong acid. It will protonate the oxygen atom in the hydroxyl group and convert it into a good leaving group. Elimination reactions are generally favored if the reaction conditions includes heat. This rules out both the substitution reactions. And as the hydroxyl group will be protonated by the strong acid, in the next step, it will be removed to generate a resonance stabilized secondary carbocation. Thus, the E1 reaction is highly favored. In the next step, the hydrogen atom adjacent to the carbocation is removed to form an alkene. The most substituted alkene formed would be the major product.

The complete, detailed mechanism is shown below:

Conclusion

The outcome of any given reaction can be predicted by considering the factors like nature of the leaving group, substrate, strength of the reagent used, solvent, and temperature.

Interpretation Introduction

(d)

Interpretation:

The complete, detailed mechanism for the reaction is to be provided and the products are to be predicted including stereochemistry where appropriate. It is be determined whether the reaction will yield exclusively one product or a mixture of products. If the reaction will yield a mixture of products, then the major product is to be determined.

Concept introduction:

In order to predict the outcome of the given reaction, four factors are considered.

First is to determine if the substrate molecule has a suitable leaving group. If the leaving group is suitable, then evaluate the type of carbon bonded to the leaving group. For an ${\text{S}}_{\text{N}}{\text{1, S}}_{\text{N}}\text{2, E1, or }\text{}\text{E2}$ reaction to take place readily, the carbon atom needs to be ${\text{sp}}^3$ hybridized. If the leaving group is on a primary carbon atom, then ${\text{S}}_{\text{N}}\text{1, and E1}$ reactions are not feasible unless the resulting carbocation is resonance stabilized. If the leaving group is on a tertiary carbon atom, then ${\text{S}}_{\text{N}}\text{2}$ reactions are not feasible. If it is attached to a secondary carbon atom, then all four reactions are equally likely. The next step is to examine the influence of the factors like the strength of the attacking species as a nucleophile and as a base, the concentration of the attacking species, the leaving group ability, and the effect of the solvent. If both substitution and elimination reactions appear to be favored, then the influencing factor is heat.

Answer to Problem 9.63P

The complete, detailed mechanism for the reaction and the products (major) including the stereochemistry are given below:

Explanation of Solution

The given reaction is

The substrate has one site for substitution or elimination as chlorine is a moderately good leaving group. The chlorine atom is attached to an ${\text{sp}}^{\text{3}}$ hybridized secondary carbon atom. The reagent and the source of nucleophile is ${{\text{(CH}}_{\text{3}}\text{)}}_3\text{CONa}$, which dissociates into

the corresponding ions: ${{\text{(CH}}_{\text{3}}\text{)}}_3{\text{CO}}^{-}{\text{ and Na}}^{+}$.

Here, ${{\text{(CH}}_{\text{3}}\text{)}}_3{\text{CO}}^{-}$ is a weak nucleophile but a strong base. This favors ${\text{S}}_{\text{N}}\text{2, and E2}$ mechanism pathways. However, as it is a weak nucleophile, an E2 reaction will be favored over ${\text{S}}_{\text{N}}\text{2}$ reaction. The solvent used in the above reaction is DMF, which is a polar aprotic solvent, which favors ${\text{S}}_{\text{N}}\text{2, and E2}$ reactions. Thus, E2 reaction will be favored considering all the reaction conditions.

For the E2 product, the base abstracts the alpha hydrogen atom which would result in the formation of a mixture of products. But since the base is a bulky base, the proton, which is least sterically hindered, will be abstracted to form the major product.

The complete, detailed mechanisms is shown below:

Conclusion

The outcome of any given reaction can be predicted by considering the factors like nature of the leaving group, substrate, strength of the reagent used, solvent, and temperature.

Interpretation Introduction

(e)

Interpretation:

The complete, detailed mechanism for the reaction is to be provided and the products are to be predicted including stereochemistry where appropriate. It is be determined whether the reaction will yield exclusively one product or a mixture of products. If the reaction will yield a mixture of products, then the major product is to be determined.

Concept introduction:

In order to predict the outcome of the given reaction, four factors are considered.

First is to determine if the substrate molecule has a suitable leaving group. If the leaving group is suitable, then evaluate the type of carbon bonded to the leaving group. For an ${\text{S}}_{\text{N}}{\text{1, S}}_{\text{N}}\text{2, E1, or }\text{}\text{E2}$ reaction to take place readily, the carbon atom needs to be ${\text{sp}}^3$ hybridized. If the leaving group is on a primary carbon atom, then ${\text{S}}_{\text{N}}\text{1, and E1}$ reactions are not feasible unless the resulting carbocation is resonance stabilized. If the leaving group is on a tertiary carbon atom, then ${\text{S}}_{\text{N}}\text{2}$ reactions are not feasible. If it is attached to a secondary carbon atom, then all four reactions are equally likely. The next step is to examine the influence of the factors like the strength of the attacking species as a nucleophile and as a base, the concentration of the attacking species, the leaving group ability, and the effect of the solvent. If both substitution and elimination reactions appear to be favored, then the influencing factor is heat.

Answer to Problem 9.63P

The complete, detailed mechanism for the reaction and the products (major) including the stereochemistry are given below:

Explanation of Solution

The given reaction is

The substrate has one site for substitution or elimination as the ${\text{TfO}}^{-}$ is a good leaving group. The ${\text{TfO}}^{-}$ group is attached to an ${\text{sp}}^{\text{3}}$ hybridized primary carbon atom. This rules out the possibility of ${\text{S}}_{\text{N}}\text{1, and E1}$ reactions. The attacking species NaN3 is a strong nucleophile but a weak base. This rules out the possibility of $\text{E2}$ reaction mechanism. Thus, ${\text{S}}_{\text{N}}\text{2}$ reaction will take place predominantly.

The ${\text{S}}_{\text{N}}\text{2}$ reaction is a single step reaction in which the nucleophile attacks the carbon to which the leaving group is attached from the side opposite the leaving group. This results in the inversion of the configuration if the reaction occurs at the chiral center.

The complete, detailed mechanism is shown below:

As the reaction yields a single product, it is the major product.

Conclusion

The outcome of any given reaction can be predicted by considering the factors like nature of the leaving group, substrate, strength of the reagent used, solvent, and temperature.

Interpretation Introduction

(f)

Interpretation:

The complete, detailed mechanism for the reaction is to be provided and the products are to be predicted including stereochemistry where appropriate. It is be determined whether the reaction will yield exclusively one product or a mixture of products. If the reaction will yield a mixture of products, then the major product is to be determined.

Concept introduction:

In order to predict the outcome of the given reaction, four factors are considered.

First is to determine if the substrate molecule has a suitable leaving group. If the leaving group is suitable, then evaluate the type of carbon bonded to the leaving group. For an ${\text{S}}_{\text{N}}{\text{1, S}}_{\text{N}}\text{2, E1, or }\text{}\text{E2}$ reaction to take place readily, the carbon atom needs to be ${\text{sp}}^3$ hybridized. If the leaving group is on a primary carbon atom, then ${\text{S}}_{\text{N}}\text{1, and E1}$ reactions are not feasible unless the resulting carbocation is resonance stabilized. If the leaving group is on a tertiary carbon atom, then ${\text{S}}_{\text{N}}\text{2}$ reactions are not feasible. If it is attached to a secondary carbon atom, then all four reactions are equally likely. The next step is to examine the influence of the factors like the strength of the attacking species as a nucleophile and as a base, the concentration of the attacking species, the leaving group ability, and the effect of the solvent. If both substitution and elimination reactions appear to be favored, then the influencing factor is heat.

Answer to Problem 9.63P

The complete, detailed mechanism for the reaction and the products (major) including the stereochemistry are given below:

Explanation of Solution

The given reaction is

The substrate has one site for substitution or elimination, which is the ${\text{MsO}}^{-}$ group, and it is a good leaving group. The leaving group is attached to an ${\text{sp}}^{\text{3}}$ hybridized primary carbon atom, which is further attached to the benzene ring. Thus, the primary carbon is at the benzylic carbon atom, which would form a resonance stabilized carbocation intermediate.

The attacking species and the solvent for the reaction is the same, that is, ethanol, which serves as a weak nucleophile and a polar protic solvent. This strongly favors ${\text{S}}_{\text{N}}\text{1}$ and E1 reaction pathways. There is no hydrogen atom on the carbon atom adjacent to the carbon atom with the leaving group. This rules out the possibility of E1 and E2 mechanisms. So, overall ${\text{S}}_{\text{N}}\text{1}$ reaction will be favored.

The first step in the ${\text{S}}_{\text{N}}\text{1}$ reaction is heterolysis where a bond between carbon and the ${\text{MsO}}^{-}$ breaks, thus yielding a resonance stabilized carbocation. In the second coordination step, the nucleophile attacks the carbocation, and thus the oxygen atom in ethanol will be attached to the carbocation. In the third step, which is a proton transfer reaction, another ethanol molecule will deprotonate the oxygen atom.

The complete, detailed mechanism is shown below:

Conclusion

The outcome of any given reaction can be predicted by considering the factors like nature of the leaving group, substrate, strength of the reagent used, solvent, and temperature.