Chapter 8, Problem 8.65P

Interpretation Introduction

(a)

Interpretation:

The mechanism for the given elimination reaction including carbocation rearrangement is to be drawn.

Concept introduction:

The $\text{E1}$ reaction proceeds with the formation of carbocation intermediate by elimination of the leaving group in the first step. This is the rate-determining step; thus, the stability of the carbocation formed increases the rate of reaction. The electron donating group (alkyl substituents) adjacent to the carbocation increases its stability. The order of stability of carbocation is ${\text{CH}}_{\text{3}}\text{}{\text{< 1}}^{\text{o}}{\text{ < 2}}^{\text{o}}{\text{ < 3}}^{\text{o}}$. The carbocation can be rearranged by $\text{1, 2-}$ hydride or $\text{1, 2-}$ methyl shift to form a more stable carbocation.

Answer to Problem 8.65P

The $\text{E1}$ mechanism for the given reaction involving carbocation rearrangement is

Explanation of Solution

The given reaction equation is

In first step, the leaving group $\text{Br}$ detaches from the carbon and forms the secondary carbocation.

The carbocation formed is rearranged by $\text{1, 2-}$ hydride shift and forms the more stable tertiary carbocation.

The water molecule acts as a base and abstracts a proton from the carbon adjacent to the carbocation, forming $\text{C=C}$ bond.

Conclusion

The mechanism for the given elimination reaction is drawn to show the carbocation rearrangement by $\text{1, 2-}$ hydride shift.

Interpretation Introduction

(b)

Interpretation:

The mechanism for the given elimination reaction without carbocation rearrangement is to be drawn.

Concept introduction:

The $\text{E1}$ reaction proceeds with the formation of carbocation intermediate by elimination of the leaving group in the first step. This is the rate-determining step; thus, the stability of the carbocation formed increases the rate of reaction. The electron donating group (alkyl substituents) adjacent to the carbocation increases its stability. The order of stability of carbocation is ${\text{CH}}_{\text{3}}\text{}{\text{< 1}}^{\text{o}}{\text{ < 2}}^{\text{o}}{\text{ < 3}}^{\text{o}}$. The carbocation can be rearranged by $\text{1, 2-}$ hydride or $\text{1, 2-}$ methyl shift to form a more stable carbocation. Sometimes the carbocation formed in the first step, after and before rearrangement, gives the same product. More substituted products are stable and hence form preferably.

Answer to Problem 8.65P

The $\text{E1}$ mechanism for the given reaction without carbocation rearrangement is

Explanation of Solution

The given reaction equation is

In the first step, the leaving group $\text{Br}$ detaches from the carbon and forms secondary carbocation.

In the second step, without rearrangement, the proton is eliminated by the base ${\text{H}}_{\text{2}}\text{O}$ from the carbon adjacent to the carbocation, forming $\text{C=C}$ bond. The more substituted double bond is more stable and hence is formed as a major product.

Conclusion

The mechanism for the given elimination reaction is drawn to without rearrangement step, indicating that the same product is formed with or without rearrangement.

Interpretation Introduction

(c)

Interpretation:

It is to be explained how the $\text{C}$ isotope labeling is useful to determine whether the rearrangement occurred in the given $\text{E1}$ reaction.

Concept introduction:

The $\text{E1}$ reaction proceeds with the formation of carbocation intermediate by elimination of the leaving group in the first step. This is the rate-determining step; thus, the stability of the carbocation formed increases the rate of reaction. The electron donating group (alkyl substituents) adjacent to the carbocation increases its stability. The order of stability of carbocation is ${\text{CH}}_{\text{3}}\text{}{\text{< 1}}^{\text{o}}{\text{ < 2}}^{\text{o}}{\text{ < 3}}^{\text{o}}$. The carbocation can be rearranged by $\text{1, 2-}$ hydride or $\text{1, 2-}$ methyl shift to form a more stable carbocation. Sometimes the carbocation formed in the first step, after and before rearrangement, gives the same product. More substituted products are stable and hence form preferably. Whether the reaction proceeds with or without rearrangement can be determined by labeling the carbon as $\text{C}$ where the leaving group is bonded.

Answer to Problem 8.65P

The reaction with carbocation rearrangement gave two products, and the reaction without carbocation rearrangement gave only one product, as shown below, indicating that the $\text{C}$ isotope labeling is useful to determine whether the rearrangement occurred in the given $\text{E1}$ reaction.

Reaction with rearrangement:

Reaction without rearrangement:

Explanation of Solution

The given reaction equation is

If the carbon bonded to the leaving group in the given substrate is labelled as $\text{C}$ and rest carbon atoms are considered as $\text{C}$, then two products are formed when the reaction occurs through carbocation rearrangement. The detailed mechanism is as follows:

In one product, one of the double bonded carbon is $\text{C}$ labelled, and in another product, both the double bonded carbons are $\text{C}$.

If the reaction proceeds without rearrangement, then only one product is formed where one of the double bonded carbon is $\text{C}$. The detailed mechanism is as follows:

As the reaction with rearrangement of carbocation formed two products, and reaction without rearrangement formed only one product, it indicates that the E1 products depend on whether the rearrangement occurred.

Conclusion

It is explained that the E1 products depend on whether the reaction includes carbocation rearrangement occurring with $\text{C}$ isotope labeling.

Interpretation Introduction

(d)

Interpretation:

How the deuterium isotope labeling is useful to determine whether the rearrangement is occurred in given $\text{E1}$ reaction, it is to be explained.

Concept introduction:

The $\text{E1}$ reaction proceeds with the formation of carbocation intermediate by elimination of the leaving group in the first step. This is the rate-determining step; thus, the stability of the carbocation formed increases the rate of reaction. The electron donating group (alkyl substituents) adjacent to the carbocation increases its stability. The order of stability of carbocation is ${\text{CH}}_{\text{3}}\text{}{\text{< 1}}^{\text{o}}{\text{ < 2}}^{\text{o}}{\text{ < 3}}^{\text{o}}$. The carbocation can be rearranged by $\text{1, 2-}$ hydride or $\text{1, 2-}$ methyl shift to form a more stable carbocation. Sometimes the carbocation formed in the first step, after and before rearrangement, gives the same product. The more substituted products are stable and hence form preferably. Whether the reaction proceeds with or without rearrangement can be determined by replacing the hydrogen that proposed to undergo migration.

Answer to Problem 8.65P

The reaction with carbocation rearrangement by migration of deuterium gave two products, and the reaction without carbocation rearrangement gave only one product as shown below, indicating that the deuterium isotope labeling is useful to determine whether the rearrangement occurred in the given $\text{E1}$ reaction.

Reaction with rearrangement:

Reaction without rearrangement:

Explanation of Solution

The given reaction equation is

If the migrating hydrogen in the given substrate is replaced with deuterium, then two products are formed when the reaction occurred through carbocation rearrangement. The detailed mechanism is as follows:

One product is formed by elimination of hydrogen atom and another by elimination of deuterium atom.

If the reaction proceeds without rearrangement, only one product is formed by elimination of deuterium atom. The detailed mechanism is as follows:

As the reaction with rearrangement of carbocation formed two products and reaction without rearrangement formed only one product, it indicates the E1 products depend on whether the rearrangement occurs.

Conclusion

It is explained on the basis of formation of different products that deuterium isotope labeling is useful to determine whether the rearrangement occurred in the given $\text{E1}$ reaction.