Chapter 8, Problem 8.68P

(a) Draw all products formed by treatment of each dibromide (A and B) with one equivalent of ${\text{NaNH}}_{\text{2}}$. (b) Label pairs of diastereomers and constitutional isomers.

                  A                                       B

Interpretation Introduction

(a)

Interpretation: All products formed by the treatment of the given dibromide (A and B) with one equivalent of ${\text{NaNH}}_{\text{2}}$ is to be drawn.

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 one-step bimolecular elimination reaction that favors the removal of a proton by a base from carbon adjacent to the leaving group that results in the formation of a carbocation is termed as $\text{E2}$ elimination reaction. The formation of a double bond takes place simultaneously through the carbocation to form an alkene as the desired product.

Answer to Problem 8.68P

The products that formed by the treatment of the given dibromide A with one equivalent of ${\text{NaNH}}_{\text{2}}$ are (I)-$\left(Z\right)-1-\left(1-\text{bromo}-2-\text{phenylvinyl}\right)-3-\text{methylbenzene}$ and (II)-$\left(Z\right)-1-\left(2-\text{bromo}-2-\text{phenylvinyl}\right)-3-\text{methylbenzene}$. The products that formed by the treatment of the given dibromide B with one equivalent of ${\text{NaNH}}_{\text{2}}$ are (III)-$\left(Z\right)-1-\left(1-\text{bromo}-2-\text{phenylvinyl}\right)-3-\text{methylbenzene}$ and (IV)-$\left(Z\right)-1-\left(2-\text{bromo}-2-\text{phenylvinyl}\right)-3-\text{methylbenzene}$.

Explanation of Solution

The reaction of dibromide A with one equivalent of ${\text{NaNH}}_{\text{2}}$ results in the elimination of one mole of $\text{HBr}$. The corresponding chemical reaction is shown below.

Figure 1

In this reaction, the given dihalide undergoes elimination reaction in the presence of one equivalent of the strong base, ${\text{NaNH}}_{\text{2}}$, that results in the formation of two alkene products, (I)-$\left(Z\right)-1-\left(1-\text{bromo}-2-\text{phenylvinyl}\right)-3-\text{methylbenzene}$ and (II)-$\left(Z\right)-1-\left(2-\text{bromo}-2-\text{phenylvinyl}\right)-3-\text{methylbenzene}$.

The reaction of dibromide B with one equivalent of ${\text{NaNH}}_{\text{2}}$ results in the elimination of one mole of $\text{HBr}$. The corresponding chemical reaction is shown as,

Figure 2

In this reaction, the given dihalide undergoes elimination reaction in the presence of one equivalent of the strong base, ${\text{NaNH}}_{\text{2}}$, that results in the formation of two alkene products, (III)-$\left(Z\right)-1-\left(1-\text{bromo}-2-\text{phenylvinyl}\right)-3-\text{methylbenzene}$ and (IV)-$\left(Z\right)-1-\left(2-\text{bromo}-2-\text{phenylvinyl}\right)-3-\text{methylbenzene}$.

Conclusion

The products that formed by the treatment of the given dibromide A with one equivalent of ${\text{NaNH}}_{\text{2}}$ are (I)-$\left(Z\right)-1-\left(1-\text{bromo}-2-\text{phenylvinyl}\right)-3-\text{methylbenzene}$ and (II)-$\left(Z\right)-1-\left(2-\text{bromo}-2-\text{phenylvinyl}\right)-3-\text{methylbenzene}$. The products that formed by the treatment of the given dibromide B with one equivalent of ${\text{NaNH}}_{\text{2}}$ are (III)-$\left(Z\right)-1-\left(1-\text{bromo}-2-\text{phenylvinyl}\right)-3-\text{methylbenzene}$ and (IV)-$\left(Z\right)-1-\left(2-\text{bromo}-2-\text{phenylvinyl}\right)-3-\text{methylbenzene}$.

Interpretation Introduction

(b)

Interpretation: The labeling of the pairs of diastereomers and constitutional isomers are to be shown.

Concept introduction: Diastereomers compounds are not mirror images of each other and are different in configuration at one or more stereocentres.

The isomers which have same molecular formula but different connectivity of atoms are constitutional isomers.

Answer to Problem 8.68P

The labeling of the pairs of diastereomers is,

The labeling of the pairs of constitutional isomers is,

Explanation of Solution

The first pair of diastereomers is shown as,

Figure 3

Thus, the products (I) and (III) are diastereomers that are not mirror images of each other and are differ in configuration.

The second pair of diastereomers is shown as,

Figure 4

Thus, the products (II) and (IV) are diastereomers that are not mirror images of each other and are differ in configuration. (I), (II) and (III),(IV) are constitutional isomers as they differ only in arrangement of bonds.

The first pair of constitutional isomers is shown as,

Figure 5

Thus, the products (I) and (II) are constitutional isomers as they possess different arrangement of bonds.

The first pair of constitutional isomers is shown as,

Figure 6

Thus, the products (III) and (IV) are constitutional isomers as they possess different arrangement of bonds.

Conclusion

The labeling of the pairs of diastereomers is shown in Figure 3 and Figure 4. The labeling of the pairs of constitutional isomers is shown in Figure 5 and Figure 6.