PKG ORGANIC CHEMISTRY
PKG ORGANIC CHEMISTRY
5th Edition
ISBN: 9781259963667
Author: SMITH
Publisher: MCG
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Chapter 8, Problem 8.68P

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

Chapter 8, Problem 8.68P, 8.68 (a) Draw all products formed by treatment of each dibromide (A and B) with one equivalent of .

                  A                                       B

Expert Solution
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Interpretation Introduction

(a)

Interpretation: All products formed by the treatment of the given dibromide (A and B) with one equivalent of NaNH2 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 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 NaNH2 are (I)-(Z)1(1bromo2phenylvinyl)3methylbenzene and (II)-(Z)1(2bromo2phenylvinyl)3methylbenzene. The products that formed by the treatment of the given dibromide B with one equivalent of NaNH2 are (III)-(Z)1(1bromo2phenylvinyl)3methylbenzene and (IV)-(Z)1(2bromo2phenylvinyl)3methylbenzene.

Explanation of Solution

The reaction of dibromide A with one equivalent of NaNH2 results in the elimination of one mole of HBr. The corresponding chemical reaction is shown below.

PKG ORGANIC CHEMISTRY, Chapter 8, Problem 8.68P , additional homework tip  1

Figure 1

In this reaction, the given dihalide undergoes elimination reaction in the presence of one equivalent of the strong base, NaNH2, that results in the formation of two alkene products, (I)-(Z)1(1bromo2phenylvinyl)3methylbenzene and (II)-(Z)1(2bromo2phenylvinyl)3methylbenzene.

The reaction of dibromide B with one equivalent of NaNH2 results in the elimination of one mole of HBr. The corresponding chemical reaction is shown as,

PKG ORGANIC CHEMISTRY, Chapter 8, Problem 8.68P , additional homework tip  2

Figure 2

In this reaction, the given dihalide undergoes elimination reaction in the presence of one equivalent of the strong base, NaNH2, that results in the formation of two alkene products, (III)-(Z)1(1bromo2phenylvinyl)3methylbenzene and (IV)-(Z)1(2bromo2phenylvinyl)3methylbenzene.

Conclusion

The products that formed by the treatment of the given dibromide A with one equivalent of NaNH2 are (I)-(Z)1(1bromo2phenylvinyl)3methylbenzene and (II)-(Z)1(2bromo2phenylvinyl)3methylbenzene. The products that formed by the treatment of the given dibromide B with one equivalent of NaNH2 are (III)-(Z)1(1bromo2phenylvinyl)3methylbenzene and (IV)-(Z)1(2bromo2phenylvinyl)3methylbenzene.

Expert Solution
Check Mark
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,

PKG ORGANIC CHEMISTRY, Chapter 8, Problem 8.68P , additional homework tip  3

The labeling of the pairs of constitutional isomers is,

PKG ORGANIC CHEMISTRY, Chapter 8, Problem 8.68P , additional homework tip  4

Explanation of Solution

The first pair of diastereomers is shown as,

PKG ORGANIC CHEMISTRY, Chapter 8, Problem 8.68P , additional homework tip  5

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,

PKG ORGANIC CHEMISTRY, Chapter 8, Problem 8.68P , additional homework tip  6

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,

PKG ORGANIC CHEMISTRY, Chapter 8, Problem 8.68P , additional homework tip  7

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,

PKG ORGANIC CHEMISTRY, Chapter 8, Problem 8.68P , additional homework tip  8

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.

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Chapter 8 Solutions

PKG ORGANIC CHEMISTRY

Ch. 8 - Prob. 8.11PCh. 8 - Problem 8.12 What alkenes are formed from each...Ch. 8 - Prob. 8.13PCh. 8 - Problem 8.14 What alkenes are formed from each...Ch. 8 - Problem 8.15 How does each of the following...Ch. 8 - Problem 8.16 Draw both the SN1 and E1 products of...Ch. 8 - Prob. 8.17PCh. 8 - Prob. 8.18PCh. 8 - Problem 8.19 Explain why...Ch. 8 - Prob. 8.20PCh. 8 - Problem 8.21 Draw the alkynes formed when each...Ch. 8 - Problem 8.22 Draw the products in each...Ch. 8 - Problem 8.23 Draw a stepwise mechanism for the...Ch. 8 - 8.24 Rank the alkenes shown in the ball-and-stick...Ch. 8 - Prob. 8.25PCh. 8 - 8.26 What is the major E2 elimination product...Ch. 8 - Prob. 8.27PCh. 8 - Prob. 8.28PCh. 8 - Prob. 8.29PCh. 8 - 8.30 Label each pair of alkenes as constitutional...Ch. 8 - Prob. 8.31PCh. 8 - Prob. 8.32PCh. 8 - Prob. 8.33PCh. 8 - For each of the following alkenes, draw the...Ch. 8 - Prob. 8.35PCh. 8 - Prob. 8.36PCh. 8 - Prob. 8.37PCh. 8 - What alkene is the major product formed from each...Ch. 8 - Prob. 8.39PCh. 8 - Prob. 8.40PCh. 8 - Pick the reactant or solvent in each part that...Ch. 8 - 8.42 In the dehydrohalogenation of...Ch. 8 - Prob. 8.43PCh. 8 - Prob. 8.44PCh. 8 - Prob. 8.45PCh. 8 - Prob. 8.46PCh. 8 - Prob. 8.47PCh. 8 - Prob. 8.48PCh. 8 - What alkyl chloride affords the following alkene...Ch. 8 - Draw the products formed when each dihalide is...Ch. 8 - Draw the structure of a dihalide that could be...Ch. 8 - Under certain reaction conditions, 2,...Ch. 8 - For which reaction mechanisms, SN1, SN2, E1 or...Ch. 8 - Draw the organic products formed in each...Ch. 8 - Prob. 8.55PCh. 8 - Draw all products, including stereoisomers, in...Ch. 8 - Draw all of the substitution and elimination...Ch. 8 - Prob. 8.58PCh. 8 - 8.59 Draw a stepwise, detailed mechanism for each...Ch. 8 - Draw the major product formed when...Ch. 8 - Draw a stepwise, detailed mechanism for the...Ch. 8 - Explain why the reaction of with gives ...Ch. 8 - Draw a stepwise detailed mechanism that...Ch. 8 - Prob. 8.64PCh. 8 - 8.65 Explain the selectivity observed in the...Ch. 8 - Prob. 8.66PCh. 8 - Prob. 8.67PCh. 8 - 8.68 (a) Draw all products formed by treatment of...
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