Chapter 11, Problem 11.53AP

Interpretation Introduction

(a)

Interpretation:

Whether the reaction of ethylene with ${\text{OsO}}_{\text{4}}$ followed by ${\text{NaHSO}}_3$ will give a racemic mixture of products that can be resolved into enantiomers under ordinary conditions is to be stated.

Concept introduction:

The reaction of alkenes with ${\text{OsO}}_{\text{4}}$ is a concerted cycloaddition reaction which gives an intermediate, osmate ester. The osmate ester on hydrolysis yields glycols. It is a syn-addition reaction. Sodium sulphite is added so that any by-product formed which contains osmium can be converted into reduced forms and then removed by filtration.

Answer to Problem 11.53AP

The reaction of ethylene with ${\text{OsO}}_{\text{4}}$ followed by ${\text{NaHSO}}_3$ will not give a racemic mixture of products that can be resolved into enantiomers under ordinary conditions. The reaction of ethylene with ${\text{OsO}}_{\text{4}}$ will yield $\text{1, 2-glycol}$ as shown below.

Explanation of Solution

Ethylene reacts with ${\text{OsO}}_{\text{4}}$ by syn-addition reaction to give osmate ester which on hydrolysis yields $\text{1, 2-glycol}$ as shown below.

Figure 1

Since, the product $\text{1, 2-glycol}$ will be formed on addition of ${\text{OsO}}_{\text{4}}$ to ethylene from either side. Therefore, the reaction of ethylene with ${\text{OsO}}_{\text{4}}$ followed by ${\text{NaHSO}}_3$ will not give a racemic mixture.

Conclusion

The reaction of ethylene with ${\text{OsO}}_{\text{4}}$ followed by ${\text{NaHSO}}_3$ will not give a racemic mixture.

Interpretation Introduction

(b)

Interpretation:

Whether the reaction of $cis\text{-2-butene}$ with ${\text{OsO}}_{\text{4}}$ followed by ${\text{NaHSO}}_3$ will give a racemic mixture of products that can be resolved into enantiomers under ordinary conditions is to be stated.

Concept introduction:

The reaction of alkenes with ${\text{OsO}}_{\text{4}}$ is a concerted cycloaddition reaction which gives an intermediate, osmate ester. The osmate ester on hydrolysis yields glycols. It is a syn-addition reaction. Sodium sulphite is added so that any by-product formed which contains osmium can be converted into reduced forms and then removed by filtration.

Answer to Problem 11.53AP

The reaction of $cis\text{-2-butene}$ with ${\text{OsO}}_{\text{4}}$ followed by ${\text{NaHSO}}_3$ will not give a racemic mixture of products. The reaction of $cis\text{-2-butene}$ with ${\text{OsO}}_{\text{4}}$ will yield meso compounds of $\text{2, 3-butanediol}$ as shown below.

Explanation of Solution

The compound $cis\text{-2-butene}$ reacts with ${\text{OsO}}_{\text{4}}$ by syn-addition reaction to give osmate ester which on hydrolysis yields meso compounds: $\left(\text{2R, 3R}\right)\text{-2, 3-butanediol}$ and $\left(\text{2R, 3S}\right)\text{-2, 3-butanediol}$ as shown below.

Figure 2

The two products of $\text{2, 3-butanediol}$ with different stereochemistry will be formed on addition of ${\text{OsO}}_{\text{4}}$ to $cis\text{-2-butene}$ from the front and the backside. But they are identical meso compounds, therefore, the reaction of $cis\text{-2-butene}$ with ${\text{OsO}}_{\text{4}}$ followed by ${\text{NaHSO}}_3$ will not give a racemic mixture.

Conclusion

The reaction of $cis\text{-2-butene}$ with ${\text{OsO}}_{\text{4}}$ followed by ${\text{NaHSO}}_3$ will not give a racemic mixture.

Interpretation Introduction

(c)

Interpretation:

Whether the reaction of $trans\text{-2-butene}$ with ${\text{OsO}}_{\text{4}}$ followed by ${\text{NaHSO}}_3$ will give a racemic mixture of products that can be resolved into enantiomers under ordinary conditions is to be stated.

Concept introduction:

The reaction of alkenes with ${\text{OsO}}_{\text{4}}$ is a concerted cycloaddition reaction which gives an intermediate, osmate ester. The osmate ester on hydrolysis yields glycols. It is a syn-addition reaction. Sodium sulphite is added so that any by-product formed which contains osmium can be converted into reduced forms and then removed by filtration.

Answer to Problem 11.53AP

The reaction of $trans\text{-2-butene}$ with ${\text{OsO}}_{\text{4}}$ followed by ${\text{NaHSO}}_3$ will give a racemic mixture of products that can be resolved into enantiomers under ordinary conditions. The reaction of $trans\text{-2-butene}$ with ${\text{OsO}}_{\text{4}}$ will yield racemic mixture of $\left(\text{2R, 3R}\right)\text{-2, 3-butanediol}$ and $\left(\text{2S, 3S}\right)\text{-2, 3-butanediol}$ as shown below.

Explanation of Solution

The compound $trans\text{-2-butene}$ reacts with ${\text{OsO}}_{\text{4}}$ by syn-addition reaction to give osmate ester which on hydrolysis yields racemic mixture of $\left(\text{2R, 3R}\right)\text{-2, 3-butanediol}$ and $\left(\text{2S, 3S}\right)\text{-2, 3-butanediol}$ as shown below.

Figure 3

The products $\left(\text{2R, 3R}\right)\text{-2, 3-butanediol}$ and $\left(\text{2S, 3S}\right)\text{-2, 3-butanediol}$ will be formed on addition of ${\text{OsO}}_{\text{4}}$ to $trans\text{-2-butene}$ from the front and the back side. Therefore, the reaction of $trans\text{-2-butene}$ with ${\text{OsO}}_{\text{4}}$ followed by ${\text{NaHSO}}_3$ will give a racemic mixture.

Conclusion

The reaction of $trans\text{-2-butene}$ with ${\text{OsO}}_{\text{4}}$ followed by ${\text{NaHSO}}_3$ will give a racemic mixture.

Interpretation Introduction

(d)

Interpretation:

Whether the reaction of $cis\text{-2-pentene}$ with ${\text{OsO}}_{\text{4}}$ followed by ${\text{NaHSO}}_3$ will give a racemic mixture of products that can be resolved into enantiomers under ordinary conditions is to be stated.

Concept introduction:

The reaction of alkenes with ${\text{OsO}}_{\text{4}}$ is a concerted cycloaddition reaction which gives an intermediate, osmate ester. The osmate ester on hydrolysis yields glycols. It is a syn-addition reaction. Sodium sulphite is added so that any by-product formed which contains osmium can be converted into reduced forms and then removed by filtration.

Answer to Problem 11.53AP

The reaction of $cis\text{-2-pentene}$ with ${\text{OsO}}_{\text{4}}$ followed by ${\text{NaHSO}}_3$ will not give a racemic mixture of products that can be resolved into enantiomers under ordinary conditions. The reaction of $cis\text{-2-pentene}$ with ${\text{OsO}}_{\text{4}}$ will yield the identical meso products: $\left(\text{2R, 3R}\right)\text{-2, 3-pentanediol}$ and $\left(\text{2R, 3S}\right)\text{-2, 3-pentanediol}$ as shown below.

Explanation of Solution

The reaction of $cis\text{-2-pentene}$ with ${\text{OsO}}_{\text{4}}$ will yield the identical meso products: $\left(\text{2R, 3R}\right)\text{-2, 3-pentanediol}$ and $\left(\text{2R, 3S}\right)\text{-2, 3-pentanediol}$ as shown below.

Figure 4

The two products: $\left(\text{2R, 3R}\right)\text{-2, 3-pentanediol}$ and $\left(\text{2R, 3S}\right)\text{-2, 3-pentanediol}$ with different stereochemistry will be formed on addition of ${\text{OsO}}_{\text{4}}$ to $cis\text{-2-pentene}$ from the front and the backside. But they are identical meso compounds, therefore, the reaction of $cis\text{-2-pentene}$ with ${\text{OsO}}_{\text{4}}$ followed by ${\text{NaHSO}}_3$ will not give a racemic mixture.

Conclusion

The reaction of $cis\text{-2-pentene}$ with ${\text{OsO}}_{\text{4}}$ followed by ${\text{NaHSO}}_3$ will not give a racemic mixture.