Chapter 5, Problem 5.56P

Draw all possible stereoisomers for each cycloalkane. Label pairs of enantiomers and diastereomers. Label any meso compound.

a. b. c.

Interpretation Introduction

(a)

Interpretation: The possible stereoisomers for the given cycloalkane are to be drawn. The labeling of any enantiomers, diastereomers or meso compound is to be stated.

Concept introduction: A carbon atom that has four nonequivalent atoms or groups attached to it is known as chiral carbon atom. Chiral carbon centers are also called as asymmetric or stereogenic centers.

The stereocenters are carbon atoms on which the interchanging of two atoms or groups results in the formation of new stereoisomers. The stereocenters are also known as stereogenic centers. $R,S$ configurations can be used to determine the relation between the two compounds. Enantiomer of a compound has opposite configuration. Diastereomers of a compound have at least one stereogenic centre with same configuration and at least one with opposite configuration.

Answer to Problem 5.56P

The possible stereoisomers for the given cycloalkane are three. Among the three stereoisomers, the labeling of two enantiomers of the given cycloalkane is $\left(1R,3R\right)-1,3-\text{dimethylcyclohexane}$ and $\left(1S,3S\right)-1,3-\text{dimethylcyclohexane}$. The labeling of one remaining meso compound is done as $\left(1R,3S\right)-1,3-\text{dimethylcyclohexane}$.

Explanation of Solution

The given cycloalkane is,

Figure 1

Due to the presence of cyclohexane ring attached with two methyl groups, the IUPAC name of this cycloalkane is $1,3-\text{dimethylcyclohexane}$. The total number of chiral centers for $1,3-\text{dimethylcyclohexane}$ is two. The total number of stereoisomers of $1,3-\text{dimethylcyclohexane}$ is calculated by the formula,

$\text{Number}\text{of}\text{stereoisomers}={\text{2}}^{\text{n}}$

Where,

• $\text{n}$ is the total number of chiral centers.

Substitute the number of chiral centers of $1,3-\text{dimethylcyclohexane}$ in the above expression.

$\begin{array}{rll}\text{Number}\text{of}\text{stereoisomers}& =& {\text{2}}^{\text{2}}\\ & =& 4\end{array}$

Thus, the total number of stereoisomers of $1,3-\text{dimethylcyclohexane}$ is $4$. This $1,3-\text{dimethylcyclohexane}$ has a plane of symmetry as shown below.

Figure 2

Thus, this compound forms only three stereoisomers instead of four. Among the three stereoisomers of $1,3-\text{dimethylcyclohexane}$, two stereoisomers are enantiomers as shown below.

Figure 3

The remaining one stereoisomer of $1,3-\text{dimethylcyclohexane}$ is a meso compound as labeled below.

Figure 4

Conclusion

The possible stereoisomers for the given cycloalkane are three. Among the three stereoisomers, the labeling of two enantiomers of the given cycloalkane is $\left(1R,3R\right)-1,3-\text{dimethylcyclohexane}$ and $\left(1S,3S\right)-1,3-\text{dimethylcyclohexane}$. The labeling of one remaining meso compound is done as $\left(1R,3S\right)-1,3-\text{dimethylcyclohexane}$.

Interpretation Introduction

(b)

Interpretation: The possible stereoisomers for the given cycloalkane are to be drawn. The labeling of any enantiomers, diastereomers or meso compound is to be stated.

Concept introduction: A carbon atom that has four nonequivalent atoms or groups attached to it is known as chiral carbon atom. Chiral carbon centers are also called as asymmetric or stereogenic centers.

The stereocenters are carbon atoms on which the interchanging of two atoms or groups results in the formation of new stereoisomers. The stereocenters are also known as stereogenic centers. $R,S$ configurations can be used to determine the relation between the two compounds. Enantiomer of a compound has opposite configuration. Diastereomers of a compound have at least one stereogenic centre with same configuration and at least one with opposite configuration.

Answer to Problem 5.56P

The possible stereoisomers for the given cycloalkane are two. Both are meso compounds and these meso compound are labeled as done as $\text{Cis}-1,4-\text{dimethylcyclohexane}$ and $\text{Trans}-1,4-\text{dimethylcyclohexane}$.

Explanation of Solution

The given cycloalkane is,

Figure 5

Due to the presence of cyclohexane ring attached with two methyl groups at first and fourth carbon atoms, the IUPAC name of this cycloalkane is $1,4-\text{dimethylcyclohexane}$. This compound does not possess any chiral carbon atom.

This $1,4-\text{dimethylcyclohexane}$ has a plane of symmetry as shown below.

Figure 6

Thus, this compound forms only two stereoisomers one is cis isomer and second is trans isomer. The given compound, $1,4-\text{dimethylcyclohexane}$ exist in the chair conformation in both cis and trans forms as shown below.

Figure 7

Thus, $1,4-\text{dimethylcyclohexane}$ possess only two sreeroisomers which are also the meso compounds.

Conclusion

The possible stereoisomers for the given cycloalkane are two. Both are meso compounds and these meso compound are labeled as done as $\text{Cis}-1,4-\text{dimethylcyclohexane}$ and $\text{Trans}-1,4-\text{dimethylcyclohexane}$.

Interpretation Introduction

(c)

Interpretation: The possible stereoisomers for the given cycloalkane are to be drawn. The labeling of any enantiomers, diastereomers or meso compound is to be stated.

Concept introduction: A carbon atom that has four nonequivalent atoms or groups attached to it is known as chiral carbon atom. Chiral carbon centers are also called as asymmetric or stereogenic centers.

The stereocenters are carbon atoms on which the interchanging of two atoms or groups results in the formation of new stereoisomers. The stereocenters are also known as stereogenic centers. $R,S$ configurations can be used to determine the relation between the two compounds. Enantiomer of a compound has opposite configuration. Diastereomers of a compound have at least one stereogenic centre with same configuration and at least one with opposite configuration.

Answer to Problem 5.56P

The possible stereoisomers for the given cycloalkane are four. It contains two pairs of enantiomers, the labeling of two pairs of enantiomers of the given cycloalkane is $\left(1S,2R\right)-1-\text{bromo}-2-\text{chlorocyclopentane}$, $\left(1R,2S\right)-1-\text{bromo}-2-\text{chlorocyclopentane}$ and $\left(1R,2R\right)-1-\text{bromo}-2-\text{chlorocyclopentane}$, $\left(1S,2S\right)-1-\text{bromo}-2-\text{chlorocyclopentane}$.

Explanation of Solution

The given cycloalkane is,

Figure 8

Due to the presence of cyclopentane ring attached with one bromine atom and one chlorine atom at first and second carbon atom respectively, the IUPAC name of this cycloalkane is $1-\text{bromo}-2-\text{chlorocyclopentane}$. The total number of chiral centers for $1-\text{bromo}-2-\text{chlorocyclopentane}$ is two. The total number of stereoisomers of $1-\text{bromo}-2-\text{chlorocyclopentane}$ is calculated by the formula,

$\text{Number}\text{of}\text{stereoisomers}={\text{2}}^{\text{n}}$

Where,

• $\text{n}$ is the total number of chiral centers.

Substitute the number of chiral centers of $1-\text{bromo}-2-\text{chlorocyclopentane}$ in the above expression.

$\begin{array}{rll}\text{Number}\text{of}\text{stereoisomers}& =& {\text{2}}^{\text{2}}\\ & =& 4\end{array}$

Thus, the total number of stereoisomers of $1-\text{bromo}-2-\text{chlorocyclopentane}$ is $4$. This $1-\text{bromo}-2-\text{chlorocyclopentane}$ compound does not possess plane of symmetry as shown below.

The four stereoisomers of $1-\text{bromo}-2-\text{chlorocyclopentane}$ are shown as,

Figure 9

The compound, $1-\text{bromo}-2-\text{chlorocyclopentane}$ possess two pairs of enantiomers which are labeled as,

Figure 10

Conclusion

The possible stereoisomers for the given cycloalkane are four. It contains two pairs of enantiomers, the labeling of two pairs of enantiomers of the given cycloalkane is $\left(1S,2R\right)-1-\text{bromo}-2-\text{chlorocyclopentane}$, $\left(1R,2S\right)-1-\text{bromo}-2-\text{chlorocyclopentane}$ and $\left(1R,2R\right)-1-\text{bromo}-2-\text{chlorocyclopentane}$, $\left(1S,2S\right)-1-\text{bromo}-2-\text{chlorocyclopentane}$.