Chapter 10.6, Problem 25E

(a)

Interpretation Introduction

Interpretation:Whether the major product formed will have like or unlike stereochemistry or whether it is chiral or meso should be determined.

  

Concept introduction:In order to assign absolute configuration of R and S, Cahn − Ingold − Prelog rules are used and the first step is to assign the priority order with the atomic number as the fundamental property. The one with the highest atomic number gest highest priority and is designated as “1” and so on.

In order to distinguish the relative stereochemistry of two chiral centers situated adjacent to one another, like or unlike notation is used.

For enantiomeric pairs that have same stereochemistry at adjacent carbon atoms, the notation “l” designates “like” and notation “u” designates“un-like”.

Four kinds of symmetry elements that may be present are tabulated as follows:

  $\begin{array}{ccc}\text{S}\text{.No}& \text{Symmetry element}& \text{Notatio}n\\ 1.& \text{Axis of symmetry}& C_n\\ 2.& \text{Plane of symmetry}& \sigma \\ 3.& \begin{array}{l}\text{Alternating axis of }\\ \text{symmetry}\end{array}& S_n\\ 4.& \text{Center of symmetry}& i\end{array}$

Presence of any symmetry element makes molecule achiral and optically inactive.

Explanation of Solution

If the groups now are arranged are read from highest towards least in clockwise fashion then R is assigned to the stereocenter, if the rotation is anticlockwise then S is assigned at the configuration.

The products formed in reaction with bromine are both R due to the clockwise priority order indicated in the reaction below.

  

Since the enantiomers have same stereochemistry on both the carbons so they are like. These enantiomers have ${\text{C}}_{\text{2}}$ axis of symmetry that makes them meso or achiral.

(b)

Interpretation Introduction

Interpretation:Whether the major product formed will have like or unlike stereochemistry or whether it is chiral or meso should be determined.

  

Concept introduction:In order to assign absolute configuration of R and S, Cahn − Ingold − Prelog rules are used and the first step is to assign the priority order with the atomic number as the fundamental property. The one with the highest atomic number gest highest priority and is designated as “1” and so on.

In order to distinguish the relative stereochemistry of two chiral centers situated adjacent to one another, like or unlike notation is used.

For enantiomeric pairs that have same stereochemistry at adjacent carbon atoms, the notation “l” designates “like” and notation “u” designates “un-like”.

Four kinds of symmetry elements that may be present are tabulated as follows:

  $\begin{array}{ccc}\text{S}\text{.No}& \text{Symmetry element}& \text{Notatio}n\\ 1.& \text{Axis of symmetry}& C_n\\ 2.& \text{Plane of symmetry}& \sigma \\ 3.& \begin{array}{l}\text{Alternating axis of }\\ \text{symmetry}\end{array}& S_n\\ 4.& \text{Center of symmetry}& i\end{array}$

Presence of any symmetry element makes molecule achiral and optically inactive.

Explanation of Solution

If the groups now are arranged are read from highest towards least in clockwise fashion the R is assigned to the stereocenter, if the rotation is anticlockwise the S is assigned at the configuration.

The products formed in the reaction of bromine are R and S indicated in the reaction below.

  

Since the enantiomers have different stereochemistry on both the carbons so they are unlike. These enantiomers have no plane or axis of symmetry so they are chiral.

(c)

Interpretation Introduction

Interpretation:Whether the major product formed will have like or unlike stereochemistry or whether it is chiral or meso should be determined.

  

Concept introduction:In order to assign absolute configuration of R and S, Cahn − Ingold − Prelog rules are used and the first step is to assign the priority order with the atomic number as the fundamental property. The one with the highest atomic number gest highest priority and is designated as “1” and so on.

In order to distinguish the relative stereochemistry of two chiral centers situated adjacent to one another, like or unlike notation is used.

For enantiomeric pairs that have same stereochemistry at adjacent carbon atoms, the notation “l” designates “like” and notation “u” designates “un-like”.

Four kinds of symmetry elements that may be present are tabulated as follows:

  $\begin{array}{ccc}\text{S}\text{.No}& \text{Symmetry element}& \text{Notatio}n\\ 1.& \text{Axis of symmetry}& C_n\\ 2.& \text{Plane of symmetry}& \sigma \\ 3.& \begin{array}{l}\text{Alternating axis of }\\ \text{symmetry}\end{array}& S_n\\ 4.& \text{Center of symmetry}& i\end{array}$

Presence of any symmetry element makes molecule achiral and optically inactive.

Explanation of Solution

If the groups now are arranged are read from highest towards least in clockwise fashion the R is assigned to the stereocenter, if the rotation is anticlockwise the S is assigned at the configuration.

The products formed in the reaction of bromine with given alkene are indicated in the reaction below.

  

Since the enantiomers have the same stereochemistry on both the carbons so they are like. Further,the presence of ${\text{C}}_{\text{2}}$ axis of symmetry results in an identical configuration that suggests that the product is meso or achiral.

(d)

Interpretation Introduction

Interpretation:Whether the major product formed will have like or unlike stereochemistry or whether it is chiral or meso should be determined.

  

Concept introduction:In order to assign absolute configuration of R and S, Cahn − Ingold − Prelog rules are used and the first step is to assign the priority order with the atomic number as the fundamental property. The one with the highest atomic number gest highest priority and is designated as “1” and so on.

In order to distinguish the relative stereochemistry of two chiral centers situated adjacent to one another, like or unlike notation is used.

For enantiomeric pairs that have same stereochemistry at adjacent carbon atoms, the notation “l” designates “like” and notation “u” designates “un-like”.

Four kinds of symmetry elements that may be present are tabulated as follows:

  $\begin{array}{ccc}\text{S}\text{.No}& \text{Symmetry element}& \text{Notatio}n\\ 1.& \text{Axis of symmetry}& C_n\\ 2.& \text{Plane of symmetry}& \sigma \\ 3.& \begin{array}{l}\text{Alternating axis of }\\ \text{symmetry}\end{array}& S_n\\ 4.& \text{Center of symmetry}& i\end{array}$

Presence of any symmetry element makes molecule achiral and optically inactive.

Explanation of Solution

If the groups now are arranged are read from highest towards least in clockwise fashion the R is assigned to the stereocenter, if the rotation is anticlockwise the S is assigned at the configuration.

The products formed in the reaction of bromine with given alkene are indicated in the reaction below.

  

Since the enantiomers have same stereochemistry on both the carbons so they are like. Further presence of ${\text{C}}_{\text{2}}$ axis of symmetry makes it meso or achiral.

(e)

Interpretation Introduction

Interpretation:Whether the major product formed will have like or unlike stereochemistry or whether it is chiral or meso should be determined.

  

Concept introduction:: In order to assign absolute configuration of R and S, Cahn − Ingold − Prelog rules are used and the first step is to assign the priority order with the atomic number as the fundamental property. The one with the highest atomic number gest highest priority and is designated as “1” and so on.

In order to distinguish the relative stereochemistry of two chiral centers situated adjacent to one another, like or unlike notation is used.

For enantiomeric pairs that have same stereochemistry at adjacent carbon atoms, the notation “l” designates “like” and notation “u” designates “un-like”.

Four kinds of symmetry elements that may be present are tabulated as follows:

  $\begin{array}{ccc}\text{S}\text{.No}& \text{Symmetry element}& \text{Notatio}n\\ 1.& \text{Axis of symmetry}& C_n\\ 2.& \text{Plane of symmetry}& \sigma \\ 3.& \begin{array}{l}\text{Alternating axis of }\\ \text{symmetry}\end{array}& S_n\\ 4.& \text{Center of symmetry}& i\end{array}$

Presence of any symmetry element makes molecule achiral and optically inactive.

Explanation of Solution

If the groups now are arranged are read from highest towards least in clockwise fashion the R is assigned to the stereocenter, if the rotation is anticlockwise the S is assigned at the configuration.

The products formed in the reaction of bromine with given alkene are indicated in the reaction below.

  

Since the major enantiomer has the same stereochemistry on both the carbons so they are like. These enantiomers have no plane or axis of symmetry so they are chiral.

(f)

Interpretation Introduction

Interpretation:Whether the major product formed will have like or unlike stereochemistry or whether it is chiral or meso should be determined.

  

Concept introduction:: In order to assign absolute configuration of R and S, Cahn − Ingold − Prelog rules are used and the first step is to assign the priority order with the atomic number as the fundamental property. The one with the highest atomic number gest highest priority and is designated as “1” and so on.

In order to distinguish the relative stereochemistry of two chiral centers situated adjacent to one another, like or unlike notation is used.

For enantiomeric pairs that have same stereochemistry at adjacent carbon atoms, the notation “l” designates “like” and notation “u” designates “un-like”.

Four kinds of symmetry elements that may be present are tabulated as follows:

  $\begin{array}{ccc}\text{S}\text{.No}& \text{Symmetry element}& \text{Notatio}n\\ 1.& \text{Axis of symmetry}& C_n\\ 2.& \text{Plane of symmetry}& \sigma \\ 3.& \begin{array}{l}\text{Alternating axis of }\\ \text{symmetry}\end{array}& S_n\\ 4.& \text{Center of symmetry}& i\end{array}$

Presence of any symmetry element makes molecule achiral and optically inactive.

Explanation of Solution

If the groups now are arranged are read from highest towards least in clockwise fashion the R is assigned to the stereocenter, if the rotation is anticlockwise the S is assigned at the configuration.

The products formed in the reaction of bromine with given alkene are R, S and S, R respectively due to the priority order indicated in the reaction below.

  

Since the enantiomers have different stereochemistry on both the carbons so they are unlike. These enantiomers havethe plane of symmetry so they are meso.