Chapter 7, Problem 35P

Interpretation Introduction

Interpretation:

Based on the structure given, the different types of stereoisomers for each of the conditions given are to be identified.

Concept introduction:

A chiral carbon atom is attached to four different atoms or groups of atoms. The number of stereoisomers is determined by using the formula $2^n$, where $n$ is the number of chiral carbon atoms present in a molecule.

Stereoisomers arise from chiral centers as well as the arrangement of atoms or groups attached to the double bonded carbon atoms.

The pair of isomers designated $\text{cis- and}\text{trans-}$ are stereoisomers of each other. They have the same constitution; but the cis isomer has both of its similar groups on the same side of the double bond, while the similar groups in the trans isomer are on opposite sides.

Enantiomers are non-superimposable mirror images of each other.

Stereoisomers that are not mirror images are diastereomers.

Answer to Problem 35P

Solution:

a) The number of stereoisomers represented by the given constitution is $8$.

b) If the substituents on the five-membered ring are cis to each other, then there are four stereoisomers represented by the given constitution.

c) If the butenyl side chain has the Z configuration of its double bond, the given constitutional isomer has four possible stereoisomers.

d) Stereochemically accurate representations of all the stereoisomers are as follows:

e) The pairs of enantiomers are as follows:

$\begin{array}{l}\left(\text{1}\right)\text{ and }\left(\text{4}\right)\hfill \\ \left(\text{2}\right)\text{ and }\left(\text{3}\right)\hfill \\ \left(\text{5}\right)\text{ and }\left(\text{8}\right)\hfill \\ \left(\text{6}\right)\text{ and }\left(\text{7}\right)\hfill \end{array}$

The pairs of diastereomers are as follows:

$\begin{array}{l}\left(\text{1}\right)\text{ and }\left(\text{2}\right)\\ \left(\text{1}\right)\text{ and }\left(\text{3}\right)\\ \left(\text{2}\right)\text{ and }\left(\text{4}\right)\\ \left(\text{3}\right)\text{ and }\left(\text{4}\right)\end{array}$

$\begin{array}{l}\left(\text{5}\right)\text{ and }\left(\text{6}\right)\\ \left(\text{5}\right)\text{ and }\left(\text{7}\right)\\ \left(\text{6}\right)\text{ and }\left(\text{7}\right)\\ \left(\text{7}\right)\text{ and }\left(\text{8}\right)\end{array}$

Explanation of Solution

a)

In the given structure, there are two chiral carbon atoms indicated by $*$. The number of stereoisomers is counted by using the formula, ${\text{2}}^{\text{n}}$, where n is the number of chiral carbon atoms in the given molecule. So the number of stereoisomers:

$\begin{array}{c}{\text{2}}^{\text{n}}=2^2\\ =4\end{array}$

Also, the double bond of butenyl side chain can form E and Z isomers, so the total number of possible stereoisomers will be:

$4\times 2=8$

Therefore, the number of stereoisomers represented by the given constitution is $8$.

b)

So, if the substituents on the five-membered ring are cis to each other, then the double bond in the butenyl side chain can be in the E or Z form. Thus, there are $4$ stereoisomers represented by this constitution, as follows:

c)

If the butenyl side chain has the Z configuration of its double bond, the two substituents on the ring can be oriented in four different ways. The number of possible stereoisomers is $4$, as shown below:

d)

There are total $8$ stereoisomers represented by the given constitution. They are as follows:

e)

Enantiomers are non-superimposable mirror images of each other. Diastereomers are the stereoisomers which are not mirror images.

The pairs of enantiomers are as follows:

$\begin{array}{l}\left(\text{1}\right)\text{ and }\left(\text{4}\right)\hfill \\ \left(\text{2}\right)\text{ and }\left(\text{3}\right)\hfill \\ \left(\text{5}\right)\text{ and }\left(\text{8}\right)\hfill \\ \left(\text{6}\right)\text{ and }\left(\text{7}\right)\hfill \end{array}$

The pairs of diastereomers are as follows:

$\begin{array}{l}\left(\text{1}\right)\text{ and }\left(\text{2}\right)\\ \left(\text{1}\right)\text{ and }\left(\text{3}\right)\\ \left(\text{2}\right)\text{ and }\left(\text{4}\right)\\ \left(\text{3}\right)\text{ and }\left(\text{4}\right)\end{array}$

$\begin{array}{l}\left(\text{5}\right)\text{ and }\left(\text{6}\right)\\ \left(\text{5}\right)\text{ and }\left(\text{7}\right)\\ \left(\text{6}\right)\text{ and }\left(\text{7}\right)\\ \left(\text{7}\right)\text{ and }\left(\text{8}\right)\end{array}$