Question

Give the stereoisomers of: (a) 2 -butene (d) 1-bromo-1-chloropropene (b) 2 -pentene (e) 2-bromo-1-chloropropene (c) 2,4 -hexadiene (f) 1 -bromo-1,2-dichloroethene Indicate for each the isomer that is expected to predominate in an equilibrium mixture. Justify your answer.

Short answer

The stereoisomers for the given compounds are as follows: a) 2-butene: (Z)-2-butene (cis) and (E)-2-butene (trans) with the (E)-isomer (trans) predominating due to lower steric hinderance. b) 2-pentene: (Z)-2-pentene (cis) and (E)-2-pentene (trans) with the (E)-isomer (trans) predominating due to lower steric hindrance. c) 2,4-hexadiene: (Z,Z)-2,4-hexadiene (cis,cis), (Z,E)-2,4-hexadiene (cis,trans), (E,Z)-2,4-hexadiene (trans,cis), and (E,E)-2,4-hexadiene (trans,trans) with the (E,E)-isomer (trans,trans) predominating due to lowest steric hindrance. d) 1-bromo-1-chloropropene: No stereoisomers for this molecule. e) 2-bromo-1-chloropropene: (Z)-2-bromo-1-chloropropene (cis) and (E)-2-bromo-1-chloropropene (trans) with the (E)-isomer (trans) predominating due to lower steric hinderance. f) 1-bromo-1,2-dichloroethene: No stereoisomers for this molecule.

Step-by-step solution

Identify stereoisomers for 2-butene

For 2-butene, we have a double bond between the second and third carbon atoms. Stereocenters exist only around the double bond. Due to this, 2-butene will have 2 possible stereoisomers: 1. (Z)-2-butene (cis-2-butene) 2. (E)-2-butene (trans-2-butene)

Evaluate which isomer will predominate

The isomer that will predominate is the one with lower steric hinderance. In the case of 2-butene, the (E)-isomer (trans) will predominate because it has lower steric hinderance due to the bulkier groups being further apart when compared to the (Z)-isomer (cis). #b) 2-pentene#

Identify stereoisomers for 2-pentene

For 2-pentene, the double bond is between the second and third carbon atoms. Stereocenters exist only around the double bond. Due to this, 2-pentene will have 2 possible stereoisomers: 1. (Z)-2-pentene (cis-2-pentene) 2. (E)-2-pentene (trans-2-pentene)

Evaluate which isomer will predominate

The isomer that will predominate in 2-pentene is the one with lower steric hindrance. In this case, the (E)-isomer (trans) will predominate due to the bulkier groups being further apart when compared to the (Z)-isomer (cis). #c) 2,4-hexadiene#

Identify stereoisomers for 2,4-hexadiene

For 2,4-hexadiene, we have two double bonds in the molecule between second and third carbon atoms and between fourth and fifth carbon atoms. Stereocenters exist only around the double bonds. Due to this, there are 4 possible stereoisomers for 2,4-hexadiene: 1. (Z,Z)-2,4-hexadiene (cis,cis-2,4-hexadiene) 2. (Z,E)-2,4-hexadiene (cis,trans-2,4-hexadiene) 3. (E,Z)-2,4-hexadiene (trans,cis-2,4-hexadiene) 4. (E,E)-2,4-hexadiene (trans,trans-2,4-hexadiene)

Evaluate which isomer will predominate

The isomer that will predominate in 2,4-hexadiene is the one with the lowest steric hindrance. In this case, the (E,E)-isomer (trans,trans) will predominate due to the bulkier groups being further apart when compared to the other isomers. #d) 1-bromo-1-chloropropene#

Identify stereoisomers for 1-bromo-1-chloropropene

For 1-bromo-1-chloropropene, the double bond is between the first and second carbon atoms. Stereocenters exist only around the double bond. The molecule has two different substituents on same carbon so we have no isomers for this molecule. #e) 2-bromo-1-chloropropene#

Identify stereoisomers for 2-bromo-1-chloropropene

For 2-bromo-1-chloropropene, we have a double bond between the first and second carbon atoms and different atoms on the second carbon. Stereocenters exist only around the double bond. Due to this, there are 2 possible stereoisomers for 2-bromo-1-chloropropene: 1. (Z)-2-bromo-1-chloropropene (cis-2-bromo-1-chloropropene) 2. (E)-2-bromo-1-chloropropene (trans-2-bromo-1-chloropropene)

Evaluate which isomer will predominate

The isomer that will predominate in 2-bromo-1-chloropropene is the one with lower steric hinderance. In this case, the (E)-isomer (trans) will predominate due to the bulkier groups being further apart when compared to the (Z)-isomer (cis). #f) 1-bromo-1,2-dichloroethene#

Identify stereoisomers for 1-bromo-1,2-dichloroethene

For 1-bromo-1,2-dichloroethene, we have a double bond between the first and second carbon atoms. Stereocenters exist only around the double bond. The molecule has different halogens on the first carbon making this molecule asymmetrical and no stereoisomers for this molecule.

Key concepts

2-butene Stereioisomers

Understanding the stereoisomers of 2-butene is fundamental in grasping isomerism in organic chemistry. 2-butene has two stereoisomers: (Z)-2-butene and (E)-2-butene, commonly referred to as cis-2-butene and trans-2-butene, respectively. The different spatial arrangements around the double bond lead to distinctive physical and chemical properties despite having the same molecular formula.

The (Z)-2-butene has the higher priority groups (according to Cahn-Ingold-Prelog rules) on the same side of the double bond, while (E)-2-butene has them on opposite sides. In comparing stability, the (E)-isomer (trans) generally predominates due to less steric hindrance, meaning the bulky groups are kept apart as opposed to being close together in the (Z)-isomer (cis). This concept illustrates the important role that spatial arrangement plays in the stability of molecules.

Stereocenters in Alkenes

In alkenes, stereocenters are points within the molecule around which the bonding can vary in space to give stereoisomers. With alkenes, the double bond itself becomes the point of interest. Unlike chiral centers, where atoms with four different substituents are present, alkenes feature planar carbon atoms that create opportunities for varying spatial arrangements of the substituents attached to the carbons involved in the double bond.

This phenomenon is inherent to compounds like 2-butene and 2-pentene, where the double bond introduces two possible configurations: (Z) and (E). The presence or absence of stereocenters will significantly influence the molecular geometry and reactivity of the alkene.

E-Z Notation in Stereochemistry

E-Z notation is an essential aspect of stereochemistry used to describe the relationship between different substituents attached to two carbon atoms that are connected by a double bond. The 'E' stands for 'Entgegen' which means 'opposite' in German while 'Z' stands for 'Zusammen', meaning 'together'.

Priority is given to the substituents based on the atomic number, where higher atomic numbers correspond to higher priority. In (Z)-configured molecules, the higher priority substituents are on the same side of the double bond, and in (E)-configured molecules, they are on opposite sides. This system provides a clear, unambiguous way to describe configurations of molecules, especially useful for complex molecules.

Steric Hindrance and Isomer Stability

Steric hindrance occurs when atoms or groups of atoms are too close to each other, causing an increase in energy and a decrease in stability due to physical crowding within the molecule. The concept of steric hindrance is key to understanding isomer stability. Typically, isomers that minimize steric repulsions are more stable and are therefore favored.

For example, in alkenes with simple substituents such as 2-butene, the (E)-isomer, with its bulkier groups on opposite sides, is usually more stable than the (Z)-isomer, where these groups are positioned on the same side. This stability influences which isomer will predominate at equilibrium.

2,4-Hexadiene Configurations

The molecule 2,4-hexadiene is a bit more complex, with two double bonds each capable of (Z) or (E) configurations. The molecule can exist in four possible stereoisomers: (Z,Z)-, (Z,E)-, (E,Z)-, and (E,E)-2,4-hexadiene. These stereoisomers have different spacial configurations and hence different physical properties.

The (E,E)-isomer is markedly more stable due to the minimized steric hindrance across both of its double bonds; the bulky alkyl groups are as far away from each other as possible. This makes the (E,E)-configuration the most likely to predominate in an equilibrium mixture.

Halogenated Alkenes Stereochemistry

Stereochemistry in halogenated alkenes adds another layer of complexity due to the presence of halogen atoms as substituents, which are substantially larger than the typical hydrogen atoms. In compounds like 1-bromo-1-chloropropene, where the halides are attached to the same carbon, there is no stereochemistry involved due to a lack of configurational isomerism around the double bond.

However, in molecules like 2-bromo-1-chloropropene, stereochemistry becomes relevant with the (Z)- and (E)- configurations due to the different atoms on the second carbon. Steric interactions between the larger halogen atoms play a significant role in determining the stability and predominance of the isomers, typically favoring the (E)-isomer where the bulky halogens are positioned across from each other, reducing steric strain.