Question

The reaction of \((\pm)-2,3\) -dibromobutane with ethoxide ion produces trans-2-bromo-2-butene while, under the same conditions, meso-2, 3 -dibromobutane produces the corresponding cis isomer. With the aid of three- dimensional representations, determine whether this is an anti (trans) elimination. Explain how it is possible for a "trans" elimination to produce both cis and trans product.

Short answer

The reaction of (±)-2,3-dibromobutane and meso-2,3-dibromobutane with ethoxide ion undergoes an E2 (bimolecular elimination) mechanism. In both cases, the leaving group (bromide ion) and the hydrogen atom being eliminated are aligned along the opposite side of the molecule, allowing for an anti (trans) elimination mechanism. This results in the formation of both trans-2-bromo-2-butene and cis-2-bromo-2-butene, demonstrating that an anti (trans) elimination can produce different isomeric products based on the relative spatial alignment of the leaving group and hydrogen atom.

Step-by-step solution

Identify the given information and the reaction conditions

The reaction conditions given are as follows: 1. (±)-2,3-dibromobutane with ethoxide ion → trans-2-bromo-2-butene 2. meso-2,3-dibromobutane with ethoxide ion → cis-2-bromo-2-butene

Visualize the 3-dimensional structures of (±)-2,3-dibromobutane and meso-2,3-dibromobutane

To better understand the reaction mechanism, let's first visualize the three-dimensional structures of both (±)-2,3-dibromobutane and meso-2,3-dibromobutane. (±)-2,3-dibromobutane exists as a racemic mixture of R,R and S,S enantiomers. They both have a chiral carbon at the 2nd and 3rd carbon atoms. In the R,R enantiomer, the two bromine atoms are on the same side of the molecule (either both wedge or both dash), whereas in the S,S enantiomer they are on the opposite side. meso-2,3-dibromobutane is the diastereomer of (±)-2,3-dibromobutane, having one chiral carbon in R configuration and the other in the S configuration. These molecules have a plane of symmetry, and due to this symmetry, the two bromine atoms are also on the opposite side of the molecule.

Determine the elimination process

The reaction of both dibromobutane isomers with ethoxide ions will undergo E2 (bimolecular elimination) reaction mechanism. In an E2 reaction, both the leaving group (bromide ion) and the hydrogen that is being eliminated are removed simultaneously in a single, concerted step leading to the formation of an alkene product. To determine if it's an anti (trans) elimination, we need to study the spatial alignment of the leaving group and the hydrogen atom being eliminated.

Visualize anti (trans) elimination in each reaction

For (±)-2,3-dibromobutanes: In both enantiomers, the hydrogen atoms that are being eliminated are on the opposite side of the molecule as the corresponding leaving group (bromide ion). This can result in an anti (trans) elimination because the leaving groups and the hydrogen atoms being eliminated are aligned along the opposite sides of the molecule, resulting in the formation of the trans-2-bromo-2-butene product. For meso-2,3-dibromobutane: The two bromine atoms are on the opposite side of the molecule, and so are the hydrogen atoms that are being eliminated. In this case, an anti (trans) elimination is also possible since the leaving groups and the hydrogen atoms being eliminated are aligned along the opposite side of the molecule, resulting in the formation of cis-2-bromo-2-butene.

Explain the formation of both cis and trans products

This problem demonstrates that an elimination reaction can be considered anti (trans) even if it results in different isomeric products, such as cis and trans. The key factor is that the relative spatial alignment of the leaving group and the hydrogen atom being eliminated is what dictates if the reaction proceeds via an anti (trans) elimination mechanism. In both reactions, the leaving group and the hydrogen atom being eliminated are aligned along the opposite side of the molecule, satisfying the condition for an anti (trans) elimination mechanism.

Key concepts

Stereochemistry

Stereochemistry is the study of how the spatial arrangement of atoms affects the properties and reactions of molecules. It plays a crucial role in understanding chemical reactions such as E2 eliminations. For instance, in the reaction of (±)-2,3-dibromobutane, stereochemistry helps us understand that this compound is a racemic mixture. This means it contains equal parts of two enantiomers, R,R and S,S, which are mirror images. In these enantiomers, stereochemical considerations dictate how atoms are arranged in 3D space.

Meso-2,3-dibromobutane, however, is different from the racemic mixture. It is a diastereomer of these enantiomers because it has a plane of symmetry. Understanding these spatial arrangements is vital to predict the product of their reaction with ethoxide ion.

Anti Elimination

Anti, or trans, elimination is a type of bimolecular E2 reaction. In such reactions, the leaving group and the hydrogen atom that gets eliminated need to be on opposite sides of the molecule. This opposite alignment is a key feature of anti elimination.

In the reaction of (±)-2,3-dibromobutane, the spatial arrangement is such that the C–H bond and the C–Br bond are placed opposite to each other in 3D space. This setup allows the ethoxide ion to abstract the hydrogen atom while the bromide ion leaves, all happening in a single concerted step.

The meso-2,3-dibromobutane also follows anti elimination, even though the products differ. The correct geometric alignment of the hydrogen and bromine determines the resulting isomer.

Racemic Mixture

A racemic mixture contains equal amounts of left- and right-handed enantiomers. In the case of (±)-2,3-dibromobutane, this means that both R,R and S,S isomers are present in equal proportions. Each of these enantiomers has chiral centers, leading to different spatial arrangements of atoms.

When such a racemic mixture reacts, both enantiomers participate in the chemical reaction, often leading to symmetrical results. While they have different spatial orientations, their transformation follows the same mechanistic pathway in E2 eliminations. This ensures the formation of the same alkene product, like trans-2-bromo-2-butene, regardless of the enantiomer involved in the reaction.

Diastereomers

Diastereomers are stereoisomers that are not mirror images of each other. A good example is the meso-2,3-dibromobutane compared to its racemic counterpart. While racemic mixtures have distinct enantiomers as their components, diastereomers differ at some but not all stereocenters.

In meso-2,3-dibromobutane, one chiral center is at R while the other is S. This difference in stereocenters introduces a plane of symmetry that is absent in the racemic form. It leads to different chemical behavior under similar conditions.

When meso compounds react in E2 eliminations, the spatial arrangement of atoms allows the reaction to produce products like cis-2-bromo-2-butene, different from what the racemic enantiomers yield. Understanding diastereomers is crucial to predict such outcomes accurately.