Question

Following are diastereomers (A) and (B) of 3 -bromo-3,4-dimethylhexane. On treatment with sodium ethoxide in ethanol, each gives 3,4 -dimethyl-3-hexene as the major product. One diastereomer gives the \(E\) alkene, and the other gives the \(Z\) alkene. Which diastereomer gives which alkene? Account for the stereoselectivity of each \(\beta\)-elimination.

Short answer

Explain the stereoselectivity of each β-elimination. Answer: Diastereomer (A) forms the E (trans) alkene, while diastereomer (B) forms the Z (cis) alkene. The stereoselectivity of each β-elimination is determined by the orientation of the β-hydrogen atom and the bromine atom in the original diastereomer. In diastereomer (A), the antiperiplanar arrangement of β-hydrogen and bromine atom leads to the formation of the E alkene. In contrast, diastereomer (B) has a synperiplanar arrangement of β-hydrogen and bromine atom, leading to the formation of the Z alkene.

Step-by-step solution

Reaction Mechanism

The diastereomers undergo β-elimination to produce an alkene with the following general mechanism: 1. Ethoxide (the conjugate base of ethanol) acts as a base and abstracts the hydrogen atom (proton) from the β-position (carbon next to the one bearing the bromine atom). 2. The newly formed alkoxide ion (the conjugate base of the alcohol formed) reforms the bond with the rest of the molecule, causing the electrons in the C-Br bond to move to the bromine atom, leading to its detachment from the molecule and forming an alkene. Now, let's look at each diastereomer:

Diastereomer A - E Alkene Formation

We will first analyze diastereomer (A) undergoing β-elimination in the presence of sodium ethoxide in ethanol: 1. Identify the hydrogen atom (proton) that can be abstracted by ethoxide at the β-position. 2. Carry out the elimination reaction and visualize the formed alkene; it will have a trans (E) configuration.

Diastereomer B - Z Alkene Formation

Next, we will analyze diastereomer (B) undergoing β-elimination in the presence of sodium ethoxide in ethanol: 1. Identify the hydrogen atom (proton) that can be abstracted by ethoxide at the β-position. 2. Carry out the elimination reaction and visualize the formed alkene; it will have a cis (Z) configuration. After carefully analyzing the β-elimination of both diastereomers, we can deduce which product is formed by each diastereomer:

Conclusion

Diastereomer (A) forms the E (trans) alkene, while diastereomer (B) forms the Z (cis) alkene when treated with sodium ethoxide in ethanol. The stereoselectivity of each β-elimination is determined by the orientation of the β-hydrogen atom and the bromine atom in the original diastereomer. Since diastereomer (A) has an antiperiplanar β-hydrogen and bromine atom, it is more likely to form the E alkene. In contrast, diastereomer (B) has a synperiplanar β-hydrogen and bromine atom, leading to the formation of the Z alkene.

Key concepts

Diastereomers

Diastereomers are a type of stereoisomers—molecules with the same molecular formula and sequence of bonded atoms (constitution), but differing in the three-dimensional orientations of their atoms in space. Unlike enantiomers, diastereomers are not mirror images of each other and often have different physical and chemical properties, such as solubility, boiling point, and reaction rate, which can be distinguished in non-chiral environments.

In the provided exercise, the diastereomers of 3-bromo-3,4-dimethylhexane have different spatial arrangements around the chiral center, which affects the outcome of the β-elimination reaction when treated with sodium ethoxide in ethanol. The difference in orientation leads to the formation of distinct E/Z alkenes as products.

Sodium Ethoxide in Ethanol

Role as a Base in Elimination Reactions

Sodium ethoxide, often used as a base in organic synthesis, is produced by the reaction of sodium metal with ethanol. In the context of β-elimination reactions, sodium ethoxide abstracts a proton from the β-carbon of a molecule, which then leads to the formation of a double bond, resulting in an alkene. The strength and steric nature of sodium ethoxide make it a good base for such transformations.

When a β-elimination occurs, the leaving group (in this case, bromine) is expelled, and the alkene product is formed. Sodium ethoxide's reactivity is crucial in promoting these reactions, leading to the formation of the diverse E/Z configuration alkenes from the corresponding diastereomers.

E/Z Alkene Configuration

Understanding E and Z Nomenclature

The terms E and Z are used in organic chemistry to describe the configuration of substituents around the double bond in alkenes. 'E' stands for 'entgegen', which means 'opposite' in German, and denotes that the higher priority substituents on each carbon of the double bond are on opposite sides. Conversely, 'Z' stands for 'zusammen', meaning 'together', indicating that the high priority substituents are on the same side.

The priority of substituents is determined by the Cahn-Ingold-Prelog priority rules. In the exercise, analyzing which diastereomer leads to an E or Z alkene is essential for understanding the stereoselectivity of the β-elimination reaction and helps predict the products of such reactions.

Stereoselectivity

Significance in Chemical Reactions

Stereoselectivity refers to the preference of a chemical reaction to yield a particular stereoisomer when more than one is possible. It is a crucial concept in organic chemistry as it affects the physical and biological properties of the compounds. For the β-elimination reaction discussed in the exercise, stereoselectivity is influenced by factors such as the steric hindrance and electronic effects of the substituents near the reactive site.

In this reaction, the diastereomers of 3-bromo-3,4-dimethylhexane will form different alkene products due to their distinct configurations. Stereoselectivity can often be rationalized by examining the stability of the transition states leading to the product alkenes. For instance, the generation of an E-alkene typically requires an antiperiplanar arrangement of the leaving group and the abstracted hydrogen, which is often associated with a more stable transition state, leading to its preference in the elimination reaction.