Question

Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene: (i) 1-Bromo-1-methylcyclohexane (ii) 2-Chloro-2-methylbutane (iii) 2.2.3-Trimethyl-3-bromopentane.

Short answer

Major alkenes: (i) 1-methyl-1-cyclohexene, (ii) 2-methyl-2-butene, (iii) 2,3,3-trimethyl-2-pentene.

Step-by-step solution

Understanding the Reaction

Dehydrohalogenation is an elimination reaction where a hydrogen halide (such as HBr or HCl) is removed, forming an alkene. Using a strong base like sodium ethoxide encourages the formation of the most substituted, and thus more stable, alkene (according to Zaitsev's rule).

Analyzing Compound (i) 1-Bromo-1-methylcyclohexane

This halide can undergo elimination to form an alkene. The most substituted position after dehydrohalogenation will be between the carbon attached to the bromine and an adjacent carbon. The hydrogen from the adjacent carbon can be removed to form either: methylcyclohexene or 1-methyl-1-cyclohexene, with 1-methyl-1-cyclohexene being the major product.

Analyzing Compound (ii) 2-Chloro-2-methylbutane

The dehydrohalogenation of 2-chloro-2-methylbutane can occur by removing a hydrogen from either carbon adjacent to the carbon holding the chlorine. This can form 2-methyl-2-butene or 2-methyl-1-butene. The more substituted and thus the more stable alkene is 2-methyl-2-butene, which will be the major product.

Analyzing Compound (iii) 2.2.3-Trimethyl-3-bromopentane

In this compound, the bromine is attached to a tertiary carbon, which can lead to the removal of a hydrogen from the adjacent secondary carbons. This forms products such as 2,3,3-trimethyl-2-pentene and 2,3,3-trimethyl-3-pentene. The major product here is 2,3,3-trimethyl-2-pentene since it is the more substituted alkene.

Summary and Conclusion

In summary, the major alkenes formed are: (i) 1-methyl-1-cyclohexene, (ii) 2-methyl-2-butene, and (iii) 2,3,3-trimethyl-2-pentene. These products follow Zaitsev's rule, where the most substituted alkene is the major product.

Key concepts

Zaitsev's Rule

Zaitsev's rule is a guiding principle in organic chemistry that helps predict the outcome of elimination reactions. It states that when a dehydrohalogenation reaction occurs, the most substituted alkene will be the major product. This is because the more substituted an alkene is, the more stable it tends to be. High stability arises from a greater degree of hyperconjugation and electron donation by alkyl groups.

Zaitsev's rule applies to reactions where a base removes a hydrogen atom adjacent to a leaving group, such as a halogen. The base pulls off a proton from the beta carbon, leading to the formation of a double bond. Among possible outcomes, the alkene that is the most substituted—that is, has the highest number of alkyl groups attached to the double-bonded carbons—is favored.

• This rule predicts the product formation in many elimination reactions.
• It's essential to remember that Zaitsev's product is not always formed due to steric hindrances or sterioelectronic effects.

Alkene Formation

Alkene formation is the process of creating alkenes, a type of hydrocarbon with one or more double bonds between carbon atoms. In the context of elimination reactions, an alkene is formed through the removal of elements, like hydrogen and a halide, from an organic compound. This transformation is called dehydrohalogenation.

The process involves:

• Formation of a double bond between carbon atoms after the removal of a hydrogen halide.
• Possibility to form multiple alkenes from a single substrate depending on the structure.
• According to Zaitsev's rule, the major product is typically the most stable alkene, which is usually the most substituted one.

Knowing the structure of the original halide helps in predicting possible alkene products and identifying the major product, providing insight into the stability and reactivity of the resulting alkenes.

Elimination Reactions

Elimination reactions are crucial in organic chemistry for synthesizing alkenes. In these reactions, elements are removed from a molecule, leading to the formation of a new double bond. Dehydrohalogenation, a common type of elimination reaction, specifically eliminates a hydrogen halide.

Key points about elimination reactions:

• They are characterized by the theta-symbol ("E") in their nomenclature, e.g., E1 and E2 mechanisms.
• In E1 reactions, the reaction involves a two-step mechanism: the loss of the leaving group followed by the deprotonation to form the double bond.
• In E2 reactions, it is a one-step concerted mechanism where the elimination of the leaving group and deprotonation occur simultaneously.

The choice between E1 and E2 paths usually depends on the structure of the substrate and the reaction conditions.

Sodium Ethoxide as a Base

Sodium ethoxide is a strong base commonly used in organic synthesis, especially in elimination reactions. Its chemical formula is NaOCH₂CH₃, where the ethoxide ion (OCH₂CH₃⁻) acts as the base that abstracts protons from substrates.

When sodium ethoxide is used,:

• It enhances the removal of hydrogen in the dehydrohalogenation process.
• It typically accelerates the E2 mechanism, favoring the formation of the most substituted product according to Zaitsev's rule.
• Its strong basic nature makes it ideal for reactions that require quick deprotonation.

Understanding how sodium ethoxide functions is crucial for manipulating reaction pathways to achieve desired outcomes in organic synthesis. It is especially useful when synthetic strategies need to rapidly and efficiently generate alkenes.