Question

Elimination of bromine from 2-bromobutane results in the formation of (a) equimolar mixture of 1 and 2 -butene (b) predominantly 2 -butene (c) predominantly 1 -butene (d) predominantly 2 -butyne

Short answer

(b) predominantly 2 -butene

Step-by-step solution

Understanding the Reaction Type

The elimination of bromine from 2-bromobutane involves an elimination reaction typically following the E2 mechanism, where a hydrogen atom is removed along with the halide, resulting in the formation of an alkene.

Identifying the Possible Products

The removal of bromine from 2-bromobutane can lead to the formation of butenes. The possible butenes formed are 1-butene and 2-butene due to the position from which the hydrogen is eliminated.

Applying Zaitsev's Rule

According to Zaitsev's Rule, in elimination reactions, the more substituted alkene is usually the major product. Therefore, 2-butene, being the more substituted alkene compared to 1-butene, is likely to be the predominant product.

Conclusion Based on the Rule

Given the application of Zaitsev's Rule, the elimination of bromine from 2-bromobutane primarily yields 2-butene, as it is the more stable and substituted product compared to 1-butene.

Key concepts

2-bromobutane

2-bromobutane is a type of organic compound categorized as a haloalkane. This molecule consists of a four-carbon alkane chain, known as butane, where a bromine atom is attached to the second carbon. This attachment makes it 'secondary', meaning it is one carbon away from the end of the chain. 2-bromobutane plays a crucial role in E2 elimination reactions.

• The presence of the bromine atom makes the carbon it is attached to electrophilic, or electron-loving.
• In an E2 reaction, this electrophilic nature facilitates the removal of the bromine atom, setting the stage for the formation of double bonds in the resulting alkene.

Understanding the structure of 2-bromobutane helps in visualizing how alkenes are formed through elimination processes.

Zaitsev's Rule

Zaitsev's Rule is a guiding principle in organic chemistry applied in elimination reactions. It states that the most substituted alkene is the preferred product of an elimination reaction. In other words, when hydrogen is removed from adjacent carbons, the resulting double bond will most likely form in a manner that results in the more stable and more substituted alkene.

• A more substituted alkene has more alkyl groups attached to the carbons of the double bond, which enhances its stability due to hyperconjugation and the dispersal of electrons.
• This rule indicates that in the case of 2-bromobutane, 2-butene is more stabilized than 1-butene because it has more substituents directly attached to the double-bonded carbons.

Consequently, when bromine is eliminated from 2-bromobutane, 2-butene, the more stable alkene, is predominantly formed according to Zaitsev's Rule.

Alkene Formation

Alkene formation is a critical outcome of elimination reactions. In the context of an E2 reaction involving 2-bromobutane, the removal of bromine and a hydrogen atom from adjacent carbons leads to the creation of a carbon-carbon double bond, resulting in an alkene. The formation of alkenes is significant because it reflects the step from saturated molecules, which have single bonds, to unsaturated molecules, which contain double bonds.

• In the conversion from 2-bromobutane, the removal of a hydrogen atom from the second or the third carbon (alongside the bromine) will provide two potential alkenes: 1-butene and 2-butene.
• However, due to Zaitsev's Rule, 2-butene is formed as the major product, as it has a more central double bond, allowing more substituent groups to stabilize it.

The alkene formation process intelligently utilizes the structure of 2-bromobutane to predict which alkene will predominantly form.

Reaction Mechanism

The reaction mechanism for the elimination of bromine from 2-bromobutane is known as the E2 mechanism. E2, short for bimolecular elimination, reflects a process where the reaction occurs in a single concerted step. During this step, the base abstracts a proton from one carbon, as the leaving group (such as bromine) simultaneously departs from an adjacent carbon.

• In the E2 mechanism, the molecule undergoes a transition state where bonds are partially formed and broken at the same time.
• This process requires a strong base that can efficiently abstract the proton, helping to facilitate the elimination of bromine and formation of the alkene.

The E2 reaction mechanism is favored when dealing with secondary haloalkanes like 2-bromobutane and results in the rearrangement of electrons to produce alkenes efficiently.