Question

Reaction of trans-2-phenyl-1-bromocyclopentane on reaction with alcoholic KOH produces (a) 4 -phenylcyclopentene (b) 2-phenylcyclopentene (c) 1-phenylcyclopentene (d) 3-phenylcyclopentene

Short answer

(b) 2-phenylcyclopentene

Step-by-step solution

Understanding the Reaction Conditions

The compound in question is trans-2-phenyl-1-bromocyclopentane, and it is reacting with alcoholic KOH. The presence of alcoholic KOH suggests that we should expect an elimination reaction, specifically an E2 mechanism, due to the strong base (KOH) and polar protic solvent (alcohol), which favors elimination over substitution.

Identifying the Possible Elimination Sites

The elimination reaction will involve the removal of a hydrogen atom and the bromine atom. For this molecule, the hydrogen to be removed must be on a carbon adjacent to the carbon holding the bromine (β-hydrogen). Trans-2-phenyl-1-bromocyclopentane has a 1-bromo and 2-phenyl group, meaning the β-hydrogen that can be removed is found on either the 3, 4, or 5 positions on the cyclopentane ring.

Determining the Major Product Based on Zaitsev's Rule

According to Zaitsev's rule, the preferred elimination will generally form the more substituted alkene. For the cyclopentane ring, removing a hydrogen from the 4-position (adjacent to the 1 and 2 carbons) will lead to the more substituted and stable alkene product. Specifically, elimination between the 1-carbon and 4-carbon results in a double bond at the 2-position of the phenyl, forming 2-phenylcyclopentene.

Writing the Reaction Mechanism

In an E2 mechanism, both the proton abstraction by the base and the loss of the leaving group (bromine) happen simultaneously. A base from the alcoholic KOH abstracts a β-hydrogen, and a double bond forms as bromine leaves, resulting in the formation of the double bond at the 2-position, giving 2-phenylcyclopentene.

Finalizing the Answer

Given the E2 reaction and considering Zaitsev's preference for more substituted alkenes, the major product of this reaction will be 2-phenylcyclopentene, aligning with option (b) in the given choices.

Key concepts

Zaitsev's Rule

Zaitsev's Rule is a guiding principle in organic chemistry that predicts the outcome of elimination reactions, such as the E2 reaction we encounter in this exercise. When a compound undergoes an elimination reaction, multiple alkenes can form. However, Zaitsev's Rule suggests that the more substituted alkene—the alkene where the double bond connects to more carbon atoms—will be the major product.

This rule stems from the stability of alkenes: more substituted alkenes are generally more stable due to hyperconjugation and the electron-donating effect of alkyl groups, which help to stabilize the double bond.

In the case of trans-2-phenyl-1-bromocyclopentane, applying Zaitsev's Rule helps us predict that the elimination of the β-hydrogen at the 4-position will form 2-phenylcyclopentene, as this position results in a more substituted alkene.

β-hydrogen

The term "β-hydrogen" refers to the hydrogen atom that is on the carbon adjacent to the carbon bonded with the leaving group, which in this case is bromine. During an E2 elimination reaction, a base abstracts this β-hydrogen, facilitating the removal of the leaving group and the formation of a double bond.

For trans-2-phenyl-1-bromocyclopentane, the relevant β-hydrogens can be found on the 3, 4, or 5 positions of the cyclopentane ring. These hydrogen atoms are crucial in the reaction because their removal leads to different possible alkene products.

The presence and location of β-hydrogens influence which double bond forms, and following Zaitsev's Rule, it becomes evident that the hydrogen at the 4-position, when removed, provides the most stable and substituted alkene, which is 2-phenylcyclopentene.

Alcoholic KOH

Alcoholic KOH, or potassium hydroxide dissolved in alcohol, plays an important role as a strong base in E2 elimination reactions. The alcohol serves as a polar protic solvent, favoring the elimination pathway over substitution.

In this reaction, KOH abstracts a β-hydrogen from trans-2-phenyl-1-bromocyclopentane, initiating the elimination of the leaving group (bromine). The choice of alcohol as the solvent aids in promoting the E2 mechanism due to its ability to stabilize the transition state through hydrogen bonding.

Thanks to the strong basic nature of KOH, it can effectively remove the β-hydrogen and facilitate the formation of a carbon-carbon double bond, ultimately resulting in the Zaitsev-favored alkene product.

Trans-2-phenyl-1-bromocyclopentane

Trans-2-phenyl-1-bromocyclopentane is the starting compound in this exercise and is crucial for understanding the elimination process. Its structure consists of a cyclopentane ring with a bromine atom and a phenyl group bonded at different positions.

The "trans" configuration indicates that these groups are on opposite sides of the ring, which can be significant in determining the stereochemistry of the product formed. This stereochemistry ensures that the β-hydrogens discussed earlier are appropriately positioned to react under the E2 mechanism.

Due to its structure, this compound features multiple potential β-hydrogens, leading to the formation of possible alkenes. However, using our understanding of Zaitsev's Rule, we determine that the most stable product, 2-phenylcyclopentene, will be the main outcome of the reaction with alcoholic KOH.