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

The product is (b) 2-phenylcyclopentene.

Step-by-step solution

Understand the Reaction Conditions

The reaction involves trans-2-phenyl-1-bromocyclopentane with alcoholic KOH. Alcoholic KOH is a common reagent used for elimination reactions that typically proceed via an E2 mechanism. In this mechanism, a hydrogen atom and a leaving group (such as a halogen) are eliminated to form a double bond.

Analyze the Structure

The substrate is trans-2-phenyl-1-bromocyclopentane. It consists of a cyclopentane ring with a bromine atom at the first position and a phenyl group at the second position in a trans configuration with respect to each other.

E2 Elimination Mechanism

In an E2 reaction, the base (alcoholic KOH) abstracts a proton from a carbon atom adjacent to the carbon bearing the leaving group (bromine). Since bromine is at position 1, the elimination can occur between carbon 1 and carbon 2. The hydrogen at position 2 can be abstracted as it is anti-periplanar to the leaving bromine atom, leading to the formation of a double bond.

Determine the Product

Consider the new double bond formed between carbons 1 and 2 after the loss of HBr. The resulting compound is a cyclopentene with the phenyl group still at position 2, leading to 2-phenylcyclopentene.

Key concepts

alcoholic KOH reagent

In the realm of organic chemistry, the reagent alcoholic KOH plays a vital role. Its importance lies in facilitating elimination reactions, specifically those that follow the E2 mechanism. Alcoholic KOH, essentially potassium hydroxide dissolved in alcohol, acts effectively as a strong base. The presence of the alcohol component is significant because it influences the reaction pathway, often ensuring the favoring of elimination over substitution.
The E2 mechanism involves two molecular entities coming together, hence the name bimolecular. This contrasts with the E1 mechanism, which is unimolecular and proceeds through a carbocation intermediate. With alcoholic KOH facilitating the E2 reaction, a proton and a leaving group are removed simultaneously. This departure creates a double bond in the substrate, resulting in an alkene product.

• Elimination via E2: Simultaneous removal of a proton and a leaving group.
• Strong base preferred: Alcoholic KOH is the choice due to its efficacy as a strong base.
• Rapid reaction: The bimolecular nature implies that reaction speed depends on both the base and substrate concentrations.

trans-2-phenyl-1-bromocyclopentane

Trans-2-phenyl-1-bromocyclopentane is the substrate at the heart of this reaction. It features a cyclopentane ring to which both a phenyl group and a bromine atom are attached. Importantly, these two substituents are *trans* to each other, meaning they occupy opposite sides of the ring plane.
This structural configuration is crucial for the reaction. In the E2 mechanism, the removal of a proton and a leaving group can occur effectively if they are anti-periplanar. The *trans* configuration ensures that the substrate is appropriately set up for the reaction to proceed efficiently.

• Trans Configuration: Phenyl and bromine are opposite each other on the ring.
• Substrate Readiness: The molecular geometry enables suitable conditions for the E2 reaction.
• Cyclopentane Core: Provides a flexible base for the substitution patterns.

double bond formation

Double bond formation is the essence of E2 elimination reactions. Once the correct conditions are met, the transition from single bonds to double bonds occurs. This transformation happens as a hydrogen atom is abstracted by the base—in this case, alcoholic KOH—from the carbon atom next to the one bonded to the bromine.
The leaving bromine concurrently departs, forming a pi bond between the involved carbon atoms. This newly formed double bond enhances molecular stability and reduces steric hindrance, creating an alkene product.

• Pi Bond Creation: Simultaneous bond-making and bond-breaking.
• Increased Stability: Double bonds generally confer greater stability compared to single bonds.
• Spatial Effect: Adjusts molecular geometry to optimize space utilization.

anti-periplanar abstraction

Anti-periplanar abstraction is a key concept that helps explain the mechanism of the E2 reaction. It emphasizes the need for the leaving group (bromine) and the hydrogen atom to be situated in opposite planes. This geometric requirement enables the simultaneous removal of the proton and the bromine, pivotal in forming the double bond.
In trans-2-phenyl-1-bromocyclopentane, this arrangement allows the E2 mechanism to proceed smoothly. The opposite orientation of bromine and hydrogen makes it straightforward for the base to abstract the proton, in line with the anti-periplanar requirement.

• Orientation Importance: Ensures the efficiency of the abstraction.
• Facilitates Bond Formation: Directly impacts the creation of a stable double bond.
• Geometric Necessity: Integral to the reaction occurring under optimal conditions.