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Chapter 24: Problem 198

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

Short Answer

Expert verified
The reaction forms 2-phenylcyclopentene.

Step by step solution

01

Identify the Reaction Type

The reaction involves the use of alcoholic KOH with trans-2-phenyl-1-bromocyclopentane. This setup typically favors an elimination reaction (E2 mechanism) because alcoholic KOH is a strong base.
02

Determine the Substrate Geometry

The substrate, trans-2-phenyl-1-bromocyclopentane, suggests a situation where the bromine atom and the hydrogen atom located on adjacent carbons are in a trans configuration. This is important for the E2 reaction, where the leaving group and the hydrogen must be anti-periplanar.
03

Identify 尾-Hydrogens for Elimination

The beta hydrogen must be on the carbon adjacent to the one attached to the bromine. Look for a hydrogen atom on either C-2 or C-3. Given the trans configuration, this hydrogen will be on C-3, opposite to the phenyl group, enabling elimination.
04

Form the Double Bond

In the E2 mechanism, the base (KOH) will remove the beta hydrogen, causing the electrons to form a double bond between C-2 and C-3 while the bromine leaves as the bromide ion.
05

Name the Alkene Product

The resulting double bond forms between C-2 and C-3, creating 2-phenylcyclopentene. This double bond location reflects the structure of the produced alkene.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Elimination Reactions
Elimination reactions are a fundamental type of organic reaction where two substituents are removed from a molecule, forming a new 蟺 bond in the process. In the case of E2 reactions, this involves a bimolecular transition state.
  • "E" in E2 refers to "elimination," highlighting the removal aspect of these reactions.
  • The "2" indicates a bimolecular reaction, meaning two molecules, typically the substrate and the base, are involved in the rate-determining step.
  • Elimination reactions often compete with substitution reactions and require specific conditions, such as a strong base, to proceed effectively.
In E2 reactions, the use of a strong base, like alcoholic KOH, is crucial because it facilitates the removal of a proton adjacent to the leaving group, leading to the formation of the double bond.
Stereochemistry in Organic Reactions
Stereochemistry plays a pivotal role in determining the outcome of organic reactions. It pertains to the spatial arrangement of atoms in molecules and influences how reactions proceed.
  • For E2 reactions, the geometric orientation of the substrate is essential. The leaving group and the hydrogen to be abstracted must align in an anti-periplanar configuration.
  • Stereochemistry determines the regiochemistry and stereoselectivity of elimination reactions, affecting which hydrogen is removed.
  • The trans configuration of molecules like trans-2-phenyl-1-bromocyclopentane often facilitates efficient elimination processes.
Thus, understanding the stereochemical arrangement in the substrate is key to predicting the product of an E2 reaction.
Alkene Formation
Alkene formation is the primary result of elimination reactions, such as the E2 type. This occurs when a hydrocarbon loses a leaving group and a 尾-hydrogen to form a carbon-carbon double bond.
  • The newly formed \( ext{C=C}\) double bond results in an unsaturated hydrocarbon, known as an alkene.
  • In our example, as the bromide ion departs from trans-2-phenyl-1-bromocyclopentane, the double bond forms between C-2 and C-3, resulting in 2-phenylcyclopentene.
  • Alkene stability is influenced by the substitution pattern, with more substituted alkenes generally being more stable.
The placement and stability of the double bond are crucial for predicting the specific alkene product of an elimination reaction.
Beta Elimination
Beta elimination is a common process in many elimination reactions, where the halogen or other leaving group and a hydrogen atom are removed from adjacent carbons.
  • The term "beta" refers to the position of the hydrogen atom relative to the leaving group; it's on a carbon next to the one bearing the leaving group.
  • In our context, the beta hydrogen is on C-3, adjacent to the bromine on C-1 of the cyclopentane ring.
  • This hydrogen is removed by the base, consequentially forming a double bond at the site vacated by the departing groups.
Beta elimination is intrinsic to E2 reactions, dictating both where the double bond forms and which product is ultimately generated.
Anti-Periplanar Geometry
Anti-periplanar geometry is an essential concept in E2 reactions, referring to the spatial relationship required between the leaving group and the hydrogen being abstracted.
  • The term "anti-periplanar" means that the leaving group and the hydrogen atom must be oriented 180 degrees apart in the same plane.
  • This alignment minimizes steric hindrance and allows optimal orbital overlap, facilitating the transfer of electrons to form the double bond.
  • In trans-2-phenyl-1-bromocyclopentane, the trans arrangement naturally predisposes the bromine and a nearby hydrogen in an anti-periplanar setup.
Such geometric considerations are vital for ensuring an E2 reaction proceeds efficiently and leads to successful alkene formation.

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Most popular questions from this chapter

Which of the following species on photolysis does give a carbene? (a) CC(C)=O (b) \(\mathrm{CH}_{2}=\mathrm{C}=\mathrm{O}\) (c) \(\mathrm{CCl}_{4}\) (d) \(\mathrm{CHCl}_{3}\)

In which case, the intermediate involved is incorrect? (a) Pinacol - pinacolone rearrangement - carbocation (b) Hofmann's bromamide reaction \(-\) Nitrene (c) Aldol condensation - carbocation (d) Anti Markonikoff's addition of HBr to propene Free radical.

Among the given compounds, the most susceptible to nucleophilic attack at the carbonyl group is (a) \(\mathrm{MeCOCl}\) (b) \(\mathrm{MeCHO}\) (c) \(\mathrm{MeCOOMe}\) (d) \(\mathrm{Me} \mathrm{CO}-\mathrm{O}-\mathrm{CO} \mathrm{Me}\)

Resonance structures of a molecule does not have (a) identical arrangement of atoms (b) nearly the same energy content (c) same number of paired electrons (d) identical bonding

The reaction intermediate in \(\mathrm{E}_{1 C b}\) is (a) Carbonium ion (b) carbon free radical (c) Carbanion (d) Carbene
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