Question

\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHI} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}\) \(\stackrel{\text { alc } \mathrm{KOH} / \Delta}{\longrightarrow}(\mathrm{A})\). The product \((\mathrm{A})\) is (a) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH} \mathrm{CH}_{3}\) (b) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}\) (d)

Short answer

Short Answer: The structure of product (A) obtained from the given reaction using alcoholic KOH and heat is (a) \(\mathrm{CH}_{3}\mathrm{CH}=\mathrm{CH}\mathrm{CH}_{2}\mathrm{CH}=\mathrm{CH}\mathrm{CH}_{3}\). This elimination reaction (E2) forms a double bond between the second and third carbons in the molecule.

Step-by-step solution

Identify the reaction

In this case, the reaction involves alcoholic KOH and heat, which is indicative of an elimination reaction. Specifically, it is an E2 (bimolecular elimination) reaction, in which both the substrate and base are involved in the rate-determining step, and a β-hydrogen and a halogen atom (iodine in this case) are eliminated, resulting in the formation of a double bond.

Locate the elimination site

In an E2 reaction, a β-hydrogen and a halogen atom are eliminated. In the given compound \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHI} \mathrm{CH}_{2}\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}\), we can locate the iodine atom on the third carbon. The β-hydrogens are present on the second and fourth carbons. Since we already have a double bond present in the molecule, it's more likely that the elimination occurs at the site with a single bond adjacent to the halogen atom - which is the second carbon.

Perform the elimination

Now that the reaction site has been identified, perform the elimination reaction by removing the iodine atom from the third carbon and a hydrogen atom from the adjacent second carbon. This creates a double bond between the second and third carbons: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHI} \mathrm{CH}_{2}\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3} \stackrel{\text { alc } \mathrm{KOH}/ \Delta}{\longrightarrow}\mathrm{CH}_{3}\mathrm{CH}=\mathrm{CH}\mathrm{CH}_{2}\mathrm{CH}=\mathrm{CH}\mathrm{CH}_{3}\)

Identify the product (A)

Based on the reaction performed, the structure of product (A) can be determined to be \(\mathrm{CH}_{3}\mathrm{CH}=\mathrm{CH}\mathrm{CH}_{2}\mathrm{CH}=\mathrm{CH}\mathrm{CH}_{3}\). Comparing the structure with the given options, the correct answer is: (a) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2}\mathrm{CH}=\mathrm{CH} \mathrm{CH}_{3}\)

Key concepts

Bimolecular Reaction

In the world of organic chemistry, bimolecular reactions are fascinating and pivotal, especially when it comes to understanding complex transformations. A bimolecular reaction refers to a reaction mechanism that involves two molecules interacting with each other as part of the rate-determining step. In the context of an E2 elimination reaction, both the attacking base and the substrate are crucial participants in this pivotal step.

E2 stands for Elimination, Bimolecular, and the reaction is characterized by a one-step process. During this process, the base removes a hydrogen atom from the β-carbon. Simultaneously, a leaving group, such as a halogen, exits from the molecule, allowing the formation of a new double bond, thereby converting an alkane to an alkene. The rate of an E2 reaction depends on the concentration of both the substrate and the base, hence the term bimolecular.

This type of reaction is particularly useful in organic synthesis because of its ability to form alkenes directly and efficiently. Here are some key characteristics of bimolecular reactions in the E2 context:

• Involves two reactants in the rate-determining step.
• Proceeds through a concerted, one-step mechanism.
• Forms alkenes by eliminating atoms from adjacent carbon atoms.

Organic Chemistry Reaction Mechanism

Reaction mechanisms in organic chemistry detail the step-by-step transformation of reactants into products, with a keen focus on electron movement and molecular changes. Understanding these mechanisms is crucial for mastering the subject, as it explains how and why particular products form.

In organic chemistry, the E2 elimination reaction mechanism offers a great example of how reactions can proceed smoothly and efficiently. The mechanism follows a concerted process, where bonds break and form simultaneously. This means that the departure of the leaving group (often a halogen like iodine) and the extraction of the hydrogen atom by the base happen in a synchronized manner.

E2 mechanisms stand apart from other elimination reactions like E1 because all changes occur in just one concerted step, without the formation of intermediates. Key points about E2 mechanisms include:

• No intermediates – a single, continuous reaction step.
• Requires a strong base to facilitate the removal of the hydrogen atom.
• Forms trans alkenes predominantly, due to favorable geometric arrangements during the reaction.

These mechanisms highlight the elegance and simplicity of organic transformations, making E2 eliminations a staple in alkene synthesis strategies.

Alkene Synthesis

Alkene synthesis is an essential topic in organic chemistry, serving as a building block for more complex molecular structures. The creation of alkenes typically involves eliminating particular atoms in a molecule to form a double bond. The E2 elimination reaction is a highly efficient method for this purpose, often utilized in both laboratory and industrial settings.

Alkenes are hydrocarbons that contain one or more carbon-carbon double bonds, and their synthesis requires careful planning and execution. In our exercise, the E2 reaction facilitates alkene formation by removing a β-hydrogen and a leaving group from adjacent carbon atoms. This reaction results in the transformation of single-bonded carbons into a double-bonded configuration, yielding the desired alkene.

Some benefits of synthesis via E2 elimination are:

• High selectivity for trans-alkenes due to spatial orientation during the reaction.
• Capability to synthesize complex alkenes from relatively simple starting materials.
• Efficient pathway requiring only one step, minimizing reaction time and resource use.

Overall, mastering alkene synthesis via the E2 mechanism equips chemists with a powerful tool for designing and creating valuable organic compounds.