Question

The major product formed when 2-Fluoropentane is treated with \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-} / \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) is (a) pent-1-ene (b) cis pent- 2 -ene (c) trans pent-2-ene

Short answer

Answer: (c) trans pent-2-ene

Step-by-step solution

Write down the starting compound and the reagent

In this case, the starting compound is 2-Fluoropentane (\(\mathrm{CH}_{3} \mathrm{CHF} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\)), and the reagent for the elimination reaction is ethoxide ion (C2H5O-) in ethanol solvent (C2H5OH).

Identify the α- and β-carbons

First, label the carbon that has the fluorine attached as the α-carbon. In this case, it's the second carbon in the chain. Then, the carbon(s) directly connected to the α-carbon are the β-carbons. Here, we have two β-carbons: one to the left and one to the right of the α-carbon.

Apply the Zaitsev's rule to analyze the possible products

According to Zaitsev's rule, when there are multiple β-carbons to choose from, the hydrogen will be removed from the β-carbon that, when forming the double bond, generates the more substituted alkene (the alkene with more carbon atoms attached to the double bond). In this case, both β-carbons have the same degree of substitution - they are each bound to two other carbon atoms.

Determine the major product formed as an alkene

Since both β-carbons are equivalent in terms of substitution, we will form a double bond between the α-carbon and one of the β-carbons. In this case, we will end up with pent-2-ene (\(\mathrm{CH}_{3} \mathrm{CH} = \mathrm{CH} \mathrm{CH}_{2} \mathrm{CH}_{3}\)).

Analyze the possibility of cis or trans configuration

The alkene we formed in step 4 has two hydrogen atoms on the α-carbon and two methyl groups on the β-carbon. The configuration can be either cis (both methyl groups on the same side of the double bond) or trans (methyl groups on opposite sides of the double bond). Since the trans isomer tends to have lower steric hindrance than the cis isomer, it is generally more stable and is thus the major product formed.

Match the major product with the answer choices

Comparing the major product, trans pent-2-ene, with the answer choices, we find that the answer is (c) trans pent-2-ene.

Key concepts

Elimination Reactions

Elimination reactions are a type of chemical reaction where elements are removed from a molecule, leading to the formation of a double bond. In the context of organic chemistry, elimination reactions are an essential mechanism for producing alkenes. These reactions typically involve the removal of a leaving group and a hydrogen atom from nearby carbon atoms.

Here's a quick breakdown of the process:

• A base, often characterized by a negatively charged ion like ethoxide \((C_2H_5O^- )\), attacks the substrate—a molecule containing a potential leaving group, like halogens (e.g., fluorine in 2-Fluoropentane).
• The base abstracts a proton from a \( eta \)-carbon (a carbon atom adjacent to the carbon with the leaving group), leading to the departure of the leaving group. This results in the formation of a double bond between the \( eta \)-carbon and the adjacent \( \alpha \)-carbon where the leaving group was attached.

This kind of elimination is called "E2" (bimolecular elimination) and occurs in a single step, without any intermediates. Understanding how these reactions occur is fundamental for predicting the structure and orientation of the resulting alkene product.

Zaitsev's Rule

Zaitsev's Rule is a guideline in organic chemistry used to predict the favored alkene product in elimination reactions. According to Zaitsev's Rule, the major product of an elimination reaction will be the more substituted alkene, meaning the double bond will form with more carbon atoms around it. This often results in a more stable configuration.

When applying Zaitsev's Rule, consider the potential hydrogens that can be removed from the adjacent \( \beta \)-carbon atoms:

• The \( \beta \)-carbon which contributes to a double bond with more alkyl groups (carbon groups) attached will form the favored product.
• This results because alkenes with more substituents are generally more stable due to hyperconjugation and the electron-releasing effect of the alkyl groups.

In scenarios, such as in the case of 2-Fluoropentane, when multiple \( \beta \)-carbons are available with equivalent substitution, either side can competitively participate in forming the double bond. However, even under these circumstances, the rule emphasizes the possibility of multiple products, with the most stable (Zaitsev) being predominant.

Alkene Formation

The process of converting a substrate to an alkene is known as alkene formation. This outcome is a result of elimination reactions, where the goal is to form a carbon-carbon double bond.

To achieve successful alkene formation:

• Choose suitable conditions such as a strong base like ethoxide ions to facilitate hydrogen abstraction.
• Identify the proper \( \beta \)-carbon for hydrogen removal to optimize the formation of a substituted alkene.

A high degree of stereospecificity might be involved, dictating whether the geometry of the newly formed double bond will be cis or trans. Trans alkenes, having substituents on opposite sides of the double bond, tend to be more stable due to less steric strain compared to cis alkenes, where substituents are on the same side.

In our case study with 2-Fluoropentane, alkene formation results in either cis or trans pent-2-ene. However, the trans configuration will prevail due to its higher stability, resulting in trans pent-2-ene being the major product. This highlights the direct relationship between Elimination Reactions, Zaitsev's Rule, and Alkene Formation in influencing organic synthesis.