Problem 45 - A Complete Resource Book in Chemistry for JEE Main

Question

The major product obtained on treatment of \(\mathrm{CH}_{3}\) \(\mathrm{CH}_{2} \mathrm{CH}(\mathrm{F}) \mathrm{CH}_{3}\) with \(\mathrm{CH}_{3} \mathrm{O} / \mathrm{CH}_{3} \mathrm{OH}\) is (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}\left(\mathrm{OCH}_{3}\right) \mathrm{CH}_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OCH}_{3}\)

Step-by-Step Solution

Answer

The major product is (b) \\mathrm{CH}_3\\mathrm{CH}=\\mathrm{CHCH}_3\.

Step 1: Understanding the Reactants and Conditions

The given compound is ethyl-2-fluorobutane treated with methanol ( CH_3OH ) and sodium methoxide ( CH_3O^- ). The key here is to recognize that this setup is conducive to an elimination reaction with fluorine as the leaving group.

Step 2: Determine the Type of Reaction

Since we have a strong base ( CH_3O^- ) and a poor leaving group ( F^- ), the most likely reaction is an E2 (bimolecular elimination) reaction. This reaction typically results in the removal of HX (in this case, HF ), leading to the formation of a double bond.

Step 3: Predict the Product Based on Zaitsev's Rule

Zaitsev's Rule states that the more substituted alkene is the preferred product in elimination reactions. In this scenario, eliminating the fluorine from the CH along with a hydrogen from an adjacent carbon atom follows this rule. The result is a more substituted alkene.

Step 4: Identify the Major Product

By removing the fluorine from the third carbon and a hydrogen from the second carbon, the most stable and substituted alkene is formed: CH_3CH=CHCH_3 , which corresponds to the option labeled (b).

Key Concepts

Zaitsev's Rule

A fundamental concept in organic chemistry, Zaitsev's Rule predicts which alkene will predominate in elimination reactions. It suggests that the more substituted alkene, one that has more alkyl groups attached to the double-bonded carbons, is typically favored as the product. In our reaction, we consider the elimination of a hydrogen atom from an adjacent carbon to the one holding the leaving group. If multiple hydrogens can be removed, Zaitsev’s Rule helps us choose the path that leads to the more stable, substituted alkene.
The reasoning behind this rule is related to the stability conferred by alkyl groups. More alkyl groups can help stabilize the double bond through hyperconjugation and the inductive effect, which enables the alkene to be of lower energy and hence more stable. For example, removing a proton from the secondary carbon in ethyl-2-fluorobutane, coupled with the fluorine leaving, results in a product with a double bond between two secondary carbons:

• This more substituted alkene structure, such as \( \text{CH}_3\text{-CH=CH-CH}_3 \), provides greater stability, leading to its predominance over less substituted isomers.

Understanding Zaitsev's Rule is critical when predicting the product distribution in elimination reactions, as it often helps chemists anticipate the most likely products.

Fluorine as a Leaving Group

In an E2 elimination reaction, the leaving group is of paramount importance. However, not all leaving groups are treated equally in terms of their ability to exit the molecule. Fluorine (\(-\mathrm{F}^-\)) is renowned for being a poor leaving group due to its small size and strong bond strength with carbon. These characteristics contribute to its reluctance to depart.
Despite this challenge, elimination reactions can still proceed with fluorine as the leaving group under certain conditions. This typically involves the presence of a strong base, such as methoxide (\(-\mathrm{OCH}_3^-\)), which can encourage the departure of the fluorine by forming a stable product (e.g., HF). To achieve successful elimination:

• There needs to be adequate driving force, such as the formation of a stable alkene, following Zaitsev's rule.
• Using a strong base enhances the elimination by effectively abstracting a proton, further promoting the creation of a double bond as fluorine leaves.

Understanding the limitations and behavior of different leaving groups such as fluorine is vital for effectively executing and predicting the outcomes of elimination reactions in the lab.

Alkene Formation in Elimination Reactions

Elimination reactions primarily occur to form alkenes by removing atoms or groups from the adjacent carbon atoms, characteristically known as dehydrohalogenation when involving halogens like fluorine. The E2 mechanism specifically involves a single concerted step where a base abstracts a proton and the leaving group exits simultaneously, leading to alkene formation.
The specific structure of the alkene is commonly dictated by factors such as Zaitsev's rule, which states the most substituted (and hence, more stable) alkene will often be the major product. In the scenario of ethyl-2-fluorobutane reacting with sodium methoxide in methanol:

• The fluorine atom leaves from the carbon it was originally attached to, facilitated by the strong base removing a neighboring hydrogen.
• The ensuing double bond formation depends on the position of the carbon where hydrogen is removed, shaped by rules predicting stability—mainly the Zaitsev's rule.

This process creates more substituted alkenes like but-2-ene (\(\text{CH}_3\text{-CH=CH-CH}_3\)).
Understanding the dynamics of elimination reactions and alkene formation helps in synthesizing specific organic molecules in chemical research and industrial applications.