Problem 33 - A Complete Resource Book in Chemistry for JEE Main

Question

The major product obtained on the 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 \(\mathrm{CH}_{3}\mathrm{CH}=\mathrm{CH}\mathrm{CH}_{3}\) (option b), formed via E2 elimination.

Step 1: Identify the Reaction Type

The substrate, \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}(\mathrm{F})\mathrm{CH}_{3}\), is treated with \(\mathrm{CH}_{3}\mathrm{O}^{-}/\mathrm{CH}_{3}\mathrm{OH}\). This is a common setup for an elimination reaction, known as the E2 mechanism, possibly forming an alkene or a substitution product.

Step 2: Analyze the Substrate

The tertiary carbon atom in \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}(\mathrm{F})\mathrm{CH}_{3}\) is attached to a fluorine. Fluorine is a weak leaving group, but the strong base (\(\mathrm{CH}_{3}\mathrm{O}^{-}\)) can facilitate elimination by abstracting a hydrogen atom.

Step 3: Determine Possible Products of E2 Elimination

The E2 mechanism will form alkenes. Assess the possible alkenes: 1. \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}\mathrm{CH}_{3} \) (option b) 2. \(\mathrm{CH}_{3}\mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2}\) (option c).

Key Concepts

Alkene Formation

In chemical reactions such as the E2 elimination mechanism, alkene formation is a prominent outcome. But what exactly are alkenes? They are hydrocarbons that contain at least one carbon-carbon double bond. The presence of this double bond is essential for the unique chemical properties alkenes display. When a substrate is treated with a strong base in the presence of a weak leaving group, the reaction's driving force is the formation of this double bond.
The E2 reaction mechanism involves the simultaneous removal of a proton (hydrogen) from one carbon and the expulsion of a leaving group from the adjacent carbon. As a result, a double bond forms between these two carbons, leading to an alkene.
The stability of the formed alkene largely depends on the substitution pattern of this double bond. Generally, alkenes with more substituents are more stable due to hyperconjugation and the delocalization of electron density over a larger area. This stability guide helps in predicting the major product of an E2 reaction.

Weak Leaving Group

In elimination reactions, especially the E2 type, the identity of the leaving group is crucial. A leaving group is a portion of the substrate that can be removed as the molecule undergoes a transformation into a new product. Commonly, strong leaving groups such as halides (like iodide and bromide) can easily depart, making them suitable for most reactions.
However, fluorine, as encountered in our exercise, acts as a weak leaving group. This is because fluorine forms a very strong bond with carbon due to its high electronegativity. It does not readily "let go" to leave with its pair of electrons.
In scenarios where a weak leaving group is involved, a strong base such as methoxide (\( ext{CH}_3 ext{O}^- \)) steps in to facilitate the process. The stronger the base, the better it can remove a proton adjacent to the carbon bonded with the leaving group, compensating for the leaving group's reluctance to exit the molecule. Understanding the nature of the leaving group is essential to predict the efficiency and outcome of E2 elimination reactions.

Strong Base Reactions

A strong base is a central player in the E2 elimination mechanism. These bases have a high affinity for protons, which makes them highly effective in deprotonating substrates. Methoxide (\( ext{CH}_3 ext{O}^- \)) is a classic example of a strong base often used in E2 reactions. Its role is to abstract a hydrogen from a substrate, enabling the formation of a double bond.
So why is a strong base necessary? In E2 reactions, both the removal of the hydrogen atom and the departure of the leaving group must happen almost simultaneously. A strong base ensures that the hydrogen elimination is favorable, which is especially important when dealing with weak leaving groups.
This base-driven reaction pathway bypasses potential substitution reactions, which might occur otherwise if the conditions weren't as stringent on elimination. Consequently, a strong base not only ensures the reaction proceeds but directs it specifically towards forming alkenes. Knowing this mechanism helps in identifying what outcomes to expect, and which conditions favor elimination over substitution.