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) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OCH}_{3}\)

Short answer

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

Step-by-step solution

Understanding the Reagents

The problem involves reacting \( \text{CH}_3 \text{CH}_2 \text{CH}(\text{F}) \text{CH}_3 \) with \( \text{CH}_3 \text{O}^- / \text{CH}_3 \text{OH} \). Here, \( \text{CH}_3 \text{O}^- \) is a strong base and a good nucleophile.

Identify Possible Reactions

Given the strong base and nucleophile \(\text{CH}_3 \text{O}^-\), there are two possible reactions: substitution or elimination. Since \(\text{F}^-\) is a poor leaving group for substitution, elimination (E2 mechanism) is more likely.

Determine the Elimination Mechanism (E2)

For an E2 reaction, the base \(\text{CH}_3 \text{O}^-\) will remove a hydrogen atom adjacent to the carbon bonded to the fluorine, resulting in the formation of a double bond. Here, a hydrogen from one of the methyl groups—specifically the one next to \(C-F\)—can be removed.

Apply Zaitsev's Rule

The major product is generally the one which is more substituted, according to Zaitsev's Rule. This means that the double bond will form between the \(C-2\) and \(C-3\) to give more substituted and stable alkene.

Identify the Major Product

Applying Zaitsev's Rule, removing a hydrogen from \(CH_3\) next to \(C-F\) gives \(\text{CH}_3\text{CH}=\text{CHCH}_3\), which is a more stable internal alkene because it is more substituted compared to other possible alkenes.

Key concepts

E2 Mechanism

The E2 mechanism, or bimolecular elimination reaction, is a fundamental concept in organic chemistry. When a strong base is present, an E2 reaction can occur. This is because the base will pull away a hydrogen atom from a carbon atom adjacent to a carbon that holds a leaving group, such as fluorine. In our exercise, the base \( \text{CH}_3 \text{O}^- \) removes a hydrogen atom from the \( \text{CH}_2 \) group next to the fluoro group. This leads to the formation of a double bond between the carbon where the hydrogen was removed and the adjacent carbon. E2 reactions are concerted processes, meaning that the base attack and the leaving group departing happen simultaneously. This behavior is distinct from other mechanisms like SN1 or SN2, where steps occur sequentially. E2 reactions typically require a strong base and a substrate that can readily form a stable alkene.

Zaitsev's Rule

Zaitsev's Rule is a guiding principle used to predict the outcome of elimination reactions like E2. It states that the major product of an elimination reaction tends to be the more substituted alkene. In simpler terms, it means that when given a choice, the double bond prefers to form in a position that results in more carbon atoms bonded to the carbons involved in this double bond. This is because more substituted alkenes are usually more stable due to hyperconjugation and electron delocalization. In the exercise we're discussing, applying Zaitsev's Rule helps us identify the major product where the double bond is located between \( C-2 \) and \( C-3 \), forming the alkene \( \text{CH}_3\text{CH}=\text{CHCH}_3 \). This product is favored because it is more substituted compared to the alternative, resulting in greater stability of the alkene.

Nucleophilic Substitution

Nucleophilic substitution is a type of reaction where a nucleophile, an electron-rich species, replaces a leaving group on a carbon atom. In the context of our exercise, nucleophilic substitution was a potential pathway for the reaction. However, as \( \text{F}^- \) is an exceptionally poor leaving group, it is unlikely that a substitution reaction would occur effectively with our reactant \( \text{CH}_3 \text{CH}_2 \text{CH}(\text{F}) \text{CH}_3 \). Strong bases such as \( \text{CH}_3 \text{O}^- \) can act as nucleophiles, but in this case, they promote an elimination (E2) rather than substitution, leading to the formation of double bonds instead. When a leaving group does not readily depart, elimination reactions are often favored. This illustrates the importance of understanding the nature of both the leaving group and the conditions under which the reaction is performed when predicting the pathway of a chemical reaction.

Alkene Stability

Alkenes, hydrocarbons containing a carbon-carbon double bond, vary in stability based on their structure. More substituted alkenes are generally more stable due to several factors, including hyperconjugation and the ability to distribute electron density better. Hyperconjugation refers to the delocalization of electrons from adjacent \( \sigma \) bonds to the \( \pi \) bond, which adds stability to the structure. Alkene stability also increases with increasing substitution because there are more opportunities for hyperconjugation, leading to better stabilization. In the context of our exercise, the formation of the product \( \text{CH}_3\text{CH}=\text{CHCH}_3 \) is favored because it results in a more substituted alkene compared to other possible outcomes. Ultimately, the stability of the resulting alkene will predict which product will majorly form in elimination reactions like E2. Thus, understanding alkene stability concepts helps chemists predict which products are more likely to form under certain reaction conditions.