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

Recognize Reaction Type

The problem involves a treatment of a haloalkane with a methoxide ion (\( \text{CH}_3 \text{O}^- \)) in methanol. This setup is typically used for elimination reactions, specifically the E2 mechanism. Let's explore this type of reaction.

Identify the Substrate and Leaving Group

In the given substrate \( \text{CH}_3 \text{CH}_2 \text{CH} (\text{F}) \text{CH}_3 \), the leaving group is fluorine (\( \text{F}^- \)). Despite being a poor leaving group, the presence of a strong base like methoxide can still facilitate elimination.

Determine the Base's Role

Methoxide ion (\( \text{CH}_3 \text{O}^- \)) acts as a strong base. In an elimination mechanism, it abstracts a proton from a beta-carbon to the leaving group, resulting in the formation of a double bond.

Assess the Beta Protons Available

Identify the beta carbons relative to the carbon containing the leaving group. Here, we have two beta carbons: the one in \( \text{CH}_3 \text{CH}_2 \) and the one in \( \text{CH}_3 \). Both have available hydrogens for elimination.

Apply Zaitsev's Rule

Zaitsev's rule states that the more substituted alkene is the preferred product in eliminations. Evaluating, abstraction of a proton from the methylene group (\( \text{CH}_2 \)) leads to the more substituted double bond, yielding \( \text{CH}_3 \text{CH}=\text{CH} \text{CH}_3 \).

Confirm the Major Product

According to the elimination mechanism and Zaitsev's rule, the major product of this reaction is \( \text{CH}_3 \text{CH}=\text{CH} \text{CH}_3 \), which corresponds to option (b).

Verify Against Other Options

Further confirm by evaluating the reaction conditions: (a) substitution unlikely due to elimination conditions, (c) less substituted alkene not favored, (d) substitution product on incorrect carbon. Thus, final choice is affirmed.

Key concepts

Haloalkane

Haloalkanes are a group of chemical compounds where one or more hydrogen atoms in an alkane have been replaced by a halogen atom, such as fluorine, chlorine, bromine, or iodine. These compounds are widely studied in organic chemistry due to their ability to participate in a variety of reactions.

In the context of elimination reactions, haloalkanes play the role of the substrate. The halogen atom acts as the leaving group. In the given exercise, the haloalkane used is a fluorine-substituted alkane \( \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH} (\mathrm{F}) \mathrm{CH}_3 \). This compound is called 2-fluorobutane. While fluorine is a poor leaving group due to its high electronegativity, the presence of a strong base can facilitate its removal during the reaction process.

Methoxide Ion

The methoxide ion, denoted as \( \mathrm{CH}_3 \mathrm{O}^- \), is a potent base commonly used in elimination reactions. It consists of a methyl group bonded to an oxygen atom carrying a negative charge. Due to its strong basic nature, methoxide can effectively abstract protons from substrates.

In E2 elimination reactions, the methoxide ion acts as a strong base rather than a nucleophile. It specifically targets protons located on beta-carbons adjacent to the leaving group in haloalkanes. By abstracting a proton, the methoxide ion facilitates the formation of a double bond between adjacent carbon atoms, leading to the generation of an alkene.

Zaitsev's Rule

Zaitsev's Rule is an important principle in organic chemistry that guides the prediction of the predominant alkene formed in an elimination reaction. It states that the more highly substituted product, possessing the greatest number of alkyl groups attached to the double-bonded carbons, will generally be the favored alkene.

In the exercise, application of Zaitsev's Rule is crucial to determining the major product. When methoxide abstracts a beta proton from 2-fluorobutane, relative to the fluoro group, two possible protons could be removed. Abstraction from the structure's methylene group leads to a more substituted and stable alkene: \( \mathrm{CH}_{3} \mathrm{CH} = \mathrm{CHCH}_{3} \). This compound, but-2-ene, is the more stable product and is thus formed preferentially according to Zaitsev's Rule.

Alkene Formation

Alkene formation is the transformation of a saturated compound into an unsaturated one by the removal of atoms, resulting in the creation of a carbon-carbon double bond. This process is central to many elimination reactions, such as the E2 mechanism observed here.

In an E2 reaction, the base abstracts a proton from a beta-carbon while the leaving group is expelled from the adjacent carbon. This simultaneous process leads to the formation of an alkene. For the exercise in question, the methoxide ion facilitates the removal of a proton from 2-fluorobutane. The expulsion of the fluorine leaving group, in conjunction with the base's action, creates a double bond, yielding the alkene \( \mathrm{CH}_3 \mathrm{CH} = \mathrm{CH} \mathrm{CH}_3 \).